A cylinder contains a fixed mass of an ideal gas at pressure $2Y$ and volume $6X$. The gas undergoes a sequence of changes from its initial state A, through states B, C and D, then finally back to its initial state A, as shown in Fig. 4.1. Fig. 4.2 shows the variation with time of the internal energy of the gas. (a) State the first law of thermodynamics. (i) Use Fig. 4.1 and Fig. 4.2 to determine the general expression for the internal energy $U$ of the gas when it has…

(a) $$\Delta U = Q + W$$ · (b)(i) $$U = \frac{3}{2}pV$$ · (b)(ii) $$U = \frac{3}{2}NkT$$ · (c)(i) $$W = 8XY$$ · (c)(ii) $$W = -20XY$$ · (d) $$Q = 12XY$$

9702/42 Oct/Nov 2025 — Physics Solved Past Paper

Q1MediumMotion in a CircleGravitational Fields

The Earth may be considered as a uniform sphere of radius $6.37 \times 10^{6}\text{ m}$ .

Cambridge is at a point on the Earth’s surface that has a latitude of $52.2^{\circ}$ north of the Equator, as shown in Fig. 1.1.

Fig. 1.1

As the Earth spins on its axis, Cambridge moves in a circle that is parallel to the Equator but with a smaller radius.

(a)

4M

(i)

Show that the radius of the circle around which Cambridge moves is $3.90 \times 10^{6}\text{ m}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

At latitude $\phi = 52.2^{\circ}$ , the radius of the circle about the Earth’s axis is

r = R\cos\phi

r = (6.37 \times 10^{6})\cos 52.2^{\circ} = 3.90 \times 10^{6}\ \text{m}

Final answer

3.90 × 10^6 m

Detailed explanation

Background Concept

A point on the rotating Earth moves in a circle around the Earth’s rotation axis. If the point is not on the Equator, that circle has a smaller radius than the Earth because the point is closer to the axis.

In a cross-section through the Earth’s centre and the point, you can form a right triangle:

the hypotenuse is the Earth’s radius $R$ (centre to the point),
the adjacent side to the latitude angle is the distance from the rotation axis, i.e. the radius $r$ of the circular path.

This leads to a simple cosine relationship.

Understanding the Question

You are told:

Earth radius $R = 6.37 \times 10^{6}\ \text{m}$ ,
Cambridge is at latitude $\phi = 52.2^{\circ}$ north.

The question asks for the radius of the circle around the Earth’s rotation axis traced out by Cambridge as the Earth spins.

Approach

Use the right-triangle geometry implied by latitude:

at latitude $\phi$ , the distance to the axis is reduced by a factor of $\cos\phi$ .
So calculate $r = R\cos\phi$ .

Step-by-Step Reasoning

Identify the relevant triangle: the line from Earth’s centre to Cambridge has length $R$ .
Latitude $\phi$ is the angle between the equatorial plane and the radius to Cambridge.
The perpendicular distance from the axis (i.e. the circular-path radius) is the “adjacent” side of that triangle.

So,

r = R\cos\phi

Substitute:

r = (6.37 \times 10^{6})\cos(52.2^{\circ})

r \approx 3.90 \times 10^{6}\ \text{m}

Key Takeaways

A point at latitude $\phi$ moves in a circle of radius $r = R\cos\phi$ about the Earth’s axis.
Latitude acts like an angle in a right-triangle projection onto the equatorial plane.

Common Mistakes

Using $\sin\phi$ instead of $\cos\phi$ .
Using $\phi$ measured from the pole instead of from the Equator (latitude is from the Equator).
Forgetting to keep $R$ in metres.

Things to Be Careful About

Calculator must be in degrees.
The radius required is the distance to the rotation axis, not the Earth’s radius and not the distance to the Earth’s centre.

Techniques used

apply trigonometry to relate latitude to the radius of a parallel circleuse the cosine relationship for adjacent and hypotenuse in a right triangle

(ii)

Calculate the speed at which Cambridge moves around the circle.

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

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Period of Earth’s rotation $T = 24.0\ \text{h} = 8.64 \times 10^{4}\ \text{s}$ .

\omega = \frac{2\pi}{T} = \frac{2\pi}{8.64 \times 10^{4}} = 7.27 \times 10^{-5}\ \text{rad s}^{-1}

Using $v = r\omega$ with $r = 3.90 \times 10^{6}\ \text{m}$ ,

v = (3.90 \times 10^{6})(7.27 \times 10^{-5}) = 2.84 \times 10^{2}\ \text{m s}^{-1}

Answer

$2.84 \times 10^{2}\ \text{m s}^{-1}$

Final answer

2.84 × 10^2 m s^-1

Detailed explanation

Background Concept

For uniform circular motion:

the angular speed $\omega$ is the rate of rotation in $\text{rad s}^{-1}$ ,
for one complete revolution, the angular displacement is $2\pi$ radians,
if the period is $T$ , then

\omega = \frac{2\pi}{T}

The linear speed $v$ of a point at radius $r$ from the axis is

v = r\omega

This is equivalent to $v = \frac{2\pi r}{T}$ .

Understanding the Question

Cambridge moves in a circle of radius $r = 3.90 \times 10^{6}\ \text{m}$ (from part (a)(i)) because Earth rotates.

You are asked to calculate the speed of this motion. The key extra piece of information is that Earth completes one rotation in about 24 hours.

Approach

Convert the Earth’s rotation period into seconds.
Find $\omega = 2\pi/T$ .
Use $v = r\omega$ .

Step-by-Step Reasoning

Convert time:

T = 24.0\ \text{h} = 24.0 \times 3600\ \text{s} = 8.64 \times 10^{4}\ \text{s}

Angular speed:

\omega = \frac{2\pi}{T} = \frac{2\pi}{8.64 \times 10^{4}} \approx 7.27 \times 10^{-5}\ \text{rad s}^{-1}

Linear speed:

v = r\omega = (3.90 \times 10^{6})(7.27 \times 10^{-5}) \approx 2.84 \times 10^{2}\ \text{m s}^{-1}

So Cambridge moves at about $284\ \text{m s}^{-1}$ .

Key Takeaways

Convert period to seconds before using $\omega = 2\pi/T$ .
Use $v = r\omega$ (or equivalently $v = 2\pi r/T$ ).

Common Mistakes

Forgetting to convert hours to seconds.
Using $r = 6.37 \times 10^{6}\ \text{m}$ (Earth’s radius) instead of the smaller radius at latitude.
Omitting units or giving $\omega$ in the wrong units.

Things to Be Careful About

Use consistent significant figures; $v$ is typically given to 3 s.f. here.
Ensure you use radians for $2\pi$ (not $360^{\circ}$ ) when calculating $\omega$ .

Techniques used

use angular speed as omega = 2 pi / Tuse v = r omega for circular motionconvert time period into seconds

(b)

A student of mass $58.6\text{ kg}$ stands on horizontal ground in Cambridge.

5M

(i)

Determine the magnitude of the resultant force that acts to cause the circular motion of the student.

$\text{resultant force}$ = ______ $\text{N}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For circular motion,

F = \frac{mv^{2}}{r}

Using $m = 58.6\ \text{kg}$ , $v = 2.84 \times 10^{2}\ \text{m s}^{-1}$ and $r = 3.90 \times 10^{6}\ \text{m}$ ,

F = \frac{58.6(2.84 \times 10^{2})^{2}}{3.90 \times 10^{6}} = 1.21\ \text{N}

Answer

$1.21\ \text{N}$

Final answer

1.21 N

Detailed explanation

Background Concept

Uniform circular motion requires an inward (centripetal) acceleration of magnitude

a = \frac{v^{2}}{r} = r\omega^{2}

By Newton’s second law, the resultant (net) force toward the centre of the circular path is

F = ma = m\frac{v^{2}}{r} = mr\omega^{2}

This resultant force is not an extra new force; it is the vector sum of real forces acting on the object.

Understanding the Question

A $58.6\ \text{kg}$ student standing in Cambridge is carried in a circle about Earth’s axis with:

radius $r = 3.90 \times 10^{6}\ \text{m}$ ,
speed $v \approx 2.84 \times 10^{2}\ \text{m s}^{-1}$ .

The question asks for the magnitude of the resultant force that produces the student’s centripetal acceleration.

Approach

Use

F = \frac{mv^{2}}{r}

because $m$ , $v$ , and $r$ are known.

Step-by-Step Reasoning

Square the speed to get $v^{2}$ .
Divide by $r$ to get the centripetal acceleration magnitude $a = v^{2}/r$ .
Multiply by $m$ to get $F = ma$ .

Numerically:

F = \frac{58.6(2.84 \times 10^{2})^{2}}{3.90 \times 10^{6}} \approx 1.21\ \text{N}

This is small compared with the student’s weight ( $\approx 58.6 \times 9.81 \approx 575\ \text{N}$ ), which is why the effect of Earth’s rotation on apparent weight is small.

Key Takeaways

Resultant force for circular motion is found from $F = mv^{2}/r$ .
Even large speeds can give small centripetal force if $r$ is very large.

Common Mistakes

Using $r = 6.37 \times 10^{6}\ \text{m}$ instead of the circle radius at latitude.
Mixing up $v$ and $\omega$ without converting properly.
Forgetting to square $v$ .

Things to Be Careful About

Keep enough significant figures from part (a)(ii) so rounding does not change the final answer noticeably.
Ensure the radius used is in metres and speed in $\text{m s}^{-1}$ so that $F$ comes out in newtons.

Techniques used

apply the centripetal force equation F = m v^2 / rsubstitute calculated circular speed into a force calculation

(ii)

On Fig. 1.2, draw an arrow to show the direction of the resultant force that acts on the student.

Fig. 1.2

1M Similar

DifficultyEasy

Worked solution

diagram-b-ii

Answer

Arrow from the student pointing towards the Earth’s axis of rotation (along the dashed line).

Final answer

Towards the Earth’s rotation axis (along the dashed line).

Detailed explanation

Background Concept

In uniform circular motion, the acceleration is always directed toward the centre of the circular path. Therefore, the resultant force (since $\vec{F} = m\vec{a}$ ) must also point toward the centre of the circle.

Understanding the Question

The student is moving in a circle around the Earth’s axis (not around the Earth’s centre). The diagram shows a dashed line from the student in the direction of the axis of rotation.

The question asks you to indicate the direction of the resultant force responsible for this circular motion.

Approach

Identify the centre of the student’s circular path: it lies on the Earth’s rotation axis.
Draw the centripetal/resultant force arrow pointing from the student toward that axis.

Step-by-Step Reasoning

Cambridge’s path is a circle whose centre is on the rotation axis.
The resultant force for circular motion is centripetal, so it points toward the centre of that circle.
On Fig. 1.2 the dashed line indicates the direction toward the axis, so the required arrow should be drawn along this dashed line, pointing inward toward the axis.

diagram-b-ii

Key Takeaways

Centripetal/resultant force always points to the centre of the circular path.
Here the centre is on the Earth’s axis, not at the Earth’s centre.

Common Mistakes

Drawing the arrow toward the Earth’s centre instead of toward the rotation axis.
Drawing the arrow tangential to the motion (tangential would correspond to changing speed, not direction).

Things to Be Careful About

The question explicitly says the student moves in a circle parallel to the Equator: that means the centre of the circle is on the axis.
Ensure the arrow direction is toward the axis (not away from it).

Techniques used

identify the centre of the circular path as the Earth’s rotation axisuse the fact that the centripetal force points toward the centre of the circular path

(iii)

On Fig. 1.3, draw labelled arrows from the student to show the directions of the forces that act on the student to cause the resultant force in (b)(ii).

Fig. 1.3

2M Similar

DifficultyMedium

Worked solution

diagram-b-iii

Answer

Forces on the student:

weight $mg$ vertically downward (towards Earth’s centre),
normal contact force $N$ from the ground perpendicular to the ground (upward),
static friction $f$ along the ground towards the Earth’s rotation axis.

These forces combine to give the resultant shown in (b)(ii).

Final answer

Weight mg down, normal N up (perpendicular to ground), and static friction along the ground towards the rotation axis.

Detailed explanation

Background Concept

The “centripetal force” is not a separate force; it is the resultant (vector sum) of the real forces acting on the object.

For a person standing on the Earth, the real forces available are:

gravitational force (weight) $mg$ toward the Earth’s centre,
contact force from the ground. The contact force can include:
- a normal reaction $N$ perpendicular to the ground,
- a static frictional force $f$ parallel to the ground.

Because the required resultant force is toward the Earth’s rotation axis (part (b)(ii)), you should include any force components needed to make the resultant point in that direction.

Understanding the Question

You are asked to draw labelled force arrows acting on a stationary student on horizontal ground in Cambridge.

From part (b)(ii), the resultant force must point toward the Earth’s rotation axis (along the dashed line). The question now asks which real forces combine to produce that resultant.

Approach

Start with the standard forces on a person standing still: $mg$ and $N$ .
Notice that the centripetal resultant is not necessarily vertical; it is toward the axis.
Include a static friction force along the ground if needed so that the vector sum of forces can point toward the axis.

Step-by-Step Reasoning

Draw weight $mg$ :

It acts toward the centre of the Earth.
On the local diagram, this is drawn vertically downward.

Draw normal reaction $N$ :

It acts perpendicular to the (horizontal) ground.
On the local diagram, this is drawn vertically upward.

Draw static friction $f$ :

Friction acts parallel to the ground surface.
Its direction should be such that when added vectorially with $mg$ and $N$ , the resultant points toward the rotation axis (the dashed-line direction).
Therefore, draw $f$ along the ground pointing toward the axis (i.e. in the direction that helps the net force align with the dashed line).

diagram-b-iii

Key Takeaways

The centripetal force is the resultant of real forces.
For an object on a surface, include both the normal reaction and (if necessary) static friction.
Always match the resultant direction (toward the centre of the circular path).

Common Mistakes

Drawing only $mg$ and $N$ even when their resultant cannot point toward the axis direction shown.
Drawing the resultant itself again instead of the real forces.
Not labelling the forces.

Things to Be Careful About

Weight $mg$ is the gravitational force (toward Earth’s centre), not “mass” and not the centripetal force.
Static friction can act even when the student is not sliding; it adjusts to whatever value is needed (up to a maximum) to prevent relative motion.
Make sure the friction arrow is parallel to the ground, while $N$ is perpendicular to the ground.

Techniques used

draw a free-body force diagram for an object on a surfaceresolve real forces to show how their vector sum produces the centripetal resultant

Q2MediumGravitational FieldsIdeal Gases

(a)

State Newton’s law of gravitation.

2M Similar

DifficultyEasy

Worked solution

Answer

Any two masses attract each other with a force

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

acting along the line joining their centres (attractive).

Final answer

Attractive force between two point masses: F = G m1 m2 / r^2 along the line joining their centres.

Detailed explanation

Background Concept

Newton’s law of gravitation describes the gravitational force between two point masses (or spherically symmetric masses treated as if their mass is concentrated at the centre). The law is an inverse-square law: the force decreases as the square of the separation increases.

The magnitude is

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

where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between their centres.

Understanding the Question

You are asked to state Newton’s law of gravitation. That means giving the correct proportionalities and/or the full formula, and indicating direction (along the line joining centres, attractive).

Approach

Write the standard equation for gravitational force, and include the statement that it is attractive and acts along the line joining the masses.

Step-by-Step Reasoning

Recall the inverse-square dependence: $F \propto 1/r^2$ .
Recall proportionality to both masses: $F \propto m_1 m_2$ .
Combine with the constant $G$ to form the standard equation and state it is attractive along the line joining centres.

Key Takeaways

Gravitational force between masses follows an inverse-square law.
The complete statement includes direction (attractive, along the line joining centres).

Common Mistakes

Writing $F = Gm_1m_2/r$ (missing the square).
Not stating the force is attractive / acts along the line joining centres.

Things to Be Careful About

$r$ is the centre-to-centre separation.
Use correct symbols: $G$ is a constant, not to be confused with $g$ (field strength near Earth).

Techniques used

recall and state Newton's law of gravitationidentify the inverse-square dependence on separation

(b)

One of the basic assumptions of the kinetic theory of gases is that there are no forces exerted between the molecules of the gas except during collisions.

State two other basic assumptions of the kinetic theory of gases.

2M Similar

DifficultyEasy

Worked solution

Answer

Any two:

Molecules move in constant random motion (in straight lines between collisions).
Collisions between molecules (and with the container walls) are perfectly elastic.

Final answer

Random straight-line motion between collisions; collisions are perfectly elastic.

