1.1 Physical Quantities and Measurement Techniques

In this section, we will discuss the different physical quantities used in measurements, such as mass, length, time, temperature, and others. The various units of measurement like SI units will also be covered.

Measurement Techniques

Scalar and Vector Quantities

Vector Calculations

1.2 Motion

In this section, we will define and calculate the key concepts of speed, velocity, acceleration, and the various graphs associated with these quantities.

Speed and Velocity

  • Speed:
    Speed is defined as the distance travelled per unit time. The formula used is: \[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
  • Velocity:
    Velocity is defined as speed in a given direction. It is a vector quantity, meaning it has both magnitude and direction.
  • Average Speed:
    The average speed is calculated as the total distance travelled divided by the total time taken. The formula is: \[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \]

Distance–Time and Speed–Time Graphs

  • Graph Interpretation:
    Sketch, plot, and interpret distance–time and speed–time graphs to understand the motion of objects. Key points to note:
    • At rest: The graph is a horizontal line.
    • Constant speed: The graph is a straight, non-horizontal line with a constant gradient.
    • Acceleration: The graph has an increasing gradient.
    • Deceleration: The graph has a decreasing gradient.
  • Calculate Speed from Distance–Time Graph:
    To calculate speed from a distance–time graph, determine the gradient of a straight-line section of the graph.
  • Calculate Distance from Speed–Time Graph:
    The area under a speed–time graph represents the distance travelled. For motion with constant speed or constant acceleration, this can be calculated easily using geometric shapes.

Acceleration

  • Acceleration of Free Fall:
    The acceleration of free fall (\(g\)) for an object near the Earth's surface is approximately constant and has a value of approximately \(9.8 \, \text{m/s}^2\).
  • Definition of Acceleration:
    Acceleration is defined as the change in velocity per unit time. The formula used is: \[ a = \frac{\Delta v}{\Delta t} \] where \(a\) is acceleration, \(\Delta v\) is the change in velocity, and \(\Delta t\) is the change in time.
  • Constant and Changing Acceleration:
    From given data or a speed–time graph, you can determine when an object is moving with constant or changing acceleration:
    • Constant acceleration: The graph has a straight line with a constant gradient.
    • Changing acceleration: The graph's gradient changes over time.
  • Calculate Acceleration from Speed–Time Graph:
    The acceleration can be calculated by finding the gradient of a speed–time graph.
  • Deceleration:
    Deceleration is a negative acceleration and can be used in calculations just like positive acceleration.

Objects Falling in a Gravitational Field

  • Motion in a Gravitational Field:
    The motion of objects falling in a uniform gravitational field can be described with or without air or liquid resistance. Without resistance, the object accelerates at \(g = 9.8 \, \text{m/s}^2\). With resistance, the object eventually reaches terminal velocity, where the force due to gravity equals the drag force, resulting in no further acceleration.

For example, the formula for speed is \( v = \frac{d}{t} \), where \( v \) is velocity, \( d \) is distance, and \( t \) is time.

Also, the following is an example of an equation displayed:

\[ a = \frac{\Delta v}{\Delta t} \]


Equations of Motion

  • Equation 1: \( v = u + at \)
    This equation describes the final velocity (\(v\)) of an object when it starts with an initial velocity (\(u\)), accelerates at a rate of \(a\) for a time \(t\).
    Example: An object starts from rest (i.e., \(u = 0\)) and accelerates at \(2 \, \text{m/s}^2\) for 5 seconds. The final velocity is: \[ v = 0 + (2 \times 5) = 10 \, \text{m/s} \]
  • Equation 2: \( S = ut + \frac{1}{2} a t^2 \)
    This equation calculates the displacement (\(S\)) of an object when it starts with an initial velocity (\(u\)), accelerates at \(a\) for a time \(t\).
    Example: A car starts from rest (\(u = 0\)) and accelerates at \(3 \, \text{m/s}^2\) for 4 seconds. The displacement is: \[ S = (0 \times 4) + 0.5 \times 3 \times 4^2 = 0 + 0.5 \times 3 \times 16 = 24 \, \text{m} \]
  • Equation 3: \( v^2 = u^2 + 2aS \)
    This equation relates the final velocity squared (\(v^2\)) to the initial velocity squared (\(u^2\)), acceleration (\(a\)), and displacement (\(S\)).
    Example: A car accelerates from \(u = 5 \, \text{m/s}\) to \(v = 15 \, \text{m/s}\) with an acceleration of \(2 \, \text{m/s}^2\). The displacement is: \[ v^2 = u^2 + 2aS \quad \Rightarrow \quad 15^2 = 5^2 + 2 \times 2 \times S \] \[ 225 = 25 + 4S \quad \Rightarrow \quad 200 = 4S \quad \Rightarrow \quad S = 50 \, \text{m} \]

Applications of Equations of Motion

  • Solving Problems:
    These equations are widely used to solve problems involving motion with constant acceleration. For example, to find the final velocity or displacement of an object in free fall or an object moving with constant acceleration, you can use these equations depending on the known quantities.

For example, the first equation of motion is \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

Also, the following are examples of the other two equations of motion:

\[ S = ut + \frac{1}{2} a t^2 \]

\[ v^2 = u^2 + 2aS \]

1.3 Mass and Weight

Mass: Mass is a measure of the quantity of matter in an object at rest relative to the observer. It is constant and does not change regardless of location.

Weight: Weight is the gravitational force acting on an object with mass. It depends on the gravitational field strength and varies based on location.

Gravitational Field Strength: Defined as the force per unit mass. It is expressed as:

\[ g = \frac{W}{m} \]

This equation is equivalent to the acceleration due to free fall, which on Earth is approximately \( 9.8 \, \text{N/kg} \).

Weights and masses can be compared using a balance. For example, a spring balance measures weight, while a beam balance measures mass by comparison to standard weights.

Weight is the effect of a gravitational field on a mass. The strength of the gravitational field (\( g \)) can be found by plotting a graph of weight against mass. The gradient of this graph gives the value of \( g \).

Graph of Weight Against Mass

Plot the graph, calculate the gradient, and interpret the result as the gravitational field strength.

Key Concepts:

  • Weight, mass, and gravitational field strength are related through the equation \( W = mg \).
  • Gravitational field strength near Earth’s surface is constant at \( 9.8 \, \text{N/kg} \).
  • All objects experience the same acceleration due to gravity in free fall, but air resistance can alter their motion.

Activity: Weight on Different Planets

Stick images of the planets with their gravitational field strength values around the classroom. Learners can use these values to calculate their weight on different planets using the formula \( W = mg \).

For example, if a learner's mass is \( 60 \, \text{kg} \):

\[ W_{\text{Mars}} = 60 \times 3.7 = 222 \, \text{N} \]

Discussion: How Sports Records Might Change

Discuss how Olympic records for weightlifting, high jump, and sprints would differ on Mars due to its lower gravitational field strength. Learners estimate the potential changes.

Advanced Topics

Gravitational Fields: A gravitational field is a region where a mass experiences a force due to gravitational attraction. The strength of this field depends on the size of the mass creating it and the distance from it.

Newton’s Law of Gravitation: For advanced learners, introduce the equation:

\[ F = G \frac{m_1 m_2}{r^2} \]

Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are masses, and \( r \) is the distance between them.

Interactive Simulation

Learners can explore gravitational attraction using the Gravity Force Lab Simulation. They calculate forces between objects and understand why gravitational forces between small masses go unnoticed.

Practice Questions

  • Q1: Calculate the weight of a \( 50 \, \text{kg} \) object on Earth.

  • Q2: If an object weighs \( 300 \, \text{N} \) on Earth, what is its mass?

  • Q3: What would be the weight of a \( 70 \, \text{kg} \) astronaut on the Moon where \( g = 1.6 \, \text{N/kg} \)?

Answers

  • A1: \[ W = mg = 50 \times 9.8 = 490 \, \text{N} \] The weight of the object is \( 490 \, \text{N} \).

  • A2: \[ m = \frac{W}{g} = \frac{300}{9.8} \approx 30.61 \, \text{kg} \] The mass of the object is approximately \( 30.61 \, \text{kg} \).

  • A3: \[ W = mg = 70 \times 1.6 = 112 \, \text{N} \] The weight of the astronaut on the Moon is \( 112 \, \text{N} \).

1.4 Density

Definition of Density: Density is defined as mass per unit volume. It indicates how closely packed the particles of a substance are. The formula for density is:

\[ \rho = \frac{m}{v} \]

Where:

  • \( \rho \) = density (kg/m3 or g/cm3)
  • \( m \) = mass (kg or g)
  • \( v \) = volume (m3 or cm3)

Determining Density

1. For a Liquid:

  1. Measure the mass of an empty measuring cylinder.
  2. Pour the liquid into the cylinder and measure its total mass.
  3. Subtract the empty cylinder’s mass to find the liquid’s mass.
  4. Record the liquid’s volume from the cylinder.
  5. Use the formula \( \rho = \frac{m}{v} \).

2. For a Regularly Shaped Solid:

  1. Measure the mass of the solid using a balance.
  2. Determine its volume using geometric formulas (e.g., \( v = \text{length} \times \text{width} \times \text{height} \) for a cuboid).
  3. Use the formula \( \rho = \frac{m}{v} \).

3. For an Irregularly Shaped Solid:

  1. Measure the mass of the solid using a balance.
  2. Fill a measuring cylinder with water and note the initial volume.
  3. Submerge the solid in the water and record the new volume.
  4. Calculate the volume of the solid as the difference between the initial and final volumes.
  5. Use the formula \( \rho = \frac{m}{v} \).

Floating and Sinking

Objects: An object will float if its density is less than the density of the liquid it is placed in. Otherwise, it will sink.

Liquids: When two immiscible liquids are mixed, the one with lower density will float on top of the denser liquid. Examples include oil on water or fresh water on saltwater.

Converting Units

To convert between g/cm3 and kg/m3:

  • 1 g/cm3 = 1000 kg/m3

Practical Activity: Investigating Density

Use the Buoyancy Simulation to explore the principles of density and floating. Try stacking liquids of different densities and observe how they layer.

Practice Questions

  • Q1: A metal block has a mass of 540 g and a volume of 200 cm3. Calculate its density.

  • Q2: If 50 ml of a liquid weighs 40 g, determine its density.

  • Q3: Will an object with a density of 800 kg/m3 float in water? Justify your answer.

  • Q4: Oil has a density of 0.92 g/cm3 and water has a density of 1.00 g/cm3. Which liquid will float on top?

Answers

  • A1: \[ \rho = \frac{m}{v} = \frac{540 \, \text{g}}{200 \, \text{cm}^3} = 2.7 \, \text{g/cm}^3 \]

  • A2: \[ \rho = \frac{m}{v} = \frac{40 \, \text{g}}{50 \, \text{ml}} = 0.8 \, \text{g/cm}^3 \]

  • A3: The object will float since its density (\( 800 \, \text{kg/m}^3 \)) is less than the density of water (\( 1000 \, \text{kg/m}^3 \)).

  • A4: Oil will float on water because its density (\( 0.92 \, \text{g/cm}^3 \)) is less than water’s density (\( 1.00 \, \text{g/cm}^3 \)).

1.5 Forces

Definition of Force: A force is a push or pull acting on an object as a result of its interaction with another object. Forces can change an object’s shape, speed, or direction of motion.

Types of Forces

Contact Forces: These forces occur when objects are physically touching each other.

  • Frictional Force: Resists motion between two surfaces in contact.
  • Tension Force: Force transmitted through a string, rope, or cable when it is pulled tight.
  • Normal Force: The support force exerted by a surface on an object resting on it.
  • Air Resistance: Force acting against a moving object in air.

Non-Contact Forces: These forces act at a distance without physical contact.

  • Gravitational Force: Attraction between two masses.
  • Magnetic Force: Attraction or repulsion between magnetic poles.
  • Electrostatic Force: Force between electrically charged objects.

Net Force

When multiple forces act on an object, the net force is the vector sum of all these forces. It determines the overall effect on the object’s motion.

Formula: \( F_{\text{net}} = F_1 + F_2 + \dots + F_n \), considering the direction of each force.

Balanced and Unbalanced Forces

Balanced Forces: If the forces acting on an object are equal in magnitude and opposite in direction, the net force is zero. This means:

  • The object remains stationary or continues to move at a constant velocity.

Condition for Balanced Forces:

\[ F_{\text{net}} = 0 \]

Unbalanced Forces: If the forces acting on an object do not cancel out, the net force is non-zero, causing the object to accelerate in the direction of the net force.

Examples of Net Force Calculations

  • Example 1: Two forces of 5 N and 7 N act in the same direction on an object. Find the net force.

  • Example 2: Two forces of 10 N and 6 N act on an object in opposite directions. Find the net force.

Practice Questions

  • Q1: A box is pulled with a force of 15 N to the right and 10 N to the left. What is the net force acting on the box?

  • Q2: An object is stationary under the influence of two forces, 20 N upwards and 20 N downwards. Is the object in equilibrium? Why?

  • Q3: Two children push a toy car with forces of 12 N and 8 N in the same direction. Calculate the net force.

Answers

  • A1: Net force = \( 15 \, \text{N} - 10 \, \text{N} = 5 \, \text{N} \) to the right.

  • A2: Yes, the object is in equilibrium because the forces are balanced (\( 20 \, \text{N} - 20 \, \text{N} = 0 \)).

  • A3: Net force = \( 12 \, \text{N} + 8 \, \text{N} = 20 \, \text{N} \) in the same direction.

1.5.1 Effects of forces

1. Resultant Forces

The resultant force is the single force that has the same effect as the combined forces acting on an object along the same straight line.

  • If forces act in the same direction, add their magnitudes.
  • If forces act in opposite directions, subtract the smaller force from the larger.

Newton’s First Law: An object remains at rest or in uniform motion unless acted upon by a resultant force. If the resultant force is zero, the object is in equilibrium.

Newton’s Second Law: A resultant force causes acceleration:
\[ F = ma \]

The direction of the force and acceleration is the same.

2. Types of Forces

Forces can be classified into two main categories:

  • Contact Forces: Forces that require physical contact. Examples include friction, air resistance, tension, and normal force.
  • Non-Contact Forces: Forces that act without physical contact. Examples include gravity, magnetic force, and electrostatic force.

3. Friction and Drag

Friction: Acts between surfaces in contact, opposing motion and producing heat.

  • Solid Friction: Between two surfaces in contact.
  • Drag (Air Resistance): Acts on objects moving through a fluid, such as air or water.

Terminal Velocity: This occurs when the force of air resistance equals the force of gravity, resulting in no net force and constant velocity.

4. Hooke’s Law

Hooke's Law: Describes the extension of a spring under a load:

\[F = kx \], where:

  • F = Force applied,
  • k = Spring constant,
  • x = Extension.

Limit of Proportionality: The point beyond which Hooke’s Law is no longer valid.

We can see that the behaviour of the graph changes from a straight line after a force of 8N. This is the limit of proportionality for this spring.


Practice Questions

1. Resultant Force

Two forces of 10 N and 15 N act on a box in opposite directions. Find the resultant force.

2. Hooke’s Law

A spring stretches by 0.05 m under a 2 N force. Calculate the spring constant.

3. Friction and Drag

Explain why a spacecraft heats up upon re-entering the Earth’s atmosphere.


Answers

1. Resultant Force

The two forces act in opposite directions. The resultant force is the difference between the magnitudes of the forces:

\[ F_{\text{resultant}} = F_1 - F_2 = 15\, \text{N} - 10 \text{N} = 5\, \text{N} \] The resultant force is 5 N in the direction of the larger force.


2. Hooke’s Law

Hooke’s Law states: \[ F = k \cdot x \]

Rearranging to find the spring constant (\( k \)):

\[ k = \frac{F}{x} = \frac{2\, \text{N}}{0.05\, \text{m}} = 40\, \text{N/m} \]

The spring constant is \( 40\, \text{N/m} \).


3. Friction and Drag

A spacecraft heats up upon re-entry because of air resistance. The drag force causes compression of air in front of the spacecraft, generating heat through friction and compression. This results in extremely high temperatures.


1.5.2 Moments of Forces

In this section, we will explore the concept of moments, their applications, and how they can be used to determine equilibrium in various scenarios.

Moment of a Force

The moment of a force is a measure of its turning effect around a pivot point. It is calculated as:

\[ \text{Moment} = \text{Force} \times \text{Perpendicular distance from the pivot} \]

Unit: Newton-metre (N·m)

Everyday Examples of Moments

  • Opening a door: The force applied on the handle creates a turning effect about the hinges (pivot).
  • Using a wrench: The force applied on the wrench handle generates a moment to turn a bolt.
  • Balancing on a seesaw: The weights of the children on either side create moments around the pivot point.

Principle of Moments

The principle of moments states that for an object in equilibrium:

\[ \text{Sum of clockwise moments} = \text{Sum of anticlockwise moments} \]

Equilibrium

An object is in equilibrium if:

  • There is no resultant force acting on it.
  • There is no resultant moment about any pivot.

Application of the Principle of Moments

  • Balancing a beam: When two weights are placed on opposite sides of a pivot, the beam remains balanced if the moments on each side are equal.
  • Multiple forces: The principle of moments can be applied to situations with more than one force on each side of the pivot.

Experiment: Demonstrating No Resultant Moment

Objective: To show that an object in equilibrium has no resultant moment.

Procedure:

  1. Place a uniform beam on a pivot and ensure it is balanced.
  2. Hang weights on either side of the beam at different distances from the pivot.
  3. Adjust the weights and their distances until the beam is level.

Conclusion: The sum of clockwise moments equals the sum of anticlockwise moments when the beam is in equilibrium.

Practice Problems

Problem 1: Balancing a Seesaw

A child of weight \( 300 \, \text{N} \) sits \( 2 \, \text{m} \) from the pivot of a seesaw. Where should a \( 400 \, \text{N} \) child sit on the other side to balance the seesaw?

Solution:

Using the principle of moments:

\[ \text{Clockwise moment} = \text{Anticlockwise moment} \]

\[ 300 \times 2 = 400 \times d \]

\[ 600 = 400d \quad \Rightarrow \quad d = \frac{600}{400} = 1.5 \, \text{m} \]

The \( 400 \, \text{N} \) child should sit \( 1.5 \, \text{m} \) from the pivot.

Problem 2: Multiple Forces

A uniform beam of length \( 5 \, \text{m} \) and weight \( 50 \, \text{N} \) is balanced on a pivot \( 2 \, \text{m} \) from one end. A \( 100 \, \text{N} \) weight is hung \( 1 \, \text{m} \) from the same end. Find the force exerted by the pivot.

