Physics 9702 / 21 — October/November 2025

(b)

A rocket is launched vertically from the surface of the Earth.

Fig. 1.1 shows the variation of the velocity of the rocket with time for the first 20s after its launch.

Fig. 1.1

3M

(i)

Determine the acceleration of the rocket.

acceleration = ______ $\text{m s}^{-2}$

DifficultyMedium-Easy

Worked solution

Working

Gradient of $v$ – $t$ graph:

a = \frac{\Delta v}{\Delta t} = \frac{320 - 0}{20 - 0} = 16\ \text{m s}^{-2}

Answer

$16\ \text{m s}^{-2}$

Final answer

16 m s^-2

Detailed explanation

Background Concept

For a velocity–time graph, the acceleration is the gradient (slope):

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

This works because acceleration is defined as the rate of change of velocity with time.

Understanding the Question

The rocket’s velocity increases linearly from $0\ \text{m s}^{-1}$ at $t=0\ \text{s}$ to $320\ \text{m s}^{-1}$ at $t=20\ \text{s}$ . You must find the acceleration from this graph.

Approach

Pick two clear points on the straight-line graph (the endpoints are easiest), calculate $\Delta v$ and $\Delta t$ , then divide.

Step-by-Step Reasoning

Read two points from the line: $(0\ \text{s}, 0\ \text{m s}^{-1})$ and $(20\ \text{s}, 320\ \text{m s}^{-1})$ .
Compute changes:

\Delta v = 320 - 0 = 320\ \text{m s}^{-1}

\Delta t = 20 - 0 = 20\ \text{s}

Gradient gives acceleration:

a = \frac{320}{20} = 16\ \text{m s}^{-2}

Key Takeaways

Acceleration equals gradient of a $v$ – $t$ graph.
Use large separations on the graph (e.g. endpoints) to reduce reading error.

Common Mistakes

Using $\Delta t / \Delta v$ instead of $\Delta v / \Delta t$ .
Reading values incorrectly from the axes (mixing up 200 and 400 gridlines, etc.).

Things to Be Careful About

Include units: gradient of $\text{m s}^{-1}$ against $\text{s}$ is $\text{m s}^{-2}$ .
Because the line goes through the origin, using endpoints is unambiguous here.

Techniques used

determine acceleration from the gradient of a velocity–time graphcalculate a gradient using change in velocity over change in timeuse units consistently when finding a gradient

(ii)

Show that the height of the rocket above the surface of the Earth at a time of 20s after launch is 3.2km.

DifficultyMedium-Easy

Worked solution

Working

Height = area under $v$ – $t$ graph (triangle):

\text{height} = \frac{1}{2} \times 20 \times 320 = 3200\ \text{m} = 3.2\ \text{km}

Answer

$3.2\ \text{km}$

Final answer

3.2 km

Detailed explanation

Background Concept

For a velocity–time graph, the displacement (change in position) is the area under the graph:

s = \int v\,dt

For straight-line segments, this is found using basic geometry (rectangles, triangles, trapezia).

Understanding the Question

At $t = 20\ \text{s}$ , the rocket’s velocity is $320\ \text{m s}^{-1}$ and the velocity increased linearly from zero. You are asked to show the rocket’s height above the surface is $3.2\ \text{km}$ . Since the rocket moves vertically upwards, the displacement equals the height.

Approach

Treat the region under the straight-line $v$ – $t$ graph from $0$ to $20\ \text{s}$ as a triangle. Find its area, giving displacement in metres, then convert to kilometres.

Step-by-Step Reasoning

The graph starts at $(0,0)$ and ends at $(20,320)$ , so the area from $0$ to $20\ \text{s}$ is a triangle.
Triangle base $= 20\ \text{s}$ and height $= 320\ \text{m s}^{-1}$ .
Area (displacement):

s = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 320 = 3200\ \text{m}

Convert to kilometres:

3200\ \text{m} = 3.2\ \text{km}

Key Takeaways

On a $v$ – $t$ graph: gradient gives acceleration; area gives displacement.
Linear increase in velocity produces a triangular area.

Common Mistakes

Using the gradient instead of the area (mixing up what each feature represents).
Forgetting the factor $\frac{1}{2}$ for a triangle.
Not converting $3200\ \text{m}$ to $3.2\ \text{km}$ .

Things to Be Careful About

Ensure the velocity stays positive; here it does, so area corresponds directly to upward displacement.
Units: $\text{s} \times \text{m s}^{-1} = \text{m}$ , which is correct for displacement.

Techniques used

use the area under a velocity–time graph to find displacementcalculate the area of a triangle from base and heightconvert metres to kilometres

(c)

The mass of the rocket in (b) is $2.9 \times 10^6\ \text{kg}$ . Assume that this mass remains constant.

For this rocket, from launch to its height at a time of 20s after launch:

6M

(i)

calculate the gain in gravitational potential energy $\Delta E_P$

$\Delta E_P$ = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

\Delta E_P = mgh = (2.9 \times 10^{6})(9.8)(3200)

\Delta E_P = 9.1 \times 10^{10}\ \text{J}

Answer

$9.1 \times 10^{10}\ \text{J}$

Final answer

9.1 × 10^10 J

Detailed explanation

Background Concept

Near the Earth’s surface, the change in gravitational potential energy is

\Delta E_P = mgh

where $m$ is mass, $g \approx 9.8\ \text{m s}^{-2}$ is gravitational field strength, and $h$ is the vertical height gained.

Understanding the Question

The rocket has mass $2.9 \times 10^{6}\ \text{kg}$ (assumed constant) and rises to a height of $3.2\ \text{km}$ in the first $20\ \text{s}$ . You must calculate its gain in gravitational potential energy over that interval.

Given/inferred:

$m = 2.9 \times 10^{6}\ \text{kg}$
$h = 3.2\ \text{km} = 3200\ \text{m}$
take $g = 9.8\ \text{m s}^{-2}$ (unless a different value is specified)

Approach

Convert height to metres, then apply $\Delta E_P = mgh$ and present the answer in joules, typically in standard form.

Step-by-Step Reasoning

Convert height:

h = 3.2\ \text{km} = 3.2 \times 10^{3}\ \text{m} = 3200\ \text{m}

Substitute:

\Delta E_P = (2.9 \times 10^{6})(9.8)(3.2 \times 10^{3})

Multiply the numbers and combine powers of ten:

2.9 \times 3.2 = 9.28

So

\Delta E_P \approx 9.28 \times 9.8 \times 10^{9} \approx 9.1 \times 10^{10}\ \text{J}

(Any close value depending on the chosen $g$ and rounding is acceptable.)

Key Takeaways

For vertical motion near Earth, $\Delta E_P = mgh$ .
Convert km to m before substitution.

Common Mistakes

Using $h = 3.2$ instead of $3200\ \text{m}$ .
Dropping the power of ten from the mass ( $2.9 \times 10^{6}$ ).
Giving units as N or W instead of J.

Things to Be Careful About

Use a consistent value of $g$ (typically $9.8$ or $9.81$ ) and round sensibly.
State the answer to an appropriate number of significant figures (often 2 s.f. here because $3.2$ is 2 s.f.).

Techniques used

use gravitational potential energy change $\Delta E = mgh$substitute numerical values with correct powers of tenexpress a result in standard form with unit

(ii)

calculate the gain in kinetic energy $\Delta E_K$

$\Delta E_K$ = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

Final speed at $20\ \text{s}$ : $v = 320\ \text{m s}^{-1}$ .

\Delta E_K = \frac{1}{2}mv^2 = \frac{1}{2}(2.9 \times 10^{6})(320)^2

\Delta E_K = 1.5 \times 10^{11}\ \text{J}

Answer

$1.5 \times 10^{11}\ \text{J}$

Final answer

1.5 × 10^11 J

Detailed explanation

Background Concept

Kinetic energy of an object of mass $m$ moving at speed $v$ is

E_K = \frac{1}{2}mv^2

If the object starts from rest, the gain in kinetic energy is simply its final kinetic energy.

Understanding the Question

From the velocity–time graph, the rocket’s velocity at $t=20\ \text{s}$ is $320\ \text{m s}^{-1}$ . The rocket starts from rest at launch, and the mass is constant at $2.9 \times 10^{6}\ \text{kg}$ . We need the increase in kinetic energy from $t=0$ to $t=20\ \text{s}$ .

Approach

Use $\Delta E_K = \frac{1}{2}m(v^2 - u^2)$ . Since $u=0$ , this becomes $\Delta E_K = \frac{1}{2}mv^2$ using $v = 320\ \text{m s}^{-1}$ .

Step-by-Step Reasoning

Initial speed $u = 0$ .
Final speed from the graph: $v = 320\ \text{m s}^{-1}$ .
Compute:

\Delta E_K = \frac{1}{2}m(v^2 - u^2) = \frac{1}{2}(2.9 \times 10^{6})(320^2)

Square the speed:

320^2 = 102400

Multiply:

\Delta E_K = 0.5 \times 2.9 \times 10^{6} \times 102400 \approx 1.48 \times 10^{11}\ \text{J} \approx 1.5 \times 10^{11}\ \text{J}

Key Takeaways

If motion starts from rest, $\Delta E_K = \frac{1}{2}mv^2$ .
Read the correct final speed from the graph.

Common Mistakes

Using the average speed or area under the graph instead of the final speed.
Forgetting the factor $\frac{1}{2}$ .
Not squaring $v$ correctly.

Things to Be Careful About

Ensure you use speed (magnitude of velocity) for kinetic energy.
Keep track of large numbers: use standard form to avoid arithmetic slips.

