Physics 9702 / 44 — October/November 2025

DifficultyEasy

Worked solution

Answer

The gravitational force between two point masses is attractive, acts along the line joining their centres, and has magnitude

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

where $r$ is the separation of the masses.

Final answer

Attractive force along line joining centres: F = G m1 m2 / r^2.

Detailed explanation

Background Concept

Newton’s law of gravitation describes the mutual force between two point masses (or spherically symmetric masses treated as if their mass were concentrated at their centres). The magnitude of the force depends on:

the product of the masses $m_1 m_2$ (bigger masses give a bigger force), and
the inverse square of their separation $r$ (doubling $r$ makes the force four times smaller).

The law is written as

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

where $G$ is the gravitational constant.

Understanding the Question

You are asked to state Newton’s law, so you need the relationship in words and/or the equation. For full credit you should include that the force acts along the line joining the masses and is attractive.

Approach

Write the standard inverse-square formula and add the direction/attractive nature, since the question asks for the law (not just the equation).

Step-by-Step Reasoning

Identify the interacting objects: two masses $m_1$ and $m_2$ separated by distance $r$ .
State the magnitude of the force using the inverse-square relationship:

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

State the direction: the force on each mass is towards the other mass, along the line joining their centres.

Key Takeaways

Gravitational force is always attractive.
Magnitude follows an inverse-square law with separation.
For spherical bodies, use centre-to-centre separation.

Common Mistakes

Omitting the $r^2$ in the denominator.
Forgetting to mention that the force is attractive / along the line joining the centres.
Using diameter or radius instead of the centre-to-centre separation.

Things to Be Careful About

$r$ is the distance between the centres of mass.
The force on each mass has the same magnitude (Newton’s third law pair) but opposite direction.

Techniques used

state a physical law in words and as an equationidentify the direction of the gravitational force

(b)

A binary star consists of star A, of mass $4.0 \times 10^{30}\ \text{kg}$ , and star B, of mass $2.0 \times 10^{30}\ \text{kg}$ , separated by a distance of $3.3 \times 10^{12}\ \text{m}$ . The stars are both in circular orbit around their common centre of gravity X, as shown in Fig. 1.1.

Fig. 1.1

The radius $R_B$ of the orbit of star B is double the radius $R_A$ of the orbit of star A.

9M

(i)

Use Newton’s law of gravitation to calculate the magnitude of the gravitational force exerted by each star on the other.

force = ______ $\text{N}$

DifficultyMedium-Easy

Worked solution

Working

F = \frac{G m_A m_B}{r^2}

F = \frac{(6.67\times 10^{-11})(4.0\times 10^{30})(2.0\times 10^{30})}{(3.3\times 10^{12})^2}

F = 4.90\times 10^{25}\ \text{N}

Answer

$4.9\times 10^{25}\ \text{N}$

Final answer

4.9 × 10^25 N

Detailed explanation

Background Concept

For two masses $m_A$ and $m_B$ separated by distance $r$ , the magnitude of the gravitational force is

F = \frac{G m_A m_B}{r^2}

This is the same force magnitude on each star (equal and opposite forces).

Understanding the Question

You are given:

$m_A = 4.0\times 10^{30}\ \text{kg}$
$m_B = 2.0\times 10^{30}\ \text{kg}$
separation $r = 3.3\times 10^{12}\ \text{m}$

You must calculate the gravitational force that each exerts on the other, i.e. the single magnitude $F$ .

Approach

Use Newton’s law of gravitation directly with the given centre-to-centre distance. Substitute carefully in standard form, then round to an appropriate number of significant figures.

Step-by-Step Reasoning

Start with

F = \frac{G m_A m_B}{r^2}

Compute the product of masses:

m_A m_B = (4.0\times 10^{30})(2.0\times 10^{30}) = 8.0\times 10^{60}

Square the separation:

r^2 = (3.3\times 10^{12})^2 = 10.89\times 10^{24} = 1.089\times 10^{25}

Now substitute:

F = \frac{(6.67\times 10^{-11})(8.0\times 10^{60})}{1.089\times 10^{25}}

Combine powers of ten:

6.67\times 8.0 = 53.36

(10^{-11})(10^{60}) = 10^{49}

So numerator is $53.36\times 10^{49} = 5.336\times 10^{50}$ . Then

F = \frac{5.336\times 10^{50}}{1.089\times 10^{25}} = 4.90\times 10^{25}\ \text{N}

Key Takeaways

Use centre-to-centre separation.
Inverse-square dependence means $r^2$ is crucial.
Equal magnitude forces act on both stars.

Common Mistakes

Forgetting to square $r$ .
Squaring $3.3\times 10^{12}$ incorrectly (especially the power of ten).
Giving units other than newtons.

Things to Be Careful About

Keep significant figures consistent (here typically 2 s.f. is fine: $4.9\times 10^{25}\ \text{N}$ ).
Don’t confuse $r$ (distance between stars) with orbital radii $R_A$ and $R_B$ (distance from centre of mass).

Techniques used

apply Newton’s law of gravitationsubstitute values with powers of ten carefullyevaluate an inverse-square calculation

(ii)

Calculate the centripetal acceleration of star A.

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

DifficultyMedium-Easy

Worked solution

Working

For star A, gravitational force provides centripetal force:

a_A = \frac{F}{m_A}

a_A = \frac{4.90\times 10^{25}}{4.0\times 10^{30}} = 1.23\times 10^{-5}\ \text{m s}^{-2}

Answer

$1.2\times 10^{-5}\ \text{m s}^{-2}$

Final answer

1.2 × 10^-5 m s^-2

Detailed explanation

Background Concept

In uniform circular motion, an object moving in a circle of radius $r$ needs a centripetal (inward) acceleration

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

This acceleration is caused by a net inward force:

F = ma

In an orbit, the centripetal force is provided by gravity.

Understanding the Question

You have already found the gravitational force between the two stars. That same force acts on star A towards the centre of its orbit (towards the other star / towards the centre of mass), so it is the centripetal force for star A. The question asks for the centripetal acceleration of star A.

Known:

$F$ from part (b)(i)
$m_A = 4.0\times 10^{30}\ \text{kg}$

Unknown:

$a_A$

Approach

Use $F = m_A a_A$ for star A, rearrange to $a_A = F/m_A$ , and substitute.

Step-by-Step Reasoning

Centripetal force on star A is the gravitational force:

F = m_A a_A

Rearrange:

a_A = \frac{F}{m_A}

Substitute values:

a_A = \frac{4.90\times 10^{25}}{4.0\times 10^{30}} = 1.225\times 10^{-5}\ \text{m s}^{-2}

Rounded to 2 s.f.:

a_A \approx 1.2\times 10^{-5}\ \text{m s}^{-2}

Key Takeaways

In orbits, gravity provides the centripetal force.
Once you know the force on a body, its centripetal acceleration is simply $a = F/m$ .

Common Mistakes

Dividing by the wrong mass (using $m_B$ instead of $m_A$ ).
Thinking the force is different on each star (it is the same magnitude).
Missing the unit $\text{m s}^{-2}$ .

Things to Be Careful About

This acceleration is the acceleration of star A towards the centre of its circular path (radially inward), not tangential acceleration (speed is constant).

Techniques used

equate gravitational force to centripetal forceapply Newton’s second law in radial directioncalculate acceleration from force and mass

(iii)

Use your answer in (b)(ii) to determine the period of the orbit of star A.

period = ______ $\text{s}$

DifficultyMedium

Worked solution

Working

Given $R_B = 2R_A$ and separation $R_A + R_B = 3.3\times 10^{12}\ \text{m}$ :

3R_A = 3.3\times 10^{12} \Rightarrow R_A = 1.1\times 10^{12}\ \text{m}

For star A,

a_A = R_A \omega^2 \Rightarrow \omega = \sqrt{\frac{a_A}{R_A}}

and $\omega = \frac{2\pi}{T}$ , so

T = 2\pi \sqrt{\frac{R_A}{a_A}}

T = 2\pi \sqrt{\frac{1.1\times 10^{12}}{1.23\times 10^{-5}}} = 1.89\times 10^{9}\ \text{s}

Answer

$1.9\times 10^{9}\ \text{s}$

Final answer

1.9 × 10^9 s

Detailed explanation

Background Concept

For uniform circular motion:

a = r\omega^2

and angular speed $\omega$ is related to period $T$ by

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

Combining these gives

a = r\left(\frac{2\pi}{T}\right)^2 = \frac{4\pi^2 r}{T^2}

So if you know $a$ and $r$ , you can find $T$ .

In a two-body system orbiting a common centre of mass, the separation of the bodies is the sum of their orbital radii about the centre of mass.

Understanding the Question

You are told the stars are separated by $3.3\times 10^{12}\ \text{m}$ and that $R_B = 2R_A$ . The diagram indicates $R_A$ and $R_B$ are the radii of the circular orbits about the common centre of gravity $X$ .

You must use the centripetal acceleration of star A from (b)(ii) and its orbital radius $R_A$ to find the period $T$ .

Approach

Use the geometry: separation $= R_A + R_B$ , and the given ratio $R_B = 2R_A$ , to find $R_A$ .
Use $a_A = R_A\omega^2$ to find $\omega$ (or directly combine into $T = 2\pi\sqrt{R_A/a_A}$ ).
Substitute values and calculate.