Solution:

Moments about the pivot:

Taking clockwise moments:

\[ 100 \times 1 = 100 \, \text{N·m} \]

Taking anticlockwise moments (beam's weight acts at its center, \( 2.5 \, \text{m} \) from one end):

\[ 50 \times (2.5 - 2) = 25 \, \text{N·m} \]

Net moment for equilibrium:

\[ \text{Clockwise moment} = \text{Anticlockwise moment} \]

\[ F_{\text{pivot}} = 100 - 25 = 75 \, \text{N} \]

1.5.3 Centre of Gravity

This section focuses on the concept of the center of gravity, its practical determination, and how it affects the stability of objects.

Definition of Centre of Gravity

The center of gravity of an object is the point where its entire weight appears to act, irrespective of its orientation.

Experiment: Determining the Centre of Gravity of an Irregularly Shaped Lamina

Objective: To determine the center of gravity of an irregularly shaped plane lamina.

Apparatus: Irregularly shaped lamina, plumb line, pin, clamp, and retort stand.

Procedure:

  1. Fix the lamina to a clamp stand using a pin so that it can freely rotate.
  2. Hang a plumb line from the pin. Mark the line traced by the plumb line on the lamina.
  3. Repeat the process by suspending the lamina from two other points and marking the plumb lines.
  4. The intersection of the three lines is the center of gravity of the lamina.

Conclusion: The point where the three lines intersect is the center of gravity of the lamina.

Effect of Centre of Gravity on Stability

The position of the center of gravity greatly influences the stability of an object:

  • Low Centre of Gravity: Objects with a lower center of gravity are more stable as their weight is concentrated closer to the base. For example, a wide-based cone resting on its wider end.
  • High Centre of Gravity: Objects with a higher center of gravity are less stable and more prone to toppling. For instance, a tall, narrow vase.

Qualitative Analysis of Stability

  • Stable Equilibrium: When an object returns to its original position after being slightly tilted. Example: A cone resting on its base.
  • Unstable Equilibrium: When an object falls over after being tilted. Example: A cone balanced on its tip.
  • Neutral Equilibrium: When an object stays in its new position after being displaced. Example: A ball on a flat surface.

Problem and Solution

Problem: Stability of a Tower

A tower has a base of \( 4 \, \text{m} \) and a height of \( 20 \, \text{m} \). Explain how lowering its center of gravity or widening its base could increase its stability.

Solution:

  • By lowering the center of gravity, the weight of the tower is distributed closer to the base, reducing the likelihood of toppling.
  • By widening the base, the tower's support area increases, making it harder for the line of action of the weight to fall outside the base, thus improving stability.

1.6 Momentum

This section covers momentum, impulse, the conservation of momentum, and the relationship between force and the rate of change of momentum.

Definition of Momentum

Momentum (p) is defined as the product of an object’s mass and its velocity:

\[ p = mv \]

Where:

  • \( p \): Momentum (\(\text{kg m/s}\))
  • \( m \): Mass (\(\text{kg}\))
  • \( v \): Velocity (\(\text{m/s}\))

Definition of Impulse

Impulse is defined as the product of force and the time duration over which the force acts. It is also equal to the change in momentum:

\[ \text{Impulse} = F\Delta t = \Delta p = \Delta(mv) \]

Where:

  • \( F \): Force (\(\text{N}\))
  • \( \Delta t \): Time interval (\(\text{s}\))
  • \( \Delta p \): Change in momentum (\(\text{kg m/s}\))

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before an interaction is equal to the total momentum after the interaction, provided no external forces act:

\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]

Where:

  • \( m_1, m_2 \): Masses of objects (\(\text{kg}\))
  • \( u_1, u_2 \): Initial velocities (\(\text{m/s}\))
  • \( v_1, v_2 \): Final velocities (\(\text{m/s}\))

Resultant Force and Momentum

A resultant force is the rate of change of momentum. This is expressed by the equation:

\[ F = \frac{\Delta p}{\Delta t} \]

Where:

  • \( F \): Resultant force (\(\text{N}\))
  • \( \Delta p \): Change in momentum (\(\text{kg m/s}\))
  • \( \Delta t \): Time interval (\(\text{s}\))

Problems and Solutions

Problem 1: Calculating Momentum

A car of mass \( 1000 \, \text{kg} \) is moving with a velocity of \( 20 \, \text{m/s} \). Calculate its momentum.

Solution:

\[ p = mv = 1000 \times 20 = 20,000 \, \text{kg m/s} \]

Problem 2: Applying Impulse

A force of \( 50 \, \text{N} \) acts on an object for \( 4 \, \text{seconds} \). Calculate the impulse.

Solution:

\[ \text{Impulse} = F\Delta t = 50 \times 4 = 200 \, \text{N·s} \]

Problem 3: Conservation of Momentum

Two trolleys collide. Trolley 1 (mass \( 2 \, \text{kg} \)) moves at \( 3 \, \text{m/s} \) and Trolley 2 (mass \( 1 \, \text{kg} \)) is stationary. After the collision, they stick together. Find their final velocity.

Solution:

Using conservation of momentum:

\[ m_1u_1 + m_2u_2 = (m_1 + m_2)v \]

\[ (2 \times 3) + (1 \times 0) = (2 + 1)v \]

\[ 6 = 3v \quad \Rightarrow \quad v = 2 \, \text{m/s} \]


1.7 Energy, Work and Power

This section introduces the fundamental concepts of energy, work, and power, which are crucial in understanding various physical phenomena.


Forms of Energy

Energy can be categorized into different forms, including:

  • Kinetic Energy (KE) – Energy due to motion.
  • Potential Energy (PE) – Energy stored due to position or configuration.
  • Heat Energy – Energy transferred due to temperature difference.
  • Light Energy – Energy carried by electromagnetic waves.
  • Electrical Energy – Energy due to the flow of electric charge.
  • Sound Energy – Energy carried by sound waves.

1.7.1 Energy Stores and Transfers

Energy Stores

Energy can be stored in various forms, including:

  • Kinetic Energy – Energy of motion.
  • Gravitational Potential Energy – Energy due to height in a gravitational field.
  • Chemical Energy – Energy stored in chemical bonds.
  • Elastic (Strain) Energy – Energy stored in stretched or compressed objects.
  • Nuclear Energy – Energy stored in the nucleus of an atom.
  • Electrostatic Energy – Energy due to electric charges.
  • Magnetic Energy – Energy in magnetic fields.
  • Internal (Thermal) Energy – Energy due to the temperature of a body.

Energy Transfers

Energy can be transferred between stores during various events and processes, including:

  • Mechanical Work – Transfer by forces, e.g., lifting a weight.
  • Electrical Work – Transfer by electric currents, e.g., in circuits.
  • Heating – Transfer due to a temperature difference.
  • Waves – Transfer via electromagnetic waves, sound waves, etc.

Principle of Conservation of Energy

Energy cannot be created or destroyed, only transferred from one store to another. The total energy in a closed system remains constant.

This principle is applied in interpreting simple flow diagrams and Sankey diagrams for complex processes.

Key Equations

  • Kinetic Energy:
  • \[ KE = \frac{1}{2}mv^2 \]

  • Change in Gravitational Potential Energy:
  • \[ \Delta PE = mg\Delta h \]

Examples

1. A pendulum converts gravitational potential energy to kinetic energy as it swings.

2. An electric motor transfers electrical energy to mechanical energy.

Application of Conservation of Energy

Consider a roller coaster: at the highest point, it has maximum gravitational potential energy, which is converted to kinetic energy as it descends.

Sankey diagrams can be used to illustrate energy efficiency, showing energy input, useful output, and wasted energy.

1.7.2 Work

1.7.2 Mechanical and Electrical Work

Understanding Work Done

Work is done when a force is applied to an object, causing it to move. The energy transferred during this process is equal to the work done.

Similarly, in electrical systems, work is done when electrical energy is transferred by a current.

Key Equation for Mechanical Work

The work done by a force is given by the equation:

\[ W = Fd = \Delta E \]

Where:

  • \( W \): Work done (\(\text{J}\))
  • \( F \): Force applied (\(\text{N}\))
  • \( d \): Distance moved in the direction of the force (\(\text{m}\))
  • \( \Delta E \): Change in energy (\(\text{J}\))

Examples

1. Lifting a box: A force is applied to lift the box, doing work equal to the increase in the box's gravitational potential energy.

2. Pushing a car: Mechanical work is done to overcome friction, transferring energy to the car's kinetic store.

Electrical Work

Electrical work is done when a current flows through a circuit. The energy transferred is equal to the product of voltage, current, and time:

\[ E = VIt \]

1.7.3 Energy Resources and Efficiency

Sources of Useful Energy

Energy can be obtained or electrical power generated from various sources:

  • Fossil Fuels (coal, oil, natural gas): Chemical energy stored in these fuels is released during combustion.
  • Biofuels: Chemical energy stored in organic material, such as plants, is released through combustion or fermentation.
  • Water:
    • Energy stored in waves is harnessed by wave power systems.
    • Energy in tides is used by tidal power stations.
    • Energy stored in water behind a hydroelectric dam is converted to electrical energy by turbines.

Advantages and Disadvantages of Energy Resources

Method Advantages Disadvantages
Fossil Fuels High energy density, reliable, widely available Non-renewable, significant environmental impact (CO2 emissions)
Biofuels Renewable, lower net CO2 emissions Can compete with food production, deforestation concerns
Hydroelectric Power Renewable, reliable, large-scale production Environmental impact on ecosystems, requires suitable locations
Wave and Tidal Power Renewable, reliable in coastal areas High initial costs, limited suitable locations

Concept of Efficiency

Efficiency measures how effectively energy is transferred from one form to another.

Key Equations:

  • \[ \text{Efficiency (%)} = \left( \frac{\text{Useful energy output}}{\text{Total energy output}} \right) \times 100 \]
  • \[ \text{Efficiency (%)} = \left( \frac{\text{Useful power output}}{\text{Total power output}} \right) \times 100 \]

Energy from the Sun

Radiation from the Sun is the primary source of energy for most energy resources, except:

  • Geothermal – Energy from Earth's internal heat.
  • Nuclear – Energy from nuclear fission reactions.
  • Tidal – Energy from gravitational interactions between Earth and the Moon.

The Sun's energy is produced by nuclear fusion, where hydrogen nuclei fuse to form helium, releasing immense energy.

Future of Nuclear Fusion

Research is underway to develop nuclear fusion as a large-scale, sustainable source of electrical energy, with minimal environmental impact.

1.7.4 Power

Definition of Power

Power is defined as the rate at which work is done or the rate at which energy is transferred.

Key Equations

  • \[ P = \frac{W}{t} \] Where:
    • \(P\) = Power (\(\text{W}\))
    • \(W\) = Work done (\(\text{J}\))
    • \(t\) = Time taken (\(\text{s}\))
  • \[ P = \frac{\Delta E}{t} \] Where:
    • \(P\) = Power (\(\text{W}\))
    • \(\Delta E\) = Energy transferred (\(\text{J}\))
    • \(t\) = Time taken (\(\text{s}\))

Example Problems

  1. Problem 1: A motor does 5000 J of work in 20 s. Calculate its power.

    Solution:

    \[ P = \frac{W}{t} = \frac{5000}{20} = 250 \text{ W} \]
  2. Problem 2: An appliance transfers 3000 J of energy in 15 s. What is its power rating?

    Solution:

    \[ P = \frac{\Delta E}{t} = \frac{3000}{15} = 200 \text{ W} \]
  3. Problem 3: A lift raises a mass of 150 kg to a height of 10 m in 30 s. Calculate the power developed by the lift (take \( g = 9.8 \, \text{m/s}^2 \)).

    Solution:

    Work done, \( W = mgh = 150 \times 9.8 \times 10 = 14700 \, \text{J} \) \[ P = \frac{W}{t} = \frac{14700}{30} = 490 \text{ W} \]
  4. Problem 4: A person pushes a box with a force of 200 N for 5 m in 10 s. Calculate the power.

    Solution:

    Work done, \( W = Fd = 200 \times 5 = 1000 \, \text{J} \) \[ P = \frac{W}{t} = \frac{1000}{10} = 100 \text{ W} \]
  5. Problem 5: A bulb consumes 1800 J of energy in 1 minute. Calculate its power.

    Solution:

    Time, \( t = 60 \, \text{s} \) \[ P = \frac{\Delta E}{t} = \frac{1800}{60} = 30 \text{ W} \]
  6. Problem 6: A machine transfers 6000 J of energy in 12 s. What is its power output?

    Solution:

    \[ P = \frac{\Delta E}{t} = \frac{6000}{12} = 500 \text{ W} \]
  7. Problem 7: A motor lifts a 250 kg object vertically by 8 m in 40 s. Find the power output of the motor (take \( g = 9.8 \, \text{m/s}^2 \)).

    Solution:

    Work done, \( W = mgh = 250 \times 9.8 \times 8 = 19600 \, \text{J} \) \[ P = \frac{W}{t} = \frac{19600}{40} = 490 \text{ W} \]

1.8 Pressure

Definition of Pressure

Pressure is defined as the force exerted per unit area.

Key Equation

  • \[ p = \frac{F}{A} \] Where:
    • \(p\) = Pressure (\(\text{Pa}\) or \(\text{N/m}^2\))
    • \(F\) = Force (\(\text{N}\))
    • \(A\) = Area (\(\text{m}^2\))

Pressure in Everyday Contexts

Pressure varies with force and area:

  • Increasing the force increases the pressure if the area remains constant.
  • Increasing the area decreases the pressure if the force remains constant.

Examples:

  • High-heeled shoes exert more pressure on the ground compared to flat shoes because of the smaller contact area.
  • A sharp knife cuts better because it concentrates force over a small area, increasing pressure.

Pressure Beneath the Surface of a Liquid

The pressure beneath the surface of a liquid increases with depth and the density of the liquid.

Key Equation

  • \[ \Delta p = \rho g \Delta h \] Where:
    • \(\Delta p\) = Change in pressure (\(\text{Pa}\))
    • \(\rho\) = Density of the liquid (\(\text{kg/m}^3\))
    • \(g\) = Gravitational field strength (\(9.8 \, \text{m/s}^2\))
    • \(\Delta h\) = Change in depth (\(\text{m}\))

Qualitative Explanation

As depth increases, more liquid is above a point, which results in a higher pressure due to the weight of the liquid. Similarly, a denser liquid results in a higher pressure for the same depth.

Example Problems

  1. Problem 1: A box exerts a force of 300 N on the ground over an area of \(2 \, \text{m}^2\). Calculate the pressure.

    Solution:

    \[ p = \frac{F}{A} = \frac{300}{2} = 150 \, \text{Pa} \]
  2. Problem 2: Calculate the pressure change at a depth of 5 m beneath the surface of water (\(\rho = 1000 \, \text{kg/m}^3\)).

    Solution:

    \[ \Delta p = \rho g \Delta h = 1000 \times 9.8 \times 5 = 49000 \, \text{Pa} \]

2.1 Kinetic Particle Model of Matter

2.1.1 States of matter

2.1.1 Properties of Solids, Liquids, and Gases

In this section, we will explore the distinguishing properties of solids, liquids, and gases, and the terms used for changes in state between these phases.

Distinguishing Properties:

  • Solids: Have a fixed shape and volume. Particles are closely packed and only vibrate in place.
  • Liquids: Have a fixed volume but no fixed shape. Particles are close together but can move past each other.
  • Gases: Have no fixed shape or volume. Particles are far apart and move freely in all directions.

Changes in State:

The terms used for the changes in state between solids, liquids, and gases are as follows:

  • Solid to Liquid: Melting
  • Liquid to Solid: Freezing
  • Liquid to Gas: Evaporation
  • Gas to Liquid: Condensation

Change of state

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2.1.2 Particle model

2.1.2 Particle Structure and Behavior of Solids, Liquids, and Gases

This section explores the particle structure of solids, liquids, and gases in terms of the arrangement, separation, and motion of particles. It also explains how temperature, pressure, and motion of microscopic particles influence the properties of matter.

Particle Structure:

The particle structure of solids, liquids, and gases can be understood through the arrangement, separation, and motion of particles:

  • Solids: Particles are closely packed in a fixed arrangement and only vibrate in place, maintaining a fixed shape and volume.
  • Liquids: Particles are close together but can move past one another, allowing liquids to flow and take the shape of their container while maintaining a fixed volume.
  • Gases: Particles are far apart and move freely at high speeds, allowing gases to expand and fill the shape and volume of their container.
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Temperature and Particle Motion:

The temperature of a substance is directly related to the kinetic energy of its particles. As temperature increases, the motion of the particles becomes more vigorous. At the lowest possible temperature, known as absolute zero (-273°C), the particles have the least possible kinetic energy, and their motion ceases.

Pressure and Gas Particles:

The pressure exerted by a gas is the result of the motion of its particles and the collisions of these particles with the surfaces of their container. The more frequent and forceful the collisions, the higher the pressure.

Brownian Motion:

Brownian motion refers to the random motion of microscopic particles suspended in a fluid (liquid or gas). This motion occurs due to random collisions between the suspended particles and the molecules of the surrounding gas or liquid. It provides evidence for the kinetic particle model of matter.

Effect of Particle Motion on Properties:

The forces and distances between particles, as well as their motion, affect the properties of solids, liquids, and gases. For example:

  • In solids, the tightly packed particles result in high density and a rigid structure.
  • In liquids, the particles can move past one another, allowing them to flow while maintaining a fixed volume.
  • In gases, the far-apart particles result in low density and the ability to expand and fill any container.

Pressure and Forces of Gas Particles:

The pressure of a gas is a result of the forces exerted by particles colliding with the surfaces of the container. This creates a force per unit area, which is the pressure. As the number of collisions increases (due to higher temperature or more gas particles), the pressure increases.

Microscopic Particles and Collisions:

The motion of microscopic particles can be influenced by collisions with fast-moving molecules (such as in the case of gases). These collisions transfer energy to the microscopic particles, causing them to move.

2.1.3 Gases and the absolute scale of temperature

2.1.3 Boyle's and Charles' Law

This section explores the relationship between pressure and volume, and temperature and volume for gases, represented by Boyle's Law and Charles' Law respectively.

Boyle's Law: Pressure vs Volume (at constant temperature)

Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional. As the volume of the gas increases, its pressure decreases, and vice versa.

Graph 1: Boyle's Law - Pressure vs Volume

Charles' Law: Temperature vs Volume (at constant pressure)

Charles' Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to the temperature (in Kelvin). As the temperature increases, the volume of the gas also increases.

Graph 2: Charles' Law - Temperature vs Volume

Boyle's Law (Mathematical Representation):

\( pV = \text{constant} \)

Charles' Law (Mathematical Representation):

\( \frac{V}{T} = \text{constant} \)

IGCSE-Type Problems

Problem 1: Boyle's Law

Question: A gas occupies a volume of \(6.0 \, \text{m}^3\) at a pressure of \(200 \, \text{kPa}\). If the volume of the gas is reduced to \(3.0 \, \text{m}^3\), calculate the new pressure, assuming the temperature is constant.