Techniques used

use kinetic energy change $\Delta E = \frac{1}{2}mv^2$read a final speed from the velocity–time graphcalculate a square and present the result in standard form

(iii)

determine the average power output of the rocket engines. Assume that resistive forces are negligible.

power = ______ $\text{W}$

DifficultyMedium

Worked solution

Working

With resistive forces negligible,

E_{\text{engines}} = \Delta E_P + \Delta E_K

E_{\text{engines}} = (9.1 \times 10^{10}) + (1.5 \times 10^{11}) = 2.4 \times 10^{11}\ \text{J}

Average power over $20\ \text{s}$ :

P = \frac{E_{\text{engines}}}{t} = \frac{2.4 \times 10^{11}}{20} = 1.2 \times 10^{10}\ \text{W}

Answer

$1.2 \times 10^{10}\ \text{W}$

Final answer

1.2 × 10^10 W

Detailed explanation

Background Concept

Power is the rate of energy transfer:

P = \frac{\Delta E}{\Delta t}

If resistive forces (air resistance, etc.) are negligible, then the work done (energy supplied) by the engines becomes the increase in the rocket’s mechanical energy:

E_{\text{engines}} = \Delta E_P + \Delta E_K

Understanding the Question

From launch to $t=20\ \text{s}$ , the rocket gains gravitational potential energy by rising and gains kinetic energy by speeding up. The question asks for the average power output of the engines over these $20\ \text{s}$ , assuming no energy is lost to resistive forces.

Known from earlier parts:

$\Delta E_P \approx 9.1 \times 10^{10}\ \text{J}$
$\Delta E_K \approx 1.5 \times 10^{11}\ \text{J}$
$\Delta t = 20\ \text{s}$

Approach

Add the two energy gains to find the total energy supplied by the engines.
Divide by the time interval to get average power.

Step-by-Step Reasoning

Neglecting resistive forces means the engines’ energy goes into:
- increasing $E_P$ (height gain), and
- increasing $E_K$ (speed gain).

So total energy transfer from engines:

E_{\text{engines}} = \Delta E_P + \Delta E_K

Substitute the values:

E_{\text{engines}} \approx 9.1 \times 10^{10} + 1.5 \times 10^{11} = 2.41 \times 10^{11}\ \text{J} \approx 2.4 \times 10^{11}\ \text{J}

Average power:

P = \frac{E_{\text{engines}}}{\Delta t} = \frac{2.4 \times 10^{11}}{20} = 1.2 \times 10^{10}\ \text{W}

Key Takeaways

With negligible resistive forces, engine work equals increase in mechanical energy.
Average power is total energy transferred divided by time.

Common Mistakes

Using only $\Delta E_P$ or only $\Delta E_K$ instead of both.
Dividing by the wrong time (e.g. using 2 s from a gridline rather than 20 s total).
Confusing power with force (giving answer in N).

Things to Be Careful About

The result is an average power over 0–20 s; instantaneous power could vary even though the velocity–time graph is linear.
Keep consistent significant figures: earlier values are about 2 s.f., so the final power should be similar.

Techniques used

add energy transfers to find total work done by the enginesuse $P = \Delta E / \Delta t$ to find average powerapply the assumption of negligible resistive forces

Q2MediumForces, Density and PressureDynamicsPhysical Quantities and Units

(a)

3M

(i)

Define pressure.

(ii)

Explain how hydrostatic pressure results in an upthrust force acting on a solid object immersed in a liquid.

DifficultyMedium-Easy

Worked solution

Answer

Hydrostatic pressure increases with depth, so the pressure on the bottom of the object is greater than on the top.

This produces a resultant upward force (difference in pressure $\times$ area), called upthrust (equal to the weight of displaced liquid).

Final answer

Pressure is greater at greater depth, so the bottom of the object experiences a larger pressure than the top, giving a net upward force (upthrust), equal to the weight of displaced liquid.

Detailed explanation

Background Concept

In a stationary liquid, pressure increases with depth:

p = \rho g h

where $\rho$ is the liquid density, $g$ is gravitational field strength, and $h$ is depth below the surface.

Forces on a surface come from pressure acting over area:

F = pA

Upthrust (buoyant force) occurs because pressure is not the same on all faces of an immersed object.

Understanding the Question

You must explain, using hydrostatic pressure, why there is an upward force on an immersed solid. The key idea is “pressure increases with depth”, so forces on the top and bottom faces differ.

Approach

State that pressure is larger deeper in the liquid.
Convert pressures on top and bottom faces into forces using $F=pA$ .
The difference between these forces is a resultant upward force (upthrust), which can also be stated as equal to the weight of displaced liquid.

Step-by-Step Reasoning

Consider the top face and bottom face of the object.
The bottom face is at a greater depth, so

p_{\text{bottom}} > p_{\text{top}}

The liquid pushes on each face with force $F=pA$ .
Therefore the upward force on the bottom face is larger than the downward force on the top face.
The side forces cancel (equal pressures at the same depth acting in opposite horizontal directions).
The net result is an upward force:

U = (p_{\text{bottom}} - p_{\text{top}})A

This net upward force is the upthrust, and it equals the weight of liquid displaced (Archimedes’ principle).

Key Takeaways

Pressure in a fluid increases with depth.
Pressure differences lead to a resultant force.
Upthrust arises because the bottom experiences a greater pressure than the top.

Common Mistakes

Saying “upthrust happens because the object is lighter” (cause and effect reversed).
Forgetting that the liquid also exerts forces on the top face (downwards).
Not mentioning pressure difference with depth.

Things to Be Careful About

Make clear that the difference in pressure (not just “pressure”) causes the net force.
Side forces cancel; only vertical pressure forces produce the resultant upthrust.

Techniques used

use the fact that pressure increases with depth in a fluiddeduce a resultant force from pressure difference on opposite facesrelate the net force to displaced fluid

(b)

A small steel ball of radius $r$ and mass $m$ falls vertically at terminal speed $v$ through oil.

The viscous drag force $D$ that acts on the ball is given by

D = 6\pi\eta rv

where $\eta$ is a property of the oil called its viscosity.

5M

(i)

On Fig. 2.1, draw labelled arrows from the ball to show the directions of the three forces that act on the ball as it falls.

Fig. 2.1

(ii)

Determine the SI base units of $\eta$ .

base units ______

DifficultyMedium-Easy

Worked solution

Working

From

D = 6\pi\eta rv

\eta = \frac{D}{6\pi rv}

Units:

[\eta] = \frac{\text{N}}{\text{m}\,(\text{m s}^{-1})} = \frac{\text{kg m s}^{-2}}{\text{m}^2\,\text{s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}

Answer

$\text{kg m}^{-1}\text{s}^{-1}$

Final answer

kg m^-1 s^-1

Detailed explanation

Background Concept

To find SI base units of a quantity, use an equation that relates it to known quantities and substitute units for each symbol. The constant $6\pi$ is dimensionless, so it does not affect units.

Force unit:

1\ \text{N} = 1\ \text{kg m s}^{-2}

Understanding the Question

You are given Stokes’ law for drag on a sphere:

D = 6\pi\eta rv

and asked for SI base units of viscosity $\eta$ .

Approach

Rearrange the equation to make $\eta$ the subject, then substitute SI units for $D$ , $r$ , and $v$ , and simplify to base units.

Step-by-Step Reasoning

Rearrange:

\eta = \frac{D}{6\pi rv}

Now substitute units:

$D$ has units N.
$r$ has units m.
$v$ has units $\text{m s}^{-1}$ .

So

[\eta] = \frac{\text{N}}{\text{m}\,(\text{m s}^{-1})}

Replace N with base units:

[\eta] = \frac{\text{kg m s}^{-2}}{\text{m}^2\,\text{s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}

Key Takeaways

Rearranging first avoids confusion.
Dimensionless constants (like $6\pi$ ) do not change units.
Viscosity $\eta$ has base units $\text{kg m}^{-1}\text{s}^{-1}$ .

Common Mistakes

Treating $6\pi$ as if it has units.
Using $\text{m s}^{-2}$ for velocity (mixing up $v$ and $a$ ).
Leaving the answer in N s m $^{-2}$ without converting to base units when asked.

Things to Be Careful About

Write velocity as $\text{m s}^{-1}$ .
Ensure final expression is in SI base units (kg, m, s).

Techniques used

rearrange an equation to isolate a target quantitysubstitute units into a formulaexpress derived units in SI base units

(c)

The oil in (b) has a density of $920\ \text{kg m}^{-3}$ and a viscosity of 4.7 in SI units.

The steel ball has a mass of $2.4 \times 10^{-3}\ \text{kg}$ and a radius of $4.2 \times 10^{-3}\ \text{m}$ .

4M

(i)

Show that the upthrust force acting on the ball is $2.8 \times 10^{-3}\ \text{N}$ .

DifficultyMedium-Easy

Worked solution

Working

Volume of ball:

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (4.2 \times 10^{-3})^3 = 3.10 \times 10^{-7}\ \text{m}^3

Upthrust:

U = \rho gV = 920 \times 9.81 \times 3.10 \times 10^{-7} = 2.8 \times 10^{-3}\ \text{N}

Answer

$2.8 \times 10^{-3}\ \text{N}$

Final answer

2.8 × 10^-3 N

Detailed explanation

Background Concept

Archimedes’ principle states that the upthrust (buoyant force) on an immersed object equals the weight of the fluid displaced:

U = \rho g V

where:

$\rho$ is the fluid density,
$g$ is gravitational field strength,
$V$ is the volume of fluid displaced (for a fully immersed object, this is the object’s volume).

For a sphere:

V = \frac{4}{3}\pi r^3

Understanding the Question

You are given oil density $\rho = 920\ \text{kg m}^{-3}$ and the ball radius $r = 4.2 \times 10^{-3}\ \text{m}$ . The ball is immersed in the oil, so it displaces a volume of oil equal to its own volume. You must show the upthrust is $2.8 \times 10^{-3}\ \text{N}$ .

Approach

Compute the sphere’s volume from its radius.
Multiply by $\rho g$ to obtain upthrust.