Step-by-Step Reasoning

1) Find $R_A$ from the separation and ratio

Given:

R_B = 2R_A

Separation between stars:

R_A + R_B = 3.3\times 10^{12}

Substitute $R_B$ :

R_A + 2R_A = 3R_A = 3.3\times 10^{12}

So

R_A = \frac{3.3\times 10^{12}}{3} = 1.1\times 10^{12}\ \text{m}

2) Relate acceleration to period

Use

a_A = R_A\omega^2

and $\omega = 2\pi/T$ :

a_A = R_A\left(\frac{2\pi}{T}\right)^2 = \frac{4\pi^2 R_A}{T^2}

Rearrange for $T$ :

T^2 = \frac{4\pi^2 R_A}{a_A}

T = 2\pi\sqrt{\frac{R_A}{a_A}}

3) Substitute numbers

With $R_A = 1.1\times 10^{12}\ \text{m}$ and $a_A = 1.23\times 10^{-5}\ \text{m s}^{-2}$ :

T = 2\pi\sqrt{\frac{1.1\times 10^{12}}{1.23\times 10^{-5}}}

Compute the inside:

\frac{1.1\times 10^{12}}{1.23\times 10^{-5}} \approx 8.98\times 10^{16}

Square root:

\sqrt{8.98\times 10^{16}} \approx 3.00\times 10^{8}

Multiply by $2\pi$ :

T \approx 6.283\times 3.00\times 10^{8} = 1.89\times 10^{9}\ \text{s}

Rounded:

T \approx 1.9\times 10^{9}\ \text{s}

Key Takeaways

For circular motion: $a = r\omega^2$ and $\omega = 2\pi/T$ .
In a binary system: separation $= R_A + R_B$ .
Use consistent SI units throughout.

Common Mistakes

Using the full separation $3.3\times 10^{12}\ \text{m}$ as $R_A$ .
Forgetting the square root when rearranging for $T$ .
Using $a = v^2/r$ without relating $v$ to $T$ correctly.

Things to Be Careful About

Ensure $R_A$ is found correctly: here $R_A$ is one third of the separation because $R_B = 2R_A$ .
Keep track of powers of ten when dividing by $10^{-5}$ (it increases the power by 5).

Techniques used

use the centre-of-mass condition to find orbital radiusrelate centripetal acceleration to angular speedrearrange for period using omega = 2 pi / T

(iv)

By placing a tick (✓) in each row, complete Table 1.1 to show how the quantities indicated for star B compare with the same quantities for star A.

Table 1.1

	B less than A	B equal to A	B greater than A
centripetal acceleration
linear speed
period

DifficultyMedium-Easy

Worked solution

Answer

Both stars have the same angular speed, so the periods are equal.

Since $a = r\omega^2$ and $v = r\omega$ with the same $\omega$ and $R_B = 2R_A$ :

centripetal acceleration: B greater than A
linear speed: B greater than A
period: B equal to A

Final answer

a_B > a_A, v_B > v_A, T_B = T_A

Detailed explanation

Background Concept

In uniform circular motion for an object at radius $r$ with angular speed $\omega$ :

v = r\omega

a = r\omega^2

The period $T$ is related to $\omega$ by

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

In a binary system orbiting a common centre of mass, both objects rotate with the same angular speed $\omega$ (they stay on opposite sides of the centre of mass, so they must complete each revolution together).

Understanding the Question

You are told $R_B$ is double $R_A$ :

R_B = 2R_A

The question asks you to compare for star B versus star A:

centripetal acceleration
linear speed
period

No calculation is required; it is about how these quantities scale with radius when angular speed is the same.

Approach

Use the fact that both stars share the same period $T$ (and so the same $\omega$ ).
Compare $a$ using $a = r\omega^2$ .
Compare $v$ using $v = r\omega$ .

Step-by-Step Reasoning

1) Period
If both stars rotate about the centre of mass together, they complete one orbit in the same time, so

T_B = T_A

2) Centripetal acceleration

a = r\omega^2

Since $\omega$ is the same for both stars, $a \propto r$ .
With $R_B = 2R_A$ :

\frac{a_B}{a_A} = \frac{R_B}{R_A} = 2

So $a_B$ is greater than $a_A$ .

3) Linear speed

v = r\omega

Again with the same $\omega$ , $v \propto r$ . Therefore:

\frac{v_B}{v_A} = \frac{R_B}{R_A} = 2

So $v_B$ is greater than $v_A$ .

Key Takeaways

In a two-body orbit, both bodies share the same angular speed and period.
With the same $\omega$ , both $v$ and $a$ increase in proportion to orbital radius $r$ .

Common Mistakes

Saying the periods are different because the radii are different (they must orbit together).
Using $a = v^2/r$ and then incorrectly concluding that larger $r$ gives smaller $a$ without accounting for the change in $v$ .

Things to Be Careful About

The relationship $a = v^2/r$ is always true, but here $v$ is not the same for both stars.
Always decide first whether $\omega$ (or $T$ ) is common to both bodies in the situation described.

Techniques used

use proportional reasoning with a = r omega^2use proportional reasoning with v = r omegarecognise that both bodies share the same angular speed and period in a binary orbit

Q2Medium-EasyIdeal Gases

(a)

The equation of state for an ideal gas may be expressed as

pV = NkT

5M

(i)

State the meaning of each of the symbols in this equation.

$p$ : ______

$V$ : ______

$N$ : ______

$k$ : ______

$T$ : ______

DifficultyEasy

Worked solution

Answer

$p$ : pressure of the gas

$V$ : volume of the gas

$N$ : number of molecules (particles) in the gas

$k$ : Boltzmann constant

$T$ : thermodynamic temperature (in kelvin)

Final answer

p: pressure; V: volume; N: number of molecules; k: Boltzmann constant; T: thermodynamic temperature (K)

Detailed explanation

Background Concept

The ideal gas equation can be written in different but equivalent forms. In A Level, two common versions are

pV = nRT

and

pV = NkT

Here, $n$ is the amount of gas in moles, $R$ is the molar gas constant, $N$ is the number of molecules (or atoms) and $k$ is the Boltzmann constant. They are related by $N = nN_A$ and $R = N_A k$ , where $N_A$ is the Avogadro constant.

Understanding the Question

You are given

pV = NkT

and asked to state what each symbol represents. No calculation is required; you just need the correct physical meaning of each letter.

Approach

List each symbol in turn and write its meaning, being careful that:

$N$ is a count of molecules/particles (not moles),
$T$ is absolute temperature (kelvin), not degrees Celsius.

Step-by-Step Reasoning

$p$ is the pressure exerted by the gas on the container walls.
$V$ is the volume occupied by the gas.
$N$ counts how many molecules (or particles) are present.
$k$ is the Boltzmann constant, the microscopic equivalent of $R$ .
$T$ is thermodynamic (absolute) temperature measured in kelvin.

Key Takeaways

$pV = NkT$ uses particle number $N$ , not moles.
Temperature must be on the kelvin scale.
$k$ links microscopic particle descriptions to macroscopic measurements.

Common Mistakes

Writing $N$ as “number of moles” (that is $n$ ).
Giving $T$ in $^{\circ}\text{C}$ instead of kelvin.
Confusing $k$ (Boltzmann constant) with $R$ (molar gas constant).

Things to Be Careful About

If units are mentioned: $p$ in $\text{Pa}$ , $V$ in $\text{m}^3$ , $T$ in $\text{K}$ .
$N$ is dimensionless (a pure count).

Techniques used

identify the physical meaning of symbols in a given equationdistinguish between number of molecules and amount of substancestate the correct unit/scale associated with thermodynamic temperature

(ii)

Using the equation of state, derive an expression for the average translational kinetic energy $E_K$ of a particle in the gas in terms of some or all of $N$ , $k$ and $T$ .

$E_K$ = ______

DifficultyMedium-Easy

Worked solution

Working

From kinetic theory,

pV = \frac{1}{3}Nm\langle c^2 \rangle

Given $pV = NkT$ ,

NkT = \frac{1}{3}Nm\langle c^2 \rangle \Rightarrow kT = \frac{1}{3}m\langle c^2 \rangle

Average translational kinetic energy per particle:

E_K = \frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT

Answer

E_K = \frac{3}{2}kT

Final answer

E_K = (3/2) kT

Detailed explanation

Background Concept

The kinetic theory of gases relates macroscopic pressure to microscopic molecular motion. For an ideal gas,

pV = NkT

Also, kinetic theory gives

pV = \frac{1}{3}Nm\langle c^2 \rangle

where $m$ is the mass of one molecule and $\langle c^2 \rangle$ is the mean of the squared molecular speeds. The root-mean-square speed is $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ .

The average translational kinetic energy per molecule is

E_K = \frac{1}{2}m\langle c^2 \rangle

Understanding the Question

You are asked to use the equation of state to derive an expression for the average translational kinetic energy $E_K$ of one particle (molecule) in the gas, in terms of $N$ , $k$ , and $T$ (though $N$ will cancel).

Approach

Use the fact that there are two valid expressions for $pV$ for an ideal gas:

from the equation of state: $pV = NkT$
from kinetic theory: $pV = \frac{1}{3}Nm\langle c^2 \rangle$

Equate them, cancel $N$ , and then rewrite in terms of $E_K = \frac{1}{2}m\langle c^2 \rangle$ .