Detailed explanation

Background Concept

The kinetic theory model treats a gas as a very large number of microscopic particles whose motion explains macroscopic quantities like pressure and temperature. To make the model solvable, several assumptions are made (ideal-gas assumptions).

Common assumptions include:

Molecules are in continuous random motion.
They move in straight lines at constant speed between collisions.
Collisions between molecules and with container walls are perfectly elastic.
The time of collision is negligible compared with the time between collisions.
The volume of the molecules is negligible compared with the volume of the container.
No intermolecular forces act except during collisions (given in the question).

Understanding the Question

One assumption has already been stated (no intermolecular forces except during collisions). You must state two different assumptions from the standard list.

Approach

Choose any two widely accepted kinetic theory assumptions that are clearly distinct from “no forces except during collisions”. The safest pair is “random motion / straight-line motion between collisions” and “perfectly elastic collisions”.

Step-by-Step Reasoning

State assumption 1 about how molecules move between collisions.
State assumption 2 about the nature of collisions.
These are both fundamental and are commonly credited.

Key Takeaways

Kinetic theory assumptions describe particle motion and simplify interactions.
“Perfectly elastic” is important because it conserves kinetic energy in collisions.

Common Mistakes

Repeating the given assumption in different words (earns no credit).
Stating a consequence rather than an assumption (e.g. “pressure is due to collisions” is an explanation, not an assumption).

Things to Be Careful About

Ensure your two statements are clearly separate assumptions.
Use precise wording: “perfectly elastic” means total kinetic energy conserved in collisions (not just momentum).

Techniques used

recall standard assumptions of the kinetic theory modelselect two valid assumptions distinct from the one given

(c)

Hydrogen gas consists of molecules that each have a mass of $3.34 \times 10^{-27}\text{ kg}$ . Hydrogen may be considered to be an ideal gas.

A spherical balloon contains $0.0160\text{ mol}$ of hydrogen gas at a temperature of $282\text{ K}$ . At this temperature, the volume of gas in the balloon is $1.87 \times 10^{-4}\text{ m}^{3}$ .

4M

(i)

Determine the pressure of the gas.

$\text{pressure}$ = ______ $\text{Pa}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using

pV = nRT

p = \frac{nRT}{V} = \frac{(0.0160)(8.31)(282)}{1.87 \times 10^{-4}}

p = 2.00 \times 10^{5}\ \text{Pa}

Answer

$2.00 \times 10^{5}\ \text{Pa}$

Final answer

2.00 × 10^5 Pa

Detailed explanation

Background Concept

For an ideal gas, the macroscopic variables pressure $p$ , volume $V$ , temperature $T$ , and amount of gas $n$ are related by the equation of state

pV = nRT

where $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ .

Understanding the Question

You are given:

$n = 0.0160\ \text{mol}$
$T = 282\ \text{K}$
$V = 1.87 \times 10^{-4}\ \text{m}^3$

and asked to determine the pressure $p$ in pascals.

Approach

Rearrange the ideal gas equation to $p = nRT/V$ , then substitute all values in SI units (they already are).

Step-by-Step Reasoning

Start from

pV = nRT

Rearrange:

p = \frac{nRT}{V}

Substitute:

p = \frac{(0.0160)(8.31)(282)}{1.87 \times 10^{-4}}

Calculate the numerator:

(0.0160)(8.31)(282) \approx 37.5

Divide by $1.87 \times 10^{-4}$ :

p \approx 2.00 \times 10^{5}\ \text{Pa}

Key Takeaways

For ideal gases, pressure is found directly from $p = nRT/V$ .
Always check units: $V$ must be in $\text{m}^3$ and $T$ in K.

Common Mistakes

Using $V$ in $\text{cm}^3$ or $\text{dm}^3$ without converting to $\text{m}^3$ .
Using $T$ in $^\circ\text{C}$ instead of kelvin.

Things to Be Careful About

Significant figures: given data are mostly 3 s.f., so $p$ to 3 s.f. is appropriate.
Do not confuse $R$ with $k$ (Boltzmann constant); $k$ would require number of molecules $N$ , not moles $n$ .

Techniques used

apply the ideal gas equation pV = nRTsubstitute given values consistently in SI unitsrearrange to solve for pressure

(ii)

Estimate the average separation of the hydrogen molecules in the gas.

$\text{average separation}$ = ______ $\text{m}$

2M Similar

DifficultyMedium

Worked solution

Working

Number of molecules:

N = nN_A = (0.0160)(6.02 \times 10^{23}) = 9.63 \times 10^{21}

Volume per molecule:

\frac{V}{N} = \frac{1.87 \times 10^{-4}}{9.63 \times 10^{21}} = 1.94 \times 10^{-26}\ \text{m}^3

Average separation $s$ :

s \approx \left(\frac{V}{N}\right)^{1/3} = (1.94 \times 10^{-26})^{1/3} = 2.68 \times 10^{-9}\ \text{m}

Answer

$2.68 \times 10^{-9}\ \text{m}$

Final answer

2.68 × 10^-9 m

Detailed explanation

Background Concept

If particles are spread roughly uniformly in a gas, a simple estimate of the average separation comes from thinking of each molecule as “owning” a small cube of volume equal to the average volume per molecule.

If the number of molecules is $N$ in a total volume $V$ , then volume per molecule is $V/N$ , and the cube side length (an estimate of separation) is

s \approx \left(\frac{V}{N}\right)^{1/3}

To find $N$ from moles:

N = nN_A

where $N_A \approx 6.02 \times 10^{23}\ \text{mol}^{-1}$ .

Understanding the Question

You are given the amount of hydrogen $n = 0.0160\ \text{mol}$ and the balloon volume $V = 1.87 \times 10^{-4}\ \text{m}^3$ . You must estimate the typical distance between neighbouring hydrogen molecules.

Approach

Convert moles to number of molecules using $N = nN_A$ .
Find average volume per molecule $V/N$ .
Take the cube root to convert a volume scale into a length scale.

Step-by-Step Reasoning

Compute $N$ :

N = (0.0160)(6.02 \times 10^{23}) = 9.63 \times 10^{21}

Average volume per molecule:

\frac{V}{N} = \frac{1.87 \times 10^{-4}}{9.63 \times 10^{21}} = 1.94 \times 10^{-26}\ \text{m}^3

Now cube root to estimate separation:

s \approx (1.94 \times 10^{-26})^{1/3}

A useful way to see the power of ten is to rewrite $1.94 \times 10^{-26}$ as $19.4 \times 10^{-27}$ :

cube root of $10^{-27}$ is $10^{-9}$
cube root of $19.4$ is about $2.7$

So

s \approx 2.7 \times 10^{-9}\ \text{m}

Key Takeaways

Convert moles to molecules with $N = nN_A$ .
Average separation in a uniform gas scales as $\left(V/N\right)^{1/3}$ .

Common Mistakes

Forgetting the cube root and using $V/N$ directly as a distance.
Mixing up $N$ (number of molecules) and $n$ (number of moles).

Things to Be Careful About

This is an estimate assuming a roughly uniform distribution.
Keep track of powers of ten carefully when cube-rooting: $10^{-27} \to 10^{-9}$ .

Techniques used

convert moles to number of molecules using the Avogadro constantcalculate volume per moleculeestimate mean separation using a cube-root approximation

(d)

3M

(i)

Use your answer in (c)(ii) to calculate the average gravitational force between adjacent molecules in hydrogen gas.

$\text{average force}$ = ______ $\text{N}$

2M Similar

DifficultyMedium

Worked solution

Working

Using

F = \frac{G m^2}{r^2}

with $m = 3.34 \times 10^{-27}\ \text{kg}$ and $r = 2.68 \times 10^{-9}\ \text{m}$ :

F = \frac{(6.67 \times 10^{-11})(3.34 \times 10^{-27})^2}{(2.68 \times 10^{-9})^2}

F = 1.04 \times 10^{-46}\ \text{N}

Answer

$1.04 \times 10^{-46}\ \text{N}$

Final answer

1.04 × 10^-46 N

Detailed explanation

Background Concept

Newton’s law of gravitation applies to any two masses. For two identical masses $m$ separated by distance $r$ :

F = \frac{Gm^2}{r^2}

This force is always attractive and acts along the line joining the masses.

Understanding the Question

You have already estimated the average separation $r$ between adjacent hydrogen molecules from (c)(ii). You are now asked to use that separation to estimate the gravitational force between two neighbouring molecules.

Given:

$m = 3.34 \times 10^{-27}\ \text{kg}$ (mass of one molecule)
$r \approx 2.68 \times 10^{-9}\ \text{m}$

Find $F$ .

Approach

Use $F = Gm^2/r^2$ , substituting the molecular mass for both objects and using the separation from (c)(ii).

Step-by-Step Reasoning

Start with

F = \frac{Gm^2}{r^2}

Substitute values:

F = \frac{(6.67 \times 10^{-11})(3.34 \times 10^{-27})^2}{(2.68 \times 10^{-9})^2}

Compute pieces:

(3.34 \times 10^{-27})^2 = 1.12 \times 10^{-53}

(2.68 \times 10^{-9})^2 = 7.18 \times 10^{-18}

So

F = \frac{(6.67 \times 10^{-11})(1.12 \times 10^{-53})}{7.18 \times 10^{-18}}

Numerator:

(6.67 \times 10^{-11})(1.12 \times 10^{-53}) \approx 7.44 \times 10^{-64}

Divide:

F \approx \frac{7.44 \times 10^{-64}}{7.18 \times 10^{-18}} = 1.0 \times 10^{-46}\ \text{N}

Key Takeaways

Gravitational attraction exists between any masses but can be extremely small for molecules.
Using standard form helps keep track of very small forces.

Common Mistakes

Using $F = Gm/r^2$ instead of $Gm^2/r^2$ (forgetting both masses contribute).
Squaring the separation incorrectly or mixing powers of ten.

Things to Be Careful About

$r$ must be in metres and $m$ in kilograms.
The separation used is an average estimate; the force is therefore also an estimate.

Techniques used

apply Newton's law of gravitation to two moleculessubstitute the separation distance from a previous parthandle very small quantities in standard form

(ii)

By considering the weight of a molecule, suggest with a reason whether your answer in (d)(i) is consistent with the assumption of the kinetic theory of gases that there are no forces exerted between molecules.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Weight of a molecule:

W = mg = (3.34 \times 10^{-27})(9.81) = 3.28 \times 10^{-26}\ \text{N}

Since $3.28 \times 10^{-26}\ \text{N} \gg 1.04 \times 10^{-46}\ \text{N}$ , the intermolecular gravitational force is negligible.

Answer

Yes, it is consistent: the force between molecules is many orders of magnitude smaller than the weight of a molecule, so it can be taken as zero.

Final answer

Yes. The intermolecular gravitational force is negligible compared with the molecule’s weight (mg).

Detailed explanation

Background Concept

A key part of judging whether a force can be neglected is comparing its magnitude with other relevant forces in the situation.

For a particle of mass $m$ near Earth, the weight is

W = mg

If an interparticle force is many orders of magnitude smaller than typical forces acting, it is reasonable in a model to treat it as effectively zero.

Understanding the Question

You found in (d)(i) that the gravitational force between neighbouring hydrogen molecules is extremely small. The kinetic theory assumption says molecules exert no forces on each other (except during collisions). You are asked to decide whether your calculated force supports that assumption by comparing with the weight of one molecule.

Approach

Compute the weight $mg$ of a hydrogen molecule.
Compare $mg$ to the gravitational force between molecules from (d)(i).
Conclude whether the intermolecular force is negligible.

Step-by-Step Reasoning

Calculate the weight:

W = mg = (3.34 \times 10^{-27})(9.81) = 3.28 \times 10^{-26}\ \text{N}

Compare with

F_{\text{between molecules}} \approx 1.0 \times 10^{-46}\ \text{N}

The ratio is about $10^{20}$ , meaning the intermolecular gravitational force is around 20 orders of magnitude smaller than the weight. Such a tiny force has no measurable effect on the gas behaviour compared with molecular collisions and is therefore negligible.

Key Takeaways

Use order-of-magnitude comparisons to justify modelling assumptions.
Molecular gravitational attraction in a gas is extraordinarily small.

Common Mistakes

Saying “not consistent because the force is not exactly zero” (models often neglect forces that are negligible).
Comparing to the wrong quantity (e.g. comparing with pressure instead of a force scale like $mg$ ).

Things to Be Careful About

You must give both a decision (consistent / not consistent) and a reason (comparison showing it is negligible).
Ensure you compare forces in the same units (newtons).

Techniques used

compare magnitudes of two forcescalculate the weight of a molecule using mgdraw a conclusion about negligibility relative to an assumption

Q3Medium-EasyTemperature

(a)

State what is meant by two objects being in thermal equilibrium.

2M Similar

DifficultyEasy

Worked solution

Answer

Two objects are in thermal equilibrium when, if they are placed in thermal contact, there is no net transfer of thermal energy between them (so they are at the same temperature).

Final answer

No net thermal energy transfer between them when in contact (same temperature).

Detailed explanation

Background Concept

Thermal equilibrium is a condition about energy transfer. When two bodies at different temperatures are put in thermal contact, thermal energy is transferred from the hotter to the cooler object. This continues until the temperatures become equal.

At that point, although microscopic energy exchanges still occur, there is no net flow of thermal energy in either direction.

Understanding the Question

You are asked for the meaning of “two objects being in thermal equilibrium”. The mark scheme typically expects two ideas:

what happens if they are put in thermal contact (net heat flow), and
the temperature condition (equal temperatures).

Approach

State the operational test: bring them into thermal contact. Then state the criterion: no net heat transfer, which implies equal temperature.

Step-by-Step Reasoning

If object A and object B are placed in thermal contact and A is hotter, energy flows A → B.
If B is hotter, energy flows B → A.
In thermal equilibrium there is no net energy transfer in either direction, which only occurs when their temperatures are equal.

Key Takeaways

Thermal equilibrium is defined by no net thermal energy transfer.
In equilibrium, the objects have the same temperature.

Common Mistakes

Saying only “they have the same temperature” and not mentioning the no-net-transfer condition (often loses a mark).
Saying “no heat transfer” (too absolute); it should be no net transfer.

Things to Be Careful About

Use the phrase “no net transfer of thermal energy” (or “no net heat flow”).
Make it clear it applies when in thermal contact.

Techniques used

use the criterion of no net thermal energy transferlink thermal equilibrium to equal temperature

(b)

Fig. 3.1 shows a type of thermometer called a constant volume gas thermometer.

Fig. 3.1

The thermometer is used to determine the thermodynamic temperature $T$ of the gas in the glass bulb.

The glass bulb is immersed in the environment for which the temperature is to be measured. The height of the movable glass tube is then adjusted so that the level of the liquid on the left-hand side aligns with the reference line X marked on the fixed glass tube. The reference line Y is marked on the side of the movable glass tube. The level of the liquid at Y is higher than at X as a result of the pressure of the gas in the glass bulb.

The difference in height $\Delta h$ between the liquid levels at X and Y is then measured using the scale. The thermodynamic temperature $T$ of the gas is directly proportional to the pressure of the gas. This pressure is directly proportional to $\Delta h$ .

7M

(i)

The value of $\Delta h$ can be used to calculate the pressure of the gas. In order to do this, the gravitational field strength is used, along with a property of the liquid.

State the property of the liquid that is used to calculate the pressure.

1M Similar

(ii)

Before the measurement of $\Delta h$ can be made, the glass bulb needs to reach thermal equilibrium with the environment for which the temperature is to be measured.

State two disadvantages of using a constant volume gas thermometer to measure temperature.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Slow response / must wait a long time for the bulb to reach thermal equilibrium with the environment.
Bulky and fragile / not convenient or portable (requires a careful set-up and adjustment to keep volume constant).

Final answer

Slow response (time to reach equilibrium) and bulky/fragile (impractical, needs careful set-up/adjustment).

Detailed explanation

Background Concept

A practical thermometer should:

reach thermal equilibrium with the object/environment reasonably quickly (short response time), and
be convenient to use (portable, robust, easy to read).

A constant-volume gas thermometer is a primary standard instrument: accurate in principle, but not designed for everyday use.

Understanding the Question

You are told that before $\Delta h$ is measured, the bulb must reach thermal equilibrium with the environment. You must give two disadvantages of using this thermometer to measure temperature. The expected answers are practical limitations stemming from the apparatus and the equilibrium requirement.

Approach

Choose two clear, distinct disadvantages:

something about time/response (equilibrium), and
something about practicality (size/fragility/complexity of adjustment/readings).