Step-by-Step Reasoning

Calculate volume:

V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (4.2 \times 10^{-3})^3

Numerically this is

V \approx 3.10 \times 10^{-7}\ \text{m}^3

Now calculate upthrust:

U = \rho g V = 920 \times 9.81 \times 3.10 \times 10^{-7}

U \approx 2.8 \times 10^{-3}\ \text{N}

which matches the required value.

Key Takeaways

Upthrust depends on fluid density and displaced volume, not on the object’s mass.
For a fully submerged sphere, use $V=\frac{4}{3}\pi r^3$ .

Common Mistakes

Using the steel density instead of the oil density.
Using diameter instead of radius.
Forgetting the factor $\frac{4}{3}\pi$ .

Things to Be Careful About

Keep powers of ten consistent: $(10^{-3})^3 = 10^{-9}$ .
Include units: $\rho$ in $\text{kg m}^{-3}$ gives $U$ in newtons when multiplied by $g$ and $V$ .

Techniques used

calculate volume of a sphereapply Archimedes' principle U = rho g Vuse standard form and correct units

(ii)

Determine the terminal speed $v$ of the ball.

$v$ = ______ $\text{m s}^{-1}$

DifficultyMedium

Worked solution

Working

At terminal speed, resultant force is zero:

mg = U + D \quad \Rightarrow \quad D = mg - U

mg = (2.4 \times 10^{-3})(9.81) = 2.35 \times 10^{-2}\ \text{N}

D = 2.35 \times 10^{-2} - 2.8 \times 10^{-3} = 2.07 \times 10^{-2}\ \text{N}

Using

D = 6\pi \eta r v \quad \Rightarrow \quad v = \frac{D}{6\pi\eta r}

v = \frac{2.07 \times 10^{-2}}{6\pi (4.7)(4.2 \times 10^{-3})} = 5.6 \times 10^{-2}\ \text{m s}^{-1}

Answer

$5.6 \times 10^{-2}\ \text{m s}^{-1}$

Final answer

5.6 × 10^-2 m s^-1

Detailed explanation

Background Concept

Terminal speed occurs when an object moving through a fluid has zero acceleration, so the resultant force is zero.

For a falling sphere in a fluid:

Weight $mg$ acts downward.
Upthrust $U$ acts upward.
Drag $D$ acts upward (opposes downward motion).

At terminal speed:

mg = U + D

Stokes’ law for viscous drag on a sphere is:

D = 6\pi\eta r v

Understanding the Question

You are given:

$m = 2.4 \times 10^{-3}\ \text{kg}$ ,
$r = 4.2 \times 10^{-3}\ \text{m}$ ,
$\eta = 4.7$ (SI units),
Upthrust from (c)(i): $U = 2.8 \times 10^{-3}\ \text{N}$ .

You must find the terminal speed $v$ .

Approach

Use terminal-speed condition (forces balance) to find drag $D$ .
Substitute $D$ into Stokes’ law and rearrange for $v$ .

Step-by-Step Reasoning

Force balance at terminal speed:

mg = U + D \Rightarrow D = mg - U

Calculate weight:

mg = (2.4 \times 10^{-3})(9.81) = 2.3544 \times 10^{-2}\ \text{N}

Now drag:

D = 2.3544 \times 10^{-2} - 2.8 \times 10^{-3} \approx 2.07 \times 10^{-2}\ \text{N}

Use Stokes’ law:

D = 6\pi\eta r v \Rightarrow v = \frac{D}{6\pi\eta r}

Substitute values:

v = \frac{2.07 \times 10^{-2}}{6\pi (4.7)(4.2 \times 10^{-3})}

This gives

v \approx 5.6 \times 10^{-2}\ \text{m s}^{-1}

Key Takeaways

At terminal speed, set resultant force to zero.
For a falling object in a fluid: $mg$ is balanced by upthrust + drag.
Stokes’ law links drag to viscosity, radius, and speed.

Common Mistakes

Setting $mg = D$ and forgetting upthrust.
Using $D = U - mg$ (wrong sign).
Substituting radius in mm instead of m.

Things to Be Careful About

Terminal speed means $a=0$ , not that forces disappear.
Keep all quantities in SI units before substitution.
Use a sensible number of significant figures (typically 2 s.f. here, matching given data).

Techniques used

apply equilibrium of forces at terminal velocitycalculate weight and use subtraction to find drag forcerearrange Stokes' law to solve for terminal speed

Q3Medium-EasyDeformation of SolidsElectricity

A wire has length $L$ and cross-sectional area $A$ . The wire is made from a metal that has Young modulus $E$ and resistivity $\rho$ .

(a)

Define the Young modulus of a material.

DifficultyEasy

Worked solution

Answer

Young modulus $E$ is the ratio of tensile stress to tensile strain, within the limit of proportionality:

E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}

Final answer

Young modulus is tensile stress divided by tensile strain (within the limit of proportionality).

Detailed explanation

Background Concept

Young modulus $E$ measures how stiff a material is when stretched (or compressed) elastically. It is defined using:

Stress: force per unit cross-sectional area

\text{stress} = \frac{F}{A}

Strain: fractional change in length

\text{strain} = \frac{\Delta L}{L}

Within the limit of proportionality (the straight-line region where Hooke’s law holds), stress is proportional to strain, so their ratio is constant.

Understanding the Question

You are asked to define Young modulus. That means you must give the relationship between stress and strain and state the condition (valid only where stress ∝ strain).

Approach

Write the definition in words and/or as an equation. Include “tensile” and “within the limit of proportionality” to match what examiners look for.

Step-by-Step Reasoning

Recognise that Young modulus uses stress and strain.
State the ratio:

E = \frac{\text{tensile stress}}{\text{tensile strain}}

If writing in terms of measured quantities for a wire:

E = \frac{F/A}{\Delta L/L}

This is enough for full credit.

Key Takeaways

$E$ is a material property that links stress and strain in the linear elastic region.
Always mention the limit of proportionality.

Common Mistakes

Defining $E$ as $F/\Delta L$ (that is spring constant $k$ for a specific object, not a material property).
Omitting “stress” and “strain” and only describing “force and extension”.

Things to Be Careful About

Young modulus applies to uniform stress and strain and to the straight-line region of the stress–strain graph.
Stress and strain are different: stress has units of Pa, strain is dimensionless.

Techniques used

state the definition of Young modulusidentify the condition for validity (limit of proportionality)

(b)

3M

(i)

State an expression, in terms of some or all of $L$ , $A$ , $E$ and $\rho$ , for the resistance $R_0$ of the wire.

$R_0$ = ______

(ii)

Show that the spring constant $k_0$ of the wire is given by

k_0 = \frac{EA}{L}

DifficultyMedium-Easy

Worked solution

Working

E = \frac{F/A}{\Delta L/L}

E = \frac{FL}{A\Delta L} \;\Rightarrow\; F = \frac{EA}{L}\,\Delta L

Since $F = k_0\Delta L$ ,

k_0 = \frac{EA}{L}

Answer

k_0 = \frac{EA}{L}

Final answer

k0 = EA/L

Detailed explanation

Background Concept

A wire behaves like a spring in the elastic (proportional) region:

F = k\,x

where $x = \Delta L$ is the extension and $k$ is the spring constant.

Young modulus $E$ connects force and extension via stress and strain:

E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}

This is essentially “Hooke’s law” written in a way that separates material property ( $E$ ) from geometry ( $L$ and $A$ ).

Understanding the Question

You must show that the spring constant of a uniform wire (original length $L$ , area $A$ , Young modulus $E$ ) is

k_0 = \frac{EA}{L}.

So you need to manipulate the Young modulus definition until it looks like $F = k_0\Delta L$ .

Approach

Write $E = (F/A)/(\Delta L/L)$ .
Rearrange to make $F$ the subject.
Compare with $F = k\Delta L$ to read off $k$ .

Step-by-Step Reasoning

Start from the definition:

E = \frac{F/A}{\Delta L/L}

Divide by a fraction by multiplying by its reciprocal:

E = \frac{F}{A}\cdot\frac{L}{\Delta L} = \frac{FL}{A\Delta L}

Rearrange for $F$ :

F = \frac{EA}{L}\,\Delta L

Compare with Hooke’s law form $F = k_0\Delta L$ . Therefore the coefficient of $\Delta L$ is the spring constant:

k_0 = \frac{EA}{L}.

Key Takeaways

For a wire, the “spring constant” depends on geometry and material:
- larger $A$ → stiffer (larger $k$ )
- larger $L$ → less stiff (smaller $k$ )
- larger $E$ → stiffer

Common Mistakes

Using $E = F/\Delta L$ (confusing $E$ with $k$ ).
Forgetting that strain is $\Delta L/L$ (missing the division by $L$ ).

Things to Be Careful About

This result holds only in the linear elastic region (limit of proportionality).
$k$ here is for the wire as a spring, so it changes if $L$ changes.

Techniques used

use stress and strain definitionsapply Young modulus relationshiprearrange to identify spring constant as force per extension

(c)

The wire is stretched, within the limit of proportionality, by a tensile force $F$ . Assume that any changes in the cross-sectional area of the wire are negligible.

2M

(i)

On Fig. 3.1, sketch the variation with $F$ of the resistance $R$ of the wire.

Fig. 3.1

DifficultyMedium-Easy

Worked solution

graph-R-vs-F

Answer

$R$ increases linearly with $F$ (straight line with positive gradient) starting at $(F=0,\,R=R_0)$ .

Final answer

Straight line increasing from R0 at F = 0.

Detailed explanation

Background Concept

Resistance of a uniform wire is

R = \frac{\rho L}{A}

If $\rho$ and $A$ are constant, then $R \propto L$ .

When a wire is stretched within the limit of proportionality, extension is proportional to force:

\Delta L \propto F

so the new length is $L' = L + \Delta L$ , which increases linearly with $F$ .