Step-by-Step Reasoning

Start with the kinetic theory form:

pV = \frac{1}{3}Nm\langle c^2 \rangle

The equation of state says for the same gas sample:

pV = NkT

Because both equal $pV$ , set them equal:

NkT = \frac{1}{3}Nm\langle c^2 \rangle

Cancel $N$ (because it is the same number of molecules on each side):

kT = \frac{1}{3}m\langle c^2 \rangle

Multiply both sides by $\frac{3}{2}$ :

\frac{3}{2}kT = \frac{1}{2}m\langle c^2 \rangle

Recognise the right-hand side as the average translational kinetic energy per molecule:

E_K = \frac{1}{2}m\langle c^2 \rangle

So,

E_K = \frac{3}{2}kT

This result is important: for an ideal gas, average translational kinetic energy depends only on absolute temperature, not on the type of gas.

Key Takeaways

Combine $pV = NkT$ with $pV = \frac{1}{3}Nm\langle c^2 \rangle$ .
The particle number $N$ cancels, so $E_K$ depends only on $T$ .
The standard result is $E_K = \frac{3}{2}kT$ (per molecule, translational only).

Common Mistakes

Forgetting the factor $\frac{1}{3}$ in the kinetic theory equation.
Using $E_K = \frac{1}{2}mc_{\text{rms}}^2$ but mixing up $c_{\text{rms}}^2$ and $\langle c^2 \rangle$ (they are equal, but you must treat them consistently).
Leaving an $N$ in the final expression (it should cancel).

Things to Be Careful About

$T$ must be in kelvin.
This is translational kinetic energy only; real molecules may also have rotational/vibrational energy, but the question explicitly asks for translational kinetic energy as used in this syllabus result.

Techniques used

use the kinetic theory relation for pressure in terms of molecular motionequate two expressions for pVrearrange to obtain mean kinetic energy per molecule

(b)

A molecule of hydrogen gas consists of two hydrogen atoms, each of nucleon number 1. A molecule of oxygen gas consists of two oxygen atoms, each of nucleon number 16.

Assume that hydrogen and oxygen both behave as ideal gases.

A sample of hydrogen gas is at the same temperature as a sample of oxygen gas.

For the two samples, determine the ratio

\frac{\text{root-mean-square (r.m.s.) speed of hydrogen molecules}}{\text{root-mean-square (r.m.s.) speed of oxygen molecules}}

ratio = ______

DifficultyMedium-Easy

Worked solution

Working

For an ideal gas,

\frac{1}{2}m c_{\text{rms}}^2 = \frac{3}{2}kT \Rightarrow c_{\text{rms}} = \sqrt{\frac{3kT}{m}}

So

\frac{c_{\text{rms,H}_2}}{c_{\text{rms,O}_2}} = \sqrt{\frac{m_{\text{O}_2}}{m_{\text{H}_2}}}

$\text{H}_2$ : nucleon number $=2$ ; $\text{O}_2$ : nucleon number $=32$ .

\frac{c_{\text{rms,H}_2}}{c_{\text{rms,O}_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

Answer

ratio $= 4$

Final answer

4

Detailed explanation

Background Concept

For an ideal gas, the r.m.s. speed is connected to temperature by

\frac{1}{2}m c_{\text{rms}}^2 = \frac{3}{2}kT

so

c_{\text{rms}} = \sqrt{\frac{3kT}{m}}

At the same temperature $T$ , the only difference between gases is the mass $m$ of a molecule: heavier molecules move more slowly on average, with speed proportional to $1/\sqrt{m}$ .

Understanding the Question

Hydrogen gas is $\text{H}_2$ (two hydrogen atoms). Each hydrogen atom has nucleon number $1$ , so the molecule has total nucleon number $2$ .

Oxygen gas is $\text{O}_2$ (two oxygen atoms). Each oxygen atom has nucleon number $16$ , so the molecule has total nucleon number $32$ .

You need the ratio:

\frac{c_{\text{rms,H}_2}}{c_{\text{rms,O}_2}}

with both samples at the same temperature.

Approach

Use $c_{\text{rms}} = \sqrt{3kT/m}$ .
Form the ratio so $3kT$ cancels.
Use nucleon number to take the ratio of molecular masses (mass (\propto) nucleon number).

Step-by-Step Reasoning

Start with

c_{\text{rms}} = \sqrt{\frac{3kT}{m}}

Form the ratio at the same $T$ :

\frac{c_{\text{rms,H}_2}}{c_{\text{rms,O}_2}} = \sqrt{\frac{3kT/m_{\text{H}_2}}{3kT/m_{\text{O}_2}}} = \sqrt{\frac{m_{\text{O}_2}}{m_{\text{H}_2}}}

Now compare molecular masses using nucleon numbers:

$m_{\text{H}_2} \propto 2$
$m_{\text{O}_2} \propto 32$

So

\frac{c_{\text{rms,H}_2}}{c_{\text{rms,O}_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

This makes sense physically: hydrogen molecules are much lighter, so they move faster at the same temperature.

Key Takeaways

For equal $T$ , $c_{\text{rms}} \propto 1/\sqrt{m}$ .
Ratios are easiest because constants ( $3kT$ ) cancel.
Use molecular mass (for $\text{H}_2$ and $\text{O}_2$ ), not atomic mass.

Common Mistakes

Using atomic masses (1 and 16) instead of molecular masses (2 and 32), giving $\sqrt{16} = 4$ coincidentally here if done inconsistently, but it can fail in other questions.
Inverting the ratio: writing $\sqrt{m_{\text{H}_2}/m_{\text{O}_2}}$ would give $1/4$ .
Forgetting the square root relationship and using $c_{\text{rms}} \propto 1/m$ .

Things to Be Careful About

Make sure both gases are stated to be at the same temperature; otherwise $T$ would not cancel.
Nucleon number is a proxy for relative mass; electron masses are negligible at this level, so the approximation is valid.

Techniques used

use the r.m.s. speed relation in terms of temperature and molecular massuse relative molecular mass (nucleon number) to compare particle massestake a square-root ratio to compare speeds at equal temperature

Q3Medium-EasyThermodynamicsIdeal Gases

(a)

With reference to molecular kinetic energy and molecular potential energy, explain what is meant by the internal energy of an ideal gas.

DifficultyMedium-Easy

Worked solution

Answer

Internal energy is the sum of the random molecular kinetic energy and the molecular potential energy.

For an ideal gas, intermolecular forces are negligible so the molecular potential energy is constant (taken as zero), hence the internal energy is due only to the random kinetic energy of the molecules.

Final answer

Internal energy is the total random molecular kinetic + potential energies; for an ideal gas the potential energy is negligible/constant so internal energy is just the total random kinetic energy.

Detailed explanation

Background Concept

The internal energy $U$ of a substance is the energy stored microscopically in its particles. For a gas this consists of:

random molecular kinetic energy (translational, and possibly rotational/vibrational in more detailed models), and
molecular potential energy due to intermolecular forces (attraction/repulsion between molecules).

An ideal gas is modelled as molecules that:

occupy negligible volume, and
exert negligible intermolecular forces on each other (except during collisions).

Because the forces are negligible, there is essentially no intermolecular potential energy change as the gas changes state, so we treat the potential energy contribution as constant (often taken as zero).

Understanding the Question

You are asked to explain what “internal energy of an ideal gas” means, explicitly referring to:

molecular kinetic energy, and
molecular potential energy.

So you must (i) state that internal energy is a sum of these, and (ii) link the ideal gas assumption to the potential energy term.

Approach

Give the definition: $U$ is total (random) molecular kinetic energy + molecular potential energy.
Apply the ideal-gas model: negligible intermolecular forces $\Rightarrow$ potential energy contribution is constant/negligible.
Conclude: for an ideal gas, internal energy depends only on molecular kinetic energy (and hence only on temperature).

Step-by-Step Reasoning

Internal energy is not about the motion of the gas as a whole; it is about the energies of molecules within it.
Molecules have random kinetic energy because they move in all directions.
If there are forces between molecules, separating/bringing them together changes potential energy.
For an ideal gas, we assume those forces are negligible, so there is no significant molecular potential energy term (or it does not change).
Therefore, the internal energy of an ideal gas is the total random kinetic energy of its molecules.

Key Takeaways

Internal energy $U$ is microscopic energy: kinetic + potential at molecular level.
For an ideal gas, potential energy is negligible/constant, so $U$ is due to kinetic energy only.

Common Mistakes

Saying internal energy is “heat energy” (heat is energy in transfer, not stored).
Forgetting to mention potential energy at all (the question explicitly asks for both).
Saying the potential energy is zero “because pressure is zero” or other unrelated statements.

Things to Be Careful About

Use the phrase random molecular kinetic energy.
Link “ideal” directly to negligible intermolecular forces, not to “no collisions” (collisions still occur).

Techniques used

state the definition of internal energy in terms of molecular energiesapply the ideal-gas assumption to neglect intermolecular potential energy

(b)

A sample of an ideal gas is initially in state A, at a pressure of $2.0 \times 10^5\ \text{Pa}$ and with a volume of $0.016\ \text{m}^3$ , as shown in Fig. 3.1.

Fig. 3.1

In state A, the temperature of the gas is $400\ \text{K}$ .

The gas undergoes two successive changes X and Y.