Step-by-Step Reasoning

Slow to reach equilibrium: The bulb and gas must come to the same temperature as the environment. This can take a long time, so it cannot measure rapidly changing temperatures well.
Impractical apparatus: It involves a bulb, connecting tubes, a liquid column and a movable tube with a scale; it is typically large, fragile and needs careful alignment to keep the gas at constant volume.

Key Takeaways

Needing thermal equilibrium usually implies slow response.
Primary-standard thermometers are often bulky/fragile/awkward to use.

Common Mistakes

Giving two versions of the same disadvantage (e.g. “slow” and “takes long time” counted as one).
Giving vague answers like “human error” without linking to a specific feature (alignment, reading $\Delta h$ , etc.).

Things to Be Careful About

Make the disadvantages clearly different.
Tie at least one disadvantage explicitly to the need for thermal equilibrium (response time).

Techniques used

link thermometer usability to response time and thermal equilibriumidentify practical limitations of a laboratory gas thermometer

(iii)

Suggest one situation in which a constant volume gas thermometer would be an appropriate type of thermometer to choose for measuring temperature.

1M Similar

(iv)

Level X aligns with $2.31\text{ cm}$ on the scale. At $0\text{ }^{\circ}\text{C}$ , level Y aligns with $8.69\text{ cm}$ .

At temperature $\theta$ , level Y aligns with $7.83\text{ cm}$ on the scale.

Determine a value for $\theta$ in $^{\circ}\text{C}$ .

$\theta$ = ______ $^{\circ}\text{C}$

3M Similar

DifficultyMedium-Easy

Worked solution

Working

At $0\,^{\circ}\text{C}$ :

\Delta h_0 = 8.69 - 2.31 = 6.38\ \text{cm}

At temperature $\theta$ :

\Delta h_{\theta} = 7.83 - 2.31 = 5.52\ \text{cm}

With $T \propto \Delta h$ :

\frac{273 + \theta}{273} = \frac{\Delta h_{\theta}}{\Delta h_0} = \frac{5.52}{6.38}

273 + \theta = 273\left(\frac{5.52}{6.38}\right) = 236\ \text{K}

\theta = 236 - 273 = -37\,^{\circ}\text{C}

Answer

$\theta = -37\,^{\circ}\text{C}$

Final answer

-37 °C

Detailed explanation

Background Concept

For a constant-volume gas thermometer, the gas pressure is proportional to the thermodynamic temperature:

p \propto T

The manometer gives a pressure difference proportional to the height difference of the liquid columns:

\Delta p = \rho g \Delta h \;\Rightarrow\; p \propto \Delta h

Combining these proportionalities (with the same gas and fixed volume) gives:

T \propto \Delta h

Thermodynamic temperature is in kelvin, related to Celsius temperature by:

T / \text{K} = 273 + \theta / ^{\circ}\text{C}

Understanding the Question

You are given scale readings for level $X$ and level $Y$ .

$X$ is fixed at $2.31\ \text{cm}$ .
At $0\,^{\circ}\text{C}$ , $Y$ is at $8.69\ \text{cm}$ .
At temperature $\theta$ , $Y$ is at $7.83\ \text{cm}$ .

The measurable quantity linked to pressure (and hence to $T$ ) is the height difference:

\Delta h = Y - X

You must find $\theta$ .

diagram-biv-1

Approach

Calculate $\Delta h_0$ at $0\,^{\circ}\text{C}$ and $\Delta h_{\theta}$ at temperature $\theta$ .
Use the proportionality $T \propto \Delta h$ to form a ratio:

\frac{T_{\theta}}{T_0} = \frac{\Delta h_{\theta}}{\Delta h_0}

Convert $T_{\theta}$ (kelvin) to $\theta$ (Celsius) using $T = 273 + \theta$ .

Step-by-Step Reasoning

Height differences:

\Delta h_0 = 8.69 - 2.31 = 6.38\ \text{cm}

\Delta h_{\theta} = 7.83 - 2.31 = 5.52\ \text{cm}

Use $T \propto \Delta h$ .
At $0\,^{\circ}\text{C}$ , $T_0 = 273\ \text{K}$ . Hence

T_{\theta} = 273\left(\frac{5.52}{6.38}\right) = 236\ \text{K}\; (\text{to }3\ \text{s.f.})

Convert to Celsius:

\theta = T_{\theta} - 273 = 236 - 273 = -37\,^{\circ}\text{C}

Key Takeaways

For a constant-volume gas thermometer, $T \propto p$ and $p \propto \Delta h$ , so $T \propto \Delta h$ .
Use kelvin in proportionality relations and only convert to Celsius at the end.

Common Mistakes

Using $\theta$ directly in the ratio (e.g. $\theta/0$ is meaningless); you must use kelvin temperatures.
Forgetting to subtract $X$ from $Y$ to get $\Delta h$ .
Mixing units or incorrectly adding/subtracting 273.

Things to Be Careful About

The proportionality is for absolute temperature: always use $T$ in kelvin.
Ensure $\Delta h$ is the difference between the two levels, not one reading alone.
Keep sensible significant figures (final $\theta$ to about 2–3 s.f. is appropriate here).

Techniques used

calculate a manometer height difference from scale readingsuse proportional reasoning with $T \propto \Delta h$convert between kelvin temperature and degrees Celsius

Q4MediumThermodynamicsIdeal Gases

A cylinder contains a fixed mass of an ideal gas at pressure $2Y$ and volume $6X$ .

The gas undergoes a sequence of changes from its initial state A, through states B, C and D, then finally back to its initial state A, as shown in Fig. 4.1.

Fig. 4.1

Fig. 4.2 shows the variation with time of the internal energy of the gas.

Fig. 4.2

(a)

State the first law of thermodynamics.

2M Similar

DifficultyEasy

Worked solution

Answer

The first law is

\Delta U = Q + W

where $\Delta U$ is the change in internal energy, $Q$ is thermal energy supplied to the gas and $W$ is the work done on the gas.

Final answer

$\Delta U = Q + W$

Detailed explanation

Background Concept

The first law of thermodynamics is an energy conservation statement for thermodynamic systems. It links:

$\Delta U$ : change in internal energy (energy in microscopic random motion / molecular interactions),
$Q$ : thermal energy transferred to the system (positive when supplied to the gas),
$W$ : work (mechanical energy transfer).

A common Cambridge convention is

\Delta U = Q + W

where $W$ is work done on the gas. (If instead you use work done by the gas, the equation becomes $\Delta U = Q - W_{\text{by}}$ .)

Understanding the Question

You are asked to state the first law, so you must write the correct equation and explain what the symbols mean, including the sign convention.

Approach

Write the standard form of the first law used in this question (work done on the gas) and define the symbols.

Step-by-Step Reasoning

Energy conservation for a thermodynamic change gives a relationship between internal energy, heating, and work.
Using the sign convention suitable for later parts (they ask for “work done on the gas”), we write

\Delta U = Q + W.

State meanings: $Q>0$ means heat supplied to the gas; $W>0$ means work done on the gas (e.g. compression).

Key Takeaways

Know the first law and be consistent with the sign convention.
For compression, work done on the gas is positive.

Common Mistakes

Writing $\Delta U = Q - W$ while still calling $W$ “work done on the gas” (sign inconsistency).
Not defining the symbols, losing explanation marks.

Things to Be Careful About

Always check whether the question uses “work done on” or “work done by” the gas.
In a cycle, total $\Delta U$ over one complete loop is zero because the gas returns to its initial state.

Techniques used

state the first law in equation formdefine the sign convention for work done on the gas

(b)

3M

(i)

Use Fig. 4.1 and Fig. 4.2 to determine the general expression for the internal energy $U$ of the gas when it has pressure $p$ and volume $V$ .

$U$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From state A: $pV = (2Y)(6X)=12XY$ and $U=18XY$ .

\frac{U}{pV} = \frac{18XY}{12XY}=\frac{3}{2}

(Also consistent with B, C and D.) Hence

U = \frac{3}{2}pV.

Answer

U = \frac{3}{2}pV

Final answer

$U = \frac{3}{2}pV$

Detailed explanation

Background Concept

For an ideal gas, internal energy $U$ depends only on temperature. If the gas is monatomic, kinetic theory gives $U = \tfrac{3}{2}nRT$ , and since $pV=nRT$ , this implies $U \propto pV$ . In this question you are not told that directly; you are expected to deduce a relationship using the given state values.

Understanding the Question

You are given a rectangular cycle on a $p$ – $V$ diagram (Fig. 4.1) and a graph of internal energy against time showing the values of $U$ at states A, B, C, D (Fig. 4.2).

You must find a general expression for $U$ when the gas has arbitrary pressure $p$ and volume $V$ .

Approach

Pick one state (A is easiest), calculate $pV$ from Fig. 4.1, read $U$ from Fig. 4.2, and see what constant relates them. Then check it matches another state to confirm the general rule.

Step-by-Step Reasoning

Read state A from Fig. 4.1: $p=2Y$ , $V=6X$ .
Compute

pV = (2Y)(6X)=12XY.

Read from Fig. 4.2 that at A,

U = 18XY.

Compare $U$ with $pV$ :

\frac{U}{pV}=\frac{18XY}{12XY}=\frac{3}{2}.

Check quickly with another point (e.g. D): $pV=(5Y)(6X)=30XY$ and $U=45XY$ , again ratio $=3/2$ . So the relationship is consistent.
Therefore the general expression is

U = \frac{3}{2}pV.

Key Takeaways

Use state-point data to infer a proportionality and a constant.
Checking a second point is a good way to confirm the deduced law.

Common Mistakes

Using $\Delta U$ between two states instead of the actual value of $U$ at a state.
Mixing up coordinates (reading $V$ from the pressure axis or vice versa).
Concluding $U=pV$ (forgetting to compute the ratio).

Things to Be Careful About

Keep track of the units: $pV$ has units of energy, so comparing $U$ to $pV$ is sensible.
The expression must be in terms of general $p$ and $V$ , not in terms of $X$ and $Y$ only.

Techniques used

extract numerical values from a p–V diagramcompare ratios to infer proportionalitydeduce an algebraic relationship from multiple state points

(ii)

An ideal gas at thermodynamic temperature $T$ contains $N$ molecules.

Use your answer in (b)(i) and the equation of state for an ideal gas to deduce an expression for $U$ in terms of $N$ and $T$ . Identify any other symbols you use.

$U$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From (b)(i),

U = \frac{3}{2}pV.

For an ideal gas,

pV = NkT.

So

U = \frac{3}{2}NkT.

Answer

U = \frac{3}{2}NkT

where $k$ is the Boltzmann constant.

Final answer

$U = \frac{3}{2}NkT$

Detailed explanation

Background Concept

The ideal gas equation of state can be written in microscopic form as

pV = NkT,

where:

$N$ is the number of molecules,
$k$ is the Boltzmann constant ( $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ ),
$T$ is thermodynamic temperature in kelvin.

If you already have an expression for $U$ in terms of $pV$ , you can immediately convert it into an expression in terms of $N$ and $T$ by substituting $pV=NkT$ .

Understanding the Question

You must use your result from (b)(i), $U=\tfrac{3}{2}pV$ , and the ideal gas equation to express internal energy $U$ purely in terms of $N$ and $T$ (and any constants such as $k$ ), and you must say what the extra symbol means.

Approach

Start from $U=\tfrac{3}{2}pV$ and replace $pV$ using $pV=NkT$ .

Step-by-Step Reasoning

Write the deduced internal-energy relationship:

U = \frac{3}{2}pV.

Use the ideal gas law in molecular form:

pV = NkT.

Substitute into the first equation:

U = \frac{3}{2}(NkT)=\frac{3}{2}NkT.

Identify $k$ as the Boltzmann constant.

Key Takeaways

$pV$ is a bridge between macroscopic variables ( $p,V$ ) and microscopic variables ( $N,T$ ).
Substitution is often all that is needed once the correct proportionality is known.

Common Mistakes

Using $pV=nRT$ but then leaving the answer in terms of $n$ and $R$ (the question asked for $N$ and $T$ ).
Forgetting to identify what $k$ is.

Things to Be Careful About

$T$ must be in kelvin.
Do not confuse $N$ (number of molecules) with $n$ (number of moles).

Techniques used

substitute the ideal gas equation of state into an expressionexpress internal energy in microscopic (molecular) variablesidentify and define physical constants used

(c)

Determine expressions, in terms of $X$ and $Y$ , for the work $W$ done on the gas during:

2M

(i)

change AB

$W$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For AB, $p=2Y$ and $\Delta V = 2X-6X=-4X$ .

Work done by gas:

W_{\text{by}} = p\Delta V = (2Y)(-4X) = -8XY.

So work done on gas:

W = -W_{\text{by}} = +8XY.

Answer

W = 8XY

Final answer

$W = 8XY$

Detailed explanation

Background Concept

On a pressure–volume diagram, the work done by a gas during a change is

W_{\text{by}} = \int p\,\mathrm{d}V.

For a constant-pressure (isobaric) change, this becomes

W_{\text{by}} = p\Delta V.

If you want the work done on the gas, then

W_{\text{on}} = -W_{\text{by}}.

Compression means $\Delta V<0$ , so $W_{\text{by}}$ is negative and $W_{\text{on}}$ is positive.

Understanding the Question

Change AB is a horizontal line at pressure $2Y$ from volume $6X$ to $2X$ (compression). You must find the work done on the gas in terms of $X$ and $Y$ .

Approach

Use $W_{\text{by}}=p\Delta V$ for constant pressure, then change sign to get work on the gas.

Step-by-Step Reasoning

Read from the $p$ – $V$ graph for AB:

$p=2Y$ (constant),
$V$ changes from $6X$ to $2X$ .

Compute the volume change:

\Delta V = V_B - V_A = 2X - 6X = -4X.

Work done by the gas:

W_{\text{by}} = p\Delta V = (2Y)(-4X)=-8XY.

Work done on the gas is the negative:

W_{\text{on}}=+8XY.

Key Takeaways

For horizontal lines on a $p$ – $V$ graph, work is $p\Delta V$ .
Compression gives positive work done on the gas.

Common Mistakes

Giving $-8XY$ without converting from “by” to “on”.
Using $\Delta V = 6X-2X$ (wrong order) after defining $\Delta V=V_B-V_A$ .

Things to Be Careful About

Always state clearly whether your $W$ is “on” or “by” the gas.
Isochoric (vertical) segments have $\Delta V=0$ so zero work (useful later).

Techniques used

read pressure and volume change from a p–V diagramuse work as the area under a p–V curveapply the sign convention for work done on the gas

(ii)

change CD.

$W$ = ______

1M Similar

(d)

Use your answers in (c) and the first law of thermodynamics to determine an expression, in terms of $X$ and $Y$ , for the net thermal energy $Q$ supplied to the gas during one full cycle ABCDA. Explain your reasoning.

$Q$ = ______

3M Similar

DifficultyMedium

Worked solution

Working

Over one full cycle, $\Delta U=0$ .

Work done on gas occurs only on AB and CD (BC and DA have $\Delta V=0$ ):

W_{\text{net,on}} = (8XY) + (-20XY) = -12XY.

First law: $\Delta U = Q + W_{\text{on}}$ :

0 = Q + (-12XY) \Rightarrow Q = 12XY.

Answer

Q = 12XY

Final answer

$Q = 12XY$

Detailed explanation

Background Concept

For any process,

\Delta U = Q + W_{\text{on}}.

For a complete cycle, the system returns to its initial state, so internal energy (a state function) returns to its initial value:

\Delta U_{\text{cycle}} = 0.

Therefore, over a full cycle,

0 = Q_{\text{net}} + W_{\text{net,on}} \quad \Rightarrow \quad Q_{\text{net}} = -W_{\text{net,on}}.

Also, on a $p$ – $V$ diagram, net work done by the gas in a cycle equals the area enclosed (with sign depending on direction). For clockwise cycles, the gas does positive net work; anticlockwise, negative.

Understanding the Question

You already found the work done on the gas during AB and CD. You must use these and the first law to find the net thermal energy $Q$ supplied to the gas over the full loop A(\to)B(\to)C(\to)D(\to)A.

Key points from Fig. 4.1:

AB and CD are horizontal (constant pressure), so work is non-zero.
BC and DA are vertical (constant volume), so $\Delta V=0$ and work is zero.

Key point from the fact it’s a cycle:

The gas ends where it started, so net $\Delta U=0$ .

Approach

Add the work done on the gas for each segment to get $W_{\text{net,on}}$ (noting isochoric segments have zero work).
Use $\Delta U_{\text{cycle}}=0$ in $\Delta U = Q + W$ to find $Q_{\text{net}}$ .