Understanding the Question

You are stretching a wire elastically with tensile force $F$ . You are told to neglect changes in cross-sectional area, so take $A$ constant. You must sketch how the wire’s resistance $R$ varies with $F$ , and the axes already show that at $F=0$ the resistance is $R_0$ .

Approach

Use $R = \rho L/A$ .
Under elastic stretching, $\Delta L \propto F$ , so $L' = L + \Delta L$ increases with $F$ .
Therefore $R$ increases with $F$ . For a 1-mark sketch, show the correct trend and intercept at $R_0$ .

Step-by-Step Reasoning

Start with

R = \frac{\rho L'}{A}

With $A$ constant and $\rho$ constant, $R \propto L'$ .
In the proportional region, $\Delta L \propto F$ , hence

L' = L + \Delta L

increases linearly with $F$ .
4. Therefore $R$ increases linearly with $F$ .
5. At $F=0$ , $L' = L$ , so $R=R_0$ , meaning the graph passes through $(0, R_0)$ .

graph-R-vs-F

Key Takeaways

If area change is negligible, stretching increases $L$ and hence increases $R$ .
In the Hooke’s law region, the increase is proportional to $F$ , giving a straight line.

Common Mistakes

Drawing the line through the origin instead of starting at $R_0$ .
Drawing $R$ decreasing with $F$ .
Assuming $A$ decreases significantly (the question tells you to neglect this).

Things to Be Careful About

The y-intercept is given as $R_0$ on the axis: the sketch must start there at $F=0$ .
Only valid within the limit of proportionality; beyond that, the trend could become non-linear.

Techniques used

relate tensile force to extension using Hooke's law behaviourlink resistance change to change in length using R = rho L / Asketch a graph from a qualitative proportionality

(ii)

On Fig. 3.2, sketch the variation with $F$ of the spring constant $k$ of the wire.

Fig. 3.2

DifficultyMedium-Easy

Worked solution

graph-k-vs-F

Answer

$k$ decreases as $F$ increases, starting at $(F=0,\,k=k_0)$ .

Final answer

k decreases from k0 as F increases.

Detailed explanation

Background Concept

For a wire acting as a spring (in the proportional region), the spring constant is

k = \frac{EA}{L}

If $E$ (material) and $A$ (cross-sectional area) are constant, then

k \propto \frac{1}{L}.

Stretching the wire increases its length $L$ , so it becomes less stiff (smaller $k$ ).

Understanding the Question

You are told the wire is stretched within the limit of proportionality by a tensile force $F$ , and changes in area are negligible. You must sketch how $k$ varies with $F$ on axes where the initial value $k_0$ at $F=0$ is shown.

Approach

As $F$ increases (still elastic), the wire length increases.
Because $k = EA/L$ , increasing $L$ makes $k$ smaller.
For a 1-mark sketch, the key is correct direction of change and correct intercept at $k_0$ .

Step-by-Step Reasoning

Initially, when $F=0$ , the wire has length $L$ and spring constant

k_0 = \frac{EA}{L}.

Under tension, the wire extends: new length $L' = L + \Delta L$ .
With $E$ and $A$ constant,

k = \frac{EA}{L'}

and since $L' > L$ , then $k < k_0$ .
4. As $F$ increases, $\Delta L$ increases, so $L'$ increases and $k$ decreases further.

graph-k-vs-F

Key Takeaways

A wire becomes less stiff as it is stretched (because stiffness is inversely proportional to length).
Always use the intercepts shown ( $k_0$ at $F=0$ ).

Common Mistakes

Drawing $k$ constant (forgetting $k$ depends on $L$ ).
Drawing $k$ increasing with $F$ .
Drawing the curve through the origin instead of starting at $k_0$ .

Things to Be Careful About

The question specifies extension is within the limit of proportionality, but even there $k$ need not be constant because the object is changing length.
Since only a sketch is required, detailed curvature is not usually assessed, but the trend must be correct.

Techniques used

use k = EA/L for a wireinfer how k changes when length increases under tensionsketch a qualitative trend on axes with given intercept

(d)

Copper has a resistivity of $1.8 \times 10^{-8}\ \Omega\text{m}$ and a Young modulus of $1.3 \times 10^{11}\ \text{Pa}$ .

A copper wire of diameter $1.6\ \text{mm}$ has a resistance of $0.034\ \Omega$ .

3M

(i)

Show that the length of the wire is $3.8\ \text{m}$ .

DifficultyMedium-Easy

Worked solution

Working

Diameter $d = 1.6\ \text{mm} = 1.6 \times 10^{-3}\ \text{m}$ , so

A = \pi\left(\frac{d}{2}\right)^2 = \pi (0.8 \times 10^{-3})^2 = 2.01 \times 10^{-6}\ \text{m}^2

Using $R = \rho L/A$ ,

L = \frac{RA}{\rho} = \frac{0.034 \times 2.01 \times 10^{-6}}{1.8 \times 10^{-8}} = 3.8\ \text{m}

Answer

$L = 3.8\ \text{m}$

Final answer

3.8 m

Detailed explanation

Background Concept

For a uniform wire,

R = \frac{\rho L}{A}

where:

$R$ is resistance ( $\Omega$ )
$\rho$ is resistivity ( $\Omega\text{ m}$ )
$L$ is length ( $\text{m}$ )
$A$ is cross-sectional area ( $\text{m}^2$ )

If the wire is circular with diameter $d$ , then

A = \pi\left(\frac{d}{2}\right)^2.

Understanding the Question

You are given for copper:

$\rho = 1.8 \times 10^{-8}\ \Omega\text{ m}$
wire diameter $d = 1.6\ \text{mm}$
resistance $R = 0.034\ \Omega$

You must show the length is $3.8\ \text{m}$ . This is a straightforward application of $R = \rho L/A$ after finding $A$ .

Approach

Convert diameter to metres.
Calculate area $A$ .
Rearrange the resistivity equation to $L = RA/\rho$ .
Substitute values carefully in SI units.

Step-by-Step Reasoning

Convert diameter:

d = 1.6\ \text{mm} = 1.6 \times 10^{-3}\ \text{m}

So radius $r = d/2 = 0.8 \times 10^{-3}\ \text{m}$ .

Area:

A = \pi r^2 = \pi (0.8 \times 10^{-3})^2 = \pi (0.64 \times 10^{-6}) \approx 2.01 \times 10^{-6}\ \text{m}^2

Rearrange $R = \rho L/A$ :

L = \frac{RA}{\rho}

Substitute:

L = \frac{0.034 \times 2.01 \times 10^{-6}}{1.8 \times 10^{-8}}

Compute powers of ten first:

\frac{10^{-6}}{10^{-8}} = 10^2

Then numbers:

\frac{0.034 \times 2.01}{1.8} \times 10^2 \approx 3.8\ \text{m}

Key Takeaways

Always convert mm to m before using SI equations.
For circular wires, area comes from $A = \pi (d/2)^2$ .

Common Mistakes

Using $A = \pi d^2$ instead of $\pi(d/2)^2$ .
Forgetting to convert $\text{mm}$ to $\text{m}$ (gives length wrong by a factor of $10^3$ ).
Rearranging incorrectly (e.g. $L = \rho A/R$ ).

Things to Be Careful About

Keep enough significant figures during intermediate steps (area) to avoid rounding error.
Check units: $\Omega \cdot \text{m}^2 /(\Omega\text{ m}) = \text{m}$ .

Techniques used

calculate cross-sectional area from diameterrearrange R = rho L / A to solve for lengthuse standard form and consistent SI units

(ii)

Use the equation in (b)(ii) to determine the spring constant of the wire.

spring constant = ______ $\text{N m}^{-1}$

DifficultyMedium-Easy

Worked solution

Working

A = 2.01 \times 10^{-6}\ \text{m}^2, \quad L = 3.8\ \text{m}

k = \frac{EA}{L} = \frac{(1.3 \times 10^{11})(2.01 \times 10^{-6})}{3.8} = 6.9 \times 10^{4}\ \text{N m}^{-1}

Answer

$k = 6.9 \times 10^{4}\ \text{N m}^{-1}$

Final answer

6.9 × 10^4 N m^-1

Detailed explanation

Background Concept

For a wire behaving elastically, the effective spring constant is

k = \frac{EA}{L}

where:

$E$ is Young modulus ( $\text{Pa} = \text{N m}^{-2}$ )
$A$ is cross-sectional area ( $\text{m}^2$ )
$L$ is length ( $\text{m}$ )

The units check works nicely:

E A \;\text{has units}\; (\text{N m}^{-2})(\text{m}^2) = \text{N}

so dividing by $L$ (m) gives $\text{N m}^{-1}$ .

Understanding the Question

You are told to use the result from (b)(ii) and the given copper values. From (d)(i) the wire length is $3.8\ \text{m}$ , and from the diameter you can use the same area $A$ calculated previously.

Approach

Use $A = \pi(d/2)^2$ (already found in (d)(i)).
Substitute $E$ , $A$ , and $L$ into $k = EA/L$ .
Give the answer in $\text{N m}^{-1}$ .

Step-by-Step Reasoning

Cross-sectional area (from diameter $1.6\ \text{mm}$ ):

A = 2.01 \times 10^{-6}\ \text{m}^2

Use $L = 3.8\ \text{m}$ and $E = 1.3 \times 10^{11}\ \text{Pa}$ .
Substitute into

k = \frac{EA}{L}

so

k = \frac{(1.3 \times 10^{11})(2.01 \times 10^{-6})}{3.8}

Multiply first:

(1.3 \times 10^{11})(2.01 \times 10^{-6}) = (1.3\times 2.01)\times 10^{5} \approx 2.61 \times 10^{5}

Then divide by $3.8$ :

k \approx \frac{2.61 \times 10^{5}}{3.8} \approx 6.9 \times 10^{4}\ \text{N m}^{-1}

Key Takeaways

A wire’s stiffness increases with area and Young modulus and decreases with length.
Always check the final unit for spring constant is $\text{N m}^{-1}$ .