Solution:

Using Boyle's Law:

\[ p_1 V_1 = p_2 V_2 \]

Substitute the known values:

\[ 200 \, \text{kPa} \times 6.0 \, \text{m}^3 = p_2 \times 3.0 \, \text{m}^3 \]

Solving for \(p_2\):

\[ p_2 = \frac{200 \, \text{kPa} \times 6.0 \, \text{m}^3}{3.0 \, \text{m}^3} = 400 \, \text{kPa} \]

Answer: The new pressure is \(400 \, \text{kPa}\).

Problem 2: Boyle's Law

Question: A gas is compressed from a volume of \(10.0 \, \text{L}\) to \(5.0 \, \text{L}\) at a pressure of \(1.5 \, \text{atm}\). Calculate the final pressure of the gas after the compression.

Solution:

Using Boyle's Law:

\[ p_1 V_1 = p_2 V_2 \]

Substitute the known values:

\[ 1.5 \, \text{atm} \times 10.0 \, \text{L} = p_2 \times 5.0 \, \text{L} \]

Solving for \(p_2\):

\[ p_2 = \frac{1.5 \, \text{atm} \times 10.0 \, \text{L}}{5.0 \, \text{L}} = 3.0 \, \text{atm} \]

Answer: The final pressure is \(3.0 \, \text{atm}\).

Problem 3: Charles' Law

Question: A gas has a volume of \(2.0 \, \text{m}^3\) at a temperature of \(300 \, \text{K}\). If the temperature is increased to \(600 \, \text{K}\), what is the new volume, assuming the pressure remains constant?

Solution:

Using Charles' Law:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Substitute the known values:

\[ \frac{2.0 \, \text{m}^3}{300 \, \text{K}} = \frac{V_2}{600 \, \text{K}} \]

Solving for \(V_2\):

\[ V_2 = \frac{2.0 \, \text{m}^3 \times 600 \, \text{K}}{300 \, \text{K}} = 4.0 \, \text{m}^3 \]

Answer: The new volume is \(4.0 \, \text{m}^3\).

Problem 4: Charles' Law

Question: A gas occupies a volume of \(4.0 \, \text{L}\) at a temperature of \(300 \, \text{K}\). If the temperature is increased to \(600 \, \text{K}\), what is the new volume, assuming the pressure is constant?

Solution:

Using Charles' Law:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

Substitute the known values:

\[ \frac{4.0 \, \text{L}}{300 \, \text{K}} = \frac{V_2}{600 \, \text{K}} \]

Solving for \(V_2\):

\[ V_2 = \frac{4.0 \, \text{L} \times 600 \, \text{K}}{300 \, \text{K}} = 8.0 \, \text{L} \]

Answer: The new volume is \(8.0 \, \text{L}\).

2.2 Thermal properties and temperature

2.2.1 Thermal expansion of solids, liquids and gases

Thermal Expansion of Solids, Liquids, and Gases at Constant Pressure

Thermal expansion refers to the increase in the size of a substance as its temperature increases. When the temperature rises, particles in a material vibrate more vigorously, causing them to move apart, leading to an increase in the material's volume.


Solids: In solids, particles are tightly packed in a fixed arrangement. As temperature increases, the vibrations of the particles become more intense, causing them to push slightly apart. This leads to a small increase in the volume of the solid. However, the expansion is usually limited due to the strong forces between particles.


Liquids: In liquids, the particles are less tightly packed than in solids and can move around each other. As the temperature increases, the kinetic energy of the particles increases, causing them to move further apart. Liquids expand more than solids but still face some constraints due to intermolecular forces.


Gases: In gases, the particles are widely spaced and move freely. When heated, the particles move even faster and push further apart, causing gases to expand the most. The expansion of gases is much more significant than that of solids and liquids at a given temperature rise, as the particles are already far apart and not strongly attracted to each other.


Thermal Expansion of a Metal Block

Thermal Expansion of a Bimetallic Strip

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Everyday Applications and Consequences of Thermal Expansion

Thermal expansion has several practical applications and consequences in everyday life:

  • Railway Tracks: Gaps are left between sections of railway tracks to allow for expansion during hot weather. If this is not done, the tracks can bend or buckle due to excessive thermal expansion.
  • Bridges: Expansion joints in bridges allow them to expand and contract as temperatures change, preventing damage to the structure.
  • Thermometers: Liquids in thermometers expand as temperature rises, allowing the temperature to be read by observing the liquid level in the tube.
  • Engine Parts: Engine components such as pistons are designed to account for thermal expansion, preventing parts from seizing or causing friction when temperatures rise during operation.
  • Electric Cables: Electric cables are made with materials that can expand and contract slightly without breaking or causing short circuits. However, excessive expansion can cause the insulation to crack.

Explanation of Thermal Expansion in Terms of Particle Motion and Arrangement

The relative order of magnitudes of thermal expansion in solids, liquids, and gases can be explained by looking at the arrangement and motion of particles:

  • Solids: In solids, particles are closely packed and only vibrate in place. Because they cannot move freely, the expansion in solids is the smallest.
  • Liquids: In liquids, particles are free to move past each other, allowing for greater expansion than in solids as the temperature increases.
  • Gases: In gases, particles are widely spaced and move freely. The kinetic energy of the particles increases significantly with temperature, causing the greatest expansion of all three states of matter.

2.2.2 Specific heat capacity

Topic 2.2.2: Internal Energy and Specific Heat Capacity

Rise in Temperature and Increase in Internal Energy

When the temperature of an object increases, its internal energy also increases. This is because the increase in temperature results in an increase in the average kinetic energy of the particles within the object. The internal energy of an object is the total energy of all its particles, including both the kinetic energy due to their motion and the potential energy due to the forces between them.

Increase in Temperature and Kinetic Energy of Particles

The temperature of an object is directly related to the average kinetic energy of its particles. When an object is heated, the particles within the object begin to move more rapidly. This increase in the motion of the particles corresponds to an increase in their kinetic energy, which in turn increases the overall internal energy of the object.

Specific Heat Capacity

Specific heat capacity is defined as the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). The equation for specific heat capacity is:

Equation:

\[ c = \frac{\Delta E}{m \Delta \theta} \]

Where:

  • c is the specific heat capacity (J/kg°C or J/kg·K)
  • ΔE is the energy added (J)
  • m is the mass of the substance (kg)
  • Δθ is the change in temperature (°C or K)

This equation allows us to calculate how much energy is required to heat a substance, depending on its mass, the temperature change, and its specific heat capacity.

Experiments to Measure Specific Heat Capacity of Solids and Liquids

Experiment to Measure Specific Heat Capacity of a Solid:

To measure the specific heat capacity of a solid, one common method is to use an electrical heater and a calorimeter. The solid is placed in thermal contact with the heater, and the temperature change is recorded over time. The amount of electrical energy supplied by the heater is calculated, and using the equation for specific heat capacity, the specific heat capacity of the solid can be determined.


Experiment to Measure Specific Heat Capacity of a Liquid:

For liquids, a similar approach is used. A known amount of liquid is placed in a calorimeter, and an electrical heater is used to heat the liquid. The temperature change of the liquid is measured, and the energy supplied by the heater is calculated. By applying the specific heat capacity equation, the specific heat capacity of the liquid can be determined.

Practice Problems: Internal Energy and Specific Heat Capacity

Problem 1: Calculating Energy Required to Heat a Solid

Question: A 0.5 kg piece of metal is heated from 20°C to 80°C. If the specific heat capacity of the metal is \( 900 \, \text{J/kg°C} \), calculate the amount of energy required to heat the metal.

Solution:

Using the formula for specific heat capacity:

\[ \Delta E = m c \Delta \theta \]

Substitute the known values:

\[ \Delta E = 0.5 \times 900 \times (80 - 20) = 0.5 \times 900 \times 60 = 27,000 \, \text{J} \]

Answer: The energy required is 27,000 J.

Problem 2: Specific Heat Capacity of a Liquid

Question: A 1.2 kg sample of water is heated from 15°C to 50°C, and the energy supplied is 42,000 J. Calculate the specific heat capacity of the water.

Solution:

Using the formula for specific heat capacity:

\[ c = \frac{\Delta E}{m \Delta \theta} \]

Substitute the known values:

\[ c = \frac{42,000}{1.2 \times (50 - 15)} = \frac{42,000}{1.2 \times 35} = \frac{42,000}{42} = 1000 \, \text{J/kg°C} \]

Answer: The specific heat capacity of the water is 1000 J/kg°C.

Problem 3: Temperature Change and Energy

Question: If 5,000 J of energy is supplied to a 2.0 kg block of copper, causing its temperature to rise from 20°C to 40°C, calculate the specific heat capacity of copper.

Solution:

Using the formula for specific heat capacity:

\[ c = \frac{\Delta E}{m \Delta \theta} \]

Substitute the known values:

\[ c = \frac{5,000}{2.0 \times (40 - 20)} = \frac{5,000}{2.0 \times 20} = \frac{5,000}{40} = 125 \, \text{J/kg°C} \]

Answer: The specific heat capacity of copper is 125 J/kg°C.

Problem 4: Energy Required to Heat a Liquid

Question: A 3.0 kg sample of oil is heated from 25°C to 75°C. If the specific heat capacity of the oil is 1,800 J/kg°C, calculate the total energy required to heat the oil.

Solution:

Using the formula for specific heat capacity:

\[ \Delta E = m c \Delta \theta \]

Substitute the known values:

\[ \Delta E = 3.0 \times 1,800 \times (75 - 25) = 3.0 \times 1,800 \times 50 = 270,000 \, \text{J} \]

Answer: The energy required is 270,000 J.

Problem 5: Increase in Internal Energy

Question: A 0.25 kg piece of aluminum is heated and its temperature increases from 30°C to 80°C. If the specific heat capacity of aluminum is 900 J/kg°C, calculate the increase in internal energy.

Solution:

Using the formula for specific heat capacity:

\[ \Delta E = m c \Delta \theta \]

Substitute the known values:

\[ \Delta E = 0.25 \times 900 \times (80 - 30) = 0.25 \times 900 \times 50 = 11,250 \, \text{J} \]

Answer: The increase in internal energy is 11,250 J.

2.2.3 Melting, boiling and evaporation

Key Concepts:

  • Melting and Boiling: Melting and boiling involve energy input without a change in temperature. In both cases, energy is used to break the bonds between particles (in solids or liquids), rather than increasing the kinetic energy of the particles. For melting, the energy is used to overcome the forces holding particles in a solid structure. For boiling, energy is used to overcome the intermolecular forces holding particles in the liquid phase.
  • Melting and Boiling Temperatures: At standard atmospheric pressure (1 atm), the melting temperature of water is 0°C, and the boiling temperature is 100°C.
  • Condensation and Solidification:
    • Condensation: When a gas loses energy, it condenses into a liquid. In terms of particles, this occurs when the gas particles lose kinetic energy, move closer together, and form intermolecular bonds to become a liquid.
    • Solidification: When a liquid loses energy, it solidifies. This happens as the particles in the liquid lose kinetic energy, slow down, and form bonds that lock the particles in a fixed structure, transitioning the liquid into a solid.
  • Evaporation: Evaporation occurs when more energetic particles at the surface of a liquid escape into the air. This happens because these particles have enough kinetic energy to overcome the forces of attraction from other liquid molecules, turning into a gas. Evaporation results in the cooling of the remaining liquid, as the particles with the most energy (and heat) escape, lowering the temperature of the liquid.
  • Differences Between Boiling and Evaporation:
    • Boiling: Boiling occurs at a specific temperature (the boiling point) and happens throughout the liquid. It involves the transition from liquid to gas when the vapor pressure of the liquid equals atmospheric pressure. It requires the input of energy to break intermolecular bonds.
    • Evaporation: Evaporation happens at any temperature and occurs only at the surface of the liquid. It doesn't require the liquid to reach its boiling point and leads to cooling of the remaining liquid.
  • Factors Affecting Evaporation:
    • Temperature: Higher temperature increases the energy of the particles, making it easier for them to escape the liquid and evaporate.
    • Surface Area: A larger surface area allows more particles to escape at once, increasing the rate of evaporation.
    • Air Movement: Increased air movement (wind) removes the evaporated particles from the surface, allowing more particles to escape and speeding up evaporation.
  • Cooling due to Evaporation: When a liquid evaporates, the more energetic particles leave, reducing the average kinetic energy of the remaining liquid. This causes the temperature of the liquid to decrease. This is why evaporation has a cooling effect. For example, sweat evaporates from the skin, helping to cool the body.

Examples of Applications and Phenomena:

  • Evaporation is used in cooling mechanisms, such as sweating and the use of cooling towers in industrial processes.
  • Melting and boiling temperatures are essential in understanding phase changes in materials, such as the use of water in heating systems (boiling point) or refrigeration (freezing point).
  • The principle of condensation is utilized in processes like distillation, where water vapor condenses to form liquid water again.

2.3 Transfer of thermal energy

There are three main modes for the transfer of thermal energy: Conduction, convection and radiation.


2.3.1 Conduction

Topic 2.3.1: Thermal Conduction and Insulation

Key Concepts:

  • Experiments to Demonstrate Thermal Conductors and Insulators:
    • Good Thermal Conductors: Materials such as metals (e.g., copper and aluminum) can be tested for thermal conductivity by heating one end of a metal rod and observing how quickly the other end becomes warm. The faster the heat is transferred, the better the thermal conductor. A simple experiment involves heating a metal rod and measuring the time it takes for the heat to reach a thermometer at the other end.
    • Bad Thermal Conductors (Thermal Insulators): Materials such as rubber, wood, and plastics can be tested by using the same setup. These materials will allow heat to transfer much more slowly, showing that they are poor conductors of heat. A practical experiment can involve wrapping the same metal rod in different insulating materials (e.g., wool, plastic) and comparing the rate of heat transfer.
  • Thermal Conduction in Solids:
    • Atomic or Molecular Lattice Vibrations: In solids, thermal conduction occurs when atoms or molecules vibrate due to thermal energy. These vibrations are passed from one particle to another, transferring heat through the material. The rate of vibration increases with temperature, leading to a more efficient transfer of heat in good conductors like metals.
    • Movement of Free Electrons in Metals: In metallic conductors, such as copper or aluminum, free (delocalized) electrons play a significant role in thermal conduction. When one part of the metal is heated, these free electrons gain kinetic energy and move faster. They then collide with other electrons and ions, transferring thermal energy across the material rapidly. This process makes metals excellent thermal conductors.
  • Why Thermal Conduction is Bad in Gases and Liquids:
    • In Gases: In gases, the particles are much farther apart compared to solids and liquids. The distance between particles is so large that the particles are less likely to collide with each other, which reduces the transfer of thermal energy. This makes gases poor conductors of heat.
    • In Liquids: Liquids have particles that are closer together than gases, but they are still far enough apart to limit effective heat transfer. The less frequent collisions between particles in liquids make them poorer conductors than solids, though they can conduct heat better than gases.
  • Thermal Conductivity of Solids:
    • Many solids conduct heat better than thermal insulators, but they do not do so as efficiently as good thermal conductors. For example, materials like wood or glass are not as effective as metals in conducting heat but still transfer thermal energy at a moderate rate. The efficiency of heat transfer in these materials depends on their atomic or molecular structure and the ability of their particles to interact and pass on energy.

Examples of Applications and Phenomena:

  • Good Thermal Conductors: Metals like copper and aluminum are used in cookware, heat exchangers, and electrical wiring because they transfer heat efficiently.
  • Bad Thermal Conductors (Thermal Insulators): Materials like fiberglass, foam, and rubber are used in insulating clothing, building materials, and thermos bottles to reduce heat transfer.
  • In Gases and Liquids: The low thermal conductivity of gases and liquids is why we use vacuum flasks to store hot beverages (to minimize heat loss) and why air is used as an insulator in double-glazed windows.

2.3.2 Convection

Topic 2.3.2: Convection in Liquids and Gases

Key Concepts:

  • Convection as a Method of Thermal Energy Transfer:
    • Convection is a method of heat transfer that occurs in liquids and gases. It involves the movement of warmer regions of a fluid (liquid or gas) to cooler areas, allowing heat to be transferred through the fluid. The process is driven by differences in temperature and, consequently, differences in density.
    • In convection, the warmer, less dense fluid rises, and the cooler, denser fluid sinks, creating a circulating flow that transfers thermal energy. This process is responsible for many everyday phenomena, such as the rising of hot air in the atmosphere and the circulation of water in a heated pot.
  • Convection in Liquids and Gases:
    • In Liquids: When a liquid is heated, the particles at the bottom gain energy, move faster, and spread out, causing the liquid to become less dense. The warmer, less dense liquid rises while the cooler, denser liquid sinks to take its place, creating a convection current.
    • In Gases: The same process occurs in gases, although the behavior of gases is less noticeable due to their lower density and higher mobility. When air is heated, it expands, becomes less dense, and rises, while cooler, denser air moves downward to replace it, forming a convection current.
  • Density Changes in Convection:
    • As a substance is heated, its particles gain energy and move faster, causing them to spread out and decrease in density. This reduced density causes the warmer fluid to rise. Conversely, as the fluid cools, its particles slow down, come closer together, and increase in density, causing the cooler fluid to sink.
  • Experiments to Illustrate Convection:
    • Simple Convection Experiment (in Water):
      1. Fill a clear glass container with water.
      2. Add a few drops of food coloring near the bottom of the container and heat the water at the bottom (e.g., using a small heater or by placing the container over a light source).
      3. Observe the movement of the food coloring as the water at the bottom warms up and rises, while the cooler water moves downward. This demonstrates convection currents in water.
    • Convection in Air (Using a Candle and Smoke):
      1. Place a lit candle on a table and hold a piece of paper above it.
      2. Observe how the smoke rises from the candle flame and spreads out as it moves through the air. This demonstrates convection in the air as warm air rises and cooler air moves down to replace it.

Examples of Applications and Phenomena:

  • Heating of Water: In a heated pot of water, convection currents allow the heat to spread throughout the water. The water at the bottom of the pot, which is heated by the heat source, becomes less dense and rises to the surface, while the cooler water moves down to replace it.
  • Atmospheric Convection: In the atmosphere, warm air rises and cool air sinks, creating wind currents and influencing weather patterns. This is why we experience breezes and storms.
  • Ocean Currents: The movement of warm water at the equator and cold water at the poles forms convection currents in the oceans, which helps regulate global climate and weather patterns.