Step-by-Step Reasoning

diagram-d-1

Identify segments with work:

AB: compression at $p=2Y$ (non-zero work).
BC: constant volume (zero work).
CD: expansion at $p=5Y$ (non-zero work).
DA: constant volume (zero work).

Use results from (c):

W_{AB,\text{on}} = 8XY,\quad W_{CD,\text{on}}=-20XY.

Net work done on the gas:

W_{\text{net,on}} = 8XY - 20XY = -12XY.

The negative sign means that overall the gas did net work on the surroundings (since work on gas is negative).

Over a cycle, internal energy returns to initial value:

\Delta U_{\text{cycle}} = 0.

Apply the first law over the cycle:

0 = Q_{\text{net}} + W_{\text{net,on}}.

So

Q_{\text{net}} = -W_{\text{net,on}} = -(-12XY)=12XY.

This positive value means net thermal energy is supplied to the gas during the cycle.

(Consistent check: the cycle is clockwise on the $p$ – $V$ diagram, so net work done by the gas is positive and hence net heat input must be positive.)

Key Takeaways

In a complete cycle, $\Delta U=0$ .
Isochoric processes do no work.
Net heat supplied equals net work done by the gas (or negative of net work done on the gas), for a cycle.

Common Mistakes

Forgetting that BC and DA have zero work and trying to include $p\Delta V$ there.
Using $\Delta U$ values from the internal energy graph and adding them incorrectly; the simplest is to use $\Delta U_{\text{cycle}}=0$ .
Sign errors: mixing “work on” and “work by” without changing sign in the first law.

Things to Be Careful About

State clearly that $\Delta U=0$ because the gas returns to state A.
Keep $W$ consistent with the version of the first law you wrote in (a).
When adding work values, include their signs (compression vs expansion).

Techniques used

sum work contributions around a p–V cycleapply that net change in internal energy over a complete cycle is zerouse the first law to relate net heat transfer to net work

Q5MediumOscillations

A steel ball on the end of a thin string oscillates with small oscillations, as shown in Fig. 5.1.

Fig. 5.1

The displacement of the centre of the ball from its equilibrium position is $x$ .

(a)

Fig. 5.2 shows the variation with $x$ of the acceleration $a$ of the ball.

Fig. 5.2

5M

(i)

Explain how Fig. 5.2 shows that the oscillations of the ball are simple harmonic.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For SHM,

a \propto -x \quad \text{(i.e. } a = -\omega^2 x\text{)}.

Fig. 5.2 is a straight line through the origin with a negative gradient, so $a$ is proportional to $x$ and always opposite in direction to $x$ ; hence the motion is simple harmonic.

Final answer

The a–x graph is a straight line through the origin with negative gradient, so a is proportional to −x (a = −ω²x), hence SHM.

Detailed explanation

Background Concept

Simple harmonic motion (SHM) is defined by the condition that the acceleration of the oscillating object is directly proportional to its displacement from equilibrium and is directed towards the equilibrium position.

Mathematically,

a = -\omega^2 x

where:

$x$ is the displacement from equilibrium,
$a$ is the acceleration,
$\omega$ is the angular frequency,
the negative sign means the acceleration is always opposite to the displacement (a restoring acceleration).

Understanding the Question

You are given a graph of acceleration $a$ against displacement $x$ for the ball as it oscillates.

The question asks how this graph demonstrates that the motion is SHM. So you must connect features of the graph (straight line, passes through origin, negative gradient) to the defining equation for SHM.

Approach

Recall the defining relation for SHM: $a = -\omega^2 x$ .
Compare this with what the graph shows:
- straight line through origin means $a \propto x$ ,
- negative gradient means $a$ has the opposite sign to $x$ (restoring).
State clearly that these two observations together satisfy the SHM condition.

Step-by-Step Reasoning

A straight line graph of $a$ against $x$ passing through $(0,0)$ shows that when $x=0$ , $a=0$ , and that $a$ is proportional to $x$ :

a \propto x.

The line slopes downwards (negative gradient). This means:
- if $x$ is positive, then $a$ is negative,
- if $x$ is negative, then $a$ is positive.

So the acceleration always points back towards equilibrium (opposite to the displacement).

That is exactly the SHM condition:

a = -\omega^2 x.

Therefore the oscillations are simple harmonic.

Key Takeaways

SHM is identified by the single condition $a = -\omega^2 x$ .
On an $a$ vs $x$ graph, SHM appears as a straight line through the origin with negative gradient.

Common Mistakes

Saying only “it is a straight line” without mentioning that it passes through the origin and has negative gradient.
Forgetting to interpret the negative gradient physically (restoring acceleration).
Confusing $a$ vs $t$ graphs (which are sinusoidal) with $a$ vs $x$ graphs (which are linear for SHM).

Things to Be Careful About

You must mention both proportionality (straight line through origin) and opposite direction (negative gradient). Either point on its own is not sufficient to establish SHM.

Techniques used

identify proportionality between acceleration and displacement from a graphuse the sign of the gradient to relate acceleration direction to displacementlink a straight line through the origin to the SHM condition a = -omega^2 x

(ii)

Determine the period $T$ of the oscillations.

$T$ = ______ $\text{s}$

3M Similar

DifficultyMedium

Worked solution

Working

Gradient of $a$ - $x$ graph:

\text{gradient} = \frac{\Delta a}{\Delta x} = \frac{-12 - (+12)}{1.5 - (-1.5)} = \frac{-24}{3.0} = -8.0\ \text{s}^{-2}.

For SHM, $a = -\omega^2 x$ , so

-\omega^2 = -8.0 \;\Rightarrow\; \omega = \sqrt{8.0} = 2.83\ \text{rad s}^{-1}.

T = \frac{2\pi}{\omega} = \frac{2\pi}{2.83} = 2.22\ \text{s}.

Answer

$T = 2.22\ \text{s}$

Final answer

2.22 s

Detailed explanation

Background Concept

In SHM, the acceleration is related to displacement by

a = -\omega^2 x.

If you plot $a$ against $x$ , this has the same form as a straight-line equation $a = mx$ , where the gradient is

m = -\omega^2.

Once $\omega$ is known, the period is found from

\omega = \frac{2\pi}{T} \quad \Rightarrow \quad T = \frac{2\pi}{\omega}.

Understanding the Question

You are given two points on the straight-line graph of acceleration $a$ (in $\text{cm s}^{-2}$ ) against displacement $x$ (in cm).

You must:

Find the gradient of this line.
Use the SHM relationship to extract $\omega$ .
Convert $\omega$ into the period $T$ .

Because both axes are in cm-based units, the gradient ends up in $\text{s}^{-2}$ (since $\text{cm s}^{-2}$ divided by cm leaves $\text{s}^{-2}$ ). This is exactly the unit expected for $\omega^2$ .

Approach

Use two clear points on the line to compute the gradient $\Delta a / \Delta x$ .
Identify gradient with $-\omega^2$ from $a=-\omega^2x$ .
Take the square root to find $\omega$ .
Use $T = 2\pi/\omega$ .

Step-by-Step Reasoning

Choose the two points shown: $(-1.5\ \text{cm}, +12\ \text{cm s}^{-2})$ and $(+1.5\ \text{cm}, -12\ \text{cm s}^{-2})$ .
Compute the gradient:

\text{gradient} = \frac{\Delta a}{\Delta x} = \frac{-12 - (+12)}{1.5 - (-1.5)} = \frac{-24}{3.0} = -8.0\ \text{s}^{-2}.

Compare with SHM form:

a = -\omega^2 x \;\Rightarrow\; \frac{a}{x} = -\omega^2.

So the gradient is $-\omega^2$ , hence

\omega^2 = 8.0\ \text{s}^{-2}, \qquad \omega = \sqrt{8.0} = 2.83\ \text{rad s}^{-1}.

Convert angular frequency to period:

T = \frac{2\pi}{\omega} = \frac{2\pi}{2.83} = 2.22\ \text{s}.

Key Takeaways

For SHM, the gradient of an $a$ vs $x$ graph equals $-\omega^2$ .
Period is obtained from $T=2\pi/\omega$ .
Unit checking helps: gradient units become $\text{s}^{-2}$ , matching $\omega^2$ .

Common Mistakes

Using $\Delta x/\Delta a$ instead of $\Delta a/\Delta x$ .
Forgetting the minus sign and writing $\omega^2 = -8$ .
Forgetting to take the square root when going from $\omega^2$ to $\omega$ .
Writing $T = \omega / 2\pi$ instead of $T = 2\pi/\omega$ .

Things to Be Careful About

Use points that lie exactly on the line to reduce reading error.
Keep enough significant figures during working; round at the end.
The period is in seconds; $\omega$ is in $\text{rad s}^{-1}$ (radian is dimensionless but should be written).

Techniques used

calculate the gradient of a straight-line graphequate the magnitude of the gradient to omega squared using a = -omega^2 xuse omega = 2 pi / T to determine the period

(b)

At time $t = 0$ , when the displacement of the ball has its maximum value, the ball is immersed in a trough containing thick oil so that the ball is just below the surface of the oil. This results in the subsequent motion of the ball being heavily damped.

5M

(i)

State what is meant by damping.

2M Similar

DifficultyEasy

Worked solution

Answer

Damping is the loss of energy from an oscillating system (due to resistive forces), causing the amplitude of the oscillations to decrease with time.

Final answer

Loss of energy from an oscillator due to resistive forces, causing its amplitude to decrease with time.

Detailed explanation

Background Concept

Real oscillators experience resistive forces (e.g. friction, air resistance, viscous drag). These forces do work against the motion, transferring mechanical energy (kinetic + potential) out of the oscillating system, usually as thermal energy in the surroundings.

This process is called damping. A key observable consequence is that the amplitude decreases with time.

Understanding the Question

The ball is placed just below the surface of thick oil, which provides a large viscous resistive force. The question asks for the meaning of “damping” in oscillations.

For full credit you need to mention both:

energy loss (or resistive force doing work), and
reduction of amplitude with time.

Approach

State damping as an energy-transfer process, then link it to the observable decay in amplitude.

Step-by-Step Reasoning

As the ball moves through a fluid, a resistive (drag) force acts opposite to the velocity.
This force does negative work on the system over each cycle, so mechanical energy decreases.
With less mechanical energy, the maximum displacement (amplitude) becomes smaller each cycle.

Thus damping is the loss of energy from the oscillating system due to resistive forces, leading to a decrease in amplitude with time.

Key Takeaways

Damping = energy removed from an oscillator by resistive forces.
The experimental sign of damping is a decreasing amplitude.

Common Mistakes

Saying “damping is friction” without stating energy loss or amplitude reduction.
Confusing damping with “frequency decreasing”; damping mainly affects amplitude (frequency may change slightly, but that is not the definition).

Things to Be Careful About

Use clear cause-and-effect wording: resistive forces → energy loss → amplitude decreases.

Techniques used

define damping in terms of energy transfer from the oscillatorlink damping to resistive forces and reduction of amplitude with time

(ii)

On Fig. 5.3, sketch a possible variation of the displacement $x$ of the ball with $t$ between $t = 0$ and $t = 2T$ .

Fig. 5.3

3M Similar

DifficultyMedium

Worked solution

diagram-5bii

Answer

Sketch starts at $x = +x_0$ at $t=0$ with zero gradient, then $x$ decreases towards $0$ without oscillating (does not cross $x=0$ ), approaching $x=0$ by $t = 2T$ .

Final answer

See sketch (heavily damped: returns to equilibrium without oscillating).

Detailed explanation

Background Concept

Damping describes energy loss from an oscillator due to resistive forces. The nature of the motion depends on the strength of damping:

Light damping (underdamped): oscillations continue but amplitude decays.
Critical damping: fastest return to equilibrium without oscillating.
Heavy damping (overdamped): returns to equilibrium without oscillating, but more slowly than critical damping.

In heavy damping there are no repeated crossings of the equilibrium position; instead, displacement decays back to zero in a non-oscillatory way.

Understanding the Question

At $t=0$ , the ball is at maximum displacement, so its instantaneous velocity is zero. Then it is immersed in thick oil, making the subsequent motion heavily damped.

You must sketch $x$ against $t$ from $0$ to $2T$ (where $T$ is the undamped SHM period found earlier), showing the qualitative effect of heavy damping.

Key required features:

starts at maximum positive displacement at $t=0$ ,
initial slope $\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)_{t=0}=0$ ,
no oscillations (no repeated maxima/minima, no crossing of $x=0$ ),
tends towards $x=0$ as time increases.

Approach

Place the starting point at $t=0$ at the top of the displacement scale.
Because it is a maximum, draw a horizontal tangent at the start.
For heavy damping, sketch a monotonic decay towards equilibrium (often drawn as an exponential-like curve), ensuring it does not overshoot through $x=0$ .

Step-by-Step Reasoning

diagram-5bii

Initial condition at $t=0$ : maximum displacement means $x(0)=+x_0$ .
Slope at $t=0$ : at a maximum in a displacement-time graph, the velocity is zero, so the tangent is horizontal.
Effect of thick oil: viscous drag is large, so mechanical energy is removed rapidly. In the heavy-damping case, the system does not have enough “inertia” relative to the damping to pass through equilibrium and continue to the other side.
Shape of curve: the displacement returns towards zero without oscillations. The curve approaches $x=0$ with the slope tending to zero as it reaches equilibrium.

Key Takeaways

Heavy damping gives a non-oscillatory return to equilibrium.
“Maximum displacement at $t=0$ ” implies a horizontal tangent at the start of the sketch.

Common Mistakes

Drawing a sinusoidal curve with decreasing amplitude (that is light damping / underdamped motion, not heavy damping).
Starting at $x=0$ instead of at maximum displacement.
Drawing the curve crossing $x=0$ and oscillating about equilibrium.
Forgetting the horizontal tangent at $t=0$ .

Things to Be Careful About

The axis is labelled up to $2T$ ; you are not required to show a period in the damped motion. $T$ is just a time scale on the axis.
“Heavily damped” in this syllabus typically means no oscillations (overdamped), not merely “quickly decreasing amplitude.”

Techniques used

use initial conditions to set the starting point and slope on an x–t graphdistinguish heavy damping from underdamping by the absence of oscillationssketch an exponential-like decay towards equilibrium

Q6MediumElectric FieldsCapacitance

(a)

Define electric field at a point.

1M Similar

DifficultyEasy

Worked solution

Answer

Electric field (strength) at a point is the force per unit positive test charge placed at that point (in the direction of the force on a positive charge).

Final answer

Force per unit positive test charge at the point (direction is that of the force on a positive charge).

Detailed explanation

Background Concept

The electric field (often meaning electric field strength) is a way to describe how a charge would experience a force in a region of space.

It is defined by

E = \frac{F}{q}

where:

$E$ is electric field strength,
$F$ is the electrostatic force on a charge,
$q$ is the charge.

The direction of $E$ is defined as the direction of the force on a small positive test charge.

Understanding the Question

You are asked to define the electric field at a point. A “point” definition means you should not talk about a whole region or about field lines; you should give the standard definition relating field to force and charge.

Approach

State the standard definition: electric field strength equals force per unit positive charge, and include the direction statement.

Step-by-Step Reasoning

Take the defining equation $E = F/q$ .
Convert it into words: “force per unit charge”.
Add the conventional direction: use a positive test charge to define the direction.

Key Takeaways

Electric field strength is defined locally (at a point).
$E$ relates force to charge: $F = qE$ .
Direction is that of the force on a positive test charge.

Common Mistakes

Saying “force per unit charge” but not specifying that it is for a positive test charge (direction can become ambiguous).
Confusing electric field with electric potential (potential is energy per unit charge).

Things to Be Careful About

Use the wording “at a point” (local definition).
Make clear it is per unit charge (not per unit mass — that is gravitational field strength).

Techniques used

state a definition in terms of force per unit chargespecify direction using a positive test charge

(b)

An isolated conducting sphere in a vacuum has a capacitance of $69\text{ pF}$ . The charge on the sphere is $+83\text{ pC}$ .

8M

(i)

On Fig. 6.1, draw field lines to represent the electric field outside the sphere due to the charge on the sphere.

Fig. 6.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

diagram-b-i

Field lines are radial, perpendicular to the surface, with arrows pointing away from the positively charged sphere (symmetrically spaced).

Final answer

Radial lines, perpendicular to the surface, arrows outwards from the + sphere.

Detailed explanation

Background Concept

Electric field lines are a diagrammatic way to show the direction and relative strength of an electric field:

The direction of the field at a point is tangent to the field line (arrow direction is the force direction on a positive test charge).
The field is stronger where the field lines are closer together.