Common Mistakes

Using diameter instead of area directly in $k = EA/L$ .
Forgetting to use SI units for $A$ (must be $\text{m}^2$ ).
Arithmetic error with powers of ten: $10^{11} \times 10^{-6} = 10^{5}$ .

Things to Be Careful About

Use the length found in (d)(i); don’t reuse the original symbolic $L$ .
Quote an appropriate number of significant figures (typically 2 s.f. is fine here, matching given data).

Techniques used

use k = EA/L for a wiresubstitute numerical values with SI unitshandle powers of ten and significant figures

Q4MediumSuperpositionWaves

(a)

State what is meant by diffraction of a wave.

DifficultyEasy

Worked solution

Answer

Diffraction is the spreading (bending) of a wave as it passes through a gap or around an obstacle.

The effect is significant when the gap/obstacle size is comparable to the wavelength.

Final answer

Diffraction is the spreading (bending) of a wave as it passes through a gap or around an obstacle, most noticeable when the gap/obstacle size is comparable to the wavelength.

Detailed explanation

Background Concept

Diffraction is a wave effect that occurs because different parts of a wavefront can interfere with each other after the wave encounters an aperture (gap) or an obstacle. When a wave passes through a narrow gap, the emerging wavefronts are no longer straight: they spread out into the region beyond the gap.

The amount of spreading depends mainly on the ratio of the aperture size to the wavelength. If the aperture is very large compared with the wavelength, the wave continues almost straight with little spreading. If the aperture size is comparable to the wavelength, the wave spreads out strongly.

Understanding the Question

The question asks for what diffraction “means”, i.e. a definition. For full credit you should mention:

spreading/bending of waves at an aperture or around an edge/obstacle, and
that it becomes significant when the aperture/obstacle size is comparable to the wavelength.

Approach

Give a concise definition (spreading around edges / through gaps), then add the condition for when it is most noticeable (gap size (\sim \lambda)).

Step-by-Step Reasoning

State the core meaning: the wave spreads out (or bends) when it encounters a gap or obstacle.
Add the standard condition: diffraction is greatest when the gap (or obstacle size) is of the same order as the wavelength.

Key Takeaways

Diffraction = spreading of a wave after an aperture/edge.
The smaller the aperture compared to (\lambda), the greater the spreading.

Common Mistakes

Saying only “bending of light” without mentioning gaps/edges (too vague).
Confusing diffraction with refraction (refraction is change of direction due to change of speed in a new medium).

Things to Be Careful About

Use wave language (spreading of the wave/wavefront).
Include the “comparable to wavelength” condition to secure the second mark when available.

Techniques used

state the definition of diffractionrelate diffraction to spreading of wavefronts at apertures/edges

(b)

A beam of vertically polarised light of wavelength $540\ \text{nm}$ is incident normally on a diffraction grating, as shown in Fig. 4.1.

Fig. 4.1

The diffraction grating has a line spacing of $5.0 \times 10^{-6}\ \text{m}$ .

The light transmitted by the diffraction grating illuminates a circular screen. The diffraction grating is at the centre X of the circle.

The central bright fringe is formed at point O on the screen and has intensity $I_0$ .

P is a point on the screen where the line XP is at a variable angle $\theta$ to the line XO. The intensity $I$ of light on the screen at P varies with $\theta$ .

6M

(i)

Show that the angle $\theta$ at which the first-order bright fringe is formed is $6.2^\circ$ .

DifficultyMedium-Easy

Worked solution

Working

For a grating,

d\sin\theta = n\lambda

First order: $n=1$ .

\sin\theta = \frac{\lambda}{d} = \frac{540\times10^{-9}}{5.0\times10^{-6}} = 0.108

\theta = \sin^{-1}(0.108) = 6.2^\circ

Answer

$\theta = 6.2^\circ$

Final answer

6.2°

Detailed explanation

Background Concept

A diffraction grating has many equally spaced slits. Light from adjacent slits interferes, producing principal maxima (bright fringes) at angles where the path difference between waves from adjacent slits is an integer number of wavelengths.

The grating equation is:

d\sin\theta = n\lambda

where:

$d$ is the slit (line) spacing,
$\theta$ is the angle from the central (straight-through) direction,
$n$ is the order number ( $0,1,2,\dots$ ),
$\lambda$ is the wavelength.

This applies for normal incidence and when measuring $\theta$ from the central maximum direction.

Understanding the Question

You are given:

wavelength $\lambda = 540\ \text{nm}$ ,
line spacing $d = 5.0\times 10^{-6}\ \text{m}$ ,
need the angle for the first-order maximum ( $n=1$ ).

“Show that” means you must demonstrate the calculation clearly.

Approach

Convert $\lambda$ into metres, use $n=1$ in $d\sin\theta=n\lambda$ , calculate $\sin\theta$ , then take $\sin^{-1}$ to obtain $\theta$ .

Step-by-Step Reasoning

Convert wavelength:

540\ \text{nm} = 540\times 10^{-9}\ \text{m}

For first order, $n=1$ , so:

\sin\theta = \frac{n\lambda}{d} = \frac{540\times10^{-9}}{5.0\times10^{-6}}

Evaluate the fraction:

\sin\theta = 0.108

Find angle:

\theta = \sin^{-1}(0.108) \approx 6.2^\circ

(Angle is small, so it is reasonable it comes out only a few degrees.)

Key Takeaways

Use $d\sin\theta=n\lambda$ for grating maxima.
Always put $\lambda$ and $d$ into consistent SI units.

Common Mistakes

Using $d = 5.0\times10^{-6}$ as if it were $5.0\times10^{-6}\ \text{nm}$ (unit mismatch).
Forgetting $n=1$ for first order.
Using $\tan\theta$ instead of $\sin\theta$ (not the grating equation).

Things to Be Careful About

$\theta$ is measured from the central direction (the $n=0$ maximum).
Keep enough significant figures in intermediate steps so the final $\theta$ rounds correctly to $6.2^\circ$ .

Techniques used

apply the diffraction grating equationsubstitute values in SI unitsuse inverse sine to find an angle

(ii)

Determine the value of $\theta$ at which the second-order bright fringe is formed.

$\theta$ = ______ $^\circ$

DifficultyMedium-Easy

Worked solution

Working

d\sin\theta = n\lambda

Second order: $n=2$ .

\sin\theta = \frac{2\lambda}{d} = 2\times 0.108 = 0.216

\theta = \sin^{-1}(0.216) = 12.5^\circ

Answer

$\theta = 12.5^\circ$

Final answer

12.5°

Detailed explanation

Background Concept

For a diffraction grating, the $n$ th order maximum occurs when the path difference between adjacent slits equals $n\lambda$ , giving:

d\sin\theta = n\lambda

As $n$ increases, $\sin\theta$ increases, so the maxima move further from the central maximum, until eventually $\sin\theta$ would exceed 1 and no higher orders are possible.

Understanding the Question

You must find the angle for the second-order bright fringe ( $n=2$ ) using the same $d$ and $\lambda$ as in part (i).

Approach

Use $n=2$ in the grating equation, calculate $\sin\theta$ , then take $\sin^{-1}$ .

Step-by-Step Reasoning

Start with:

\sin\theta = \frac{n\lambda}{d}

Substitute $n=2$ :

\sin\theta = \frac{2\lambda}{d}

Using the value from (i), $\lambda/d = 0.108$ :

\sin\theta = 2(0.108) = 0.216

Take inverse sine:

\theta = \sin^{-1}(0.216) \approx 12.5^\circ

Key Takeaways

Higher order ( $n=2$ ) gives a larger angle than first order.
You can reuse $\lambda/d$ from the first-order calculation to save time.

Common Mistakes

Doubling the angle instead of doubling $\sin\theta$ (i.e. writing $\theta \approx 2\times 6.2^\circ$ is not correct, because $\sin$ is not linear).
Forgetting that $\sin\theta$ must be (\le 1).

Things to Be Careful About

Use degrees consistently (calculator mode).
Final rounding: typically to 1 d.p. here (matching the style of part (i)).

Techniques used

apply the diffraction grating equationchange the diffraction order in the formulause inverse sine to obtain the angle

(iii)

On Fig. 4.2, sketch the variation of the intensity $I$ with $\theta$ for values of $\theta$ from $-15^\circ$ to $+15^\circ$ .

Fig. 4.2

DifficultyMedium

Worked solution

Answer

Sketch a symmetric pattern about $\theta=0$ with narrow principal maxima at:

$\theta = 0^\circ$ (central maximum) with intensity $I_0$ ,
$\theta \approx \pm 6.2^\circ$ (first order),
$\theta \approx \pm 12.5^\circ$ (second order),
with intensity very small between these maxima.

Final answer

Peaks at 0°, ±6.2°, ±12.5°; symmetric; intensity ~0 between peaks; central peak at I0.

Detailed explanation

Background Concept

A diffraction grating produces very sharp (narrow) bright fringes at angles that satisfy:

d\sin\theta = n\lambda

Each integer $n$ gives an “order” of maximum. Because the grating has many slits, the constructive interference is strong only at very specific angles, so the maxima are narrow compared with a two-slit pattern.

The pattern is symmetric: if there is a maximum at $+\theta$ there is also one at $-\theta$ , because $\sin(-\theta) = -\sin\theta$ and negative orders correspond to the other side.

Understanding the Question

You must sketch intensity $I$ against angle $\theta$ from $-15^\circ$ to $+15^\circ$ .

From earlier parts:

first order at $\pm 6.2^\circ$ ,
second order at $\pm 12.5^\circ$ ,
and the central maximum at $0^\circ$ has intensity $I_0$ (given).

So the sketch should show where the maxima are located and the overall qualitative shape.

Approach

Mark the positions of the principal maxima at $\theta = 0, \pm 6.2^\circ, \pm 12.5^\circ$ .
Draw narrow peaks at these angles (very thin compared with the $\theta$ scale).
Ensure the graph is symmetric about $\theta=0$ .
Between maxima, show intensity close to zero (or much smaller than the peaks).