2.3.3 Radiation

Topic 2.3.4: Thermal Radiation

Key Concepts:

  • Thermal Radiation and Infrared Radiation:
    • Thermal radiation refers to infrared radiation that all objects emit, regardless of their temperature. This radiation carries energy away from the object.
    • Unlike other forms of heat transfer such as conduction or convection, thermal radiation does not require a medium (like air or water) to travel through. It can even transfer energy through a vacuum, such as from the Sun to Earth.
  • Factors Affecting Radiation Emission, Absorption, and Reflection:
    • Surface Colour:
      • Dark or black surfaces are better emitters and absorbers of infrared radiation compared to light or white surfaces. They absorb more energy and also radiate it away more efficiently.
      • Light or white surfaces reflect more infrared radiation and are poor absorbers and emitters.
    • Surface Texture:
      • Dull, matte surfaces are better at emitting and absorbing infrared radiation compared to shiny surfaces. Shiny, reflective surfaces tend to reflect infrared radiation away, making them poor absorbers and emitters.
  • Energy Balance and Constant Temperature:
    • For an object to maintain a constant temperature, the rate at which it receives energy (through radiation) must equal the rate at which it loses energy (through radiation). If the rates are not equal, the object will either heat up or cool down.
  • Effects of Imbalance in Energy Transfer:
    • If an object receives more energy than it emits, it will heat up (the temperature will rise).
    • If an object emits more energy than it receives, it will cool down (the temperature will fall).
  • Experiments to Distinguish Good and Bad Emitters of Infrared Radiation:
    • Experiment 1: Surface Colour and Emission
      1. Place objects with different surface colours (e.g., a black object and a white object) under an infrared detector.
      2. Observe the infrared radiation emitted from the surfaces. The black object will emit more infrared radiation (be a better emitter) than the white object.
    • Experiment 2: Surface Texture and Emission
      1. Use objects with different surface textures, such as a shiny metal object and a dull metal object.
      2. Place these objects near a heat source and observe with an infrared thermometer. The dull object will emit more infrared radiation compared to the shiny object.
  • Experiments to Distinguish Good and Bad Absorbers of Infrared Radiation:
    • Experiment 1: Absorption and Surface Colour
      1. Place a black object and a white object near an infrared radiation source.
      2. Use a thermometer to measure the temperature change in both objects over time. The black object will absorb more infrared radiation and increase in temperature faster compared to the white object.
    • Experiment 2: Absorption and Surface Texture
      1. Place a dull object and a shiny object under the same conditions near a heat source.
      2. Measure the temperature change. The dull object will absorb more radiation and heat up more quickly than the shiny object.
  • Factors Affecting the Rate of Emission of Radiation:
    • The rate of emission of radiation depends on the object's surface temperature. As the surface temperature increases, the amount of infrared radiation emitted increases.
    • The rate of emission also increases with the surface area of the object. Larger objects with greater surface area emit more infrared radiation than smaller objects.

2.3.4 Consequences of thermal energy transfer

Topic 2.3.4: Applications and Consequences of Thermal Energy Transfer

Basic Everyday Applications and Consequences of Thermal Energy Transfer

  • Heating Objects (e.g., Kitchen Pans):
    • When a kitchen pan is placed on a stove, thermal energy is transferred to the pan primarily through conduction from the hot surface of the stove to the pan's material (often metal). The metal particles in the pan vibrate more rapidly as they gain energy, which is then passed on to the rest of the pan and the food inside.
    • The pan becomes hot as the thermal energy is conducted from the stove to the pan, and this heat is transferred into the food by further conduction, heating it up.
  • Heating a Room by Convection:
    • In a room, heating is often done using a radiator or heating system. As the air near the radiator heats up, it becomes less dense and rises, creating an upward current of warm air. This process is known as convection.
    • As the warm air rises, cooler air moves in to replace it, and the cycle repeats. This process creates a convection current that distributes heat throughout the room, warming the space efficiently.

Complex Applications and Consequences of Thermal Energy Transfer

  • Burning Wood or Coal in a Fire:
    • When wood or coal burns in a fire, multiple forms of heat transfer occur:
      • Conduction: Heat is conducted from the burning material to the surrounding air, and from the fire to any objects in contact with the flames.
      • Convection: The heated air and gases around the fire become less dense and rise, while cooler air moves in to replace it, forming convection currents. These currents spread heat throughout the room or the outdoor area.
      • Radiation: The fire also emits infrared radiation, which can be felt as heat even without direct contact. This radiation transfers energy to nearby objects and people, warming them up.
  • A Radiator in a Car:
    • In a car, the radiator plays a crucial role in cooling the engine by transferring excess heat away from the engine to the air. The process involves:
      • Conduction: The engine heat is transferred to the coolant inside the radiator through the metal parts of the engine and the radiator itself.
      • Convection: The heated coolant flows through the radiator, where air is blown over it (often by a fan) to help carry away the heat. The warm coolant loses heat to the air, and the cooled liquid returns to the engine to absorb more heat.
      • Radiation: The radiator itself also emits heat in the form of infrared radiation to its surroundings, although this is less significant than conduction and convection in this application.

3.1 General Properties of waves

Key Concepts:

Transverse and Longitudinal Waves

Practical Activities

Seismic Waves

Interactive Simulations

Extension: Superposition of Waves

Learn about the superposition of waves and how constructive and destructive interference occur. Explore what happens when two waves meet in a spring or slinky.

Research Resources

3.2 Reflection, Refraction, and Diffraction

Key Concepts:

Wave Phenomena: Reflection, Refraction, and Diffraction

Using a Ripple Tank to Demonstrate Wave Phenomena

In a ripple tank, various wave phenomena can be demonstrated:

Effects of Wavelength and Gap Size on Diffraction

Interactive Simulations

Practical Activities

Assessment and Extension

Set qualitative questions for practice, including sketching diffraction patterns. Learners can also investigate the diffraction of white light and its relation to rainbows, along with the occurrence of dark patches due to diffraction.

For advanced learners: Explain how the diffraction of light creates rainbows and why certain areas appear dark due to constructive and destructive interference.

Problem 1: Wave Speed Calculation

Question: A wave has a frequency of 20 Hz and a wavelength of 5 meters. What is the speed of the wave?

Solution:

We can use the wave equation:

\[ v = f \lambda \]

Where:

Substitute the given values:

\[ v = 20 \, \text{Hz} \times 5 \, \text{m} \]

\[ v = 100 \, \text{m/s} \]

Thus, the wave speed is 100 m/s.

Problem 2: Wavelength Calculation for Sound Wave

Question: A sound wave travels at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is the wavelength of the sound wave?

Solution:

We can use the same wave equation:

\[ v = f \lambda \]

Rearranging to solve for \(\lambda\), we get:

\[ \lambda = \frac{v}{f} \]

Substitute the given values:

\[ \lambda = \frac{340 \, \text{m/s}}{170 \, \text{Hz}} \]

\[ \lambda = 2 \, \text{m} \]

Thus, the wavelength of the sound wave is 2 meters.

Problem 3: Frequency of a Wave

Question: A wave travels with a speed of 12 m/s and has a wavelength of 3 meters. What is the frequency of the wave?

Solution:

Using the wave equation:

\[ v = f \lambda \]

Rearrange the equation to solve for frequency \(f\):

\[ f = \frac{v}{\lambda} \]

Substitute the given values:

\[ f = \frac{12 \, \text{m/s}}{3 \, \text{m}} \]

\[ f = 4 \, \text{Hz} \]

Thus, the frequency of the wave is 4 Hz.

Problem 4: Wave Energy

Question: A wave has an amplitude of 0.5 meters. How does the energy of the wave change if the amplitude is doubled?

Solution:

The energy of a wave is proportional to the square of its amplitude. The formula for wave energy is:

\[ E \propto A^2 \]

So if the amplitude is doubled, the energy will increase by a factor of 4:

\[ E_{\text{new}} = 4 \times E_{\text{original}} \]

Thus, doubling the amplitude will increase the wave's energy by a factor of 4.

Problem 5: Comparing Transverse and Longitudinal Waves

Question: A wave travels along a rope and causes the rope to move up and down. Is this wave a transverse or longitudinal wave?

Solution:

The movement of the rope is perpendicular to the direction of wave propagation, which is characteristic of a transverse wave.

In a transverse wave, the direction of vibration is at right angles to the direction of propagation, as seen in this example.

3.2 Light

3.2.1 Reflection of light

Introduction to Reflection

Reflection is the change in direction of a wave at a boundary between two different media, so that the wave returns into the original medium. In the case of light waves, this occurs when light strikes a reflective surface, such as a plane mirror.

Key Terms in Reflection

  • Normal: A line perpendicular to the surface at the point of incidence, used as a reference to measure angles.
  • Angle of Incidence: The angle between the incident ray and the normal to the surface at the point of incidence.
  • Angle of Reflection: The angle between the reflected ray and the normal to the surface at the point of reflection.

Law of Reflection

The law of reflection states that the angle of incidence is equal to the angle of reflection. This can be written as:

\[ \theta_i = \theta_r \]

Where:

  • \(\theta_i\): Angle of incidence
  • \(\theta_r\): Angle of reflection
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Formation of an Optical Image by a Plane Mirror

When light rays reflect off a plane mirror, an optical image is formed. The characteristics of this image are:

  • The image is virtual, meaning it cannot be projected onto a screen.
  • The image is the same size as the object.
  • The image is the same distance from the mirror as the object.
  • The image is laterally inverted (left and right are reversed).

Simple Experiment: Investigating the Law of Reflection

To investigate the law of reflection, learners can use the following steps:

  1. Place a plane mirror on a flat surface.
  2. Use a protractor to measure the angle of incidence as a light ray strikes the mirror.
  3. Measure the angle of reflection as the light ray bounces off the mirror.
  4. Compare the angle of incidence and the angle of reflection. They should be equal.

Image Characteristics: Practical Application

Ask learners to consider their reflection in a plane mirror. They can perform the following activities:

  • Stand in front of the mirror and raise their right arm. They will see the left arm of their reflection raised.
  • Notice that the image is the same size and distance as their actual self but appears flipped horizontally (lateral inversion).

Learn about real and virtual images: A virtual image cannot be projected onto a screen, while a real image can. A plane mirror always forms a virtual image.

Uses of Reflection

Reflection is useful in many applications, such as:

  • Periscope: A device that uses mirrors to allow observation from a hidden or protected position.
  • Pepper's Ghost: A technique used in theater and entertainment to create ghostly illusions using reflected light.
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Hands-on Activity: Building a Periscope

Students can create a simple periscope using mirrors and cardboard to observe the law of reflection in action. This will reinforce the concept of how light reflects off mirrors to provide images from different viewpoints.

Extended Assessment

To further investigate the law of reflection, learners can use the following assessment:

  • Use a protractor to measure angles of incidence and reflection for different light rays.
  • Draw ray diagrams to show the behavior of light reflecting off a plane mirror.
  • Use optical pins to determine the position and characteristics of an image formed by a plane mirror.

3.2.2 Refraction of light

Introduction to Refraction

Refraction is the change in direction of light as it passes from one transparent material into another, caused by a change in the speed of light. The bending of light depends on the angle at which it enters the new medium and the refractive indices of the materials involved.

Key Terms in Refraction

  • Normal: A line perpendicular to the surface at the point of incidence used to measure angles.
  • Angle of Incidence (i): The angle between the incident ray and the normal to the surface.
  • Angle of Refraction (r): The angle between the refracted ray and the normal to the surface.
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Experiment to Demonstrate Refraction

To demonstrate refraction, learners can use a ray box and a transparent block (e.g., Perspex or glass). Here's a simple experiment:

  1. Shine a light ray at an angle onto a transparent block placed on a piece of paper.
  2. Use a protractor to measure the angle of incidence and the angle of refraction.
  3. Trace the incident and refracted rays on the paper and measure the angles formed with the normal line.

From this experiment, learners will observe how the light ray bends when it passes from air into a different material (e.g., glass, water). The light slows down as it enters a denser medium and bends towards the normal.

Critical Angle and Total Internal Reflection

The critical angle is the angle of incidence above which total internal reflection occurs. This happens when light cannot pass out of a denser medium into a less dense one, but instead reflects entirely within the denser medium.

The equation for the critical angle (c) is:

\[ n = \frac{1}{\sin c} \]

Where:

  • n: Refractive index of the medium
  • c: Critical angle
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Refractive Index

The refractive index (n) of a material is the ratio of the speed of light in a vacuum (c) to the speed of light in the material (v). It can also be calculated using the equation:

\[ n = \frac{\sin i}{\sin r} \]

Where:

  • i: Angle of incidence
  • r: Angle of refraction

The refractive index is dimensionless (no units) and varies depending on the material.

Applications of Refraction

Refraction is used in various applications, including:

  • Optical Fibres: Optical fibres use the principle of total internal reflection to transmit light signals, especially in telecommunications.
  • Magnifying Glasses: Convex lenses use refraction to magnify objects by bending light rays to a focal point.
  • Mirages: Refraction in air layers of different temperatures causes the illusion of water on the road on hot days.
Descriptive Alt Text

Problem 1: Calculate the Critical Angle

A ray of light travels from glass (refractive index \( n = 1.5 \)) into air (\( n = 1.0 \)). Calculate the critical angle for total internal reflection at the glass-air boundary.

Solution:

The formula for the critical angle is:

\[ \sin c = \frac{n_2}{n_1} \]

Here, \( n_1 = 1.5 \) (glass) and \( n_2 = 1.0 \) (air).

Substitute the values:

\[ \sin c = \frac{1.0}{1.5} = 0.6667 \]

Take the inverse sine to find \( c \):

\[ c = \sin^{-1}(0.6667) \approx 41.8^\circ \]

Answer: The critical angle is approximately \( 41.8^\circ \).

Problem 2: Find the Refractive Index

A light ray passes from air into water. The angle of incidence is \( 45^\circ \), and the angle of refraction is \( 30^\circ \). Calculate the refractive index of water.

Solution:

The refractive index can be calculated using Snell's law:

\[ n = \frac{\sin i}{\sin r} \]

Here, \( i = 45^\circ \) and \( r = 30^\circ \).

Substitute the values:

\[ n = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{0.7071}{0.5} = 1.414 \]

Answer: The refractive index of water is approximately \( 1.414 \).

Problem 3: Speed of Light in a Medium

The refractive index of a transparent material is \( 1.6 \). Calculate the speed of light in the material. The speed of light in a vacuum is \( 3.0 \times 10^8 \, \text{m/s} \).

Solution:

The speed of light in a medium is given by:

\[ v = \frac{c}{n} \]

Here, \( c = 3.0 \times 10^8 \, \text{m/s} \) and \( n = 1.6 \).

Substitute the values:

\[ v = \frac{3.0 \times 10^8}{1.6} = 1.875 \times 10^8 \, \text{m/s} \]

Answer: The speed of light in the material is \( 1.875 \times 10^8 \, \text{m/s} \).

Problem 4: Critical Angle for Water

The refractive index of water is \( 1.33 \). Calculate the critical angle for a light ray moving from water to air.

Solution:

Use the formula for the critical angle:

\[ \sin c = \frac{n_2}{n_1} \]

Here, \( n_1 = 1.33 \) (water) and \( n_2 = 1.0 \) (air).

Substitute the values:

\[ \sin c = \frac{1.0}{1.33} \approx 0.7519 \]

Take the inverse sine to find \( c \):

\[ c = \sin^{-1}(0.7519) \approx 48.8^\circ \]

Answer: The critical angle is approximately \( 48.8^\circ \).

Problem 5: Total Internal Reflection

A glass prism with a refractive index of \( 1.5 \) is surrounded by air. Will total internal reflection occur if the angle of incidence is \( 45^\circ \)?

Solution:

First, calculate the critical angle for the glass-air interface:

\[ \sin c = \frac{n_2}{n_1} = \frac{1.0}{1.5} = 0.6667 \]

\[ c = \sin^{-1}(0.6667) \approx 41.8^\circ \]

Since the angle of incidence (\( 45^\circ \)) is greater than the critical angle (\( 41.8^\circ \)), total internal reflection will occur.

Answer: Yes, total internal reflection will occur.

3.2.3 Thin lens

Lenses

Action of Thin Converging and Diverging Lenses

  • Converging Lens (Convex): Causes parallel rays of light to converge at the principal focus.
  • Diverging Lens (Concave): Spreads parallel rays outward, as if they originated from a point (virtual focus).

Key Terms

  • Focal Length (\( f \)): The distance between the lens center and the principal focus.
  • Principal Axis: A straight line passing through the optical center and the focal points of the lens.
  • Principal Focus: The point where rays parallel to the principal axis converge or appear to diverge.

Ray Diagrams

Steps to draw ray diagrams for a converging lens:

  1. Draw a ray parallel to the principal axis. After passing through the lens, it passes through the principal focus (\( F \)).
  2. Draw a ray passing through the optical center of the lens. It remains undeviated.
  3. Mark the point where the two rays intersect to locate the image.

The nature of the image depends on the object's position relative to the focal point:

  • Beyond \( 2f \): Image is real, inverted, and diminished.
  • At \( 2f \): Image is real, inverted, and same size.
  • Between \( f \) and \( 2f \): Image is real, inverted, and magnified.
  • At \( f \): No image is formed.
  • Within \( f \): Image is virtual, upright, and magnified.

For a diverging lens: The rays appear to diverge from the virtual principal focus. The image is always virtual, upright, and diminished.

Uses of Lenses

  • Magnifying Glass: Uses a converging lens to produce an enlarged virtual image.
  • Corrective Lenses:
    • Long-sightedness: Corrected using a converging lens to focus light on the retina.
    • Short-sightedness: Corrected using a diverging lens to spread light before it reaches the eye.

IGCSE Practice Questions

  1. A converging lens has a focal length of 15 cm. An object is placed 30 cm from the lens.
    • Calculate the image distance (\( v \)) using the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
    • State the nature of the image (real or virtual, inverted or upright, magnified or diminished).

    Solution:

    • Given: \( f = 15 \, \text{cm}, \, u = 30 \, \text{cm} \)
    • Lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
    • \[ \frac{1}{15} = \frac{1}{v} - \frac{1}{30} \implies \frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} \]
    • \( v = 10 \, \text{cm} \)
    • Nature of the image: The image is real, inverted, and diminished.
  2. A diverging lens forms an image 10 cm from the lens when an object is placed 25 cm away.
    • Determine the focal length (\( f \)) of the lens.
    • State the characteristics of the image.

    Solution:

    • Given: \( u = 25 \, \text{cm}, \, v = -10 \, \text{cm} \) (negative because the image is virtual).
    • Lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
    • \[ \frac{1}{f} = \frac{1}{-10} - \frac{1}{25} = -\frac{5}{50} - \frac{2}{50} = -\frac{7}{50} \]
    • \( f = -7.14 \, \text{cm} \) (negative because it is a diverging lens).
    • Characteristics: The image is virtual, upright, and diminished.
  3. Draw a ray diagram for a converging lens where:
    • The object is placed at a distance greater than twice the focal length (\( 2f \)).
    • Label the principal axis, focal points, and the position of the image.
    • Describe the size, orientation, and type of the image.