In change X, it is heated at constant volume to a pressure of $4.0 \times 10^5\ \text{Pa}$ . At the end of change X, the gas is in state B.

In change Y, it is then allowed to expand at constant temperature back to its original pressure. At the end of change Y, the gas is in state C.

7M

(i)

Determine the internal energy of the gas in state A.

internal energy = ______ $\text{J}$

DifficultyMedium-Easy

Worked solution

Working

For an ideal gas,

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

U = \frac{3}{2}(2.0 \times 10^5)(0.016) = 4.8 \times 10^3\ \text{J}

Answer

$4.8 \times 10^3\ \text{J}$

Final answer

4.8 × 10^3 J

Detailed explanation

Background Concept

For an ideal gas, internal energy depends only on temperature. In the simplest model used at this level (ideal monatomic gas),

U = \frac{3}{2}nRT

Using the equation of state,

pV = nRT

we can also write

U = \frac{3}{2}pV

This is useful because $p$ and $V$ are often given directly.

Understanding the Question

State A has:

pressure $p_A = 2.0 \times 10^5\ \text{Pa}$
volume $V_A = 0.016\ \text{m}^3$
temperature $T_A = 400\ \text{K}$

You are asked to find the internal energy of the gas in state A.

Approach

Use the ideal-gas internal energy formula. Since $p$ and $V$ are given, the quickest route is:

U = \frac{3}{2}pV

Then substitute the state A values.

Step-by-Step Reasoning

Start from

U = \frac{3}{2}pV

Calculate $pV$ first:

pV = (2.0 \times 10^5)(0.016) = 3.2 \times 10^3

Now multiply by $\frac{3}{2}$ :

U = \frac{3}{2}(3.2 \times 10^3) = 4.8 \times 10^3\ \text{J}

Check units: $\text{Pa} \cdot \text{m}^3 = \text{N m}^{-2} \cdot \text{m}^3 = \text{N m} = \text{J}$ .

Key Takeaways

For an ideal gas (at this level), $U$ can be found using $U = \frac{3}{2}pV$ .
$pV$ naturally has joule units.

Common Mistakes

Using $U = pV$ instead of $U = \frac{3}{2}pV$ .
Forgetting that $p$ must be in pascals and $V$ in $\text{m}^3$ .
Giving $U$ in units of $\text{Pa}$ or $\text{m}^3$ instead of joules.

Things to Be Careful About

Keep powers of ten clear: $2.0 \times 10^5 \times 0.016 = 3.2 \times 10^3$ , not $3.2 \times 10^4$ .
Quote the final answer to a sensible number of significant figures (here 2 s.f. matches the data).

Techniques used

use the ideal-gas equation to relate state variablessubstitute into the internal energy relation for an ideal gasevaluate a numerical result with correct unit

(ii)

Determine the temperature of the gas in state B.

temperature = ______ $\text{K}$

DifficultyEasy

Worked solution

Working

At constant volume, $p \propto T$ .

T_B = T_A\frac{p_B}{p_A} = 400\frac{4.0 \times 10^5}{2.0 \times 10^5} = 800\ \text{K}

Answer

$800\ \text{K}$

Final answer

800 K

Detailed explanation

Background Concept

From the ideal gas equation $pV = nRT$ , if the amount of gas $n$ is fixed and the volume is constant, then

\frac{p}{T} = \text{constant} \quad \Rightarrow \quad p \propto T

So doubling the absolute temperature doubles the pressure (and vice versa).

Understanding the Question

Change X takes the gas from state A to state B:

volume stays at $0.016\ \text{m}^3$
pressure increases from $2.0 \times 10^5\ \text{Pa}$ to $4.0 \times 10^5\ \text{Pa}$
initial temperature is $400\ \text{K}$

You must find the temperature in state B.

Approach

Use the constant-volume form of the ideal-gas law:

\frac{p_A}{T_A} = \frac{p_B}{T_B}

and rearrange for $T_B$ .

Step-by-Step Reasoning

Write

\frac{p_A}{T_A} = \frac{p_B}{T_B}

Rearrange:

T_B = T_A \frac{p_B}{p_A}

Substitute values:

T_B = 400 \times \frac{4.0 \times 10^5}{2.0 \times 10^5} = 400 \times 2 = 800\ \text{K}

Key Takeaways

At constant volume for a fixed mass of gas: $p/T$ is constant.
Use ratios to avoid unnecessary calculation of $n$ .

Common Mistakes

Using $pV=\text{constant}$ (that is for constant temperature, not constant volume).
Using degrees Celsius instead of kelvin in proportional relationships.

Things to Be Careful About

Temperature must be in kelvin for $p \propto T$ .
Check that the process is explicitly “constant volume” before using $p/T$ constant.

Techniques used

use proportional reasoning from the ideal-gas equation at constant volumecalculate the new temperature from a pressure ratio

(iii)

Determine the volume of the gas in state C.

volume = ______ $\text{m}^3$

DifficultyEasy

Worked solution

Working

Change Y is at constant temperature, so

p_BV_B = p_CV_C

V_C = \frac{p_BV_B}{p_C} = \frac{(4.0 \times 10^5)(0.016)}{2.0 \times 10^5} = 0.032\ \text{m}^3

Answer

$0.032\ \text{m}^3$

Final answer

0.032 m^3

Detailed explanation

Background Concept

For a fixed mass of ideal gas at constant temperature (isothermal process), the ideal gas equation gives

pV = nRT = \text{constant}

So

p_1V_1 = p_2V_2

This is Boyle’s law.

Understanding the Question

State B is reached after change X:

$p_B = 4.0 \times 10^5\ \text{Pa}$
$V_B = 0.016\ \text{m}^3$

Change Y is then an isothermal expansion from state B until the pressure returns to its original value:

$p_C = 2.0 \times 10^5\ \text{Pa}$

You must find the volume $V_C$ .

Approach

Because temperature is constant during change Y, use

p_BV_B = p_CV_C

Solve for $V_C$ .

Step-by-Step Reasoning

Start with the isothermal condition:

p_BV_B = p_CV_C

Rearrange:

V_C = \frac{p_BV_B}{p_C}

Substitute values:

V_C = \frac{(4.0 \times 10^5)(0.016)}{2.0 \times 10^5}

The ratio $\frac{4.0 \times 10^5}{2.0 \times 10^5} = 2$ , so

V_C = 2 \times 0.016 = 0.032\ \text{m}^3

Key Takeaways

Isothermal change $\Rightarrow$ $pV$ constant.
If pressure halves, volume doubles (for isothermal conditions).

Common Mistakes

Using $p/T$ constant (that is for constant volume, not constant temperature).
Using state A temperature ( $400\ \text{K}$ ) in calculations for change Y; only the constancy of temperature matters here.

Things to Be Careful About

Make sure you use the correct states: the isothermal starts at B and ends at C.
Keep units consistent (Pa and $\text{m}^3$ ).

Techniques used

use Boyle's law for an isothermal changecalculate an unknown volume from pressure and volume ratios

(iv)

On Fig. 3.1, draw two lines, one to represent change X and one to represent change Y. Label your lines X and Y respectively.

DifficultyMedium-Easy

Worked solution

diagram-3biv

Answer

X: vertical line at $V = 0.016\ \text{m}^3$ from $2.0 \times 10^5\ \text{Pa}$ to $4.0 \times 10^5\ \text{Pa}$ .

Y: isothermal curve (rectangular hyperbola) from B to C ending at $p = 2.0 \times 10^5\ \text{Pa}$ and $V = 0.032\ \text{m}^3$ .

Final answer

X: vertical isochore at V = 0.016 m^3 from 2.0×10^5 Pa to 4.0×10^5 Pa; Y: isothermal curve from B to C ending at (2.0×10^5 Pa, 0.032 m^3).

Detailed explanation

Background Concept

On a pressure–volume ( $p$ – $V$ ) graph for a fixed mass of ideal gas:

Constant volume (isochoric): $V$ fixed, so the graph is a vertical line.
Constant temperature (isothermal): $pV = \text{constant}$ , so the graph is a rectangular hyperbola; as $V$ increases, $p$ decreases.

Understanding the Question

You start at state A: $(V, p) = (0.016\ \text{m}^3, 2.0 \times 10^5\ \text{Pa})$ .

Change X: heated at constant volume to reach state B at pressure $4.0 \times 10^5\ \text{Pa}$ .

Change Y: then expands at constant temperature until pressure returns to $2.0 \times 10^5\ \text{Pa}$ , reaching state C.

You must draw the two lines on the given $p$ – $V$ axes and label them X and Y.

Approach

Plot/identify point B directly above A at the same volume (because X is constant volume).
Draw change X as a vertical line from A to B.
For change Y, use $pV=\text{constant}$ starting from B. Find $V_C$ (it is $0.032\ \text{m}^3$ from part (iii)).
Draw a smooth decreasing hyperbola from B to C and label it Y.

Step-by-Step Reasoning

Change X: volume does not change, so you move straight up (pressure increases) at $V=0.016\ \text{m}^3$ until you reach $p=4.0 \times 10^5\ \text{Pa}$ . That endpoint is B.
Change Y: temperature stays constant, so $pV$ stays constant. Starting at B, $p$ falls back to $2.0 \times 10^5\ \text{Pa}$ while $V$ increases to keep $pV$ constant. The endpoint is C at $V=0.032\ \text{m}^3$ .
The shape for an isotherm is a curve (not a straight line) that flattens as $V$ increases.

diagram-3biv

Key Takeaways

Isochoric process $\Rightarrow$ vertical line on a $p$ – $V$ graph.
Isothermal process $\Rightarrow$ hyperbola ( $pV=\text{constant}$ ) on a $p$ – $V$ graph.