For a conducting sphere in electrostatic equilibrium:

All excess charge resides on the surface.
The field inside the conductor is zero.
The field just outside the surface is perpendicular (normal) to the surface.

Understanding the Question

You have a positively charged isolated conducting sphere in a vacuum. You must sketch the electric field outside it using field lines on the given diagram.

Approach

Use the known field pattern for a positively charged sphere:

Draw several straight lines radiating out from the sphere.
Ensure lines meet the surface at right angles.
Put arrowheads pointing away from the sphere.
Make the pattern symmetric (uniformly spaced around the sphere).

Step-by-Step Reasoning

Because the sphere is positively charged, a positive test charge would be repelled, so arrows point outwards.
A sphere has spherical symmetry, so the field magnitude depends only on distance from the centre, giving a radial pattern.
A conductor’s surface is an equipotential, so field lines must be perpendicular to it.

diagram-b-i

Key Takeaways

Positive charges have field lines pointing away; negative charges have field lines pointing toward.
For a conducting sphere, field lines are radial and normal to the surface.

Common Mistakes

Drawing circular lines around the sphere (that would be more like magnetic field lines).
Forgetting arrowheads.
Drawing lines that are not perpendicular to the conducting surface.
Drawing lines inside the conducting sphere (field inside is zero).

Things to Be Careful About

Keep the line pattern symmetric.
Field lines should not cross.
If you vary density, it should be denser near the surface and spread out with distance.

Techniques used

use rules for drawing electric field lines around a conductorshow radial symmetry and correct arrow directionrepresent relative field strength using line density

(ii)

Calculate the electric potential at the surface of the sphere.

$\text{electric potential}$ = ______ $\text{V}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

V = \frac{Q}{C} = \frac{83 \times 10^{-12}}{69 \times 10^{-12}} = 1.20

Answer

$V = 1.2\ \text{V}$

Final answer

1.2 V

Detailed explanation

Background Concept

Capacitance $C$ relates the charge stored $Q$ to the potential difference $V$ :

C = \frac{Q}{V}

So

V = \frac{Q}{C}

For an isolated charged conductor, the “potential at the surface” is the potential of the conductor relative to zero potential at infinity.

Understanding the Question

You are told:

capacitance of the isolated sphere: $C = 69\ \text{pF}$
charge on the sphere: $Q = +83\ \text{pC}$

You must calculate the electric potential at the surface of the sphere.

Approach

Use the capacitance definition $V = Q/C$ . Convert pC and pF into SI units, substitute, then quote the answer in volts.

Step-by-Step Reasoning

Convert to SI:
- $Q = 83\ \text{pC} = 83 \times 10^{-12}\ \text{C}$
- $C = 69\ \text{pF} = 69 \times 10^{-12}\ \text{F}$
Use

V = \frac{Q}{C}

Substitute:

V = \frac{83 \times 10^{-12}}{69 \times 10^{-12}} = 1.20\ \text{V}

The $10^{-12}$ factors cancel neatly, leaving a value close to $1\ \text{V}$ .

Key Takeaways

For any conductor/capacitor: $Q = CV$ .
Always convert prefixes (pico $=10^{-12}$ ) before substituting.

Common Mistakes

Forgetting to convert pF or pC to SI, leading to a factor-of- $10^{12}$ error.
Writing $V = C/Q$ instead of $V = Q/C$ .

Things to Be Careful About

Units: $\text{F} = \text{C V}^{-1}$ , so $Q/C$ correctly gives volts.
Significant figures: typically 2 s.f. is appropriate here (e.g. $1.2\ \text{V}$ ).

Techniques used

apply the definition of capacitance Q = CVconvert pico-units to SI unitsrearrange and substitute to find potential

(iii)

Determine the radius of the sphere.

$\text{radius}$ = ______ $\text{m}$

2M Similar

DifficultyMedium

Worked solution

Working

For an isolated sphere,

C = 4\pi \varepsilon_0 r \;\;\Rightarrow\;\; r = \frac{C}{4\pi \varepsilon_0}

r = \frac{69 \times 10^{-12}}{4\pi (8.85 \times 10^{-12})} = 0.62

Answer

$\text{radius} = 0.62\ \text{m}$

Final answer

0.62 m

Detailed explanation

Background Concept

An isolated conducting sphere in a vacuum behaves like a capacitor whose other “plate” is effectively at infinity.

Its capacitance depends only on its radius $r$ :

C = 4\pi \varepsilon_0 r

where $\varepsilon_0$ is the permittivity of free space.

Understanding the Question

You are given the sphere’s capacitance $C = 69\ \text{pF}$ and asked to find its radius. No charge value is needed for this part because $C$ depends only on geometry for an isolated sphere.

Approach

Use $C = 4\pi \varepsilon_0 r$ and rearrange to $r = C/(4\pi \varepsilon_0)$ . Substitute $C$ in farads and $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}$ .

Step-by-Step Reasoning

Start with

C = 4\pi \varepsilon_0 r

Rearrange:

r = \frac{C}{4\pi \varepsilon_0}

Convert $C$ :

C = 69\ \text{pF} = 69 \times 10^{-12}\ \text{F}

Substitute:

r = \frac{69 \times 10^{-12}}{4\pi (8.85 \times 10^{-12})}

The $10^{-12}$ cancels, giving a value of order $1\ \text{m}$ .

Calculate to obtain

r \approx 0.62\ \text{m}

Key Takeaways

For an isolated sphere, capacitance is proportional to radius: $C \propto r$ .
The constant of proportionality is $4\pi\varepsilon_0$ .

Common Mistakes

Using the parallel-plate formula $C = \varepsilon_0 A/d$ (not applicable here).
Forgetting that $\varepsilon_0$ has units of $\text{F m}^{-1}$ .
Missing the factor of $4\pi$ .

Things to Be Careful About

Use SI units: $C$ in farads.
Keep enough significant figures in intermediate steps; round only at the end.

Techniques used

use the capacitance of an isolated sphere C = 4\pi\varepsilon_0 rrearrange to make the radius the subjectsubstitute constants and evaluate

(iv)

Calculate the electric field strength $E$ at the surface of the sphere. Give a unit with your answer.

$E$ = ______ $\text{unit}$ ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At the surface,

E = \frac{Q}{4\pi \varepsilon_0 r^2}

E = \frac{83 \times 10^{-12}}{4\pi (8.85 \times 10^{-12})(0.62)^2} = 1.9

Answer

$E = 1.9\ \text{V m}^{-1}$ (or $1.9\ \text{N C}^{-1}$ )

Final answer

1.9 V m^-1

Detailed explanation

Background Concept

For a spherically symmetric charge distribution (such as a charged conducting sphere), the field outside is the same as if all the charge were concentrated at the centre.

So at distance $r$ from the centre,

E = \frac{Q}{4\pi \varepsilon_0 r^2}

The unit of $E$ can be written as $\text{N C}^{-1}$ (from $E=F/q$ ) or equivalently $\text{V m}^{-1}$ (from the potential gradient).

Understanding the Question

You have already found the radius $r$ of the sphere in part (iii), and you know the charge $Q$ from the stem. You are asked for the electric field strength at the surface, i.e. at $r$ equal to the sphere’s radius.

Approach

Use the spherical-field formula $E = Q/(4\pi\varepsilon_0 r^2)$ with:

$Q = 83 \times 10^{-12}\ \text{C}$ ,
$r = 0.62\ \text{m}$ ,
$\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F m}^{-1}$ .
Then state the answer with a correct unit.

Step-by-Step Reasoning

Write the formula:

E = \frac{Q}{4\pi \varepsilon_0 r^2}

Substitute values:

E = \frac{83 \times 10^{-12}}{4\pi (8.85 \times 10^{-12})(0.62)^2}

Note that $10^{-12}$ cancels, so the field will be a few $\text{V m}^{-1}$ .
Calculate to get:

E \approx 1.9\ \text{V m}^{-1}

Unit check: $\text{V m}^{-1}$ is acceptable, and is equivalent to $\text{N C}^{-1}$ .

Key Takeaways

Outside a charged sphere: treat it like a point charge at the centre.
Field at the surface uses $r$ equal to the sphere radius.
Remember acceptable units: $\text{N C}^{-1}$ or $\text{V m}^{-1}$ .

Common Mistakes

Using $E = Q/(4\pi\varepsilon_0 r)$ (missing the square).
Using diameter instead of radius.
Giving an incorrect unit (e.g. $\text{N}$ or $\text{V}$ ).

Things to Be Careful About

Keep consistent SI units (metres, coulombs).
Quote the final answer to 2 s.f. unless the question/data suggest otherwise.

Techniques used

use the surface field of a spherical charge distributionsubstitute values with correct powers of tenquote an appropriate unit for electric field strength

(c)

The sphere in (b) is discharged by connecting it to earth ( $0\text{ V}$ ) through a resistor of resistance $120\text{ M}\Omega$ .

Calculate the time taken for the charge to fall to $26\text{ pC}$ .

$\text{time}$ = ______ $\text{s}$

2M Similar

DifficultyMedium

Worked solution

Working

Time constant:

\tau = RC = (120 \times 10^{6})(69 \times 10^{-12}) = 8.28 \times 10^{-3}\ \text{s}

For discharge,

Q = Q_0 e^{-t/\tau} \;\;\Rightarrow\;\; t = \tau \ln\left(\frac{Q_0}{Q}\right)

t = (8.28 \times 10^{-3}) \ln\left(\frac{83}{26}\right) = 9.6 \times 10^{-3}\ \text{s}

Answer

$\text{time} = 9.6 \times 10^{-3}\ \text{s}$

Final answer

9.6 × 10^-3 s

Detailed explanation

Background Concept

When a capacitor discharges through a resistor, the charge on the capacitor decreases exponentially:

Q = Q_0 e^{-t/RC}

where:

$Q_0$ is the initial charge at $t=0$ ,
$R$ is the resistance,
$C$ is the capacitance,
$\tau = RC$ is the time constant.

After one time constant $\tau$ , the charge has fallen to $Q_0/e$ .

Understanding the Question

You have the same sphere from part (b), so the capacitance is still $C = 69\ \text{pF}$ . It is discharged to earth through a resistor $R = 120\ \text{M}\Omega$ .

The charge falls from $Q_0 = 83\ \text{pC}$ to $Q = 26\ \text{pC}$ .

You must find the time $t$ for that change.

Approach

Compute the time constant $\tau = RC$ using SI units.
Use the discharge equation $Q = Q_0 e^{-t/\tau}$ .
Rearrange for $t$ by taking natural logs.

Step-by-Step Reasoning

Convert units:
- $R = 120\ \text{M}\Omega = 120 \times 10^{6}\ \Omega$
- $C = 69\ \text{pF} = 69 \times 10^{-12}\ \text{F}$
Time constant:

\tau = RC = (120 \times 10^{6})(69 \times 10^{-12}) = 8.28 \times 10^{-3}\ \text{s}

Discharge law:

Q = Q_0 e^{-t/\tau}

Divide by $Q_0$ :

\frac{Q}{Q_0} = e^{-t/\tau}

Take $\ln$ :

\ln\left(\frac{Q}{Q_0}\right) = -\frac{t}{\tau}

So

t = \tau \ln\left(\frac{Q_0}{Q}\right)

Substitute $Q_0 = 83\ \text{pC}$ and $Q = 26\ \text{pC}$ (the prefix cancels in the ratio):

t = (8.28 \times 10^{-3})\ln\left(\frac{83}{26}\right)

\ln\left(\frac{83}{26}\right) \approx 1.16

t \approx 9.6 \times 10^{-3}\ \text{s}

Key Takeaways

Capacitor discharge is exponential, characterised by $\tau = RC$ .
To find time for a given change in $Q$ , rearrange using $\ln$ .

Common Mistakes

Using base-10 log instead of natural log.
Forgetting to convert $\text{M}\Omega$ and pF into SI units when calculating $RC$ .
Using $t = RC (Q_0/Q)$ (missing the logarithm).

Things to Be Careful About

Use $\ln$ , not $\lg$ .
The ratio $Q_0/Q$ is dimensionless, so you can use pC directly in that ratio, but $RC$ must be in seconds using SI units.

Techniques used

apply the exponential discharge equation Q = Q0 e^{-t/RC}rearrange using natural logarithmscalculate the time constant RC from given R and C

Q7MediumAlternating Currents

An alternating voltage $V$ varies with time $t$ according to

V = 18\cos 40\pi t

where $V$ is in $\text{V}$ and $t$ is in $\text{s}$ .

(a)

2M

(i)

show that the period is $0.050\text{ s}$

1M Similar

DifficultyEasy

Worked solution

Working

From $V = 18\cos 40\pi t$ , angular frequency $\omega = 40\pi\ \text{rad s}^{-1}$ .

T = \frac{2\pi}{\omega} = \frac{2\pi}{40\pi} = \frac{1}{20} = 0.050\ \text{s}

Answer

$T = 0.050\ \text{s}$

Final answer

0.050 s

Detailed explanation

Background Concept

A sinusoidal a.c. quantity can be written in the form

V = V_0 \cos(\omega t) \quad \text{or} \quad V = V_0 \sin(\omega t)

where:

$V_0$ is the peak (maximum) value,
$\omega$ is the angular frequency in $\text{rad s}^{-1}$ ,
$t$ is time.

The period $T$ is the time for one complete cycle. For sinusoidal motion,

\omega = \frac{2\pi}{T}

Understanding the Question

You are given an a.c. voltage:

V = 18\cos 40\pi t

and asked to show that its period is $0.050\ \text{s}$ . The key information is the coefficient of $t$ inside the cosine, which is $\omega$ .

Approach

Read off $\omega$ from $\cos(\omega t)$ .
Use $T = 2\pi/\omega$ .

Step-by-Step Reasoning

From

V = 18\cos 40\pi t

compare with $V = V_0\cos(\omega t)$ to get

\omega = 40\pi\ \text{rad s}^{-1}.

Then

T = \frac{2\pi}{\omega} = \frac{2\pi}{40\pi} = \frac{1}{20} = 0.050\ \text{s}.

Key Takeaways

The angular frequency is the factor multiplying $t$ inside the sine/cosine.
Period and angular frequency are linked by $\omega = 2\pi/T$ .

Common Mistakes

Using $T = 1/\omega$ instead of $T = 2\pi/\omega$ .
Treating $40\pi$ as the frequency in Hz (it is angular frequency in $\text{rad s}^{-1}$ ).

Things to Be Careful About

Always include $2\pi$ when converting between $\omega$ and $T$ .
Quote $T$ with unit seconds and appropriate significant figures.

Techniques used

identify angular frequency from a sinusoidal equationuse the relationship between angular frequency and periodconvert between period and frequency

(ii)

determine the root-mean-square (r.m.s.) voltage.

$\text{r.m.s. voltage}$ = ______ $\text{V}$

1M Similar

DifficultyEasy

Worked solution

Working

Peak voltage $V_0 = 18\ \text{V}$ .

V_{\text{r.m.s.}} = \frac{V_0}{\sqrt{2}} = \frac{18}{\sqrt{2}} = 12.7\ \text{V}

Answer

$V_{\text{r.m.s.}} \approx 13\ \text{V}$

Final answer

13 V

Detailed explanation

Background Concept

The root-mean-square (r.m.s.) value of an alternating voltage is the d.c. voltage that would produce the same mean power in a resistor as the a.c. voltage.

For a sinusoidal voltage

V = V_0 \cos(\omega t)

the r.m.s. value is

V_{\text{r.m.s.}} = \frac{V_0}{\sqrt{2}}.

This result comes from averaging $V^2$ over a full cycle (since power in a resistor depends on $V^2$ ).

Understanding the Question

The given wave is

V = 18\cos 40\pi t.

So the peak voltage (maximum magnitude) is the amplitude, $18\ \text{V}$ . The question asks for the r.m.s. voltage.

Approach

Identify $V_0$ from the coefficient in front of the cosine.
Apply $V_{\text{r.m.s.}} = V_0/\sqrt{2}$ .

Step-by-Step Reasoning

From $V = 18\cos 40\pi t$ :

V_0 = 18\ \text{V}.

Then

V_{\text{r.m.s.}} = \frac{18}{\sqrt{2}} = 12.7\ \text{V} \approx 13\ \text{V}.

Key Takeaways

For a sine/cosine wave: $V_{\text{r.m.s.}} = V_0/\sqrt{2}$ .
$V_0$ is simply the amplitude of the voltage-time equation.

Common Mistakes

Using $V_{\text{r.m.s.}} = V_0\sqrt{2}$ (inverted).
Confusing peak-to-peak voltage ( $2V_0$ ) with peak voltage ( $V_0$ ).