    Solution:

    • Ray diagram: The image will be located between \( f \) and \( 2f \), on the opposite side of the lens.
    • Characteristics: The image is real, inverted, and diminished.
  4. Explain how a converging lens is used to correct long-sightedness. Support your explanation with a ray diagram showing the corrected path of light rays.

    Solution:

    • Long-sightedness occurs when light rays focus behind the retina.
    • A converging lens refracts the light rays inward, causing them to focus directly on the retina.
    • Ray diagram: Show parallel rays converging after passing through the lens.
  5. White light passes through a triangular prism and produces a spectrum of colors.
    • Explain why this happens, referencing the concept of dispersion.
    • List the colors in the spectrum in order of increasing frequency.
    • What is the term for light that consists of only one wavelength?

    Solution:

    • Dispersion occurs because different wavelengths of light refract by different amounts when passing through a prism, separating into colors.
    • Order of colors (increasing frequency): Red, Orange, Yellow, Green, Blue, Indigo, Violet.
    • The term for light consisting of one wavelength is "monochromatic."

The lens formula relates the object distance (\( u \)), image distance (\( v \)), and focal length (\( f \)):

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

The magnification (\( M \)) is given by:

\[ M = \frac{\text{Image Height}}{\text{Object Height}} = \frac{v}{u} \]

3.2.4 Dispersion of light

What is Dispersion?

Dispersion is the splitting of white light into its component colors (spectrum) due to varying degrees of refraction. A prism is commonly used to demonstrate dispersion.

The Visible Spectrum

  • Colors: Red, Orange, Yellow, Green, Blue, Indigo, Violet (ROYGBIV).
  • Red: Longest wavelength, lowest frequency.
  • Violet: Shortest wavelength, highest frequency.

Monochromatic Light

Light of a single frequency is called monochromatic light, e.g., laser light.

Fun Experiments

Newton's Disc: Combine primary colors (red, green, and blue) to produce white light using a rotating color wheel.

Prism and Ray Box: Use a prism to split light into its spectrum by shining light through it at an angle.

3.3 Electromagnetic spectrum

Key Properties

  • The electromagnetic spectrum consists of all electromagnetic waves, which share the following properties: \[ \text{1. All electromagnetic waves travel at the same speed in a vacuum, approximately } 3.0 \times 10^8 \, \text{m/s}. \] \[ \text{2. They can propagate in a vacuum (do not require a medium).} \]
  • The speed of electromagnetic waves in air is approximately the same as in a vacuum.

Main Regions of the Electromagnetic Spectrum

The electromagnetic spectrum is ordered by wavelength (\( \lambda \)) and frequency (\( f \)): \[ \boxed { \text{Radio waves } \rightarrow \text{Microwaves } \rightarrow \text{Infrared } \rightarrow \text{Visible light } \rightarrow \text{Ultraviolet } \rightarrow \text{X-rays } \rightarrow \text{Gamma rays}. } \]

  • \( \text{Frequency increases as you move to the right, while wavelength decreases.} \)

Uses of Electromagnetic Waves

  • Radio waves: \[ \text{Applications: Radio and TV broadcasting, radio-frequency identification (RFID), and astronomy.} \]
  • Microwaves: \[ \text{Applications: Satellite communication, mobile phones, and microwave ovens.} \]
  • Infrared: \[ \text{Applications: Remote controls, thermal imaging, optical fibers, and intruder alarms.} \]
  • Visible light: \[ \text{Applications: Vision, photography, and illumination.} \]
  • Ultraviolet: \[ \text{Applications: Sterilization, security marking, and detecting fake banknotes.} \]
  • X-rays: \[ \text{Applications: Medical scanning and security scanners.} \]
  • Gamma rays: \[ \text{Applications: Cancer treatment, food sterilization, and medical equipment sterilization.} \]

Harmful Effects of Electromagnetic Radiation

  • Microwaves: Internal heating of body cells.
  • Infrared: Skin burns.
  • Ultraviolet: Damage to surface cells and eyes, causing skin cancer or cataracts.
  • X-rays and Gamma rays: Ionization causing cell mutation or damage to DNA.

Communication and Electromagnetic Waves

  • Microwaves: \[ \text{Used for satellite communication, mobile phones, and satellite TV.} \]
  • Radio waves: \[ \text{Used for Bluetooth, passing through walls with signal attenuation.} \]
  • Optical fibers (using visible light and infrared): \[ \text{Used for high-speed broadband and cable TV as glass is transparent to these wavelengths.} \]

Digital vs. Analogue Signals

  • Analogue signals vary continuously, while digital signals use discrete values (0 and 1).
  • Advantages of digital signals:
    \( \text{1. Increased data transmission rate.} \)
    \( \text{2. Accurate signal regeneration over long distances, resulting in less signal degradation.} \)
  • Examples: Digital watches, mobile communication, and high-speed internet.

Additional Resources

Explore the scale of the electromagnetic spectrum at: The Scale of the Universe.

3.4 Sound

Production of Sound and Nature of Sound Waves

  1. Sound is produced by vibrating sources. Examples include:

    \(\text{Tuning forks, guitar strings, vocal cords}\).

  2. Longitudinal Nature of Sound Waves

    Sound waves are longitudinal waves consisting of:

    • Compression: Regions of high pressure where particles are closer together.
    • \(\text{Compression: High particle density.}\)

    • Rarefaction: Regions of low pressure where particles are farther apart.
    • \(\text{Rarefaction: Low particle density.}\)

  3. Hearing Range of Humans

    The frequency range audible to humans is:

    \[20 \, \text{Hz} \leq f \leq 20,000 \, \text{Hz}\].

  4. Requirement of Medium

    Sound waves require a medium (solid, liquid, or gas) to propagate. In a vacuum:

    \[\(v_{\text{sound}} = 0\).\]

Speed of Sound

  1. Speed of sound in different media:

    \[\(v_{\text{solids}} > v_{\text{liquids}} > v_{\text{gases}}\].

    For example, in air:

    \[v \approx 330 \, \text{m/s to } 350 \, \text{m/s}\].

  2. Determining the speed of sound using distance and time:

    \(v = \frac{d}{t}\).

  3. Measuring speed of sound using echoes:

    \(v = \frac{2d}{t_{\text{echo}}}\), where \(t_{\text{echo}}\) is the time for the sound to travel to the reflecting surface and back.

Amplitude, Frequency, and Sound Properties

  • Loudness is determined by the amplitude \((A)\):

    \(A \uparrow \implies \text{Louder sound}\).

  • Pitch is determined by the frequency \((f)\):

    \(f \uparrow \implies \text{Higher pitch}\).

  • Oscilloscope Traces:

    • Amplitude corresponds to loudness.
    • Frequency corresponds to pitch.

Ultrasound

  1. Definition: Ultrasound is sound with a frequency:

    \(f > 20,000 \, \text{Hz}\).

  2. Uses of Ultrasound:

    • Non-destructive testing of materials.
    • Medical scanning of soft tissues.
    • Sonar for calculating depth:
    • \(d = \frac{v \cdot t}{2}\), where \(v\) is the speed of sound and \(t\) is the time taken for the sound to return.

Reflection of Sound: Echo

  1. Definition of Echo: An echo is the reflection of sound waves off a surface.

  2. Application of Echo:

    Used in sonar and echolocation. For example:

    \[v = \frac{2d}{t_{\text{echo}}}\].

Key Points

  • Sound waves are longitudinal and require a medium for propagation:
  • \(\text{Solids, liquids, and gases support sound propagation.}\)

  • Speed of sound varies depending on the medium:
  • \[v_{\text{solids}} > v_{\text{liquids}} > v_{\text{gases}}\].

  • Sound properties:
    • Loudness depends on amplitude.
    • Pitch depends on frequency.
  • Ultrasound has practical applications in medical imaging, sonar, and cleaning.

4.1 Simple phenomena of magnetism

Forces Between Magnetic Poles and Materials

Magnetic poles interact through forces of attraction and repulsion:

Induced magnetism: When a magnetic material (e.g., iron) is placed near a magnet, its domains align temporarily, inducing magnetism in it.

Differences between magnetic and non-magnetic materials:

Magnetic Fields

A magnetic field is the region around a magnet where a magnetic pole experiences a force.

Key characteristics:

To plot magnetic field lines:

  1. Use iron filings to visualize the field.
  2. Use a small plotting compass to determine the direction of the field.

Magnetic forces arise due to interactions between magnetic fields. For example:

Temporary and Permanent Magnets

Temporary magnets: Made of soft iron, easy to magnetise, and lose magnetism quickly.

Permanent magnets: Made of steel, harder to magnetise, and retain magnetism for a long time.

Examples:

Magnetic Fields Due to Current

The magnetic field around a current-carrying wire forms concentric circles:

\[ B \propto \frac{I}{r} \]

For solenoids:

Changing the direction of current reverses the direction of the magnetic field.

4.2 Electrical quantities

Electricity is the form of energy that is most widely used.
It is a convenient form of energy, and it can easily be converted to other forms of energy.

The equation for velocity is given by:

4.2.1 Electric charge

Introduction to Electric Charge

  • There are two types of electric charges:
    • Positive charge \((+)\)
    • Negative charge \((-)\)
  • Like charges repel each other:
    • Positive \((+)\) repels Positive \((+)\)
    • Negative \((-)\) repels Negative \((-)\)
  • Opposite charges attract each other:
    • Positive \((+)\) attracts Negative \((-)\)

Experiment: Static Electricity by Friction

Materials: Plastic rod, cloth, small scraps of paper.

Procedure:

  • Rub the plastic rod with a cloth to build up static charge.
  • Bring the rod close to small scraps of paper.

Observation: The rod attracts the paper due to static charge.

Electrostatic Charges by Friction

  • Charging by friction involves the transfer of negative charges (electrons).
  • Positive charges (protons) remain in the nucleus and are not transferred.

Unit of Charge

  • Electric charge is measured in coulombs (\(C\)).

Electric Fields

  • An electric field is a region where an electric charge experiences a force.
  • The direction of the electric field is the direction of the force on a positive charge.

Electric Field Patterns

  • Around a point charge:
    • Field lines radiate outwards from a positive charge and inwards towards a negative charge.
  • Between two oppositely charged plates:
    • Field lines are straight, parallel, and evenly spaced.

4.2.2 Electric current

Definition of Electric Current

  • Electric current is the rate of flow of charge.
  • It is measured in amperes (\(A\)).
  • The equation for current is: \[ I = \frac{Q}{t} \] where:
    • \(I\) = Current in amperes (\(A\))
    • \(Q\) = Charge in coulombs (\(C\))
    • \(t\) = Time in seconds (\(s\))

Conventional Current vs. Electron Flow

  • Conventional current flows from positive to negative.
  • Electron flow (actual flow of electrons) is from negative to positive.

Conductors and Insulators

Definition

  • Conductors: Materials that allow the flow of electric charge (e.g., metals like copper and aluminum).
  • Insulators: Materials that do not allow the flow of electric charge (e.g., plastic, wood).

Experiment: Testing Conductors and Insulators

Materials: Battery, wires, bulb, various materials to test.

Procedure:

  • Connect the materials in a circuit with a battery and bulb.
  • Observe if the bulb lights up.

Observation: If the bulb lights up, the material is a conductor; if not, it is an insulator.

Electrical Conduction in Metals

  • In metals, electrical conduction occurs due to the movement of free electrons.
  • These free electrons move through the metal lattice when a potential difference is applied.

Series and Parallel Circuits

Current in Series Circuits

  • The current is the same at every point in a series circuit.

Current in Parallel Circuits

  • The total current entering a junction equals the total current leaving the junction.

Direct Current (DC) vs. Alternating Current (AC)

  • Direct Current (DC): Current flows in one direction.
  • Alternating Current (AC): Current changes direction periodically.

Practice Questions

  1. What are the two types of electric charges?
  2. Describe the forces between like and unlike charges.
  3. Explain how a plastic rod can become charged by friction.
  4. Define electric current and state its unit.
  5. Distinguish between conductors and insulators with examples.
  6. Describe the current flow in a series and parallel circuit.

4.2.3 Electromotive force and potential difference

Electromotive Force (e.m.f)

  • Definition: The electromotive force (e.m.f) is the electrical work done by a source in moving a unit charge around a complete circuit.
  • It is measured in volts (\(V\)).
  • The equation for electromotive force: \[ E = \frac{W}{Q} \] where:
    • \(E\) = electromotive force in volts (\(V\))
    • \(W\) = work done in joules (\(J\))
    • \(Q\) = charge in coulombs (\(C\))

Potential Difference (p.d)

  • Definition: Potential difference (p.d) is the work done by a unit charge passing through a component.
  • It is also measured in volts (\(V\)).
  • The equation for potential difference: \[ V = \frac{W}{Q} \] where:
    • \(V\) = potential difference in volts (\(V\))
    • \(W\) = work done in joules (\(J\))
    • \(Q\) = charge in coulombs (\(C\))

Using a Voltmeter

  • Voltmeters are used to measure e.m.f and p.d in a circuit.
  • They can be analogue or digital and have different ranges depending on the voltage being measured.
  • To measure potential difference, the voltmeter is connected in parallel across the component.

Series and Parallel Circuits

Potential Difference in Series Circuits

  • The total potential difference across the components in a series circuit is equal to the sum of the individual p.d.s across each component: \[ V_{\text{total}} = V_1 + V_2 + V_3 + \cdots \]

Potential Difference in Parallel Circuits

  • The potential difference across each branch of a parallel circuit is the same and equal to the e.m.f of the source: \[ V_{\text{total}} = V_1 = V_2 = V_3 = \cdots \]

Combined e.m.f of Sources

  • For several sources in series, the combined e.m.f is the sum of individual e.m.f.s: \[ E_{\text{total}} = E_1 + E_2 + E_3 + \cdots \]

Experiment: Investigating e.m.f and p.d

Objective: Measure the e.m.f around a power supply and the p.d across components in series and parallel circuits.

Procedure:

  • Set up circuits with components in series and parallel.
  • Use a voltmeter to measure the e.m.f and p.d at various points.

Observation: In a series circuit, the total p.d is the sum of the p.d.s across individual components. In a parallel circuit, the p.d is the same across all branches.

Extension: Kirchhoff's Second Law

  • Kirchhoff's second law states that the sum of the e.m.f.s in a closed loop is equal to the sum of the potential differences across the components.
  • This law relates to the conservation of energy in a circuit. Some energy may be used to heat the battery due to internal resistance.

Simulations

4.2.4 Resistance

Equation for Resistance

The equation for resistance is:

\[ R = \frac{V}{I} \]
  • \(R\) = resistance in ohms (\(\Omega\))
  • \(V\) = potential difference in volts (V)
  • \(I\) = current in amperes (A)

Experiment to Determine Resistance

To determine the resistance of a resistor:

  • Connect a resistor in a circuit with a voltmeter in parallel and an ammeter in series.
  • Measure the voltage (\(V\)) across the resistor and the current (\(I\)) flowing through it.
  • Use the formula \(R = \frac{V}{I}\) to calculate the resistance.

Factors Affecting Resistance

  • Resistance is directly proportional to the length of the wire: \[ R \propto L \]
  • Resistance is inversely proportional to the cross-sectional area of the wire: \[ R \propto \frac{1}{A} \]

Current–Voltage (I–V) Graphs

  • Resistor of constant resistance: The I–V graph is a straight line through the origin, indicating that resistance remains constant.
  • Filament lamp: The I–V graph is a curve, indicating that resistance increases as the lamp heats up.
  • Diode: The I–V graph shows current flowing in one direction only, with a sharp increase in current once the threshold voltage is reached.

4.3.2 Combined Resistance

Resistors in Series

  • The combined resistance of resistors in series is the sum of their individual resistances: \[ R_{\text{total}} = R_1 + R_2 + R_3 + \cdots \]

Resistors in Parallel

  • The combined resistance of two resistors in parallel is less than the resistance of either resistor alone: \[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Simulation Resources

4.2.5 Electrical energy and electrical power

Electricity is a convenient and versatile form of energy used in various applications. Understanding the concepts of electrical energy and electrical power is crucial in determining the efficiency and cost of using electrical devices.

Electrical Power (\( P \))

Electrical power is the rate at which electrical energy is transferred or converted by an electrical circuit. It is measured in Watts (W), where 1 W = 1 Joule/second.

Equations for Electrical Power

The power (\( P \)) consumed by an electrical device depends on the voltage (\( V \)), current (\( I \)), and resistance (\( R \)) of the circuit. The following equations can be used to calculate power:

  • Using voltage and current: \[ P = VI \]
  • Using current and resistance (Ohm's Law: \( V = IR \)): \[ P = I^2R \]
  • Using voltage and resistance: \[ P = \frac{V^2}{R} \]

Explanation of Equations

  • \( P = VI \): Power is the product of voltage and current.
  • \( P = I^2R \): Power is proportional to the square of the current and the resistance.
  • \( P = \frac{V^2}{R} \): Power is proportional to the square of the voltage and inversely proportional to the resistance.

Electrical Energy (\( E \))

Electrical energy is the total amount of energy transferred by an electrical device over time. It is measured in Joules (J).

Equation for Electrical Energy

The electrical energy (\( E \)) transferred by a device over a time period \( t \) is given by:

\[ E = Pt = VIt \]

  • \( E \) = Electrical energy (in Joules, J)
  • \( P \) = Power (in Watts, W)
  • \( t \) = Time (in seconds, s)
  • \( V \) = Voltage (in Volts, V)
  • \( I \) = Current (in Amperes, A)

The Kilowatt-Hour (kWh)

In practical applications, especially in household and industrial contexts, electrical energy is often measured in kilowatt-hours (kWh), rather than Joules. One kilowatt-hour is the amount of energy consumed when a device with a power rating of 1 kilowatt operates for 1 hour.

Conversion Between kWh and Joules

To convert between kilowatt-hours and Joules, use the following relationships:

  • \( 1 \, \text{kWh} = 1000 \, \text{W} \times 3600 \, \text{s} = 3.6 \times 10^6 \, \text{J} \)
  • \( 1 \, \text{J} = \frac{1}{3.6 \times 10^6} \, \text{kWh} \)

Example Conversions

  • Convert 2 kWh to Joules: \[ 2 \, \text{kWh} = 2 \times 3.6 \times 10^6 \, \text{J} = 7.2 \times 10^6 \, \text{J} \]
  • Convert 5 million Joules to kWh: \[ 5 \times 10^6 \, \text{J} = \frac{5 \times 10^6}{3.6 \times 10^6} \, \text{kWh} \approx 1.39 \, \text{kWh} \]

Cost of Electrical Energy

The cost of electrical energy consumption is typically calculated based on the number of kilowatt-hours used and the rate charged by the electricity provider. The total cost (\( C \)) is given by:

\[ C = \text{Energy (kWh)} \times \text{Rate (Cost per kWh)} \]

Example Problems

Problem 1: Power Calculation

A light bulb operates at 240 V and draws a current of 0.5 A. Calculate the power consumed by the light bulb.