Step-by-Step Reasoning

Central maximum: at $\theta=0^\circ$ by definition; label its height as $I_0$ .
First orders: at $\theta=\pm 6.2^\circ$ from (b)(i). Add two narrow peaks there.
Second orders: at $\theta=\pm 12.5^\circ$ from (b)(ii). Add two narrow peaks there.
There are no other orders within $\pm 15^\circ$ because the next would be $n=3$ :

\sin\theta = \frac{3\lambda}{d} = 3(0.108)=0.324 \Rightarrow \theta\approx 18.9^\circ

which is outside the plotted range.

Between these angles, the grating produces minima and very low intensity, so the sketch should drop close to the axis between the sharp peaks.

(Exact relative heights of different orders are not usually determinable without extra information about the slit width/envelope, so the key marks are typically for correct positions and symmetry and indicating sharp maxima.)

Key Takeaways

Grating maxima occur only at discrete angles given by $d\sin\theta=n\lambda$ .
The pattern is symmetric about the central maximum.
A grating gives narrow, well-separated bright fringes.

Common Mistakes

Placing peaks at the wrong angles (e.g. at $\pm 5^\circ$ because of the grid markings).
Drawing broad sinusoidal oscillations like a two-source interference pattern instead of sharp grating maxima.
Forgetting the negative-angle side (pattern must be symmetric).

Things to Be Careful About

Put the first-order peaks at $\pm 6.2^\circ$ (not $\pm 6.0^\circ$ unless your sketch resolution forces it; aim clearly between $5$ and $10$ ).
Put the second-order peaks at $\pm 12.5^\circ$ (between $10$ and $15$ ).
Central maximum height should be labelled $I_0$ as stated in the question.

Techniques used

identify the angular positions of principal maxima from the grating equationuse symmetry of the diffraction pattern about the central maximumsketch intensity as sharp maxima separated by near-zero intensity

(c)

A polarising filter is placed in the path of the light beam that is incident on the diffraction grating in Fig. 4.1. The transmission axis of the filter is at $45^\circ$ to the vertical.

Suggest how the variation of intensity with $\theta$ for the light on the screen compares with the answer in (b)(iii).

DifficultyMedium

Worked solution

Answer

The filter reduces the incident intensity by Malus’ law:

I' = I\cos^2 45^\circ = \frac{1}{2}I

So the maxima/minima occur at the same $\theta$ values as in (b)(iii), but the whole $I$ – $\theta$ graph is reduced in height by a factor $\frac{1}{2}$ (central maximum becomes $\frac{1}{2}I_0$ ).

Final answer

Same fringe angles as (b)(iii), but all intensities are halved (central becomes 0.5 I0).

Detailed explanation

Background Concept

For plane-polarised light passing through a polarising filter, the transmitted intensity is given by Malus’s law:

I = I_{\text{in}}\cos^2\phi

where $\phi$ is the angle between the incident polarisation direction and the transmission axis of the polariser.

A diffraction grating’s angular positions of maxima depend on wavelength and grating spacing via:

d\sin\theta = n\lambda

These angles do not depend on the intensity of the incoming light; changing input intensity simply scales the output intensity at all angles.

Understanding the Question

Initially the light is vertically polarised. A polarising filter is inserted with its transmission axis at $45^\circ$ to the vertical.

You are asked how the intensity pattern $I(\theta)$ on the screen compares to your sketch in (b)(iii). In other words: do the peaks move, and do their heights change?

Approach

Use Malus’s law with $\phi = 45^\circ$ to find the factor by which the incident intensity is reduced.
Argue that the grating equation (and hence the $\theta$ values for maxima) is unchanged because $d$ and $\lambda$ are unchanged.
Therefore the whole $I(\theta)$ graph is the same shape but scaled vertically.

Step-by-Step Reasoning

The incident light is vertically polarised, so its electric field oscillates vertically.
The polariser axis is at $45^\circ$ to vertical, so $\phi = 45^\circ$ .
Apply Malus’s law:

I_{\text{after filter}} = I_{\text{before filter}}\cos^2 45^\circ

and since $\cos 45^\circ = \frac{1}{\sqrt{2}}$ ,

\cos^2 45^\circ = \frac{1}{2}

so

I_{\text{after filter}} = \frac{1}{2} I_{\text{before filter}}

The diffraction grating still sends light into the same orders at the same angles because $d\sin\theta=n\lambda$ is unchanged.
Therefore at every angle $\theta$ (including the maxima and minima), the observed intensity is half what it was in (b)(iii):

I'(\theta) = \frac{1}{2} I(\theta)

In particular, the central maximum changes from $I_0$ to $\frac{1}{2}I_0$ .

Key Takeaways

Malus’s law gives an intensity scaling factor of $\cos^2\phi$ through a polariser.
Diffraction angles depend on geometry and wavelength, not on intensity.
Adding a polariser before the grating changes peak heights, not peak positions.

Common Mistakes

Saying the peaks move to different $\theta$ values (they do not, because $\lambda$ and $d$ are unchanged).
Using $\cos 45^\circ$ instead of $\cos^2 45^\circ$ .
Thinking the output intensity becomes zero because the axis is not vertical (at $45^\circ$ it transmits half, not none).

Things to Be Careful About

The light becomes polarised along the transmission axis after the filter, but that does not affect the grating equation in this syllabus context.
Make sure you compare to (b)(iii): “same shape, scaled down” is the key comparison.

Techniques used

apply Malus's law for transmission through a polarisercompare two intensity-angle graphs by scaling the vertical axisidentify unchanged angular positions of diffraction maxima

Q5MediumD.C. CircuitsElectricity

(a)

State Kirchhoff’s first law.

DifficultyEasy

Worked solution

Answer

At any junction, the total current entering equals the total current leaving (algebraic sum of currents at a junction is zero).

Final answer

At any junction, the total current entering equals the total current leaving (algebraic sum of currents at a junction is zero).

Detailed explanation

Background Concept

Kirchhoff’s first law (often called KCL) is a direct consequence of conservation of charge. Charge cannot build up at a junction in a steady d.c. circuit, so the rate at which charge arrives at the junction must equal the rate at which charge leaves.

Current is defined as charge flow rate:

I = \frac{\Delta Q}{\Delta t}

So if charge is conserved at a junction, the currents must balance.

Understanding the Question

You are asked to state Kirchhoff’s first law, so you should give a clear sentence definition. No calculation or circuit diagram is required.

Approach

Write the law in words (and optionally as an equation) emphasising that it applies at a junction and it is about currents in and out.

Step-by-Step Reasoning

In a steady circuit, no net charge accumulates at a junction.
Therefore, charge per unit time entering = charge per unit time leaving.
Since current is charge per unit time, total current into the junction = total current out.

Equivalent mathematical statement:

\sum I_{\text{in}} = \sum I_{\text{out}}

(or algebraic sum of currents at a node is zero).

Key Takeaways

Kirchhoff’s first law expresses conservation of charge in circuit form.
It is applied at junctions where currents split or combine.

Common Mistakes

Stating Kirchhoff’s second law (about voltages around a loop) instead.
Forgetting to specify “at a junction”.

Things to Be Careful About

Use correct wording: currents entering and leaving a junction.
You may say “sum of currents into a node equals sum out” or “algebraic sum is zero”; both are acceptable when clearly expressed.

Techniques used

state Kirchhoff’s first law in terms of charge conservation at a junctionrelate currents entering and leaving a node

(b)

Fig. 5.1 shows a circuit containing a thermistor T that has a negative temperature coefficient.

Fig. 5.1

5M

(i)

The thermistor has resistance $R_0$ at a temperature of $0^\circ\text{C}$ .

On Fig. 5.2, sketch a possible variation of the resistance of the thermistor with temperature between $0^\circ\text{C}$ and $100^\circ\text{C}$ .

Fig. 5.2

DifficultyMedium-Easy

Worked solution

Answer

diagram-b-i-1

A decreasing curve starting at $R_0$ at $0^\circ\text{C}$ and falling to a lower resistance at $100^\circ\text{C}$ (non-linear, typical NTC shape).

Final answer

Resistance decreases with increasing temperature from R0 at 0°C to a lower value at 100°C (non-linear NTC curve).

Detailed explanation

Background Concept

A thermistor is a resistor whose resistance changes significantly with temperature. A negative temperature coefficient (NTC) thermistor has resistance that decreases as temperature increases.

Physically, increasing temperature increases the number of charge carriers available in the semiconducting material, so conductivity increases and resistance decreases. The relationship is typically non-linear (often close to exponential over wide ranges), so the graph is not a straight line.

Understanding the Question

You are told the thermistor has resistance $R_0$ at $0^\circ\text{C}$ . You must sketch how its resistance might vary from $0^\circ\text{C}$ to $100^\circ\text{C}$ .

Key requirements implied by the wording/axes:

The curve must pass through $(0^\circ\text{C}, R_0)$ .
Because it is NTC, the resistance must be lower at $100^\circ\text{C}$ than at $0^\circ\text{C}$ .
“Possible variation” means the exact numbers are not needed, but the overall trend and a sensible non-linear shape are.

Approach

Draw axes as given. Place the starting point at $R_0$ when $T=0^\circ\text{C}$ . Then sketch a smooth curve that decreases as temperature increases, typically steep at first and then flattening somewhat (a common NTC behaviour).

Step-by-Step Reasoning

diagram-b-i-1

Mark the point $T=0^\circ\text{C}$ on the x-axis and $R_0$ on the y-axis; start the curve there.
Move to the right (increasing temperature): because it is NTC, resistance must go down.
Make it non-linear: a typical NTC thermistor shows a rapid fall at lower temperatures and then the rate of decrease becomes smaller at higher temperatures (curve becomes less steep).