Things to Be Careful About

Significant figures: $V_0 = 18\ \text{V}$ is 2 s.f., so $V_{\text{r.m.s.}}$ should be given to 2 s.f. (about $13\ \text{V}$ ).
Include the unit V.

Techniques used

use the r.m.s.–peak relationship for a sinusoidcalculate a numerical value and round appropriatelystate the result with correct units

(b)

On Fig. 7.1, sketch the variation of $V$ with $t$ for values of $t$ from $t = 0$ to $t = 100\text{ ms}$ .

Fig. 7.1

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Amplitude $= 18\ \text{V}$ and $T = 0.050\ \text{s} = 50\ \text{ms}$ .

Key points:

$t=0$ : $V=+18\ \text{V}$
$t=12.5\ \text{ms}$ : $V=0$
$t=25\ \text{ms}$ : $V=-18\ \text{V}$
$t=37.5\ \text{ms}$ : $V=0$
$t=50\ \text{ms}$ : $V=+18\ \text{V}$

Repeat for $50$ to $100\ \text{ms}$ .

Answer

Sketch a cosine wave of peak $\pm 18\ \text{V}$ completing two cycles between $0$ and $100\ \text{ms}$ , starting at $+18\ \text{V}$ at $t=0$ .

Final answer

Two-cycle cosine wave, peak ±18 V, period 50 ms, starting at +18 V at t = 0.

Detailed explanation

Background Concept

A cosine waveform

V = V_0 \cos(\omega t)

has these useful features:

Peak value: $V_0$ (at $t = 0$ if there is no phase shift).
Period: $T = 2\pi/\omega$ .
Over one cycle starting at $t=0$ $t = 0$ :
- at $t=0$ : $\cos(0)=1$ so $V=+V_0$ ,
- at $t=T/4$ : $\cos(\pi/2)=0$ so $V=0$ ,
- at $t=T/2$ : $\cos(\pi)=-1$ so $V=-V_0$ ,
- at $t=3T/4$ : $V=0$ ,
- at $t=T$ : back to $+V_0$ .

Understanding the Question

You need to sketch $V$ from $t=0$ to $t=100\ \text{ms}$ for

V = 18\cos(40\pi t).

This is a cosine wave of amplitude $18\ \text{V}$ with period $T=0.050\ \text{s}=50\ \text{ms}$ , so between $0$ and $100\ \text{ms}$ there are exactly two full cycles.

Approach

Convert the period into ms to match the axis.
Mark the standard cosine key points every $T/4$ .
Draw a smooth sinusoidal curve through these points, repeating after each period.

Step-by-Step Reasoning

Peak (amplitude): the coefficient of the cosine is $18$ , so the graph reaches $+18\ \text{V}$ and $-18\ \text{V}$ .
Period: $T=50\ \text{ms}$ .
Quarter period:

\frac{T}{4} = 12.5\ \text{ms}.

So the key points in the first cycle are:

$0\ \text{ms}$ : $+18\ \text{V}$
$12.5\ \text{ms}$ : $0\ \text{V}$
$25\ \text{ms}$ : $-18\ \text{V}$
$37.5\ \text{ms}$ : $0\ \text{V}$
$50\ \text{ms}$ : $+18\ \text{V}$

Then the same pattern repeats from $50\ \text{ms}$ to $100\ \text{ms}$ .

diagram-b-1

Key Takeaways

Read $V_0$ directly as the amplitude.
Use $T/4$ spacing to place zeros and peaks accurately.
A cosine with no phase shift starts at its maximum.

Common Mistakes

Drawing a sine wave starting at $0$ instead of a cosine starting at $+18\ \text{V}$ .
Using the wrong period (e.g. $25\ \text{ms}$ instead of $50\ \text{ms}$ ).
Forgetting the negative half-cycle (must go down to $-18\ \text{V}$ ).

Things to Be Careful About

The time axis is in ms: ensure consistent units when locating $T/4$ , $T/2$ , etc.
The sketch should be smooth and periodic, not triangular or piecewise straight lines.
Ensure the peaks align at $t=0, 50, 100\ \text{ms}$ for a cosine wave here.

Techniques used

extract amplitude and period from a trigonometric time equationidentify key points of a cosine wave over multiple cyclessketch a sinusoidal graph with correct scaling

(c)

The alternating voltage is rectified to produce an output voltage across a load resistor R, as shown in Fig. 7.2.

Fig. 7.2

Fig. 7.3 shows the variation with $t$ of the power $P$ in the load resistor.

Fig. 7.3

State three conclusions that can be drawn from Fig. 7.3. The conclusions may be qualitative or quantitative. Use the space for any working.

3M Similar

DifficultyMedium

Worked solution

Answer

From Fig. 7.3:

$P$ is always $\ge 0$ (never negative).
The pulses repeat every $25\ \text{ms}$ (so power frequency $= 40\ \text{Hz}$ , i.e. twice the a.c. frequency).
The maximum power is about $27\ \text{W}$ (so if $V_\text{peak}=18\ \text{V}$ across $R$ , then $R \approx 18^2/27 \approx 12\ \Omega$ ).

Final answer

P is always non-negative; pulses every 25 ms (40 Hz, double input); peak power ≈ 27 W (so R ≈ 12 Ω if Vpeak = 18 V).

Detailed explanation

Background Concept

For a resistor $R$ , instantaneous electrical power is

P = VI = \frac{V^2}{R} = I^2R.

Key consequences:

Power in a resistor is never negative because $V^2 \ge 0$ .
If a voltage is rectified (made all one polarity), the voltage waveform changes, and because $P \propto V^2$ , the power waveform often has a different period from the original voltage.

Rectification:

Half-wave rectification produces one output pulse per input cycle.
Full-wave rectification produces two output pulses per input cycle, so the output (and power) repeats twice as often.

Understanding the Question

You are shown a graph (Fig. 7.3) of the power $P$ in a load resistor after an a.c. voltage has been rectified.

You must state three conclusions from the graph. Valid conclusions include observations about:

sign of $P$ ,
period / frequency of the pulses,
maximum value (and hence possible deductions about $R$ if peak voltage is known).

Approach

Read three distinct features directly from the graph:

whether the graph goes negative,
the time between repeating pulses (period of $P$ ),
the peak value of $P$ .

Optionally, combine peak power with $P_{\max} = V_{\max}^2/R$ to estimate $R$ if the peak voltage is known/assumed unchanged by the rectifier.

Step-by-Step Reasoning

Sign of power: The curve lies on/above the time axis only, so $P \ge 0$ always. This is consistent with a resistor and with rectification (no negative power shown).
Period of the power waveform: The pulses repeat at intervals of about $25\ \text{ms}$ . Therefore

T_P \approx 25\ \text{ms} \Rightarrow f_P \approx \frac{1}{0.025} = 40\ \text{Hz}.

The original a.c. had $T=50\ \text{ms}$ ( $20\ \text{Hz}$ ), so the power repeats twice as often, indicating two similar power pulses per input cycle. This is what you expect from full-wave rectification (or any process that makes the output depend on $|V|$ or $V^2$ ).

Peak power: From the vertical axis, the maximum is about $27\ \text{W}$ .

If the peak output voltage across the resistor is $18\ \text{V}$ (ideal rectifier, ignoring any diode voltage drops), then

P_{\max} = \frac{V_{\max}^2}{R} \Rightarrow R = \frac{V_{\max}^2}{P_{\max}} = \frac{18^2}{27} \approx 12\ \Omega.

diagram-c-1

Key Takeaways

In a resistor, $P = V^2/R$ , so power is always non-negative.
Rectification can change the frequency of the output waveform; full-wave rectification produces pulses at twice the original frequency.
Peak power values can be used with $P_{\max} = V_{\max}^2/R$ to estimate resistance (given the peak voltage across the load).

Common Mistakes

Saying power is alternating (it is not negative here).
Using the original period ( $50\ \text{ms}$ ) instead of the observed power period ( $25\ \text{ms}$ ).
Mixing up peak and r.m.s. values when using $P = V^2/R$ .

Things to Be Careful About

The graph is of power, not voltage: the period you read is $T_P$ , which may differ from the voltage period.
If real diodes are involved, the peak voltage across $R$ could be slightly less than $18\ \text{V}$ ; the resistance calculation assumes ideal rectification unless stated otherwise.
Make sure your three conclusions are distinct (e.g. don’t give three different ways of stating the same period/frequency fact).

Techniques used

interpret key features from a graph of power against timerelate changes in period to frequency doubling after rectificationapply the power relationship for a resistor

Q8MediumQuantum Physics

Fig. 8.1 shows the three lowest-frequency lines in the part of the emission spectrum for hydrogen that relates to electron transitions to the ground state (level $n = 1$ ).

Fig. 8.1

The numbers represent the frequencies, in $10^{15}\text{ Hz}$ , associated with the spectral lines.

(a)

Use the photon model of electromagnetic radiation to explain how the existence of spectral lines in the emission spectrum provides evidence for discrete electron energy levels in the hydrogen atom.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

In the photon model, electromagnetic radiation is emitted as photons with energy

E = hf.

An emitted photon is produced when an electron drops from a higher level to a lower level, so

hf = \Delta E.

Because the spectrum contains only certain (discrete) frequencies, only certain values of $\Delta E$ occur, so the electron energies must be in discrete levels (not a continuum).

Final answer

Discrete spectral lines imply only certain photon energies (E = hf), so only certain energy differences occur and hence electron energy levels are discrete.

Detailed explanation

Background Concept

In the photon model, electromagnetic radiation is quantised into photons. Each photon has energy

E = hf

where $h$ is Planck’s constant and $f$ is the frequency.

In an atom, an electron can move between allowed energy levels. If it moves from a higher energy level to a lower one, it emits a photon. Conservation of energy gives

E_{\text{photon}} = E_{\text{upper}} - E_{\text{lower}} = \Delta E.

So each possible electron transition produces a photon whose energy (and hence frequency) is fixed.

Understanding the Question

The figure shows an emission spectrum of hydrogen for transitions ending at the ground state $n=1$ (a series of lines). You are asked to use the photon model to explain why seeing separate lines (rather than a continuous band) is evidence that electron energies in hydrogen are discrete.

Approach

Connect three ideas in a short chain:

Emission occurs as photons with $E = hf$ .
A photon’s energy equals the difference between two electron energy levels, $\Delta E$ .
Only certain frequencies are present (lines), so only certain photon energies and hence only certain energy differences occur, implying discrete energy levels.

Step-by-Step Reasoning

Because radiation is emitted as photons, each photon has a definite energy given by $E = hf$ .
When the electron drops to a lower level, it loses energy; that lost energy appears as a photon:

hf = \Delta E.

The spectrum shows only a few distinct frequencies (separate lines), not all possible frequencies.
Therefore only certain values of $hf$ occur, so only certain values of $\Delta E$ occur.
The only way for energy differences to take only specific values is for the electron energies themselves to be restricted to specific (discrete) levels.

Key Takeaways

Line spectra arise because atomic energy levels are quantised.
Each spectral line corresponds to one transition with fixed $\Delta E$ .
Photon energy links spectra to energy level spacings via $E = hf$ .

Common Mistakes

Saying “electrons emit photons of any energy” (that would give a continuous spectrum).
Missing the link $hf = \Delta E$ and only stating “levels are discrete” with no justification.
Confusing absorption and emission (here it is emission: electron loses energy and photon is produced).

Things to Be Careful About

Make it explicit that discrete frequencies mean discrete photon energies, and those equal energy differences between levels.
Use the photon equation $E = hf$ somewhere; otherwise the argument is incomplete.

Techniques used

apply the photon model using E = hflink photon energy to electron transition energy differencesinfer discrete energy levels from discrete emitted frequencies

(b)

The energy of the ground state (level $n = 1$ ) in a hydrogen atom is $-13.6\text{ eV}$ .

7M

(i)

Calculate the energy, in $\text{J}$ , of the ground state.

$\text{energy}$ = ______ $\text{J}$

1M Similar

DifficultyEasy

Worked solution

Working

E = -13.6\,\text{eV} = -13.6 \times 1.60 \times 10^{-19}\,\text{J}

E = -2.18 \times 10^{-18}\,\text{J}

Answer

$-2.18 \times 10^{-18}\,\text{J}$

Final answer

-2.18 × 10^−18 J

Detailed explanation

Background Concept

The electronvolt (eV) is a unit of energy often used in atomic physics. The conversion is

1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}.

So to convert eV to J, multiply by $1.60 \times 10^{-19}$ .

Understanding the Question

You are told the ground state energy of hydrogen is $-13.6\,\text{eV}$ and asked to express this energy in joules (J). The negative sign indicates the electron is bound (energy must be supplied to remove it to zero energy at infinity).

Approach

Multiply the value in eV by the conversion factor $1.60 \times 10^{-19}\,\text{J eV}^{-1}$ , keeping the negative sign.

Step-by-Step Reasoning

Start with

E_1 = -13.6\,\text{eV}.

Convert units:

E_1 = -13.6 \times (1.60 \times 10^{-19})\,\text{J}.

Calculate the numeric part:

13.6 \times 1.60 = 21.76.

So

E_1 = -21.76 \times 10^{-19}\,\text{J} = -2.176 \times 10^{-18}\,\text{J}.

To 3 significant figures:

E_1 = -2.18 \times 10^{-18}\,\text{J}.

Key Takeaways

Use $1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}$ .
Keep the sign of the energy when converting units.

Common Mistakes

Dividing by $1.60 \times 10^{-19}$ instead of multiplying.
Dropping the negative sign.
Writing $-2.18 \times 10^{18}\,\text{J}$ (power of ten sign error).

Things to Be Careful About

Scientific notation: $21.76 \times 10^{-19} = 2.176 \times 10^{-18}$ .
Significant figures: the given 13.6 has 3 s.f., so the result should be 3 s.f.

Techniques used

convert electronvolts to joules using 1 eV = 1.60 × 10^−19 Juse appropriate significant figures

(ii)

Show that the energy difference between levels $n = 1$ and $n = 2$ is $10.2\text{ eV}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For the lowest-frequency line (transition $n=2 \to 1$ ),

\Delta E = hf = (6.63 \times 10^{-34})(2.47 \times 10^{15}) = 1.64 \times 10^{-18}\,\text{J}

Convert to eV:

\Delta E = \frac{1.64 \times 10^{-18}}{1.60 \times 10^{-19}} = 10.2\,\text{eV}

Answer

$\Delta E_{1\to 2} = 10.2\,\text{eV}$

Final answer

10.2 eV

Detailed explanation

Background Concept

When an electron drops between two energy levels, the atom emits a photon. The photon energy equals the energy difference between the two levels:

\Delta E = hf.

To express this energy difference in electronvolts, use

1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}.

So

\text{energy in eV} = \frac{\text{energy in J}}{1.60 \times 10^{-19}}.

Understanding the Question

The spectrum shown is for transitions to the ground state $n=1$ (Lyman series). The three lowest frequencies are given. The smallest energy difference to $n=1$ corresponds to the transition from the closest higher level, $n=2 \to 1$ , so we use the lowest frequency line, $f = 2.47 \times 10^{15}\,\text{Hz}$ .

You need to calculate the energy of that photon and show it corresponds to $10.2\,\text{eV}$ .

Approach

Take the lowest frequency $f$ because it corresponds to the smallest energy gap.
Compute $\Delta E = hf$ in joules.
Convert joules to eV by dividing by $1.60 \times 10^{-19}$ .

Step-by-Step Reasoning

Identify the correct line: for transitions ending at $n=1$ , the smallest gap is $n=2 \to 1$ , giving the lowest frequency. Hence take

f = 2.47 \times 10^{15}\,\text{Hz}.

Photon energy:

\Delta E = hf = (6.63 \times 10^{-34})(2.47 \times 10^{15})\,\text{J}.

Multiply the numbers and combine powers of ten:

6.63 \times 2.47 = 16.38, \qquad 10^{-34} \times 10^{15} = 10^{-19}.

So

\Delta E = 16.38 \times 10^{-19}\,\text{J} = 1.638 \times 10^{-18}\,\text{J}.

Convert to eV:

\Delta E = \frac{1.638 \times 10^{-18}}{1.60 \times 10^{-19}}\,\text{eV} = 10.24\,\text{eV} \approx 10.2\,\text{eV}.

This matches the stated value.

Key Takeaways

Lowest frequency corresponds to the smallest energy difference because $\Delta E \propto f$ .
Use $\Delta E = hf$ then convert J to eV.