Solution: \[ P = VI = 240 \, \text{V} \times 0.5 \, \text{A} = 120 \, \text{W} \]

Problem 2: Energy Consumption

An electric heater with a power rating of 1500 W operates for 3 hours. Calculate the energy consumed in kilowatt-hours and Joules.

Solution: \[ E = Pt = 1500 \, \text{W} \times 3 \times 3600 \, \text{s} = 16.2 \times 10^6 \, \text{J} \] \[ E = \frac{1500 \, \text{W}}{1000} \times 3 \, \text{h} = 4.5 \, \text{kWh} \]

Problem 3: Cost of Electricity

If the rate of electricity is $0.15 per kWh, calculate the cost of operating the heater in Problem 2.

Solution: \[ C = 4.5 \, \text{kWh} \times 0.15 \, \text{\$/kWh} = 0.675 \, \text{\$} \]

Summary

  • Electrical power is the rate of energy transfer and can be calculated using different formulas depending on the known variables.
  • Electrical energy is the total energy transferred over time, often measured in Joules or kilowatt-hours.
  • One kilowatt-hour is equivalent to \( 3.6 \times 10^6 \) Joules.
  • The cost of electricity depends on the energy consumed and the rate charged per kilowatt-hour.

4.3 Electric circuits

4.3.1 Circuit diagrams and circuit components

Drawing and Interpreting Circuit Diagrams

  • Learn and use standard symbols for components such as cells, batteries, switches, resistors, and lamps.
  • Build circuits by interpreting circuit diagrams containing components like diodes, LEDs, and thermistors.

4.3.2 Series and parallel circuits

Advantages of Parallel Circuits

  • Lamps in parallel maintain the same brightness as the total current is distributed.
  • If one lamp is removed or “blows,” the others continue to work.

4.3.3 Action and use of circuit components

Action of a Potential Divider

A potential divider can be used to obtain a specific voltage from a larger voltage source.

Equation for Two Resistors in a Potential Divider

\[ \frac{R_1}{R_2} = \frac{V_1}{V_2} \]
  • \(R_1\), \(R_2\): resistances of the two resistors
  • \(V_1\), \(V_2\): voltages across the two resistors

Kirchhoff’s Laws

  • Kirchhoff’s voltage law: The sum of the e.m.f.s in a closed loop is equal to the sum of potential differences.

4.4 Electrical safety

4.4.1 Hazards of Electrical Systems

Learning Activities:

4.4.4 and 4.4.5 Fuse Ratings and Safety

Learning Activities:

4.5 Electromagnetic effects

4.5.1 Electromagnetic induction

Definition

Electromagnetic induction is the process by which an electric current is induced in a conductor when it is exposed to a changing magnetic field. This phenomenon was discovered by Michael Faraday in 1831 and is the fundamental principle behind many electrical devices such as generators, transformers, and inductors.

Faraday's Law of Electromagnetic Induction

Faraday's law states that the induced electromotive force (emf) in a closed circuit is proportional to the rate of change of magnetic flux through the circuit.

The mathematical expression for Faraday's law is:

\[ \mathcal{E} = -\frac{d\Phi_B}{dt} \]

  • \(\mathcal{E}\) = Induced electromotive force (emf) in volts (V)
  • \(\Phi_B\) = Magnetic flux through the circuit in Weber (Wb)
  • \(\frac{d\Phi_B}{dt}\) = Rate of change of magnetic flux
  • The negative sign indicates that the induced emf opposes the change in magnetic flux, in accordance with Lenz's Law.

Magnetic Flux (\( \Phi_B \))

Magnetic flux is the measure of the total magnetic field passing through a given area. It is given by:

\[ \Phi_B = B \cdot A \cdot \cos \theta \]

  • \( B \) = Magnetic field strength (in Tesla, T)
  • \( A \) = Area of the surface (in square meters, m²)
  • \( \theta \) = Angle between the magnetic field and the normal to the surface

Lenz's Law

Lenz's law states that the direction of the induced current or emf is such that it opposes the change in magnetic flux that caused it. This law is a consequence of the conservation of energy.

In mathematical form, Lenz's law is already represented in Faraday's law by the negative sign:

\[ \mathcal{E} = -\frac{d\Phi_B}{dt} \]

Factors Affecting Electromagnetic Induction

The magnitude of the induced emf depends on several factors:

  • The rate of change of magnetic flux (\( \frac{d\Phi_B}{dt} \))
  • The number of turns in the coil (\( N \)): For a coil with \( N \) turns, the induced emf is given by: \[ \mathcal{E} = -N \frac{d\Phi_B}{dt} \]
  • The strength of the magnetic field (\( B \))
  • The area of the loop (\( A \)) exposed to the magnetic field
  • The angle (\( \theta \)) between the magnetic field and the loop

Applications of Electromagnetic Induction

Electromagnetic induction is used in various practical applications, including:

  • Generators: Convert mechanical energy into electrical energy by rotating a coil within a magnetic field.
  • Transformers: Transfer electrical energy between two circuits through electromagnetic induction, often to step up or step down voltage.
  • Inductors: Store energy in a magnetic field and oppose changes in current.
  • Electric motors: Operate on the principle of electromagnetic induction and the motor effect to convert electrical energy into mechanical energy.
  • Wireless charging: Transfers energy from a charging pad to a device using electromagnetic induction.

Example Problem

Problem: A circular loop with an area of 0.2 m² is placed in a uniform magnetic field of 0.5 T. The magnetic field is perpendicular to the plane of the loop and decreases uniformly to 0 T in 2 seconds. Calculate the induced emf in the loop.

Solution:

Step 1: Calculate the initial magnetic flux: \[ \Phi_{B_{\text{initial}}} = B \cdot A = 0.5 \, \text{T} \times 0.2 \, \text{m}^2 = 0.1 \, \text{Wb} \]

Step 2: The final magnetic flux is: \[ \Phi_{B_{\text{final}}} = 0 \, \text{Wb} \]

Step 3: Calculate the rate of change of magnetic flux: \[ \frac{d\Phi_B}{dt} = \frac{\Phi_{B_{\text{final}}} - \Phi_{B_{\text{initial}}}}{\Delta t} = \frac{0 - 0.1 \, \text{Wb}}{2 \, \text{s}} = -0.05 \, \text{Wb/s} \]

Step 4: The induced emf is: \[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -(-0.05 \, \text{V}) = 0.05 \, \text{V} \]

Summary

  • Electromagnetic induction is the process of generating an emf by changing the magnetic flux through a conductor.
  • Faraday's law quantifies the induced emf as proportional to the rate of change of magnetic flux.
  • Lenz's law states that the induced emf opposes the change in flux, ensuring energy conservation.
  • Electromagnetic induction has many practical applications, including generators, transformers, and motors.

4.5.2 The a.c. generator

Basic Structure of an AC Generator

  • Components:
    • Coil: A rectangular coil of wire rotates in a magnetic field.
    • Magnet: Provides the magnetic field through which the coil rotates. This can be a permanent magnet or an electromagnet.
    • Slip Rings: Metal rings connected to the ends of the rotating coil. They rotate with the coil and provide continuous contact.
    • Brushes: Stationary conductive contacts that press against the slip rings and allow current to flow from the rotating coil to the external circuit.
  • Principle of Operation: Electromagnetic induction. As the coil rotates in the magnetic field, a changing magnetic flux induces an electromotive force (e.m.f.) in the coil according to Faraday's Law of Electromagnetic Induction.

Faraday's Law of Electromagnetic Induction

According to Faraday's Law, the induced e.m.f. \[ \mathcal{E} \] is proportional to the rate of change of magnetic flux \( \Phi \) through the coil:

\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

Lenz's Law

Lenz’s Law states that the induced current will flow in a direction such that its magnetic field opposes the change in flux that caused it:

\[ \mathcal{E} = -N \frac{d\Phi}{dt} \]

  • \( N \) = Number of turns in the coil
  • \( \Phi = B \cdot A \cdot \cos(\theta) \)
  • \( B \) = Magnetic field strength
  • \( A \) = Area of the coil
  • \( \theta \) = Angle between the magnetic field and the normal to the coil

How the AC Generator Works

As the coil rotates in the magnetic field, the angle \( \theta \) changes continuously, causing the magnetic flux through the coil to vary. This varying flux induces an alternating current (a.c.) in the coil.

The induced e.m.f. follows a sinusoidal pattern:

\[ \mathcal{E}(t) = \mathcal{E}_{\text{max}} \sin(\omega t) \]

  • \( \mathcal{E}(t) \) = Instantaneous e.m.f.
  • \( \mathcal{E}_{\text{max}} \) = Maximum e.m.f.
  • \( \omega \) = Angular velocity of the rotating coil

Graph of e.m.f. Against Time

The graph of induced e.m.f. against time for a simple a.c. generator is a sine wave:

  • Peaks: Occur when the plane of the coil is parallel to the magnetic field (\( \theta = 90^\circ \) or \( 270^\circ \)). The induced e.m.f. is maximum.
  • Troughs: Occur when the plane of the coil is parallel to the magnetic field in the opposite direction.
  • Zeros: Occur when the plane of the coil is perpendicular to the magnetic field (\( \theta = 0^\circ \) or \( 180^\circ \)). The induced e.m.f. is zero.
Graph of e.m.f. vs. time

Sketch of e.m.f. vs. Time

The sinusoidal graph can be described mathematically as:

\[ \mathcal{E}(t) = \mathcal{E}_{\text{max}} \sin(\omega t) \]

Slip Rings and Brushes

Slip rings allow continuous transfer of the alternating current generated in the coil to the external circuit without reversing the connections, maintaining the alternating nature of the output.

Comparison with a DC Generator

  • AC Generator: Uses slip rings and produces alternating current.
  • DC Generator: Uses a commutator to produce direct current.

Practice and Investigation

  • Qualitative Questions:
    • What happens to the induced e.m.f. if the speed of rotation of the coil is doubled?
    • How does the graph of e.m.f. vs. time change if the magnetic field strength is increased?
  • Interactive Simulation: Explore how generators work using the following simulation:

    PhET Generator Simulation

Extended Learning: Faraday's and Lenz's Laws

Using Faraday’s Law and Lenz’s Law, we can predict and explain the behavior of more complex generators:

  • Calculate the induced e.m.f. for different shapes of coils and magnetic fields.
  • Predict the effects of reversing the magnetic field or the direction of rotation.

4.5.3 Magnetic effect of a current

The magnetic effect of a current refers to the phenomenon where an electric current flowing through a conductor produces a magnetic field around it. This effect is the basis of electromagnetism and has various applications in devices like electromagnets, electric motors, and transformers.

Oersted's Experiment

The magnetic effect of current was first discovered by the Danish physicist Hans Christian Oersted in 1820. His experiment demonstrated that an electric current produces a magnetic field, establishing a crucial link between electricity and magnetism.

Setup of Oersted's Experiment

The experimental setup consisted of the following components:

  • A straight wire connected to a power source (battery).
  • A magnetic compass placed near the wire to detect the presence of a magnetic field.

Procedure

  1. The wire was positioned horizontally, and a compass was placed directly below it.
  2. No current was initially flowing through the wire, and the compass needle aligned itself with Earth's magnetic field, pointing in the north-south direction.
  3. When an electric current was passed through the wire, the compass needle deflected from its original position.
  4. The direction of deflection changed when the direction of the current in the wire was reversed.

Observations

  • A current-carrying conductor produces a magnetic field around it.
  • The strength of the magnetic field depends on the magnitude of the current.
  • The direction of the magnetic field depends on the direction of the current in the conductor.

Conclusion

Oersted's experiment proved that an electric current generates a magnetic field, establishing the foundation of electromagnetism.

Right-Hand Thumb Rule

The direction of the magnetic field produced by a straight current-carrying conductor can be determined using the Right-Hand Thumb Rule:

"If you point the thumb of your right hand in the direction of the current, the curled fingers of your right hand will point in the direction of the magnetic field."

Magnetic Field Around a Straight Conductor

The magnetic field around a straight conductor carrying current forms concentric circles. The magnitude of the magnetic field (\( B \)) at a distance \( r \) from the conductor is given by:

\[ B = \frac{\mu_0 I}{2 \pi r} \]

  • \( B \) = Magnetic field (in Tesla)
  • \( \mu_0 \) = Permeability of free space (\( 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \))
  • \( I \) = Current in the conductor (in Amperes)
  • \( r \) = Distance from the conductor (in meters)

Applications of Magnetic Effect of Current

  • Electromagnets: Devices where a magnetic field is produced by passing current through a coil of wire wound around a soft iron core.
  • Electric Motors: Machines that convert electrical energy into mechanical energy using the magnetic effect of current.
  • Relays: Electromagnetic switches used to control circuits.
  • Loudspeakers: Devices that convert electrical signals into sound using the magnetic effect of current.

Practice Questions

  • Quantitative Questions:
    • A straight conductor carrying a current of 5 A produces a magnetic field at a distance of 0.1 m. Calculate the magnitude of the magnetic field.
    • What happens to the magnetic field if the current in the conductor is doubled?
  • Qualitative Questions:
    • Explain how Oersted's experiment demonstrated the relationship between electricity and magnetism.
    • Describe the Right-Hand Thumb Rule and its application.

Further Investigation

4.5.4 Force on a current-carrying conductor

  • Describe an experiment to show that a force acts on a current-carrying conductor in a magnetic field, including the effect of reversing:
    • The current
    • The direction of the magnetic field

Learning Activities:

  • Introduce the motor effect, which occurs when a current-carrying wire in a magnetic field experiences a force.
  • Explain that the force, magnetic field, and current directions are all perpendicular to each other.
  • Use Fleming’s Left-Hand Rule to predict the direction of movement:

    \[ \text{Thumb} \rightarrow \text{Force (Motion)} \]

    \[ \text{First Finger} \rightarrow \text{Magnetic Field} \]

    \[ \text{Second Finger} \rightarrow \text{Current} \]

4.5.4.2 Relative Directions of Force, Magnetic Field, and Current

  • Recall and use the relative directions of force, magnetic field, and current.

4.5.5 The d.c. motor

Introduction to DC Motors

A DC (Direct Current) motor is an electrical machine that converts electrical energy into mechanical energy. It operates on the principle of the motor effect, where a current-carrying conductor placed in a magnetic field experiences a force.

Principle of Operation

The DC motor works on the motor effect, which can be explained by Fleming's Left-Hand Rule. According to this rule:

  • Thumb: Direction of Force (Motion)
  • First Finger: Direction of Magnetic Field
  • Second Finger: Direction of Current

The three directions are perpendicular to each other.

Construction of a DC Motor

  • Stator: The stationary part of the motor that provides a magnetic field.
  • Rotor (Armature): The rotating part of the motor where current flows through a coil.
  • Commutator: A split ring that reverses the direction of current in the coil after every half-turn, ensuring continuous rotation in one direction.
  • Brushes: Carbon or metal contacts that maintain electrical contact with the rotating commutator.

Working of a DC Motor

  1. When current flows through the armature, a magnetic field is created around it.
  2. This magnetic field interacts with the field from the stator, producing a force on the armature according to Fleming's Left-Hand Rule.
  3. The commutator ensures that the direction of current in the armature reverses after each half-turn, maintaining continuous rotation in the same direction.

Equations for Torque and Power

Torque

The torque (\( T \)) produced in a DC motor is given by:

\[ T = B I L r \sin \theta \]

  • \( B \) = Magnetic flux density (in Tesla)
  • \( I \) = Current flowing through the armature (in Amperes)
  • \( L \) = Length of the conductor in the magnetic field (in meters)
  • \( r \) = Radius of the armature (in meters)
  • \( \theta \) = Angle between the current direction and the magnetic field

Electrical Power Input

The electrical power input (\( P_{\text{in}} \)) to the motor is given by:

\[ P_{\text{in}} = V I \]

  • \( V \) = Voltage applied to the motor (in Volts)
  • \( I \) = Current through the motor (in Amperes)

Mechanical Power Output

The mechanical power output (\( P_{\text{out}} \)) is given by:

\[ P_{\text{out}} = T \omega \]

  • \( T \) = Torque (in Newton-meters)
  • \( \omega \) = Angular velocity (in radians per second)

Efficiency of a DC Motor

The efficiency (\( \eta \)) of a DC motor is the ratio of mechanical power output to electrical power input:

\[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \]

Types of DC Motors

  • Shunt-Wound DC Motor: The field windings are connected in parallel (shunt) with the armature.
  • Series-Wound DC Motor: The field windings are connected in series with the armature.
  • Compound-Wound DC Motor: Combines features of both shunt and series motors.

Applications of DC Motors

  • Electric vehicles
  • Elevators and hoists
  • Electric trains
  • Fans and blowers
  • Household appliances

Advantages of DC Motors

  • Simple construction and design
  • High starting torque
  • Precise speed control

Disadvantages of DC Motors

  • Requires regular maintenance due to brushes and commutator wear
  • Limited lifespan of brushes and commutator

Practice Questions

  • Quantitative Questions:
    • A DC motor with an armature current of 5 A and a magnetic flux density of 0.1 T produces a torque of 2 Nm. Calculate the length of the armature conductor if the radius of the armature is 0.05 m.
    • Calculate the efficiency of a DC motor if the electrical power input is 500 W and the mechanical power output is 400 W.
  • Qualitative Questions:
    • Explain how a commutator works in a DC motor.
    • Discuss the advantages and disadvantages of using DC motors in electric vehicles.

Further Investigation

4.5.6 The transformer

Construction of a Simple Transformer

  • Core: A soft iron core that provides a continuous magnetic path, enhancing the magnetic field between the coils.
  • Primary Coil: The coil connected to the input voltage source, where an alternating current (AC) flows, creating a changing magnetic field.
  • Secondary Coil: The coil connected to the output, where the induced voltage is generated due to the changing magnetic flux.
  • Working Principle: Transformers work on the principle of electromagnetic induction. A changing magnetic field in the primary coil induces an alternating voltage in the secondary coil.

Types of Transformers

  • Step-Up Transformer:
    • Increases the voltage.
    • Number of turns in the secondary coil (\( N_s \)) is greater than in the primary coil (\( N_p \)).
    • Decreases the current in the secondary coil.
  • Step-Down Transformer:
    • Decreases the voltage.
    • Number of turns in the secondary coil (\( N_s \)) is less than in the primary coil (\( N_p \)).
    • Increases the current in the secondary coil.