Common Mistakes

Drawing the isothermal as a straight line.
Drawing change X as horizontal (horizontal would mean constant pressure, not constant volume).
Making Y end at the wrong pressure or not matching the correct final volume.

Things to Be Careful About

Label both lines clearly with X and Y as requested.
Ensure the isothermal curve goes through the correct endpoints (B and C).
Keep the curve smooth and decreasing; it should not dip below the final point or turn upward.

Techniques used

identify an isochoric process on a p-V graphidentify an isothermal process as a rectangular hyperbola on a p-V graphuse given state points to sketch the correct curve and label it

Q4MediumOscillations

(a)

State what is meant by the frequency of the oscillations of an oscillating object.

(b)

An object is oscillating.

Fig. 4.1 shows the variation of the acceleration $a$ of the object with its displacement $x$ from the equilibrium position.

Fig. 4.2 shows the variation of the kinetic energy $E_K$ of the object with time $t$ .

Fig. 4.1

Fig. 4.2

9M

(i)

Explain how Fig. 4.2 shows that the period of the oscillations is $0.80\ \text{s}$ .

DifficultyMedium-Easy

Worked solution

Answer

Successive maxima of $E_K$ are at $t=0.2\ \text{s}$ and $t=0.6\ \text{s}$ , so the period of the $E_K$ graph is $0.4\ \text{s}$ . Since $E_K$ has two maxima per oscillation, the oscillation period is $2\times 0.4 = 0.80\ \text{s}$ .

Final answer

Maxima of EK are 0.4 s apart; EK repeats twice per cycle, so T = 0.80 s.

Detailed explanation

Background Concept

For simple harmonic motion, speed (and hence kinetic energy) varies during the motion:

Kinetic energy is

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

In SHM, the object passes through equilibrium twice per cycle (once in each direction) and is at a turning point twice per cycle.
Therefore $v^2$ (and so $E_K$ ) has two identical peaks per oscillation, meaning the kinetic-energy graph has half the period of the displacement (or velocity) graph.

Understanding the Question

You are shown a graph of $E_K$ against $t$ . The question asks how this graph indicates that the oscillation period (time for one full back-and-forth motion) is $0.80\ \text{s}$ .

Approach

Read the time between repeated features on the $E_K$ graph (e.g. peak-to-peak).
Recognise that this time corresponds to half an oscillation period because $E_K$ peaks twice per oscillation.
Double the time found in step 1.

Step-by-Step Reasoning

From Fig. 4.2, successive maxima occur at $t=0.2\ \text{s}$ and $t=0.6\ \text{s}$ .

\Delta t_{\text{between peaks}} = 0.6 - 0.2 = 0.4\ \text{s}

These peaks correspond to the object passing through equilibrium with maximum speed; it does that twice per oscillation.
Hence $0.4\ \text{s}$ is half the oscillation period:

T = 2(0.4\ \text{s}) = 0.80\ \text{s}

Key Takeaways

$E_K \propto v^2$ , so it is always positive and repeats twice per cycle.
Peak-to-peak time on an $E_K$ graph corresponds to $T/2$ for SHM.

Common Mistakes

Taking the $0.4\ \text{s}$ peak-to-peak time as the oscillation period (forgetting the “twice per cycle” idea).
Using zero-to-zero spacing incorrectly without checking whether the zeros repeat every half-cycle.

Things to Be Careful About

Make sure you compare same features (peak-to-peak or zero-to-zero).
State explicitly why doubling is needed: $E_K$ has two maxima in one oscillation.

Techniques used

read time intervals from a graphuse the fact that kinetic energy repeats twice per oscillation in SHMrelate the period of kinetic energy variation to the period of oscillation

(ii)

Calculate the angular frequency $\omega$ of the oscillations.

$\omega$ = ______ $\text{rad s}^{-1}$

(iii)

Apart from the period, frequency and angular frequency of the oscillations, determine three other conclusions about the object and its oscillations that may be drawn from Fig. 4.1 and Fig. 4.2. The conclusions may be qualitative or quantitative. Use the space below for any working.

1 ______

2 ______

3 ______

DifficultyMedium

Worked solution

Answer

Fig. 4.1 shows $a \propto -x$ (straight line through origin with negative gradient), so the motion is simple harmonic with acceleration always towards equilibrium.
From Fig. 4.2, the maximum kinetic energy is $E_{K\,\text{max}} = 7.0 \times 10^{-4}\ \text{J}$ (so the total energy is $7.0 \times 10^{-4}\ \text{J}$ ).
From Fig. 4.2, $E_K=0$ at $t=0,\ 0.4,\ 0.8\ \text{s}$ so the object is at turning points (speed zero) at these times, and $E_K$ is maximum at $t=0.2$ and $0.6\ \text{s}$ so it passes through equilibrium with maximum speed at these times.

Final answer

a ∝ −x so SHM with restoring acceleration; 2) EK,max = 7.0×10^-4 J (total energy same); 3) EK=0 at turning points and EK max at equilibrium (times read from graph).

Detailed explanation

Background Concept

Two key SHM ideas connect to the given graphs:

Acceleration–displacement relation

For SHM,

a = -\omega^2 x

So a graph of $a$ against $x$ is a straight line through the origin with negative gradient. The negative sign means acceleration is always directed back towards equilibrium.

Energy in SHM

Kinetic energy:

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

Potential energy (for SHM system) is greatest at extreme displacement, least at equilibrium.
Total energy is constant (ignoring damping):

E_{\text{total}} = E_K + E_P = \text{constant}

Understanding the Question

You must give three different conclusions (not including period/frequency/angular frequency) that can be read from:

Fig. 4.1: $a$ vs $x$ .
Fig. 4.2: $E_K$ vs $t$ .

Conclusions can be qualitative (what kind of motion, where acceleration points) or quantitative (a maximum energy value, specific times when something happens).

Approach

Pick distinct information sources:

Use the shape/sign of the $a$ – $x$ graph to identify SHM and the direction of acceleration.
Use the peaks/zeros on the $E_K$ – $t$ graph to infer when the object is at equilibrium vs turning points.
Read a numerical maximum from the $E_K$ graph (this gives total energy too).

Step-by-Step Reasoning

Conclusion 1 (from Fig. 4.1): SHM and restoring acceleration

The graph is a straight line through the origin with negative gradient.
This matches $a = -\omega^2 x$ .
Therefore acceleration is proportional to displacement and opposite in direction, i.e. always towards equilibrium.

Conclusion 2 (from Fig. 4.2): maximum kinetic energy (and total energy)

The highest point on the $E_K$ graph is $7.0 \times 10^{-4}\ \text{J}$ .
In SHM without damping, total energy is constant and equals the maximum kinetic energy (because at equilibrium the potential energy is minimum).
So $E_{\text{total}} = 7.0 \times 10^{-4}\ \text{J}$ .

Conclusion 3 (from Fig. 4.2): where the object is in its cycle at certain times

When $E_K = 0$ , then $v=0$ (since $E_K = \tfrac{1}{2}mv^2$ ), which happens at turning points (maximum displacement).
The graph shows $E_K=0$ at $t=0,\ 0.4,\ 0.8\ \text{s}$ , so those are turning points.
When $E_K$ is maximum, speed is maximum, which occurs at equilibrium position. The maxima occur at $t=0.2$ and $0.6\ \text{s}$ , so the object passes through equilibrium then.

Key Takeaways

A straight-line $a$ – $x$ graph with negative gradient is the signature of SHM.
$E_K=0$ means speed is zero (turning points).
$E_K$ maximum occurs at equilibrium; its maximum value equals total energy (if no damping).

Common Mistakes

Saying “negative acceleration” rather than “acceleration opposite to displacement” (sign depends on which side of equilibrium the object is on).
Forgetting that $E_K$ being maximum implies equilibrium (not maximum displacement).
Using the $E_K$ peak spacing to restate the period (the question excludes period/frequency/ $\omega$ ).

Things to Be Careful About

Keep the three conclusions distinct; do not repeat the same idea in different words.
When quoting numerical values from graphs, include the correct power of ten from the axis label (here $E_K / 10^{-4}\ \text{J}$ ).
If mentioning “total energy”, it is only constant if damping is negligible (the graph here suggests no decay in peak height).

Techniques used

infer SHM from a linear a–x relationship with negative gradientextract qualitative motion information from energy-time featuresread maximum values directly from a graph

(iv)

Describe the interchange between kinetic energy and potential energy during the oscillations. Numerical values are not required.

DifficultyMedium-Easy

Worked solution

Answer

At the equilibrium position, speed is maximum so kinetic energy is maximum and potential energy is minimum.

As the object moves away from equilibrium, kinetic energy decreases and is converted into potential energy.

At the extreme displacement (turning points), speed is zero so kinetic energy is zero and potential energy is maximum; on returning, potential energy is converted back into kinetic energy. Total energy remains constant (neglecting damping).

Final answer

KE max at equilibrium and PE min; moving away KE converts to PE; at extremes KE=0 and PE max, then PE converts back to KE (total energy constant).