Common Mistakes

Using the highest frequency line instead of the lowest for the $n=2 \to 1$ transition.
Multiplying by $1.60 \times 10^{-19}$ when converting J to eV (you must divide).
Forgetting that frequencies are given in units of $10^{15}\,\text{Hz}$ .

Things to Be Careful About

Powers of ten: $10^{-34} \times 10^{15} = 10^{-19}$ .
Rounding: intermediate value $10.24\,\text{eV}$ rounds to $10.2\,\text{eV}$ consistent with the data precision.

Techniques used

use E = hf to find a transition energyconvert joules to electronvoltsidentify the lowest-frequency line as the smallest energy transition to n = 1

(iii)

Complete Table 8.1 to show the energy differences from the ground state, and the energies of the levels up to $n = 4$ , in the hydrogen atom. Use the space for any working.

Table 8.1

level	(energy difference from $n = 1$ )/eV	energy/eV
$n = 4$
$n = 3$
$n = 2$	10.2
$n = 1$	0.0	–13.6

4M Similar

DifficultyMedium

Worked solution

Working

For transitions to $n=1$ , energy difference from ground state is $\Delta E = hf$ .

For $n=3 \to 1$ ( $f = 2.92 \times 10^{15}\,\text{Hz}$ ):

\Delta E_{1\to 3} = \frac{(6.63 \times 10^{-34})(2.92 \times 10^{15})}{1.60 \times 10^{-19}} = 12.1\,\text{eV}

For $n=4 \to 1$ ( $f = 3.09 \times 10^{15}\,\text{Hz}$ ):

\Delta E_{1\to 4} = \frac{(6.63 \times 10^{-34})(3.09 \times 10^{15})}{1.60 \times 10^{-19}} = 12.8\,\text{eV}

Level energies: $E_n = E_1 + \Delta E_{1\to n}$ with $E_1 = -13.6\,\text{eV}$ .

E_2 = -13.6 + 10.2 = -3.4\,\text{eV}

E_3 = -13.6 + 12.1 = -1.5\,\text{eV}

E_4 = -13.6 + 12.8 = -0.8\,\text{eV}

Answer

Completed Table 8.1:

level	energy difference from $n=1$ / eV	energy / eV
$n=4$	$12.8$	$-0.8$
$n=3$	$12.1$	$-1.5$
$n=2$	$10.2$	$-3.4$
$n=1$	$0.0$	$-13.6$

Final answer

n=4: 12.8 eV, −0.8 eV; n=3: 12.1 eV, −1.5 eV; n=2: 10.2 eV, −3.4 eV

Detailed explanation

Background Concept

Hydrogen emission lines for transitions to the ground state arise when an electron falls from an excited level $n$ down to $n=1$ and emits a photon. The photon energy equals the energy gap:

\Delta E_{1\to n} = hf.

If $E_1$ is the ground state energy, then the energy of level $n$ is

E_n = E_1 + \Delta E_{1\to n}.

Here $E_1 = -13.6\,\text{eV}$ .

Understanding the Question

You are given three frequencies (in $10^{15}\,\text{Hz}$ ) for the three lowest-frequency lines associated with transitions to $n=1$ . These correspond to transitions $n=2 \to 1$ , $n=3 \to 1$ , and $n=4 \to 1$ .

You must complete a table listing:

the energy differences from the ground state to levels $n=2,3,4$ (in eV), and
the energies of those levels (also in eV).

The table already includes $\Delta E_{1\to 2} = 10.2\,\text{eV}$ and $E_1 = -13.6\,\text{eV}$ .

Approach

For each frequency, calculate the photon energy using $\Delta E = hf$ .
Convert joules to eV.
Use $E_n = -13.6 + \Delta E_{1\to n}$ to get each level energy.

Step-by-Step Reasoning

1) Match frequency to transition

For transitions ending at $n=1$ , higher starting $n$ means a larger energy drop, so a higher frequency (since $\Delta E \propto f$ ). Therefore:

lowest frequency $2.47 \times 10^{15}\,\text{Hz}$ is $2 \to 1$ (already used to get $10.2\,\text{eV}$ ),
next $2.92 \times 10^{15}\,\text{Hz}$ is $3 \to 1$ ,
highest of the three $3.09 \times 10^{15}\,\text{Hz}$ is $4 \to 1$ .

2) Compute energy differences in eV

For $n=3$ :

\Delta E_{1\to 3} = hf = (6.63 \times 10^{-34})(2.92 \times 10^{15}) = 1.94 \times 10^{-18}\,\text{J}.

Convert to eV:

\Delta E_{1\to 3} = \frac{1.94 \times 10^{-18}}{1.60 \times 10^{-19}} = 12.1\,\text{eV}.

For $n=4$ :

\Delta E_{1\to 4} = (6.63 \times 10^{-34})(3.09 \times 10^{15}) = 2.05 \times 10^{-18}\,\text{J}.

\Delta E_{1\to 4} = \frac{2.05 \times 10^{-18}}{1.60 \times 10^{-19}} = 12.8\,\text{eV}.

3) Compute the level energies

Using $E_n = E_1 + \Delta E_{1\to n}$ with $E_1 = -13.6\,\text{eV}$ :

E_2 = -13.6 + 10.2 = -3.4\,\text{eV}.

E_3 = -13.6 + 12.1 = -1.5\,\text{eV}.

E_4 = -13.6 + 12.8 = -0.8\,\text{eV}.

These values become less negative as $n$ increases, approaching $0\,\text{eV}$ for very large $n$ (the ionisation limit).

Key Takeaways

For emission to the ground state, each line corresponds to one gap $\Delta E_{1\to n}$ .
Calculate gaps via $\Delta E = hf$ and convert to eV.
Level energies follow from $E_n = E_1 + \Delta E_{1\to n}$ .

Common Mistakes

Adding the ground state energy with the wrong sign (e.g. $-13.6 - 12.1$ ).
Mixing up which frequency corresponds to which transition.
Forgetting the factor of $10^{15}$ in the given frequencies.
Giving excited level energies as positive (they should be negative for bound states).

Things to Be Careful About

Keep consistent rounding (typically to 0.1 eV here, matching the given $10.2\,\text{eV}$ ).
Use the same constants each time: $h = 6.63 \times 10^{-34}\,\text{J s}$ and $1\,\text{eV} = 1.60 \times 10^{-19}\,\text{J}$ .
The energy difference “from the ground state” is a positive number, while the level energies themselves are negative relative to the zero at infinity.

Techniques used

use E = hf to determine transition energy differencesconvert photon energies between joules and electronvoltsobtain excited-state energies by adding energy differences to the ground-state energy

Q9MediumNuclear PhysicsAstronomy and Cosmology

(a)

State what is meant by the mass defect of a nucleus.

2M Similar

DifficultyEasy

Worked solution

Answer

The mass defect is the difference between the total mass of the separate (free) nucleons and the mass of the nucleus formed from those nucleons (same $Z$ and $A$ ).

Final answer

Difference between the total mass of the separate nucleons and the mass of the nucleus.

Detailed explanation

Background Concept

Nuclei are bound systems: when nucleons (protons and neutrons) bind together, energy is released and the final bound nucleus has a lower total energy than the separated nucleons.

Einstein’s mass–energy relation,

E = mc^2

means that a decrease in energy corresponds to a decrease in mass.

The mass defect $\Delta m$ is the “missing mass”:

\Delta m = (\text{sum of masses of free nucleons}) - (\text{mass of nucleus}).

This mass defect corresponds to the binding energy:

E_b = \Delta m\,c^2.

Understanding the Question

You are asked to state what is meant by mass defect. So you need a clear definition, mentioning (1) the sum of masses of separate nucleons, (2) the mass of the nucleus, and (3) that it is the difference between them.

Approach

Write the definition as a comparison between:

the total mass of the individual nucleons if they were separated, and
the mass of the nucleus when they are bound.

Optionally (for completeness), note that it relates to binding energy via $E = mc^2$ .

Step-by-Step Reasoning

Consider a nucleus with given proton number $Z$ and nucleon number $A$ .
If you take the same numbers of protons and neutrons as separate particles, their masses add to a total “free nucleon mass”.
The actual nucleus mass is smaller than this sum.
The difference (free total minus nucleus mass) is the mass defect.

Key Takeaways

Mass defect is a difference in mass between free nucleons and the bound nucleus.
It arises because binding releases energy, reducing the total mass of the system.

Common Mistakes

Defining it as the difference between “reactants and products” in a reaction (that’s a reaction mass change, not the definition of mass defect of a nucleus).
Saying “mass defect is the energy released” without mentioning it is a mass difference.

Things to Be Careful About

Make clear it is the mass of the same nucleons when separated compared with when bound.
Use “difference” with the correct sense (sum of free nucleon masses is larger).

Techniques used

define mass defect in terms of constituent nucleon massesstate the comparison between separated nucleons and the bound nucleuslink missing mass to binding energy via mass-energy equivalence

(b)

The nuclear fusion reaction for the formation of helium-4 from deuterium is represented by

{}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He}.

Table 9.1 shows the masses of the nuclides involved in this reaction.

Table 9.1

nuclide	nuclide mass /u
${}^{2}_{1}\text{H}$	2.013553
${}^{4}_{2}\text{He}$	4.001505

Calculate the energy released in the formation of $1.00\text{ mol}$ of helium-4.

$\text{energy}$ = ______ $\text{J}$

4M Similar

DifficultyMedium

Worked solution

Working

Mass defect per reaction:

\Delta m = 2(2.013553\ \text{u}) - 4.001505\ \text{u} = 0.025601\ \text{u}

\Delta m = 0.025601\times 1.66\times 10^{-27} = 4.25\times 10^{-29}\ \text{kg}

Energy per nucleus formed:

E = \Delta mc^2 = (4.25\times 10^{-29})(3.00\times 10^8)^2 = 3.83\times 10^{-12}\ \text{J}

Energy for $1.00\ \text{mol}$ ( $N_A = 6.02\times 10^{23}\ \text{mol}^{-1}$ ):

E_{\text{mol}} = (3.83\times 10^{-12})(6.02\times 10^{23}) = 2.31\times 10^{12}\ \text{J}

Answer

$2.31\times 10^{12}\ \text{J}$

Final answer

2.31 × 10^12 J

Detailed explanation

Background Concept

In a nuclear reaction, the energy released comes from a decrease in mass between the initial and final nuclear products.

1 atomic mass unit is a mass unit used for nuclides:

1\ \text{u} = 1.66\times 10^{-27}\ \text{kg}

The energy equivalent of a mass change $\Delta m$ is given by:

E = \Delta m\,c^2

For one mole, you multiply the energy per reaction by Avogadro’s constant:

N_A = 6.02\times 10^{23}\ \text{mol}^{-1}

Understanding the Question

The reaction is:

{}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He}

You are given the nuclide masses in u for deuterium and helium-4. You must:

find the mass difference between reactants and product for one helium nucleus formed,
convert that mass difference to energy,
scale up to $1.00\ \text{mol}$ of helium-4 (i.e. $N_A$ helium nuclei).

Approach

Compute total initial mass: two deuterium nuclei.
Subtract final mass: one helium-4 nucleus.
Convert u to kg.
Use $E = \Delta mc^2$ to get energy per reaction.
Multiply by $N_A$ to get energy for 1 mol.

Step-by-Step Reasoning

1) Mass defect in u

Initial mass:

m_i = 2\times 2.013553\ \text{u} = 4.027106\ \text{u}

Final mass:

m_f = 4.001505\ \text{u}

Mass defect:

\Delta m = m_i - m_f = 4.027106 - 4.001505 = 0.025601\ \text{u}

2) Convert to kg

\Delta m = 0.025601\times 1.66\times 10^{-27} = 4.25\times 10^{-29}\ \text{kg}

3) Energy per helium nucleus formed

E = \Delta mc^2 = (4.25\times 10^{-29})(3.00\times 10^8)^2

E = (4.25\times 10^{-29})(9.00\times 10^{16}) = 3.83\times 10^{-12}\ \text{J}

4) Energy for 1.00 mol helium-4

$1.00\ \text{mol}$ contains $N_A$ helium nuclei, so:

E_{\text{mol}} = (3.83\times 10^{-12})(6.02\times 10^{23}) = 2.31\times 10^{12}\ \text{J}

Key Takeaways

Energy released in nuclear reactions comes from a mass decrease: $E = \Delta mc^2$ .
Nuclide masses often come in u, so unit conversion is essential.
“Per mole” means multiply by $N_A$ .

Common Mistakes

Forgetting the factor of 2 for the two deuterium nuclei.
Using $c = 3.0\times 10^8$ incorrectly (e.g. not squaring it).
Converting u to kg wrongly by using $10^{-26}$ instead of $10^{-27}$ .
Multiplying by $1.00\ \text{mol}$ as if it were a number instead of using $N_A$ .

Things to Be Careful About

The reaction shown produces one helium nucleus per reaction, so the number of reactions for 1 mol He is $N_A$ .
Keep track of significant figures; typical final answers are 2–3 s.f.
Always include the unit (J).

Techniques used

calculate mass change from nuclide massesconvert atomic mass units to kilogramsapply mass-energy equivalencescale energy per reaction up using the Avogadro constant

(c)

The star Sirius has a radius of $1.19 \times 10^{9}\text{ m}$ and loses mass due to nuclear fusion at a rate of $1.09 \times 10^{11}\text{ kg s}^{-1}$ . Assume that the power of the radiation emitted by the star is equal to the power released by this process.

4M

(i)

Determine a value for the luminosity of Sirius. Give a unit with your answer.

$\text{luminosity}$ = ______ $\text{unit}$ ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

L = \frac{\Delta E}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right)c^2

L = (1.09\times 10^{11})(3.00\times 10^8)^2 = (1.09\times 10^{11})(9.00\times 10^{16}) = 9.81\times 10^{27}\ \text{W}

Answer

$9.81\times 10^{27}\ \text{W}$

Final answer

9.81 × 10^27 W

Detailed explanation

Background Concept

A star’s luminosity $L$ is the total energy it emits per unit time:

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

If the star’s energy comes from nuclear fusion, the source is mass being converted to energy. Einstein’s relation gives:

\Delta E = \Delta mc^2

So the power (energy per second) released by a mass-loss rate $\frac{\Delta m}{\Delta t}$ is:

L = \left(\frac{\Delta m}{\Delta t}\right)c^2

Understanding the Question

You are told Sirius loses mass due to fusion at a known rate, and you should assume all the power released becomes radiation power. That means the luminosity equals the rate of mass-energy conversion.

Given:

$\frac{\Delta m}{\Delta t} = 1.09\times 10^{11}\ \text{kg s}^{-1}$
$c = 3.00\times 10^8\ \text{m s}^{-1}$

Find $L$ and give its unit.

Approach

Use:

L = \left(\frac{\Delta m}{\Delta t}\right)c^2

Then substitute the numbers and quote the unit as watts (W), since it is power.

Step-by-Step Reasoning

Start from $L = \frac{\Delta E}{\Delta t}$ .
With $\Delta E = \Delta mc^2$ , divide by $\Delta t$ :

L = \left(\frac{\Delta m}{\Delta t}\right)c^2

Substitute:

L = (1.09\times 10^{11})(3.00\times 10^8)^2

Square the speed of light:

(3.00\times 10^8)^2 = 9.00\times 10^{16}

Multiply:

L = (1.09\times 10^{11})(9.00\times 10^{16}) = 9.81\times 10^{27}\ \text{W}

Key Takeaways

Luminosity is power.
Fusion power can be found from mass-loss rate using $L = \dot{m}c^2$ .

Common Mistakes

Using $L = mc^2$ without dividing by time.
Forgetting to square $c$ .
Giving the unit as J instead of W.

Things to Be Careful About

$\text{kg s}^{-1} \times \text{m}^2\text{s}^{-2} = \text{kg m}^2\text{s}^{-3} = \text{W}$ , so the unit checks out.
Keep powers of ten consistent when multiplying in standard form.

Techniques used

relate mass loss rate to power using mass-energy equivalencetreat luminosity as energy emitted per unit timeuse standard form and correct SI units

(ii)

Use your answer in (c)(i) to determine the surface temperature of Sirius.

$\text{surface temperature}$ = ______ $\text{K}$

2M Similar

DifficultyMedium

Worked solution

Working

L = 4\pi r^2\sigma T^4 \Rightarrow T = \left(\frac{L}{4\pi r^2\sigma}\right)^{1/4}

T = \left(\frac{9.81\times 10^{27}}{4\pi(1.19\times 10^9)^2(5.67\times 10^{-8})}\right)^{1/4}

T = \left(9.72\times 10^{15}\right)^{1/4} = 9.95\times 10^3\ \text{K}

Answer

$1.00\times 10^{4}\ \text{K}$

Final answer

1.00 × 10^4 K

Detailed explanation

Background Concept

A star radiates approximately like a black body, so its luminosity is related to its surface temperature by the Stefan–Boltzmann law:

L = 4\pi r^2\sigma T^4

where

$L$ is luminosity in $\text{W}$ ,
$r$ is radius in $\text{m}$ ,
$\sigma = 5.67\times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$ ,
$T$ is the surface temperature in $\text{K}$ .