Transformer Equations

The voltage and the number of turns in the primary and secondary coils are related by:

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]

  • \( V_p \) = Voltage across the primary coil
  • \( V_s \) = Voltage across the secondary coil
  • \( N_p \) = Number of turns in the primary coil
  • \( N_s \) = Number of turns in the secondary coil

Equation for 100% Efficiency

For an ideal transformer with 100% efficiency, the input power equals the output power:

\[ I_p V_p = I_s V_s \]

  • \( I_p \) = Current in the primary coil
  • \( I_s \) = Current in the secondary coil

Power Loss in Transmission

Power loss in transmission cables is given by:

\[ P_{\text{loss}} = I^2 R \]

  • \( P_{\text{loss}} \) = Power lost due to resistance in the cables
  • \( I \) = Current through the cables
  • \( R \) = Resistance of the transmission cables

To minimize power loss, the current is reduced by using step-up transformers to increase the voltage during transmission.

Advantages of High-Voltage Transmission

  • Reduces Power Loss: By increasing the voltage and reducing the current, power loss (\( I^2 R \)) is minimized.
  • Improves Efficiency: Less power is wasted as heat in the transmission lines.
  • Allows Long-Distance Transmission: High-voltage transmission makes it feasible to transport electricity over long distances without significant loss.

Functioning of a Transformer

  1. An alternating current in the primary coil creates a changing magnetic field.
  2. The changing magnetic field is concentrated in the soft iron core and passes through the secondary coil.
  3. This changing magnetic flux induces an alternating voltage in the secondary coil.

Practice Questions

  • Quantitative Questions:
    • If a step-up transformer has 100 turns on the primary coil and 500 turns on the secondary coil, and the input voltage is 230 V, what is the output voltage?
    • Calculate the current in the secondary coil if the power input to a transformer is 500 W and the output voltage is 110 V.
  • Qualitative Questions:
    • Why is a soft iron core used in a transformer?
    • Explain why high-voltage transmission is more efficient than low-voltage transmission.

Further Investigation

5.1 The nuclear model of atom

5.1.1 The atom

The atom consists of three subatomic particles: protons, neutrons, and electrons. The structure of the atom can be described as follows:

  • The nucleus is located at the center of the atom and contains:
    • Protons, which have a positive charge of \( +1 \).
    • Neutrons, which have no charge (neutral).
  • Electrons are negatively charged particles (\( -1 \)) that orbit the nucleus in specific energy levels or shells.

The nucleus is very small compared to the overall size of the atom but contains most of the atom's mass.

Ions and Their Formation

An atom can gain or lose electrons to form ions:

  • Positive ions (cations) are formed when an atom loses electrons.
  • Negative ions (anions) are formed when an atom gains electrons.

The overall charge of an ion is calculated as:

\[ \text{Ion Charge} = \text{Number of Protons} - \text{Number of Electrons} \]

Proton Number, Nucleon Number, and Isotopes

  • The proton number (\( Z \)) represents the number of protons in the nucleus. This is also the atomic number of the element.
  • The nucleon number (\( A \)) represents the total number of protons and neutrons in the nucleus.

The nuclide notation is written as:

\[ ^{A}_{Z}X \]
  • \( A \): Nucleon number (protons + neutrons).
  • \( Z \): Proton number (number of protons).
  • \( X \): Symbol of the element.

Isotopes are atoms of the same element that have the same proton number but different nucleon numbers due to varying numbers of neutrons.

Examples:

\[ ^{12}_{6}\text{C}, \, ^{13}_{6}\text{C}, \, ^{14}_{6}\text{C} \]

Alpha-Particle Scattering Experiment

The alpha-particle scattering experiment, conducted by Rutherford, provided evidence for the nuclear model of the atom:

  • Alpha (\( \alpha \)) particles were directed at a thin sheet of gold foil.
  • The observations included:
    • Most \( \alpha \)-particles passed straight through the foil, indicating that the atom is mostly empty space.
    • A few \( \alpha \)-particles were deflected at large angles, indicating the presence of a small, dense, positively charged nucleus.
    • A very small number were reflected back, suggesting that the nucleus contains most of the atom's mass.

The key conclusions from the experiment were: \[ \text{1. The nucleus is very small and positively charged.} \] \[ \text{2. The atom is mostly empty space.} \] \[ \text{3. The nucleus contains most of the mass of the atom.} \]

5.1.2 The nucleus

1. Composition of the Nucleus

The nucleus of an atom is composed of protons and neutrons:

  • Protons are positively charged particles.
  • Neutrons are uncharged particles.

The proton number, denoted by \( Z \), represents the number of protons in the nucleus, while the nucleon number (mass number), denoted by \( A \), represents the total number of protons and neutrons. The number of neutrons can be calculated by:

\[ \text{Number of neutrons} = A - Z \]

5.2 Radioactivity

Detection of Radioactivity

1. Definition of Radioactivity

Radioactivity is the spontaneous emission of particles or electromagnetic radiation from unstable atomic nuclei. This occurs because the nucleus tries to become more stable by releasing energy.

2. Types of Radioactive Emissions

  • Alpha (\( \alpha \)) Particles: Helium nuclei consisting of two protons and two neutrons. They have a charge of \( +2 \) and are heavily ionizing.
  • Beta (\( \beta \)) Particles: High-speed electrons (\( \beta^- \)) or positrons (\( \beta^+ \)) emitted from the nucleus. They have a charge of \( -1 \) (electron) or \( +1 \) (positron) and are moderately ionizing.
  • Gamma (\( \gamma \)) Rays: High-energy electromagnetic radiation with no charge or mass. They are weakly ionizing but highly penetrating.

3. Methods of Detecting Radioactivity

Several devices are used to detect and measure radioactivity. These include:

3.1 Geiger-Müller (GM) Tube

The Geiger-Müller tube is a commonly used detector for ionizing radiation. It works as follows:

  • Ionizing radiation enters the tube and ionizes the gas inside.
  • The ions are attracted to the electrodes, creating a small electric current.
  • The current is amplified and registered as a "click" or a reading on a digital display.

The count rate (\( N \)) is proportional to the intensity of radiation:

\[ N \propto I \]

3.2 Cloud Chamber

A cloud chamber allows visualization of radiation tracks. It works by:

  • Supercooling a vaporized substance (e.g., alcohol) in a sealed chamber.
  • Ionizing radiation passes through the vapor, causing ionization.
  • The ions act as condensation nuclei, forming visible tracks that indicate the path of the radiation.

3.3 Scintillation Counter

A scintillation counter detects radiation by the following process:

  • Ionizing radiation interacts with a scintillating material, causing it to emit light.
  • The emitted light is detected by a photomultiplier tube, which converts it into an electrical signal.
  • The electrical signal is amplified and recorded as a count.

3.4 Photographic Film

Radiation darkens photographic film in proportion to the intensity of radiation exposure. This method is used in dosimeters for monitoring radiation exposure levels.

4. Ionization and Penetration Abilities of Radiation

The ionizing and penetration abilities of the three types of radiation differ significantly:

Type of Radiation Ionization Ability Penetration Ability Stopped By
\( \alpha \) Strong Weak Paper or skin
\( \beta \) Moderate Intermediate Thin metal (e.g., aluminum)
\( \gamma \) Weak Strong Thick lead or concrete

5. Practical Applications of Radioactivity Detection

  • Medical Imaging and Treatment: Detection of radiation in radiography, cancer treatment, and PET scans.
  • Industrial Uses: Monitoring thickness of materials, detecting leaks, and sterilizing equipment.
  • Research: Studying radioactive decay, nuclear physics experiments, and carbon dating.

5.2.2 The three types of emission

Charges Involved in Radioactive Decay

  • \( \alpha \)-Decay: In \( \alpha \)-decay, an unstable nucleus emits an \( \alpha \)-particle, which is a helium nucleus (\( ^4_2He \)) consisting of two protons and two neutrons. The \( \alpha \)-particle has a charge of \( +2 \) because it contains two protons.
  • \( \beta^- \)-Decay: In \( \beta^- \)-decay, a neutron in the nucleus converts into a proton and emits a \( \beta^- \)-particle (electron) and an antineutrino (\( \overline{\nu} \)). The \( \beta^- \)-particle has a charge of \( -1 \).
  • \( \beta^+ \)-Decay: In \( \beta^+ \)-decay, a proton in the nucleus converts into a neutron and emits a \( \beta^+ \)-particle (positron) and a neutrino (\( \nu \)). The \( \beta^+ \)-particle has a charge of \( +1 \).
  • \( \gamma \)-Decay: In \( \gamma \)-decay, the nucleus transitions from a higher energy state to a lower energy state, emitting a \( \gamma \)-ray (high-energy photon). \( \gamma \)-rays have no charge (\( 0 \)).

Examples of Radioactive Decays

  1. \( \boxed{^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He} \) (Alpha decay, \( \alpha \)-particle charge: \( +2 \)).

  2. \( \boxed{^{14}_6C \rightarrow ^{14}_7N + \beta^- + \overline{\nu}} \) (Beta-minus decay, \( \beta^- \)-particle charge: \( -1 \)).

  3. \( \boxed{^{22}_{11}Na \rightarrow ^{22}_{10}Ne + \beta^+ + \nu} \) (Beta-plus decay, \( \beta^+ \)-particle charge: \( +1 \)).

  4. \( \boxed{^{60}_{27}Co \rightarrow ^{60}_{28}Ni + \gamma} \) (Gamma decay, \( \gamma \)-ray charge: \( 0 \)).

  5. \( \boxed{^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^{4}_{2}He} \) (Alpha decay, \( \alpha \)-particle charge: \( +2 \)).

  6. \( \boxed{^{3}_{1}H \rightarrow ^{3}_{2}He + \beta^- + \overline{\nu}} \) (Beta-minus decay, \( \beta^- \)-particle charge: \( -1 \)).

  7. \( \boxed{^{11}_{6}C \rightarrow ^{11}_{5}B + \beta^+ + \nu} \) (Beta-plus decay, \( \beta^+ \)-particle charge: \( +1 \)).

  8. \( \boxed{^{99}_{43}Tc \rightarrow ^{99}_{44}Ru + \gamma} \) (Gamma decay, \( \gamma \)-ray charge: \( 0 \)).

  9. \( \boxed{^{210}_{84}Po \rightarrow ^{206}_{82}Pb + ^{4}_{2}He} \) (Alpha decay, \( \alpha \)-particle charge: \( +2 \)).

  10. \( \boxed{^{32}_{15}P \rightarrow ^{32}_{16}S + \beta^- + \overline{\nu}} \) (Beta-minus decay, \( \beta^- \)-particle charge: \( -1 \)).

  11. \( \boxed{^{13}_{7}N \rightarrow ^{13}_{6}C + \beta^+ + \nu} \) (Beta-plus decay, \( \beta^+ \)-particle charge: \( +1 \)).

  12. \( \boxed{^{137}_{55}Cs \rightarrow ^{137}_{56}Ba + \gamma} \) (Gamma decay, \( \gamma \)-ray charge: \( 0 \)).

  13. \( \boxed{^{222}_{86}Rn \rightarrow ^{218}_{84}Po + ^{4}_{2}He} \) (Alpha decay, \( \alpha \)-particle charge: \( +2 \)).

  14. \( \boxed{^{24}_{11}Na \rightarrow ^{24}_{12}Mg + \beta^- + \overline{\nu}} \) (Beta-minus decay, \( \beta^- \)-particle charge: \( -1 \)).

  15. \( \boxed{^{18}_{9}F \rightarrow ^{18}_{8}O + \beta^+ + \nu} \) (Beta-plus decay, \( \beta^+ \)-particle charge: \( +1 \)).

5.2.3 Radioactive decay

Emission of Radiation and Its Properties

1. Random and Spontaneous Emission of Radiation

The emission of radiation from an unstable nucleus is both spontaneous and random:

  • It occurs without external influence.
  • It is random in direction and time, meaning it is impossible to predict when or where a nucleus will decay.

2. Types of Radiation

The three types of radiation are:

  • Alpha (\( \alpha \)) Radiation: Helium nuclei consisting of two protons and two neutrons.
  • Beta (\( \beta^- \)) Radiation: High-speed electrons emitted during the decay of a neutron into a proton.
  • Gamma (\( \gamma \)) Radiation: High-frequency electromagnetic waves emitted by the nucleus after a decay process.

2.1 Nature, Ionizing Effects, and Penetrating Abilities

Type of Radiation Nature Relative Ionizing Effect Relative Penetrating Power
\( \alpha \) Helium nucleus (\( 2 \, \text{protons} + 2 \, \text{neutrons} \)) High Low (stopped by paper or skin)
\( \beta^- \) High-speed electron Moderate Intermediate (stopped by thin aluminum)
\( \gamma \) Electromagnetic wave Low High (stopped by thick lead or concrete)

3. Deflection in Electric and Magnetic Fields

  • \( \alpha \)-particles (\( +2 \) charge) are deflected slightly towards the negative plate in an electric field.
  • \( \beta^- \)-particles (\( -1 \) charge) are deflected strongly towards the positive plate in an electric field due to their lower mass.
  • \( \gamma \)-radiation (neutral) is not deflected in electric or magnetic fields.

The deflection depends on the relative charge-to-mass ratio. In magnetic fields, the direction of deflection follows the right-hand rule for positive charges and the left-hand rule for negative charges.

4. Ionizing Effects of Radiation

The ionizing ability of radiation is influenced by:

  • Kinetic Energy: Higher energy results in greater ionization.
  • Electric Charge: Radiation with greater charge (e.g., \( \alpha \)-particles) has a stronger ionizing effect.

Highly ionizing radiation like \( \alpha \)-particles loses energy rapidly and has low penetration. Weakly ionizing radiation like \( \gamma \)-rays penetrates deeply without much ionization.

5. Radioactive Decay and Nuclide Notation

Radioactive decay transforms an unstable nucleus into a more stable one, often forming a new element. This is represented using nuclide notation:

\[ ^A_Z X \rightarrow ^{A-4}_{Z-2} Y + ^4_2 \alpha \quad \text{(Alpha decay)} \]

\[ ^A_Z X \rightarrow ^A_{Z+1} Y + \beta^- \quad \text{(Beta decay)} \]

\[ ^A_Z X \rightarrow ^A_Z X + \gamma \quad \text{(Gamma emission)} \]

6. Practical Demonstration of Penetrating Powers

To illustrate the relative penetrating powers of \( \alpha \), \( \beta^- \), and \( \gamma \) radiation:

  • Use a radioactive source at a safe distance.
  • Place materials (paper, aluminum, lead) between the source and detector.
  • Observe how radiation intensity changes with different materials.

Background Radiation and Its Detection

Introduction to Radiation

  • Radiation is present all around us, even without a radioactive source being nearby.
  • A Geiger-Müller (GM) tube and counter can sporadically detect radiation from the environment, showing that background radiation is common and largely harmless.
  • The count rate, as shown on a Geiger-Müller counter, is measured in counts per minute (\( \text{counts/min} \)) or counts per second (\( \text{counts/s} \)).

Sources of Background Radiation

  • Common sources of background radiation include:
    • \( \textbf{Radon gas:} \) Found in the air, originating from the decay of uranium in rocks.
    • \( \textbf{Rocks and buildings:} \) Certain rocks, like granite, contain radioactive isotopes.
    • \( \textbf{Food and drink:} \) Contain trace amounts of radioactive isotopes such as \( ^{40}K \).
    • \( \textbf{Cosmic rays:} \) High-energy particles from the Sun and outer space.
    • \( \textbf{Nuclear weapons testing:} \) Residual fallout from past tests.
    • \( \textbf{Nuclear power:} \) Low-level emissions from nuclear power plants.
    • \( \textbf{Medical sources:} \) Diagnostic tools such as X-rays and radioactive tracers.
  • Radon gas contributes the largest share to background radiation in most areas.

Detection of Radiation

  • Geiger-Müller Counter:
    • Measures ionising radiation in terms of counts per minute (\( \text{counts/min} \)) or counts per second (\( \text{counts/s} \)).
    • Background radiation is measured first, and this value is subtracted to calculate the corrected count rate.
  • Cloud Chamber:
    • Detects ionising particles by observing condensation trails in a supersaturated vapor.
    • \( \alpha \)-particles produce thick, short tracks due to their high ionisation power.
  • Spark Counter:
    • Used to detect \( \alpha \)-particles by observing sparks created as the particles ionise air between electrodes.

Practical Measurements

  • To measure the corrected count rate:
    1. Measure the background radiation count rate (\( R_{\text{background}} \)).
    2. Measure the total count rate (\( R_{\text{total}} \)) with the radioactive source.
    3. Calculate the corrected count rate: \[ R_{\text{corrected}} = R_{\text{total}} - R_{\text{background}} \]

Extension: Observing Fundamental Particles

  • The cloud chamber can also be used to observe fundamental particles, such as electrons and muons.
  • Examples of tracks in a cloud chamber:
    • \( \textbf{Electrons:} \) Thin, curved tracks due to their small mass and charge.
    • \( \textbf{Muons:} \) Straight tracks, as they are more massive and less easily deflected.

Additional Resources

8. Practice Questions

Test your understanding with these questions:

Questions and Answers

  1. Explain why \( \alpha \)-particles are more ionizing but less penetrating than \( \gamma \)-rays.

    Answer: \( \alpha \)-particles are more ionizing because they are large, carry a +2 charge, and interact strongly with atoms in their path, causing significant ionization. However, due to their size and charge, they lose energy quickly and cannot penetrate deeply into materials. In contrast, \( \gamma \)-rays are uncharged electromagnetic waves, which interact less with matter, resulting in weaker ionization but greater penetration.

  2. Describe the deflection of \( \beta^- \)-particles in an electric field.

    Answer: \( \beta^- \)-particles (high-speed electrons) carry a negative charge. In an electric field, they are strongly deflected towards the positive plate due to their low mass and negative charge. The degree of deflection is greater compared to \( \alpha \)-particles because \( \beta^- \)-particles have a much smaller mass.

  3. Identify the type of radiation emitted in this decay equation:

    \[ ^{14}_6 C \rightarrow ^{14}_7 N + \beta^- \]

    Answer: This is an example of \( \beta^- \) decay. In this process, a neutron in the carbon-14 nucleus converts into a proton, emitting a \( \beta^- \)-particle (electron) and an antineutrino. The proton increases the atomic number by 1, transforming carbon into nitrogen.

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5.2.4 Half-life

Definition of Half-Life

  • The half-life of an isotope is defined as the time taken for half the nuclei of that isotope in any sample to decay.
  • This concept can be visualized using decay curves or calculated using data from tables.
  • Half-life calculations typically exclude background radiation.

Calculating Half-Life

  • To calculate the half-life:
    1. Measure the time it takes for the count rate or the number of radioactive nuclei to reduce to half its original value.
    2. Use data from decay curves, which plot the activity or number of nuclei over time.
  • Example:
    • If the activity of a sample drops from \( 100 \, \text{counts/min} \) to \( 50 \, \text{counts/min} \) in 4 minutes, the half-life is \( 4 \, \text{minutes} \).