Detailed explanation

Background Concept

In SHM (with negligible damping), mechanical energy continually swaps between:

Kinetic energy

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

Potential energy of the system (e.g. elastic potential energy for a spring)

Total energy stays constant:

E_{\text{total}} = E_K + E_P = \text{constant}

Key physical links:

Speed is greatest at equilibrium, zero at turning points.
Displacement magnitude is greatest at turning points, zero at equilibrium.

Understanding the Question

You must describe (no numbers needed) how kinetic and potential energy change during one oscillation: what happens at equilibrium, as it moves outwards, at the turning point, and as it returns.

Approach

Describe the energy changes over a cycle using the two special positions:

equilibrium ( $x=0$ )
turning points ( $x=\pm x_0$ )

and state the conversion between energies as the object moves.

Step-by-Step Reasoning

At equilibrium ( $x=0$ ): the object is moving fastest, so $v$ is maximum.
Therefore $E_K$ is maximum. At the same moment, the potential energy of the oscillating system is minimum.
Moving away from equilibrium: the restoring acceleration slows it down, so $v$ decreases. Hence $E_K$ decreases. The object is moving to larger $|x|$ , so the system stores more potential energy; $E_P$ increases. This is energy conversion from kinetic to potential.
At a turning point ( $x=\pm x_0$ ): the object momentarily stops before reversing direction, so $v=0$ and $E_K=0$ . The displacement is maximum, so the potential energy is maximum.
Returning towards equilibrium: the restoring force accelerates it back, increasing $v$ . So $E_P$ decreases and is converted back into $E_K$ .
Over the whole oscillation, if there is no damping, the total mechanical energy remains constant.

Key Takeaways

Equilibrium: $E_K$ maximum, $E_P$ minimum.
Turning points: $E_K=0$ , $E_P$ maximum.
Energy is transferred back and forth while the total stays constant (undamped).

Common Mistakes

Saying both $E_K$ and $E_P$ are maximum at the same time.
Claiming $E_P$ is zero at equilibrium in all cases (it is minimum there, but not necessarily zero depending on choice of reference).
Forgetting to mention that the object stops briefly at the turning points ( $E_K=0$ ).

Things to Be Careful About

Use the words converted or transferred between $E_K$ and $E_P$ .
If you mention conservation of energy, implicitly assume damping is negligible; otherwise total energy would decrease with time.

Techniques used

link kinetic energy to speed and potential energy to displacementdescribe the energy changes at equilibrium and at turning pointsstate conservation of total energy for undamped oscillations

Q5MediumElectric Fields

(a)

Explain why the electric potential near an isolated proton is positive.

DifficultyMedium-Easy

Worked solution

Answer

Electric potential $V$ at a point is the work done per unit positive charge in bringing a test charge from infinity.

Near a proton, a positive test charge is repelled, so external work must be done against the repulsive force.

Hence the work done per unit charge is positive, so the potential is positive (equivalently $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$ and $Q>0$ ).

Final answer

Electric potential is positive because bringing a positive test charge from infinity requires positive work against repulsion from the proton (Q>0).

Detailed explanation

Background Concept

Electric potential $V$ at a point is defined as the work done per unit positive test charge by an external agent in bringing the test charge from infinity to that point (with no change in kinetic energy).

Mathematically, for a point charge $Q$ ,

V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}

The sign of $V$ matches the sign of $Q$ (because $r>0$ and the constant is positive). Electric potential energy of a charge $q$ at potential $V$ is $U = qV$ .

Understanding the Question

You are asked specifically about the region near an isolated proton (a single positive charge). The task is to justify why the potential there is positive, using the meaning of potential (work done per unit charge) or the standard expression for $V$ .

Approach

Use either (or both) of these equivalent arguments:

Definition argument: decide whether bringing a positive test charge from infinity requires positive or negative external work.
Formula argument: quote $V \propto Q/r$ and use $Q>0$ .

Step-by-Step Reasoning

A proton has charge $+e$ , so it creates an electric field pointing radially outward.
Consider a positive test charge $+q$ brought from infinity towards the proton.
The electric force on $+q$ is repulsive (it acts outward), so to move the charge inward slowly an external force must pull inward.
Because the external force is in the same direction as the displacement (both inward), the external work done is positive.
Potential is work done per unit positive charge, so $V$ is positive.

Equivalently, substituting $Q=+e$ into $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$ shows $V>0$ for any finite $r$ .

Key Takeaways

Electric potential is tied to work done per unit positive charge.
Near a positive source charge, a positive test charge must be pushed against repulsion, so $V$ is positive.
The expression $V \propto Q/r$ is a fast sign check: sign of $V$ equals sign of $Q$ .

Common Mistakes

Saying “the potential is positive because the charge is positive” with no link to definition (often loses explanation marks).
Confusing electric potential $V$ with electric field $E$ (field has direction; potential does not).
Discussing potential energy $U$ instead of potential $V$ (they are related but not the same).

Things to Be Careful About

The definition uses a positive test charge; the sign conclusion depends on that convention.
“Work done by the field” would be negative here; the definition typically refers to work done by an external agent (or minus the work done by the field). State your wording clearly.

Techniques used

use the definition of electric potential as work done per unit chargereason about the sign of work done when moving a positive test charge in an electric fieldrelate the sign of potential to the sign of the source charge using V proportional to Q

(b)

An isolated metal sphere is positively charged and has radius $R$ , as shown in Fig. 5.1.

Fig. 5.1

Line XY passes through the centre of the sphere.
Point P lies on line XY at a variable displacement $x$ from the centre of the sphere.
Point Q is at a fixed position that is not on line XY.

The electric field strength at the surface of the sphere is $E_0$ .

4M

(i)

On Fig. 5.1, draw an arrow at point Q to show the direction of the electric field at that point.

DifficultyEasy

Worked solution

diagram-b-i

Answer

At $Q$ , the electric field is directed radially away from the centre of the positively charged sphere (arrow pointing away from the sphere along the line joining the centre to $Q$ ).

Final answer

Radially outward from the sphere’s centre at Q.

Detailed explanation

Background Concept

The electric field direction at a point is defined as the direction of the force on a positive test charge placed at that point.

For a spherically symmetric charge distribution (such as an isolated charged conducting sphere), the field outside behaves as if all charge were concentrated at the centre, so field lines are radial.

Understanding the Question

A metal sphere is positively charged. Point $Q$ is somewhere outside the sphere but not on the marked diameter line. You must show the direction of $\vec{E}$ at $Q$ by drawing an arrow.

Approach

Use symmetry: outside a charged sphere, $\vec{E}$ points along the radius line through the centre and the observation point. Since the sphere is positively charged, the field points away from the centre.

Step-by-Step Reasoning

Imagine placing a tiny positive test charge at $Q$ .
A positively charged sphere repels a positive test charge.
Therefore the force (and hence $\vec{E}$ ) at $Q$ is away from the sphere.
Because of spherical symmetry, “away from the sphere” means along the line from the centre of the sphere through $Q$ .

So the arrow at $Q$ must point directly away from the centre.

Key Takeaways

$\vec{E}$ direction = force on a positive test charge.
Outside a charged sphere: field is radial.
Positive charge: field points outward.

Common Mistakes

Drawing the arrow tangent to the sphere (that would suggest circular field lines, not electrostatic).
Drawing the arrow “away from the surface” but not aligned with the centre-to- $Q$ line.
Reversing direction (inward) which would correspond to a negative sphere.

Things to Be Careful About

The arrow must be at point $Q$ (not on the surface).
Direction is radial from the centre, not necessarily horizontal/vertical on the page.

Techniques used

use the fact that the electric field direction is the force direction on a positive test chargeuse spherical symmetry to infer a radial field directioninfer field direction around a positively charged conductor

(ii)

On Fig. 5.2, sketch the variation of the electric field $E$ at point P with $x$ for values of $x$ between $x = -3R$ and $x = 3R$ . Do not include the region inside the sphere between $x = -R$ and $x = R$ .

Fig. 5.2

DifficultyMedium

Worked solution

diagram-b-ii

Answer

For $|x|>R$ , the field is as for a point charge at the centre so $E \propto \dfrac{1}{x^2}$ .

At $x=R$ , $E=+E_0$ and decreases towards $0$ as $x$ increases (e.g. at $x=2R$ , $E=+\dfrac{1}{4}E_0$ ; at $x=3R$ , $E=+\dfrac{1}{9}E_0$ ).

At $x=-R$ , $E=-E_0$ and increases towards $0$ from below as $x$ becomes more negative (e.g. at $x=-2R$ , $E=-\dfrac{1}{4}E_0$ ; at $x=-3R$ , $E=-\dfrac{1}{9}E_0$ ).

No graph drawn for $-R < x < R$ (region inside the sphere omitted).

Final answer

Two inverse-square curves: negative from x = −3R to −R reaching −E0 at −R; positive from x = R to 3R starting at +E0 at R and decreasing towards 0.

Detailed explanation

Background Concept

For an isolated charged conducting sphere:

All excess charge resides on the surface.
The electric field inside the conducting material (and throughout the interior cavity, for a symmetrical isolated sphere) is zero in electrostatic equilibrium.
Outside the sphere, the field is the same as that of a point charge $Q$ at the centre:

E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}

The direction is radial. If we choose an $x$ -axis through the centre, the sign of $E$ depends on whether the field points in the $+x$ or $-x$ direction.