Any smooth decreasing non-linear curve passing through $R_0$ at $0^\circ\text{C}$ gains full credit.

Key Takeaways

NTC thermistor: $R$ decreases as temperature increases.
Sketch questions award marks for correct trend, correct anchoring at given points, and a realistic (non-linear) curve.

Common Mistakes

Drawing an increasing graph (that would be a positive temperature coefficient).
Drawing a straight line when a curved relation is expected.
Not passing through the given point $R_0$ at $0^\circ\text{C}$ .

Things to Be Careful About

Ensure the curve is clearly below $R_0$ by $100^\circ\text{C}$ .
Keep the sketch smooth (not jagged) and clearly non-linear unless the question explicitly indicates linear behaviour.

Techniques used

interpret negative temperature coefficient behavioursketch a qualitative graph with correct trend and curvatureuse a given reference point on axes to anchor the sketch

(ii)

With reference to the current in the cell, explain why the current in resistor R decreases with increasing temperature of the thermistor.

DifficultyMedium

Worked solution

Answer

As temperature increases, the thermistor resistance decreases, so the parallel combination has a smaller equivalent resistance and the current in the cell increases.

The larger cell current causes a larger p.d. drop across internal resistance $r$ ( $= Ir$ ), so the terminal p.d. across the parallel network (and across $R$ ) decreases:

V = E - Ir

With $R$ constant, $I_R = V/R$ decreases.

Final answer

Thermistor resistance falls, so cell current increases; increased Ir drop across internal resistance reduces terminal p.d. across R, so I_R = V/R decreases.

Detailed explanation

Background Concept

In a circuit with a cell of e.m.f. $E$ and internal resistance $r$ , the terminal potential difference across the external circuit is

V = E - Ir

where $I$ is the current delivered by the cell. If $I$ increases, the voltage lost inside the cell across $r$ (the “lost volts” $Ir$ ) increases, so the terminal p.d. $V$ decreases.

For resistors in parallel, both branches have the same potential difference across them. If the p.d. across the parallel network changes, the current in a branch with fixed resistance $R$ changes according to

I_R = \frac{V}{R}

An NTC thermistor’s resistance decreases as temperature increases.

Understanding the Question

The circuit has a cell (with internal resistance $r$ ) supplying a parallel pair: a fixed resistor $R$ and a thermistor $T$ .

You must explain why the current in resistor $R$ decreases when the thermistor’s temperature increases, and you must do this with reference to the current in the cell.

So your explanation must mention:

what happens to thermistor resistance with temperature,
what that does to the total (cell) current,
how the cell current affects terminal p.d. because of internal resistance,
why that makes the current in $R$ smaller.

Approach

Link the chain of cause and effect:

Temperature up (\Rightarrow) thermistor resistance down.
Parallel equivalent resistance down (\Rightarrow) cell current increases.
Cell current up (\Rightarrow) bigger internal drop (Ir) (\Rightarrow) terminal p.d. down.
Terminal p.d. across $R$ down (\Rightarrow) $I_R = V/R$ down.

Step-by-Step Reasoning

The thermistor is NTC, so increasing temperature decreases its resistance $R_T$ .
$R$ and the thermistor are in parallel. When one branch resistance decreases, the equivalent resistance of the parallel combination decreases:

\frac{1}{R_{\text{eq}}} = \frac{1}{R} + \frac{1}{R_T}

If $R_T$ decreases, $1/R_T$ increases, so $1/R_{\text{eq}}$ increases and therefore $R_{\text{eq}}$ decreases.
3. The external resistance seen by the cell becomes smaller, so the current drawn from the cell increases (consistent with $I = E/(r + R_{\text{eq}})$ ).
4. Because the cell has internal resistance, a larger cell current produces a larger internal p.d. drop $Ir$ . Therefore the terminal p.d. across the parallel network is reduced:

V = E - Ir

Resistor $R$ has this same (reduced) p.d. across it (parallel branches share the same voltage), so

I_R = \frac{V}{R}

decreases since $V$ decreases while $R$ is constant.

That is why the current in $R$ decreases even though the thermistor branch current increases.

Key Takeaways

NTC thermistor: temperature up (\Rightarrow) resistance down.
Reduced equivalent resistance in parallel (\Rightarrow) increased supply (cell) current.
Internal resistance causes terminal voltage to fall as cell current rises.
Branch current through a fixed resistor depends on the shared branch voltage.

Common Mistakes

Saying “thermistor resistance decreases so current in $R$ decreases” with no link to internal resistance (in an ideal cell with $r=0$ , $V$ would not drop and $I_R$ would stay the same).
Claiming the p.d. across $R$ increases because “current increases” (it is the internal drop that increases).
Treating the circuit as if $R$ is in series with the thermistor (they are in parallel).

Things to Be Careful About

The question explicitly asks for reference to current in the cell; include the step that cell current increases.
Use the correct expression $V = E - Ir$ (lost volts). If you write $V = E + Ir$ you reverse the effect.
Remember: in parallel, the p.d. across each branch is the same as the terminal p.d. across the combination.

Techniques used

relate change in thermistor resistance to change in total circuit currentuse internal resistance to link cell current to terminal potential differenceapply V = IR for a fixed resistor branch in parallel

(c)

The electromotive force (e.m.f.) $E$ of the cell in Fig. 5.1 is $1.50\text{V}$ . The internal resistance $r$ of the cell is $0.12\ \Omega$ .

Resistor R has a resistance of $6.00\ \Omega$ .

At a particular temperature of the thermistor, the current in R is $0.200\text{A}$ .

For this temperature of the thermistor, determine:

4M

(i)

the current in the cell

current = ______ $\text{A}$

DifficultyMedium-Easy

Worked solution

Working

Potential difference across $R$ (and across the parallel network):

V = I_R R = 0.200 \times 6.00 = 1.20\ \text{V}

Using $V = E - Ir$ :

I = \frac{E - V}{r} = \frac{1.50 - 1.20}{0.12} = 2.50\ \text{A}

Answer

$2.50\ \text{A}$

Final answer

2.50 A

Detailed explanation

Background Concept

With internal resistance $r$ , a cell of e.m.f. $E$ delivers a terminal p.d. $V$ given by

V = E - Ir

where $I$ is the current in the cell. The external circuit here is a parallel combination, so the terminal p.d. $V$ is the same p.d. across each parallel branch.

For a resistor, Ohm’s law gives

V = IR

Understanding the Question

Given:

$E = 1.50\ \text{V}$
$r = 0.12\ \Omega$
$R = 6.00\ \Omega$
current in resistor $R$ : $I_R = 0.200\ \text{A}$

Find: the current in the cell, $I$ .

Key idea: because $R$ is in parallel with the thermistor, the p.d. across $R$ equals the terminal p.d. of the cell.

Approach

Use $V = I_R R$ to find the p.d. across resistor $R$ .
Treat this as the terminal p.d. $V$ across the external circuit.
Use $V = E - Ir$ to solve for $I$ .

Step-by-Step Reasoning

Compute the p.d. across $R$ :

V = I_R R = 0.200\ \text{A} \times 6.00\ \Omega = 1.20\ \text{V}

This is also the terminal p.d. across the parallel network.
Use the internal resistance relation:

V = E - Ir

Rearrange for $I$ :

I = \frac{E - V}{r}

Substitute:

I = \frac{1.50 - 1.20}{0.12} = \frac{0.30}{0.12} = 2.50\ \text{A}

Key Takeaways

In parallel, each branch has the same p.d. as the terminal p.d.
Internal resistance means terminal p.d. is less than e.m.f. when current flows.
Combine $V = IR$ with $V = E - Ir$ to connect a branch current to the supply current.

Common Mistakes

Using $V = E$ directly (ignoring internal resistance).
Treating $I_R$ as the cell current (it is only the current in one branch).
Using $V = E + Ir$ instead of $V = E - Ir$ .

Things to Be Careful About

Keep units consistent: volts, ohms, amps.
Recognise that the $1.20\ \text{V}$ found from $I_R R$ is the terminal p.d., not the e.m.f.
Quote the final current to appropriate significant figures (here $2.50\ \text{A}$ matches the given data).

Techniques used

use V = IR to find terminal potential difference from a branch currentapply V = E - Ir to relate terminal p.d. to cell currentrearrange and substitute to solve for current

(ii)

the resistance of the thermistor.

resistance = ______ $\Omega$

DifficultyMedium-Easy

Worked solution

Working

Terminal p.d. across the parallel network:

V = I_R R = 0.200 \times 6.00 = 1.20\ \text{V}

From (c)(i), cell current $I = 2.50\ \text{A}$ , so thermistor current:

I_T = I - I_R = 2.50 - 0.200 = 2.30\ \text{A}

Thermistor resistance:

R_T = \frac{V}{I_T} = \frac{1.20}{2.30} = 0.522\ \Omega

Answer

$0.522\ \Omega$

Final answer

0.522 Ω

Detailed explanation

Background Concept

In a parallel circuit, currents split at a junction. Kirchhoff’s first law states

I = I_R + I_T

where $I$ is the current from the cell and $I_R$ , $I_T$ are the branch currents in resistor $R$ and thermistor $T$ .

Also, the p.d. across parallel branches is the same, so the voltage across the thermistor equals the voltage across $R$ . Once you know the branch voltage $V$ and the thermistor current $I_T$ , the thermistor resistance is

R_T = \frac{V}{I_T}

Understanding the Question

At the stated temperature, you already know:

$I_R = 0.200\ \text{A}$ and $R = 6.00\ \Omega$ so you can find the branch voltage $V$ .
From part (c)(i) you can find the cell current $I$ .

You must determine the thermistor’s resistance $R_T$ .

Approach

Calculate the terminal p.d. $V$ using $V = I_R R$ .
Use Kirchhoff’s first law to find $I_T = I - I_R$ .
Use $R_T = V/I_T$ .