Understanding the Question

You are given the radius of Sirius and, from part (c)(i), the luminosity. You must find the star’s surface temperature using the Stefan–Boltzmann relationship.

Given:

$r = 1.19\times 10^9\ \text{m}$
$L = 9.81\times 10^{27}\ \text{W}$ (from (c)(i))

Find: $T$ .

Approach

Rearrange the Stefan–Boltzmann law for $T$ :

T = \left(\frac{L}{4\pi r^2\sigma}\right)^{1/4}

Then substitute and evaluate carefully in standard form, remembering the final step is a fourth root.

Step-by-Step Reasoning

Rearrange:

T = \left(\frac{L}{4\pi r^2\sigma}\right)^{1/4}

Calculate the surface area term $4\pi r^2$ :

r^2 = (1.19\times 10^9)^2 = 1.416\times 10^{18}\ \text{m}^2

4\pi r^2 = 4\pi(1.416\times 10^{18}) \approx 1.78\times 10^{19}\ \text{m}^2

Multiply by $\sigma$ :

4\pi r^2\sigma \approx (1.78\times 10^{19})(5.67\times 10^{-8}) \approx 1.01\times 10^{12}

(Unit check: $\text{m}^2\times \text{W m}^{-2}\text{K}^{-4} = \text{W K}^{-4}$ .)

Form the ratio:

\frac{L}{4\pi r^2\sigma} \approx \frac{9.81\times 10^{27}}{1.01\times 10^{12}} \approx 9.72\times 10^{15}

So:

T = (9.72\times 10^{15})^{1/4}

Take the fourth root. Split into number and power of ten:

(9.72\times 10^{15})^{1/4} = (9.72)^{1/4}\times 10^{15/4}

10^{15/4} = 10^{3.75} = 5.62\times 10^3

and

(9.72)^{1/4} \approx 1.77

So:

T \approx 1.77\times (5.62\times 10^3) \approx 9.95\times 10^3\ \text{K} \approx 1.00\times 10^4\ \text{K}

Key Takeaways

Use $L = 4\pi r^2\sigma T^4$ to connect luminosity and surface temperature.
Rearranging gives a fourth-root dependence: temperature changes relatively slowly with luminosity.

Common Mistakes

Forgetting the $4\pi$ in the surface area.
Using $L = \sigma T^4$ (missing the area factor).
Taking a square root instead of a fourth root.
Mixing up radius and diameter.

Things to Be Careful About

Use SI units: $r$ in m, $\sigma$ in $\text{W m}^{-2}\text{K}^{-4}$ .
Handle powers of ten carefully when taking $1/4$ powers: $10^{15}$ becomes $10^{3.75}$ .
Quote the final temperature to appropriate significant figures (typically 2–3 s.f.).

Techniques used

apply the Stefan-Boltzmann lawrearrange to solve for temperaturecompute a fourth root carefully in standard form

(d)

Explain how cosmologists use standard candles to estimate the distance of a galaxy from the Earth.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

A standard candle has known luminosity $L$ (absolute brightness). Measure its apparent brightness / flux $F$ at Earth. Use the inverse-square law

F = \frac{L}{4\pi d^2}

to calculate

d = \sqrt{\frac{L}{4\pi F}}

so obtain the galaxy’s distance.

Final answer

Use known luminosity of a standard candle, measure flux, apply F = L/(4πd^2) to find d.

Detailed explanation

Background Concept

The key idea is that some astronomical objects have a known luminosity (power output), so they can act as standard candles.

Luminosity $L$ : total power radiated by the source (W).
Flux $F$ : power received per unit area at Earth (W m $^{-2}$ ), sometimes called apparent brightness.

For radiation spreading uniformly in space, the inverse-square relation applies:

F = \frac{L}{4\pi d^2}

where $d$ is the distance to the source.

Understanding the Question

You must explain how standard candles allow cosmologists to estimate the distance to a galaxy. That means describing:

what makes an object a standard candle (known $L$ ),
what is measured (flux $F$ ),
how distance is calculated (inverse-square law and rearrangement).

Approach

State that the standard candle’s luminosity is known from calibration (e.g. Cepheid period–luminosity relation or Type Ia supernova peak luminosity). Then explain measuring its apparent brightness and using $F = L/(4\pi d^2)$ to solve for $d$ .

Step-by-Step Reasoning

Identify a standard candle within the galaxy (e.g. Cepheid variable or Type Ia supernova).
Determine (or use known) luminosity $L$ of that object. This is the “standard” reference value.
Measure the flux $F$ received at Earth using a telescope/detector.
Assume isotropic emission so the power spreads over a sphere of area $4\pi d^2$ .
Use:

F = \frac{L}{4\pi d^2}

Rearrange:

d = \sqrt{\frac{L}{4\pi F}}

The computed $d$ gives the distance to the galaxy.

Key Takeaways

Standard candles work because their absolute luminosity is known.
Measuring flux and using an inverse-square law turns that into a distance.

Common Mistakes

Confusing luminosity $L$ with flux $F$ (luminosity is intrinsic; flux depends on distance).
Using $F = L/d$ instead of $F \propto 1/d^2$ .
Omitting the idea that the luminosity must be known/calibrated first.

Things to Be Careful About

The method assumes light is not significantly absorbed/scattered; in practice cosmologists correct for interstellar extinction and redshift effects.
Distinguish “apparent brightness” (measured at Earth) from “absolute brightness/luminosity” (property of the source).

Techniques used

use an object of known luminosity as a calibration sourcemeasure apparent brightness (flux) at Earthapply the inverse-square law to determine distance

Q10MediumMedical Physics

(a)

State what is meant by contrast in an X-ray image.

1M Similar

(b)

X-rays of intensity $I_{0}$ are incident normally on a structure, as shown in Fig. 10.1.

Fig. 10.1

Material P has a linear attenuation coefficient of $0.35\text{ cm}^{-1}$ .
The X-rays emerging from the structure in region A have an intensity of $0.053I_{0}$ .

5M

(i)

Show that the intensity of the X-rays emerging in region B is $0.13I_{0}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For region B (through material P of thickness $x = 5.8\ \text{cm}$ ):

I = I_0 e^{-\mu x}

\frac{I}{I_0} = e^{-0.35 \times 5.8} = e^{-2.03} = 0.13

Answer

$I = 0.13 I_0$

Final answer

0.13 I0

Detailed explanation

Background Concept

X-rays are attenuated (reduced in intensity) as they pass through matter. For a material of linear attenuation coefficient $\mu$ and thickness $x$ , the transmitted intensity is

I = I_0 e^{-\mu x}

where:

$I_0$ is the incident intensity,
$I$ is the emerging intensity,
$\mu$ has unit $\text{cm}^{-1}$ (if $x$ is in cm).

Understanding the Question

You are told material P has $\mu = 0.35\ \text{cm}^{-1}$ . From the diagram, region B corresponds to X-rays passing through thickness $5.8\ \text{cm}$ of material P (with no extra insert on that path). You must calculate the transmitted intensity in region B as a fraction of $I_0$ .

Approach

Use the attenuation law with $\mu = 0.35\ \text{cm}^{-1}$ and $x = 5.8\ \text{cm}$ , then compute $I/I_0$ .

Step-by-Step Reasoning

Start from

\frac{I}{I_0} = e^{-\mu x}

Substitute values:

\frac{I}{I_0} = e^{-0.35 \times 5.8}

Multiply in the exponent:

0.35 \times 5.8 = 2.03

Evaluate:

e^{-2.03} \approx 0.13

So $I \approx 0.13 I_0$ .

Key Takeaways

X-ray attenuation through a uniform thickness is exponential.
The ratio $I/I_0$ is found directly from $e^{-\mu x}$ .

Common Mistakes

Using a linear decrease (e.g. $I = I_0 - \mu x$ ), which is incorrect.
Mixing units (e.g. using $x$ in m with $\mu$ in $\text{cm}^{-1}$ ).

Things to Be Careful About

Ensure $x$ is in cm because $\mu$ is given in $\text{cm}^{-1}$ .
Quote the transmitted intensity as a fraction of $I_0$ as requested.

Techniques used

apply exponential attenuation of X-rayssubstitute thickness into an attenuation equationevaluate an exponential to obtain an intensity ratio

(ii)

Determine the linear attenuation coefficient $\mu$ of material Q.

$\mu$ = ______ $\text{cm}^{-1}$

3M Similar

DifficultyMedium

Worked solution

Working

For region A, thickness of Q is $2.1\ \text{cm}$ so thickness of P is $5.8-2.1=3.7\ \text{cm}$ .

\frac{I_A}{I_0}=e^{-\mu_P(3.7)}\,e^{-\mu(2.1)}

0.053 = e^{-0.35(3.7)}e^{-2.1\mu}

Take $\ln$ :

\ln(0.053)= -0.35(3.7) -2.1\mu

\mu = \frac{-\ln(0.053)-0.35(3.7)}{2.1}=\frac{2.94-1.30}{2.1}=0.78\ \text{cm}^{-1}

Answer

$\mu = 0.78\ \text{cm}^{-1}$

Final answer

0.78 cm^-1

Detailed explanation

Background Concept

If X-rays pass through more than one material, the attenuation happens successively. If the beam passes through thickness $x_1$ of material 1 (coefficient $\mu_1$ ) and then thickness $x_2$ of material 2 (coefficient $\mu_2$ ), then

I = I_0 e^{-\mu_1 x_1} e^{-\mu_2 x_2}

This works because after the first layer the intensity becomes $I_0 e^{-\mu_1 x_1}$ , and that reduced intensity then gets attenuated by the second layer.

Understanding the Question

You are given:

material P has $\mu_P = 0.35\ \text{cm}^{-1}$ ,
total thickness across the structure is $5.8\ \text{cm}$ ,
the insert material Q has thickness $2.1\ \text{cm}$ along the relevant path,
in region A, the emerging intensity is $I_A = 0.053 I_0$ .

You must find $\mu$ for material Q.

diagram-bii-1

Approach

Work out how much of the path is through P and how much is through Q.
Write the combined attenuation equation for region A.
Substitute the known numbers.
Take natural logs to solve for $\mu$ .

Step-by-Step Reasoning

Thickness through Q is $2.1\ \text{cm}$ , so thickness through P in that same line is

x_P = 5.8 - 2.1 = 3.7\ \text{cm}

Use successive attenuation:

\frac{I_A}{I_0} = e^{-\mu_P x_P} e^{-\mu x_Q}

So

0.053 = e^{-0.35(3.7)} e^{-\mu(2.1)}

A good way to isolate $\mu$ is to take logs:

\ln(0.053) = -0.35(3.7) - 2.1\mu

Evaluate the known terms:

$0.35(3.7)=1.295 \approx 1.30$
$-\ln(0.053)=2.94$

Then

2.1\mu = 2.94 - 1.30 = 1.64

and

\mu = \frac{1.64}{2.1} = 0.78\ \text{cm}^{-1}

Key Takeaways

Multiple layers multiply the attenuation factors: $e^{-\mu_1 x_1}e^{-\mu_2 x_2}$ .
Natural logs turn products/exponentials into linear equations you can rearrange.

Common Mistakes

Using $5.8\ \text{cm}$ as the thickness of P in region A (forgetting that some of it is replaced by Q).
Adding attenuation coefficients directly (e.g. $\mu_P+\mu$ ) without weighting by thickness.
Using $\log_{10}$ without handling conversion (the scheme expects $\ln$ ).

Things to Be Careful About

Keep thicknesses in cm to match $\mu$ in $\text{cm}^{-1}$ .
When taking logs, remember $\ln(e^x)=x$ and that $\ln(0.053)$ is negative.
Quote $\mu$ to a sensible number of significant figures (typically 2 s.f. here).

Techniques used

model sequential attenuation through multiple materialsform a product of exponentials for layered mediatake natural logarithms to linearise an exponential equationrearrange to solve for an attenuation coefficient

(iii)

Use the information in (b)(i) to suggest why the X-rays emerging from the structure form an image that has poor contrast.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The transmitted intensities are quite similar ( $0.053I_0$ and $0.13I_0$ ), so the difference in detector signal/brightness between the two regions is small, giving poor contrast.

Final answer

Because the transmitted intensities (0.053 I0 and 0.13 I0) are not very different, the brightness difference is small so contrast is poor.

Detailed explanation

Background Concept

In X-ray imaging, contrast depends on how different the detected X-ray intensity is between two neighbouring regions. If the transmitted intensities are close, the detector produces similar grey levels and it is difficult to distinguish structures.

Understanding the Question

You are asked to use the result from (b)(i) (region B intensity $0.13I_0$ ) together with the given region A intensity $0.053I_0$ to explain why the image contrast is poor.

Approach

Compare the two transmitted intensities. If the difference (or ratio) is small, the image will show only a small difference in grey level.

Step-by-Step Reasoning

Region A transmits $0.053I_0$ .
Region B transmits $0.13I_0$ .
These are both small fractions of $I_0$ and are within the same order of magnitude.
Therefore the detector signals (and hence grey levels) for A and B will be relatively close, so the boundary is not strongly distinguished.

Key Takeaways

High contrast requires a large difference in transmitted intensity between regions.
If intensities are close, the image appears “flat” with similar grey shades.

Common Mistakes

Saying “poor contrast because intensity is low” without comparing the two regions.
Confusing contrast with sharpness (edge blurring) or resolution.

Things to Be Careful About

The question explicitly says to use (b)(i), so you must mention the two values $0.053I_0$ and $0.13I_0$ (or their small difference) in your reasoning.

Techniques used

compare transmitted intensities from different regionslink intensity differences to image contrastinfer qualitative image quality from quantitative attenuation

(c)

Explain how X-rays are used in computed tomography (CT) scanning to produce a three-dimensional image of an internal structure.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

A narrow (fan) beam of X-rays passes through the body to detectors.
Source and detectors rotate around the patient so intensities are measured for many directions/paths.
A computer uses the attenuation data to reconstruct cross-sectional ‘slice’ images, and many slices are combined to form a 3D image.

Final answer

Rotate X-ray source/detectors to take many attenuation measurements from different angles; computer reconstructs cross-sectional slices and combines slices to form a 3D image.

Detailed explanation

Background Concept

A CT scan measures how much X-rays are attenuated by tissue. The attenuation depends on the material and thickness, and can be described by the exponential law. In CT, instead of taking a single projection image, the system measures many projections from different angles and uses computation to reconstruct the internal structure.

Understanding the Question

The question asks how CT scanning uses X-rays to produce a three-dimensional image. The key ideas are:

many measurements from different angles,
reconstruction of a cross-section (a “slice”),
combining slices to build a 3D image.

diagram-c-1

Approach

Describe:

the rotating X-ray source and detector array,
the collection of transmitted intensity data for many angles,
the computer reconstruction into a cross-sectional image,
repeating along the body (or using multiple detector rows) to form a 3D model.

Step-by-Step Reasoning

The scanner produces a narrow beam (often fan-shaped) of X-rays directed through the patient.
Detectors on the opposite side measure the transmitted intensity after the beam passes through the body.
The source and detectors rotate around the patient, so for a single slice the system obtains transmitted intensities for many different directions (many “projections”).
A computer algorithm uses these projections to calculate the X-ray attenuation at many points in that slice, reconstructing a 2D cross-sectional image.
By scanning many adjacent slices (moving the patient through the ring, or using a multi-row detector), the set of 2D slices is stacked/combined to generate a 3D image of the internal structure.

Key Takeaways

CT differs from a normal X-ray because it uses many angles and computer reconstruction.
The basic measurement is still attenuation of X-rays through tissue.
3D comes from combining many reconstructed slices.

Common Mistakes

Describing a single X-ray image only (no rotation, no reconstruction).
Omitting the computer reconstruction step.
Forgetting to explain how 3D is obtained (multiple slices).

Things to Be Careful About

Mention rotation around the patient (multiple directions) for the key CT idea.
State clearly that the output is cross-sectional slices which are then combined into 3D.

Techniques used

describe data collection using multiple X-ray projectionsexplain reconstruction of a cross-sectional image from attenuation measurementsexplain how stacking slices forms a three-dimensional image

Physics 9702/42 — October/November 2025

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