Understanding the Question

Point $P$ lies on the diameter line XY at displacement $x$ from the centre. You must sketch $E$ against $x$ from $-3R$ to $3R$ , but you must omit the inside region $-R < x < R$ .

You are told the field at the surface has magnitude $E_0$ . So at $x=R$ , $E=+E_0$ (to the right, outward), and at $x=-R$ , $E=-E_0$ (to the left, outward which is the negative direction).

Approach

Use the “point charge at centre” model for the outside region, giving an inverse-square dependence on distance from the centre.
Put the correct signs:

For $x>R$ the field points to the right (positive).
For $x<-R$ the field points to the left (negative).

Anchor the sketch using the given reference value $E_0$ at $x=\pm R$ , then show the curve approaching $0$ as $|x|$ increases.

Step-by-Step Reasoning

Outside the sphere, distance from centre is $r=|x|$ along the line XY.
Since $E \propto 1/r^2$ , you can write relative to the surface value:

\frac{E}{E_0} = \left(\frac{R}{|x|}\right)^2

Now include direction (sign). For a positive sphere, the field is always outward:
- At positions on the right ( $x>0$ ), outward is toward $+x$ , so $E$ is positive.
- At positions on the left ( $x<0$ ), outward is toward $-x$ , so $E$ is negative.

So a compact signed form for the sketch is:

E(x)=\begin{cases} +E_0\left(\frac{R}{x}\right)^2, & x>R \\ -E_0\left(\frac{R}{x}\right)^2, & x<-R \end{cases}

(Here the second line works because $x^2$ is positive but we explicitly place the minus sign.)

Use key points to shape the curve:
- At $x=R$ , $E=+E_0$ .
- At $x=2R$ , $E=+E_0/4$ .
- At $x=3R$ , $E=+E_0/9$ .
- At $x=-R$ , $E=-E_0$ .
- At $x=-2R$ , $E=-E_0/4$ .
- At $x=-3R$ , $E=-E_0/9$ .
Since the question says “Do not include the region inside the sphere”, you leave a gap between $x=-R$ and $x=R$ (no line drawn there).

diagram-b-ii

Key Takeaways

Outside a charged sphere, $E \propto 1/r^2$ like a point charge at the centre.
The sign of $E$ on an axis comes from direction relative to the chosen positive axis.
Electrostatic conductors have no electric field inside (but here you are told to omit that region anyway).

Common Mistakes

Drawing $E$ as constant outside the sphere (confusing with a uniform field between plates).
Putting the same sign on both sides (forgetting that outward on the left is the negative direction).
Making the graph linear in $x$ instead of inverse-square curvature.
Drawing a line through the interior region when instructed not to.

Things to Be Careful About

At $x=\pm R$ the magnitude is exactly $E_0$ (your curve should meet those axis markings).
As $|x|$ increases, $E$ must approach $0$ (never increase in magnitude).
The curve should be smooth and asymptotic toward $E=0$ for large $|x|$ within the plotted range.

Techniques used

model the field outside a charged conducting sphere as that of a point charge at its centreuse inverse-square dependence to sketch magnitude variation with distanceuse sign convention along an axis to assign positive and negative field directions

(c)

The proton and the electron in a hydrogen atom are separated by a distance of $5.3 \times 10^{-11}\ \text{m}$ .

Calculate the electric potential energy of the proton and the electron.

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

DifficultyMedium-Easy

Worked solution

Working

U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}

U = (8.99 \times 10^9)\frac{(+1.60\times 10^{-19})(-1.60\times 10^{-19})}{5.3\times 10^{-11}}

U = -4.3 \times 10^{-18}\ \text{J}

Answer

$-4.3 \times 10^{-18}\ \text{J}$

Final answer

-4.3 × 10^-18 J

Detailed explanation

Background Concept

The electric potential energy $U$ of two point charges separated by distance $r$ (taking $U=0$ at infinite separation) is

U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}

If $q_1 q_2 > 0$ (like charges), then $U$ is positive (you must do work to bring them together).
If $q_1 q_2 < 0$ (unlike charges), then $U$ is negative (energy is released when they come together).

Understanding the Question

A hydrogen atom has a proton ( $+e$ ) and an electron ( $-e$ ) separated by $r = 5.3 \times 10^{-11}\ \text{m}$ . You must calculate their electric potential energy using the point-charge formula and give the answer in joules.

Approach

Use $U = k\,q_1 q_2/r$ with $k = 1/(4\pi\varepsilon_0) = 8.99 \times 10^9\ \text{N m}^2\text{C}^{-2}$ .
Substitute $q_1 = +e$ and $q_2 = -e$ , so the product is negative.
Compute the magnitude and attach the negative sign.

Step-by-Step Reasoning

Identify charges:
- Proton: $q_1 = +1.60 \times 10^{-19}\ \text{C}$
- Electron: $q_2 = -1.60 \times 10^{-19}\ \text{C}$
Separation: $r = 5.3 \times 10^{-11}\ \text{m}$
Use the formula:

U = (8.99 \times 10^9)\frac{(+1.60\times 10^{-19})(-1.60\times 10^{-19})}{5.3\times 10^{-11}}

Multiply charges:

q_1 q_2 = -(1.60)^2 \times 10^{-38} = -2.56 \times 10^{-38}

Multiply by $k$ :

(8.99 \times 10^9)(2.56 \times 10^{-38}) = 23.0 \times 10^{-29} = 2.30 \times 10^{-28}

Divide by $r$ :

|U| = \frac{2.30 \times 10^{-28}}{5.3 \times 10^{-11}} = 0.434 \times 10^{-17} = 4.34 \times 10^{-18}

Put the sign back (opposite charges):

U = -4.34 \times 10^{-18}\ \text{J} \approx -4.3 \times 10^{-18}\ \text{J}

Key Takeaways

Use $U = k q_1 q_2 / r$ for two point charges.
Opposite charges give negative potential energy.
Always check powers of ten carefully when dealing with atomic distances.

Common Mistakes

Forgetting the negative sign (very common).
Using $r^2$ instead of $r$ (that would be for force/field, not potential energy).
Mixing units (distance must be in metres, charge in coulombs).

Things to Be Careful About

Give the final answer to a sensible number of significant figures (limited by $5.3 \times 10^{-11}$ , typically 2 s.f.).
Do not confuse electric potential energy (J) with electric potential (V).

Techniques used

use the Coulomb potential energy equation for two point chargessubstitute charge values with correct signcalculate and express the result in standard form with appropriate significant figures

Q6MediumAlternating CurrentsCapacitance

Fig. 6.1 shows part of a bridge rectifier circuit that can be used for rectification of an alternating input voltage $V_{IN}$ .

Fig. 6.1

The circuit contains four diodes, one of which is shown.
The rectified output voltage $V_{OUT}$ is applied across load resistor R.

(a)

4M

(i)

State what is meant by rectification.

(ii)

State the name of the type of rectification produced by a bridge rectifier circuit.

(iii)

Complete the circuit in Fig. 6.1 by drawing the three missing diodes inside the dashed circles.

DifficultyMedium-Easy

Worked solution

diagram-6a-iii

Answer

Draw the three missing diodes so that the two diodes connected to the positive output node have their cathodes at the positive node, and the two diodes connected to the negative output node have their anodes at the negative node (bridge rectifier).

Final answer

Three diodes added to form a standard bridge: cathodes meet at +VOUT, anodes meet at −VOUT.

Detailed explanation

Background Concept

A diode conducts (ideally) only when it is forward biased (anode at a higher potential than cathode). A bridge rectifier uses four diodes so that:

when one input terminal is positive, one pair of diodes conducts and drives current through the load in one direction;
when the input reverses, the other pair conducts, but the load current direction stays the same.

A good way to remember the bridge arrangement:

At the positive output node, the cathodes of two diodes meet (so current can flow into the + node from either a.c. terminal).
At the negative output node, the anodes of two diodes meet (so current can flow out of the − node to either a.c. terminal).

Understanding the Question

The diagram shows the bridge layout with one diode already drawn and three dashed circles where the other diodes must be added. The load resistor (R) is across (V_{OUT}). You must orient the remaining diodes so that (V_{OUT}) has the same polarity for both halves of (V_{IN}).

Approach

Think about the two half-cycles:

Left input terminal positive relative to right.
Right input terminal positive relative to left.

In each case, choose a path through two forward-biased diodes and (R) that makes the same end of (R) positive.

Step-by-Step Reasoning

diagram-6a-iii

Label the bridge nodes: left and right are the a.c. input terminals; top is (+V_{OUT}) and bottom is (-V_{OUT}).
Ensure there is a diode from each a.c. terminal to the + output node that conducts when that terminal is positive (anode at the a.c. terminal, cathode at + output).
Ensure there is a diode from the − output node to each a.c. terminal that conducts to complete the circuit (anode at − output, cathode at the a.c. terminal).

This guarantees current through the load is always from + output to − output.

Key Takeaways

Bridge rectifier: two diodes conduct each half-cycle.
Cathodes join at the + output; anodes join at the − output.

Common Mistakes

Reversing one diode so that one half-cycle produces a negative output.
Drawing a diode so that both halves are blocked (no conduction path).

Things to Be Careful About

Diode symbol direction: conventional current flows from anode to cathode when forward biased.
Make sure the output is taken across the correct diagonal of the bridge (across (R)).