Applications of Half-Life

  • Household Fire (Smoke) Alarms:
    • Use isotopes like \( \text{Americium-241} \), which emits \( \alpha \)-particles and has a long half-life.
    • A long half-life ensures the source does not need frequent replacement.
    • The radiation is harmless to humans due to shielding and the nature of \( \alpha \)-radiation.
  • Food Irradiation:
    • Gamma rays are used to kill bacteria, extending the shelf life of food.
  • Sterilization of Medical Equipment:
    • Gamma radiation sterilizes equipment by killing microbes without high temperatures.
  • Thickness Control:
    • Radiation (usually \( \beta \)-rays) is used to measure and control the thickness of materials such as paper or metal sheets.
    • The choice of radiation depends on its penetrating power and absorption by the material.
  • Medical Diagnosis and Treatment:
    • Gamma rays are used to treat cancer by targeting and killing cancerous cells.
    • Radioisotopes like \( \text{Technetium-99m} \) are used in diagnostic imaging.
  • Carbon Dating:
    • Uses the isotope \( \text{Carbon-14} \) to date objects that were once living.
    • \( \text{Carbon-14} \) is absorbed by living organisms and decays after death, allowing the determination of age.

Experiments and Resources

Practice Questions

  • Calculate the half-life from a given decay curve.
  • Explain why a long half-life is suitable for a smoke alarm.
  • Describe how gamma rays are used to sterilize medical equipment.
  • Determine the age of a fossil using carbon dating data.

Extended Research

  • Investigate the use of isotopes in:
    • Household fire alarms (\( \text{Americium-241} \)).
    • Food irradiation.
    • Medical equipment sterilization.
    • Thickness control in manufacturing.
    • Medical imaging and cancer treatment (\( \text{Technetium-99m} \) and gamma rays).

5.2.5 Safety precautions

6.1 Earth and the solar system

6.1.1 The earth

1. Earth's Rotation

The Earth rotates on its axis, which is tilted at an angle of approximately 23.5°, once every 24 hours. This rotation explains:

  • The apparent daily motion of the Sun across the sky. The Sun appears to rise in the east and set in the west due to Earth's rotation.
  • The periodic cycle of day and night. As the Earth rotates, different parts of its surface move into and out of sunlight.

2. Earth's Revolution

The Earth orbits the Sun once every 365.25 days. This orbital motion explains:

  • The periodic nature of the seasons. The tilt of Earth's axis causes different hemispheres to receive varying amounts of sunlight throughout the year.
  • The North and South Poles experience 24 hours of darkness during winter and 24 hours of light during summer due to the tilt of Earth's axis.

3. Moon's Orbit

The Moon orbits the Earth once every month (approximately 27.3 days). This explains:

  • The periodic nature of the Moon’s phases (new moon, first quarter, full moon, last quarter).

4. Orbital Speed

The average orbital speed of an object can be calculated using the equation:

\[ v = \frac{2\pi r}{T} \]

Where:

  • \( v \): Orbital speed
  • \( r \): Average radius of the orbit
  • \( T \): Orbital period (time for one complete orbit)

For example, Earth's average orbital speed around the Sun can be calculated using its average radius (\( r \approx 1.496 \times 10^{11} \, \text{m} \)) and orbital period (\( T \approx 365.25 \, \text{days} \)).

5. Practical Activities

  • Use balls and a lamp to model Earth's rotation and revolution. Rotate a ball representing Earth on its axis and move it in an elliptical orbit around the lamp representing the Sun.
  • Use simulations to show how Earth's rotation and revolution result in day-night cycles and seasons.
  • Investigate orbital motion further with tools like Google Earth or NASA’s Earth-Now app.

6. Practice Problems

  1. Calculate the average orbital speed of a satellite orbiting Earth at a radius of \( r = 42,000 \, \text{km} \) with an orbital period of \( T = 24 \, \text{hours} \).
  2. Explain why the poles experience extended periods of light or darkness during certain seasons.
  3. Determine the speed of the hour hand, minute hand, and second hand of a clock using the orbital speed formula.

6.1.2 The solar system

Components of the Solar System

The Solar System consists of:

  • One star: the Sun.
  • Eight named planets in the following order from the Sun:
    1. Mercury
    2. Venus
    3. Earth
    4. Mars
    5. Jupiter
    6. Saturn
    7. Uranus
    8. Neptune
  • Minor planets that orbit the Sun, such as:
    • Dwarf planets (e.g., Pluto).
    • Asteroids in the asteroid belt.
  • Moons that orbit planets.
  • Smaller Solar System bodies, such as comets and natural satellites.

Comparison of Planets

The planets are classified into two categories:

  • The four planets closest to the Sun (Mercury, Venus, Earth, Mars) are rocky and small.
  • The four planets furthest from the Sun (Jupiter, Saturn, Uranus, Neptune) are gaseous and large.

This distinction is explained by the accretion model, where gravitational forces caused lighter elements to migrate outward during the Solar System's formation.

Key Equations

The average orbital speed of a planet is given by:

\[ v = \frac{2\pi r}{T} \]
  • \( v \): Orbital speed (m/s).
  • \( r \): Average radius of the orbit (m).
  • \( T \): Orbital period (s).

Elliptical Orbits

Planets, minor planets, and comets follow elliptical orbits. The Sun is not at the center of these ellipses, except when the orbits are approximately circular. For example:

  • Comets travel faster when closer to the Sun due to the conservation of energy.
  • The orbital speed decreases as the distance from the Sun increases, due to the weaker gravitational force.

Gravitational Field Strength

The strength of the gravitational field depends on:

  • The mass of the planet or celestial body.
  • The distance from the planet or body.

The Sun contains most of the mass in the Solar System, which explains why it exerts the strongest gravitational force and holds planets in orbit.

Speed of Light and Astronomical Distances

Light travels at a speed of approximately \( 3.0 \times 10^8 \, \text{m/s} \). For example:

  • The time for light to travel from the Sun to the Earth is approximately: \[ t = \frac{d}{c} \approx \frac{1.5 \times 10^{11}}{3.0 \times 10^8} \approx 500 \, \text{s} \]
  • One light-year is the distance light travels in one year and is used to measure astronomical distances.

Formation of the Solar System

The Solar System formed through the following process:

  • The interstellar cloud of gas and dust collapsed under its own gravity.
  • The cloud began to rotate, forming an accretion disk with the Sun at its center.
  • Planets formed through the gradual clumping of material in the disk.

6.2 Stars and the universe

6.2.1 The Sun as a star

Definition of Stars

A star is a massive celestial object composed mostly of hydrogen and helium, undergoing nuclear fusion reactions that release energy in the form of light and heat.

Structure of a Star

Stars are composed of several layers:

  • Core: The innermost region where nuclear fusion occurs, primarily converting hydrogen to helium.
  • Radiative Zone: The layer where energy is transferred outward through radiation.
  • Convective Zone: The outer layer where energy is transferred through convection currents.
  • Photosphere: The visible surface of the star, emitting most of the star's light.
  • Chromosphere and Corona: The outermost layers, emitting ultraviolet and X-ray radiation.

Energy Production in Stars

Stars produce energy through nuclear fusion:

\[ 4 \, ^1H \rightarrow \, ^4He + 2e^+ + 2\nu_e + \text{energy} \]
  • \( ^1H \): Hydrogen nuclei (protons).
  • \( ^4He \): Helium nucleus.
  • \( e^+ \): Positron.
  • \( \nu_e \): Neutrino.

This process releases vast amounts of energy according to Einstein's equation:

\[ E = mc^2 \]
  • \( E \): Energy (J).
  • \( m \): Mass converted during fusion (kg).
  • \( c \): Speed of light (\( 3.0 \times 10^8 \, \text{m/s} \)).

Life Cycle of a Star

The evolution of a star depends on its initial mass:

  1. Nebula: A cloud of gas and dust collapses under gravity.
  2. Protostar: The collapsing material heats up, forming a dense core.
  3. Main Sequence: Stable nuclear fusion begins, balancing gravitational collapse.
  4. Post-Main Sequence:
    • Low-mass stars expand into red giants and end as white dwarfs.
    • High-mass stars become red supergiants, leading to supernovae, and form neutron stars or black holes.

Star Classification

Stars are classified based on their temperature and luminosity into spectral types:

  • O, B, A, F, G, K, M (from hottest to coolest).

The relationship between a star's luminosity (\( L \)) and its temperature (\( T \)) is given by the Stefan-Boltzmann law:

\[ L = 4 \pi R^2 \sigma T^4 \]
  • \( R \): Radius of the star (m).
  • \( \sigma \): Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \)).
  • \( T \): Temperature (K).

Galaxies and Stars

Stars are grouped into galaxies, each containing billions of stars. Our Sun is part of the Milky Way galaxy.

Distances to stars are measured using light-years:

\[ 1 \, \text{light-year} = 9.46 \times 10^{15} \, \text{m} \]

6.2.2 Stars

Astronomical Distances

The universe is vast, and distances between celestial objects are enormous. The Sun, for example, is much closer to the Earth than most stars in the universe.

To measure such large distances, we use the unit of a light-year. A light-year is defined as the distance light travels in a vacuum in one year:

\[ 1 \, \text{light-year} = 9.5 \times 10^{15} \, \text{m} \]

Exploring Astronomical Objects

Images captured by the Hubble Space Telescope provide detailed views of stars, galaxies, and other celestial objects. Learners can explore such images from the archive at Hubble Top 100 Images.

Constellations and Navigation

Throughout history, humans have interpreted star groupings to form constellations. These constellations have been used for navigation and storytelling across cultures.

Life Cycle of a Star

The life cycle of a star involves several stages, influenced by its mass and composition:

(a) Formation

Stars are formed from interstellar clouds of gas and dust, primarily hydrogen. These clouds collapse under their own gravity.

(b) Protostar

A collapsing interstellar cloud becomes a protostar. As it collapses, the temperature increases due to gravitational attraction.

(c) Stable Star

A protostar becomes a stable star when the inward gravitational force is balanced by the outward force created by high temperatures from nuclear fusion.

(d) Hydrogen Fuel Depletion

Over time, all stars run out of hydrogen as the primary fuel for nuclear reactions.

(e) Red Giant

When hydrogen in the star's core is depleted, the star expands to form a red giant. The core begins converting helium into heavier elements.

(f) Low-Mass Stars

A red giant from a less massive star sheds its outer layers, forming a planetary nebula with a white dwarf at the center.

(g) High-Mass Stars

A red supergiant undergoes a supernova explosion. This explosion creates a nebula with heavier elements and leaves behind either a neutron star or a black hole at the center.

(h) Formation of New Stars

The nebula from a supernova may collapse to form new stars, sometimes with orbiting planetary systems.

Energy Balances in a Star

The balance between forces in a stable star is crucial:

  • Inward force: Gravitational attraction pulling matter toward the center.
  • Outward force: Pressure generated by high temperatures from nuclear fusion.

Summary

Stars evolve over billions of years, creating and distributing elements throughout the universe. The process of star formation and death is critical to the formation of new stars, planets, and life as we know it.

6.2.3 The Universe

1. The Milky Way and Galaxies

Know that the Milky Way is one of many billions of galaxies making up the Universe and that the diameter of the Milky Way is approximately 100,000 light-years.

2. Redshift

Redshift is described as an increase in the observed wavelength of electromagnetic radiation emitted from receding stars and galaxies.

The light emitted from distant galaxies appears redshifted in comparison to light emitted on Earth.

This redshift is evidence that:

  • The Universe is expanding.
  • It supports the Big Bang theory.

3. Light-Year

A light-year is the distance travelled by light in one year in a vacuum. Its value is approximately:

\[ 1 \, \text{light-year} = 9.5 \times 10^{15} \, \text{m}. \]

4. Evidence of an Expanding Universe

Hubble's law relates the speed \( v \) at which a galaxy is moving away to its distance \( d \) from Earth:

\[ v = H_0 \cdot d, \] where \( H_0 \) is the Hubble constant. The current estimate for \( H_0 \) is approximately: \[ H_0 = 2.2 \times 10^{-18} \, \text{per second}. \]

5. Age of the Universe

The equation for the age of the Universe is given as:

\[ d = \frac{1}{H_0}. \] This equation provides evidence that all matter in the Universe was once concentrated at a single point, supporting the Big Bang theory.

6. Cosmic Microwave Background Radiation (CMBR)

Microwave radiation of a specific frequency is observed at all points in space, known as the Cosmic Microwave Background Radiation (CMBR).

The CMBR was produced shortly after the Universe was formed and has since been expanded into the microwave region of the electromagnetic spectrum as the Universe expanded.

7. Doppler Shift

The Doppler shift for sound and light explains the change in observed frequency due to relative motion between the source and the observer. The formula for frequency is:

\[ f_o = f_s \cdot \frac{v}{v \pm v_s}, \] where:
  • \( f_o \) is the observed frequency,
  • \( f_s \) is the source frequency,
  • \( v \) is the speed of the wave in the medium, and
  • \( v_s \) is the speed of the source relative to the observer.

The \( + \) or \( - \) in the denominator depends on whether the source is moving towards or away from the observer.

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Formula Sheet

Mechanics

\[\text{Average speed (ms}^{-1}) = \frac{\text{distance (m)}}{\text{time (s)}}\]
\[\text{Average velocity (ms}^{-1}) = \frac{\text{displacement (m)}}{\text{time (s)}}\]
\[v = \frac{s}{t}\]
\[\text{Period of a pendulum (s)} = \frac{\text{total time (s)}}{\text{number of swings}}\]
\[T = \frac{t}{\text{number}}\]
\[\text{Acceleration (ms}^{-2}) = \frac{\text{final velocity (ms}^{-1}) - \text{initial velocity (ms}^{-1})}{\text{time (s)}}\]
\[a = \frac{v - u}{t}\]
\[\text{Weight (N)} = \text{mass (kg)} \times \text{gravitational field strength (ms}^{-2})\]
\[F = mg \quad \text{(Earth’s gravitational field strength } g = 10 \text{ ms}^{-2})\]
\[\text{Force (N)} = \text{mass (kg)} \times \text{acceleration (ms}^{-2})\]
\[F = ma\]
\[\text{Density (kgm}^{-3}) = \frac{\text{mass (kg)}}{\text{volume (m}^{3})}\]
\[\rho = \frac{M}{V}\]
\[\text{Hooke’s law: Force (N)} = \text{constant (Nm}^{-1}) \times \text{extension (m)}\]
\[F = kx\]
\[\text{Momentum (p)} = \text{mass (kg)}) \times \text{Velocity (ms^{-1})}\]
\[p = mv\]
\[\text{Impulse (I)} = \text{Force (F)}) \times \Delta\text{t (Ns)}\]
\[I = F \Delta t \]
Impulse is defined as the change in momentum:
\[ \Delta(mv) = m \Delta v = m \cdot (v - u) \]
Also, impulse can be related to force and time:
\[ (mv) = m \Delta v = m \cdot (v - u) = F \cdot \Delta t \]
\[\text{Pressure (Pa)} = \frac{\text{force (N)}}{\text{area (m}^{2})}\]
\[P = \frac{F}{A}\]
\[\text{Fluid Pressure (Pa)} = \rho \cdot g \cdot h\]
\[\text{Work (J)} = \text{force (N)} \times \text{distance moved (m)}\]
\[\Delta E = Fd\]
\[\text{Power (W)} = \frac{\text{work (J)}}{\text{time (s)}}\]
\[P = \frac{\Delta E}{t}\]

Thermal Physics

\[\text{Boyle’s Law:} \quad P_1V_1 = P_2V_2\]
\[\text{Energy (J)} = \text{mass (kg)} \times \text{specific heat capacity (Jkg}^{-1}\text{°C}^{-1}) \times \text{temperature change (°C)}\]
\[E = mc\Delta T\]

Electricity

\[\text{Current (A)} = \frac{\text{charge (C)}}{\text{time (s)}}\]
\[I = \frac{Q}{t}\]
\[\text{Voltage (V)} = \frac{\text{energy transferred (J)}}{\text{charge (C)}}\]
\[V = \frac{E}{Q}\]
\[\text{Voltage (V)} = \text{current (A)} \times \text{resistance (Ω)}\]
\[V = IR\]
\[\text{Power (W)} = \text{current (A)} \times \text{voltage (V)}\]
\[P = IV\]

Waves

\[\text{Wave speed (ms}^{-1}) = \text{frequency (Hz)} \times \text{wavelength (m)}\]
\[c = f\lambda\]
\[\text{Frequency (Hz)} = \frac{1}{\text{Period (s)}}\]
\[F = \frac{1}{T}\]

Nuclear Physics

Alpha Decay (\(\alpha\)-decay)

In alpha decay, a nucleus emits an alpha particle (two protons and two neutrons), represented as:

\[ {}^{A}_{Z}X \rightarrow {}^{A-4}_{Z-2}Y + {}^{4}_{2}\alpha \]

Example: Uranium-238 decays into Thorium-234 by emitting an alpha particle:

\[ {}^{238}_{92}U \rightarrow {}^{234}_{90}Th + {}^{4}_{2}\alpha \]

Beta Minus Decay (\(\beta^{-}\)-decay)

In beta-minus decay, a neutron is converted into a proton, emitting an electron and an antineutrino:

\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + \beta^{-} + \bar{\nu}_e \]

Example: Carbon-14 decays into Nitrogen-14:

\[ {}^{14}_{6}C \rightarrow {}^{14}_{7}N + \beta^{-} + \bar{\nu}_e \]

Beta Plus Decay (\(\beta^{+}\)-decay)

In beta-plus decay, a proton is converted into a neutron, emitting a positron and a neutrino:

\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z-1}Y + \beta^{+} + \nu_e \]

Example: Fluorine-18 decays into Oxygen-18:

\[ {}^{18}_{9}F \rightarrow {}^{18}_{8}O + \beta^{+} + \nu_e \]

Gamma Decay (\(\gamma\)-decay)

In gamma decay, an excited nucleus releases energy by emitting a gamma ray (\(\gamma\)), without changing the number of protons or neutrons:

\[ {}^{A}_{Z}X^* \rightarrow {}^{A}_{Z}X + \gamma \]

Example: Cobalt-60 emits gamma radiation:

\[ {}^{60}_{27}Co^* \rightarrow {}^{60}_{27}Co + \gamma \]

Mass-Energy Equivalence

The energy released in nuclear reactions can be calculated using Einstein's equation:

\[ E = mc^2 \]

Where:

  • \( E \) = energy (J)
  • \( m \) = mass defect (kg)
  • \( c \) = speed of light (\( 3 \times 10^8 \, \text{ms}^{-1} \))

Cosmology

Hubble's Law

Hubble's Law states the relationship between the recession velocity of a galaxy and its distance from Earth:

\[ v = H_0 \cdot d \]

Where:

  • \( v \) = recession velocity of the galaxy (ms\(^{-1}\))
  • \( H_0 \) = Hubble constant (kms\(^{-1}\)Mpc\(^{-1}\))
  • \( d \) = distance to the galaxy (Mpc)