Step-by-Step Reasoning

Voltage across resistor $R$ :

V = I_R R = 0.200\ \text{A} \times 6.00\ \Omega = 1.20\ \text{V}

This same $V$ is across the thermistor because of the parallel connection.

The current in the cell (from part (c)(i)) is $I = 2.50\ \text{A}$ . Apply Kirchhoff’s first law at the junction where current splits:

I = I_R + I_T \quad \Rightarrow \quad I_T = I - I_R

So

I_T = 2.50 - 0.200 = 2.30\ \text{A}

Now find thermistor resistance:

R_T = \frac{V}{I_T} = \frac{1.20\ \text{V}}{2.30\ \text{A}} = 0.5217\ \Omega \approx 0.522\ \Omega

Key Takeaways

Use KCL to connect total current to branch currents in parallel.
In parallel circuits, the p.d. across each branch is the same.
Resistance can be found directly from $R = V/I$ once $V$ and $I$ for that component are known.

Common Mistakes

Using $I_T = I_R - I$ (wrong sign/order).
Using $V = E$ instead of the actual branch voltage (here the correct branch voltage is $1.20\ \text{V}$ ).
Assuming the thermistor current equals the cell current.

Things to Be Careful About

Ensure you use the terminal p.d. across the parallel network, not the e.m.f.
Keep enough significant figures in intermediate steps; rounding too early can shift the final resistance.
The thermistor resistance comes out very small because the thermistor branch current is large; this is consistent with an NTC thermistor at high temperature (low resistance).

Techniques used

use Kirchhoff’s first law at a junction to relate branch currentsuse V = IR to determine resistance from branch voltage and currentsubtract branch current from total current to find the other branch current

Q6Medium-EasyParticle Physics

The nuclide $^3_1\text{H}$ is an isotope of hydrogen that is called tritium.

(a)

4M

(i)

Determine the numbers of protons, neutrons and electrons in a neutral atom of tritium.

number of protons = ______
number of neutrons = ______
number of electrons = ______

DifficultyEasy

Worked solution

Working

For $^3_1\text{H}$ : proton number $Z = 1$ , nucleon number $A = 3$ .

Protons $= Z = 1$ .

Neutrons $= A - Z = 3 - 1 = 2$ .

Neutral atom: electrons $=$ protons $= 1$ .

Answer

number of protons $= 1$

number of neutrons $= 2$

number of electrons $= 1$

Final answer

protons 1, neutrons 2, electrons 1

Detailed explanation

Background Concept

Nuclide notation is written as $^A_Z\text{X}$ , where:

$Z$ is the proton (atomic) number = number of protons.
$A$ is the nucleon (mass) number = number of protons + number of neutrons.

For a neutral atom, the total charge is zero, so:

number of electrons = number of protons.

Understanding the Question

You are told the nuclide is tritium, $^3_1\text{H}$ . You must find how many protons, neutrons and electrons are in a neutral atom (so it includes the electron(s) outside the nucleus).

Approach

Read $Z$ to get the number of protons.
Use $A - Z$ to get the number of neutrons.
Use neutrality to get the number of electrons.

Step-by-Step Reasoning

From $^3_1\text{H}$ :

$Z = 1$ , so there is $1$ proton.
$A = 3$ , meaning total nucleons in the nucleus is $3$ .

Neutrons are the remaining nucleons after counting the protons:

\text{neutrons} = A - Z = 3 - 1 = 2

Because the atom is neutral, total positive charge from $1$ proton is balanced by $1$ electron:

electrons $= 1$ .

Key Takeaways

$Z$ gives protons directly.
$A - Z$ gives neutrons.
Neutral atom: electrons = protons.

Common Mistakes

Using $A$ as the number of neutrons (it is total nucleons, not neutrons).
Forgetting the word neutral and giving electrons as $0$ .

Things to Be Careful About

Distinguish nucleus (protons + neutrons) from the whole atom (includes electrons).
Always subtract in the correct order: $A - Z$ , not $Z - A$ .

Techniques used

interpret nuclide notation to identify proton number and nucleon numbercalculate neutron number from nucleon number minus proton numberuse charge neutrality to relate electron number to proton number

(ii)

Draw a labelled diagram to represent a simple model of the arrangement of the protons, neutrons and electrons in a tritium atom.

DifficultyMedium-Easy

Worked solution

diagram-6-a-ii-1

Answer

Nucleus containing $1$ proton and $2$ neutrons, with $1$ electron shown outside the nucleus (on a shell/orbit), all particles labelled.

Final answer

Diagram showing nucleus (1p, 2n) with 1 electron outside

Detailed explanation

Background Concept

A simple atomic model for A-Level credit usually shows:

A small central nucleus containing protons and neutrons.
Electrons occupying regions outside the nucleus (often drawn as shells/orbits in a simplified Bohr-style diagram).

The key marking points are typically correct placement (nucleons in nucleus, electrons outside) and correct labels and numbers.

Understanding the Question

You must draw a labelled diagram for tritium, $^3_1\text{H}$ , showing how many protons, neutrons and electrons it has and where they are found.

Approach

Use $^3_1\text{H}$ to identify nucleus contents: $1$ proton, $2$ neutrons.
For a neutral atom, add $1$ electron outside the nucleus.
Label each particle type clearly.

Step-by-Step Reasoning

Tritium has $Z=1$ so its nucleus contains one proton.
It has $A=3$ , so total nucleons are $3$ and therefore neutrons are $3-1=2$ .
Neutral atom means one electron outside the nucleus.

A correct diagram can be very simple: a central circle labelled “nucleus” containing symbols like $p$ and $n$ with correct counts (e.g. one $p$ and two $n$ ), and one electron $e^-$ on an outer circle/ellipse.

diagram-6-a-ii-1

Key Takeaways

Protons and neutrons belong in the nucleus.
Electrons are outside the nucleus.
Include correct numbers and labels.

Common Mistakes

Putting electrons inside the nucleus.
Drawing the correct shape but not labelling $p$ , $n$ , and $e^-$ .
Drawing the wrong number of neutrons (e.g. $3$ neutrons instead of $2$ ).

Things to Be Careful About

The question asks for a simple model: do not overcomplicate; clear labels matter most.
Make sure the diagram indicates one electron (not two, which would correspond to helium).

Techniques used

represent the nucleus and electron(s) in a simple atomic modellabel subatomic particles clearly and place them in correct regions

(b)

Tritium is radioactive and undergoes $\beta^-$ decay to form an isotope of helium (He). Gamma radiation is not emitted during this decay.

3M

(i)

Complete the equation to represent the radioactive decay of tritium.

^3_1\text{H} \rightarrow \text{}^{......}_{......}\text{He} + \text{}^{......}_{......}\beta + ^0_0\text{X}

DifficultyMedium-Easy

Worked solution

Working

In $\beta^-$ decay, nucleon number $A$ is unchanged and proton number $Z$ increases by $1$ .

So $^3_1\text{H} \rightarrow {}^3_2\text{He}$ .

Beta particle: $^0_{-1}\beta$ .

To conserve lepton number, an antineutrino is emitted: $^0_0\text{X}$ .

Answer

^3_1\text{H} \rightarrow {}^3_2\text{He} + {}^0_{-1}\beta + {}^0_0\text{X}

Final answer

^3_1H -> ^3_2He + ^0_-1β + ^0_0X

Detailed explanation

Background Concept

In nuclear equations you must conserve:

Nucleon number $A$ (total number of protons + neutrons).
Proton number $Z$ (total charge in units of $+e$ ).

For $\beta^-$ decay:

A neutron in the nucleus changes into a proton.
An electron (the $\beta^-$ particle) is emitted.
An electron antineutrino is also emitted.

This is commonly summarised as:

n \rightarrow p + e^- + \bar{\nu}_e

So $A$ stays the same, but $Z$ increases by $1$ .

Understanding the Question

You are told tritium ( $^3_1\text{H}$ ) undergoes $\beta^-$ decay to form an isotope of helium, and no gamma is emitted. You must fill in the missing nuclide numbers for helium and for the beta particle, and keep the extra particle $X$ .

Approach

Use $\beta^-$ decay rule: $A$ unchanged, $Z \to Z+1$ .
Write the beta particle as $^0_{-1}\beta$ (or equivalent).
Keep $X$ as $^0_0\text{X}$ as shown, meaning it has zero nucleon number and zero charge.

Step-by-Step Reasoning

Start with tritium:

Left side has $A=3$ , $Z=1$ .

After $\beta^-$ decay:

$A$ remains $3$ (no nucleons are created/destroyed).
$Z$ increases by $1$ : $1 \to 2$ .

So the daughter nucleus is:

{}^3_2\text{He}

A $\beta^-$ particle is an electron, written in nuclear notation as:

{}^0_{-1}\beta

To account for the weak interaction process, an additional neutral particle with $A=0$ and $Z=0$ is emitted; it is written as $^0_0X$ and will be identified in part (ii).

Putting it all together:

^3_1\text{H} \rightarrow {}^3_2\text{He} + {}^0_{-1}\beta + {}^0_0\text{X}

Key Takeaways

In $\beta^-$ decay: $A$ constant, $Z$ increases by $1$ .
The beta particle has $A=0$ and $Z=-1$ .

Common Mistakes

Writing $^3_1\text{He}$ (forgetting $Z$ increases).
Using $\beta^+$ instead of $\beta^-$ (wrong sign on $Z$ for the beta particle).
Changing $A$ (it should not change for $\beta$ decay).

Things to Be Careful About

The question explicitly says gamma is not emitted, so do not add $\gamma$ .
Ensure the daughter is helium, so the symbol must be He with $Z=2$ .

Techniques used

apply conservation of nucleon number in a nuclear equationapply conservation of charge/proton number in a nuclear equationidentify the emitted beta particle in beta-minus decay

(ii)

State the name of particle X.

(c)

Determine the quark composition of a tritium nucleus.