Techniques used

apply the conduction direction of a diode to choose diode orientationuse symmetry of the bridge rectifier to place the remaining diodesensure load current direction is the same for both half-cycles

(b)

The input voltage varies with time $t$ according to the equation

V_{IN} = 34 \sin 18t

where $V_{IN}$ is in V and $t$ is in s.

6M

(i)

Show that the period of the input voltage is $0.35\ \text{s}$ .

DifficultyMedium-Easy

Worked solution

Working

Given

V_{IN} = 34\sin(18t)

so angular frequency

\omega = 18\ \text{rad s}^{-1}

T = \frac{2\pi}{\omega} = \frac{2\pi}{18} = 0.349\ \text{s} \approx 0.35\ \text{s}

Answer

$T = 0.35\ \text{s}$

Final answer

0.35 s

Detailed explanation

Background Concept

A sinusoidal voltage can be written as

V = V_0\sin(\omega t)

where:

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

The period (T) is related to (\omega) by

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

Understanding the Question

You are given (V_{IN} = 34\sin(18t)). The coefficient of (t) inside the sine function is (\omega). You must calculate the time for one complete cycle and show it is (0.35\ \text{s}).

Approach

Read (\omega) directly from the expression.
Use (T = 2\pi/\omega).
Round to (0.35\ \text{s}) as required.

Step-by-Step Reasoning

Compare (34\sin(18t)) with (V_0\sin(\omega t)). Therefore (\omega = 18\ \text{rad s}^{-1}).
Substitute into the period formula:

T = \frac{2\pi}{18}

Calculate:

T = 0.349\ \text{s}

Rounding to two decimal places gives (0.35\ \text{s}), matching the value to be shown.

Key Takeaways

In (\sin(\omega t)), the coefficient of (t) is (\omega).
Convert between (\omega) and (T) using (\omega = 2\pi/T).

Common Mistakes

Treating (18) as the frequency (f) in hertz (it is angular frequency).
Using (T = 1/18) (wrong because (18 \neq f)).

Things to Be Careful About

(\omega) is in (\text{rad s}^{-1}), not (\text{Hz}).
Keep enough significant figures during calculation before rounding.

Techniques used

identify angular frequency from a sinusoidal equationuse the relation between angular frequency and periodevaluate a numerical result to appropriate significant figures

(ii)

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

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

(iii)

On Fig. 6.2, sketch the variation of $V_{OUT}$ with $t$ from $t = 0$ to $t = 0.35\ \text{s}$ .

Fig. 6.2

DifficultyMedium

Worked solution

diagram-6b-iii

Answer

Sketch (V_{OUT}) as a full-wave rectified sine wave (all positive), with peak (34\ \text{V}); zeros at (t=0), (0.175\ \text{s}) and (0.35\ \text{s}); peaks at (t=0.0875\ \text{s}) and (0.2625\ \text{s}).

Final answer

Full-wave rectified sine: VOUT = |34 sin(18t)| from 0 to 0.35 s.

Detailed explanation

Background Concept

A bridge rectifier produces full-wave rectification, meaning the output voltage across the load is the magnitude of the input (ideal diodes assumed):

V_{OUT}(t) = \left|V_{IN}(t)\right|

So the negative half-cycles are inverted to become positive. This also means the output waveform repeats every half input period.

Understanding the Question

You are given (V_{IN} = 34\sin(18t)) and you must sketch (V_{OUT}) from (t=0) to (t=0.35\ \text{s}) (one full input cycle). Because the circuit is a bridge rectifier, the output is always (\ge 0) (ignoring diode drops).

Approach

Identify key times in one cycle of the sine wave: zeros at (0), (T/2), (T); maxima at (T/4) and (3T/4).
Apply full-wave rectification: reflect the negative half-cycle above the time axis.
Keep the amplitude at (34\ \text{V}) and ensure the output is never negative.

Step-by-Step Reasoning

From part (b)(i), (T = 0.35\ \text{s}).
For (\sin) wave:
- Zero crossings occur at (t = 0), (T/2), (T):

0,\ 0.175\ \text{s},\ 0.35\ \text{s}

Positive peak occurs at (t = T/4 = 0.0875\ \text{s}) with value (+34\ \text{V}).
Negative peak occurs at (t = 3T/4 = 0.2625\ \text{s}) with value (-34\ \text{V}).
Full-wave rectification makes that negative peak into (+34\ \text{V}).

So the output consists of two identical positive “humps” between 0 and 0.35 s.

diagram-6b-iii

Key Takeaways

Bridge rectifier output is the absolute value of the input (ideal).
Output frequency is doubled (period halved).

Common Mistakes

Sketching half-wave rectification (output zero for half the time).
Keeping the negative half-cycle negative.
Using the wrong peak value or wrong timing of peaks.

Things to Be Careful About

The sketch is only from 0 to 0.35 s, so you should show exactly two peaks.
If diode drops were considered, peaks would be slightly less than 34 V, but unless stated, assume ideal diodes.

Techniques used

apply full-wave rectification as an absolute-value transformationuse key times in a sinusoid to locate zeros and maximasketch a waveform with correct period and amplitude

(c)

Resistor R has a resistance of $56\ \text{k}\Omega$ . A capacitor of capacitance $12\ \mu\text{F}$ is connected into the circuit of Fig. 6.1 in order to smooth the output voltage.

5M

(i)

On Fig. 6.1, draw the capacitor correctly connected into the circuit.

(ii)

Calculate the time constant of the smoothing circuit.

time constant = ______ $\text{s}$

DifficultyMedium-Easy

Worked solution

Working

\tau = RC

R = 56\ \text{k}\Omega = 5.6\times 10^{4}\ \Omega,\qquad C = 12\ \mu\text{F} = 1.2\times 10^{-5}\ \text{F}

\tau = (5.6\times 10^{4})(1.2\times 10^{-5}) = 0.672\ \text{s} \approx 0.67\ \text{s}

Answer

(0.67\ \text{s})

Final answer

0.67 s

Detailed explanation

Background Concept

For a capacitor discharging through a resistor, the time constant is

\tau = RC

The time constant sets how quickly the capacitor voltage changes: after a time (\tau), the voltage has fallen to (1/e) of its initial value during discharge.

Understanding the Question

You are given (R = 56\ \text{k}\Omega) and (C = 12\ \mu\text{F}). The time constant of the smoothing circuit is the product (RC), in seconds.

Approach

Convert (\text{k}\Omega) to (\Omega) and (\mu\text{F}) to (\text{F}).
Multiply (R) and (C).

Step-by-Step Reasoning

Convert units:

56\ \text{k}\Omega = 56\times 10^{3}\ \Omega = 5.6\times 10^{4}\ \Omega

12\ \mu\text{F} = 12\times 10^{-6}\ \text{F} = 1.2\times 10^{-5}\ \text{F}

Multiply:

\tau = RC = (5.6\times 10^{4})(1.2\times 10^{-5}) = 6.72\times 10^{-1}\ \text{s} = 0.672\ \text{s}

Round appropriately: (\tau \approx 0.67\ \text{s}).

Key Takeaways

(\tau = RC) and must be in seconds.
Always convert prefixes (k, (\mu)) before multiplying.

Common Mistakes

Using (56) instead of (56\times 10^3).
Using (12) instead of (12\times 10^{-6}).
Giving (\tau) in the wrong unit.

Things to Be Careful About

Check powers of ten: (10^{4}\times 10^{-5} = 10^{-1}).
Give a sensible number of significant figures (usually 2–3).

Techniques used

convert prefixed units to base unitsapply the time constant relation tau = RCmultiply and round to appropriate significant figures

(iii)

During each discharge cycle, the time for which the capacitor is discharging is $0.14\ \text{s}$ .

Determine the minimum value of the smoothed output voltage.

minimum voltage = ______ $\text{V}$

DifficultyMedium

Worked solution

Working

Assume ideal diodes so capacitor charges to peak

V_{\max} = 34\ \text{V}

During discharge,

V = V_{\max} e^{-t/\tau}

with (t = 0.14\ \text{s}) and (\tau = 0.672\ \text{s}):

V_{\min} = 34\, e^{-0.14/0.672} = 27.6\ \text{V} \approx 28\ \text{V}

Answer

(28\ \text{V})

Final answer

28 V

Detailed explanation

Background Concept

In a smoothed rectifier output, the capacitor:

charges rapidly to near the peak output voltage when the rectified supply rises above the capacitor voltage;
discharges through the load resistor between peaks.

For a discharge through a resistor, the capacitor voltage falls exponentially:

V(t) = V_0 e^{-t/RC} = V_0 e^{-t/\tau}

where (V_0) is the voltage at the start of discharge and (\tau = RC).

Understanding the Question

You are told that for each discharge cycle, the capacitor is discharging for (0.14\ \text{s}). You already have (R = 56\ \text{k}\Omega) and (C = 12\ \mu\text{F}), so (\tau) can be found (or taken from part (c)(ii)). The minimum smoothed output voltage is the capacitor voltage at the end of the discharge interval.

Approach

Take the maximum voltage (V_{\max}) to be the peak of the rectified waveform (ideal bridge: same as input peak).
Use exponential discharge over time (t = 0.14\ \text{s}) with time constant (\tau).

Step-by-Step Reasoning

Peak of input is (34\ \text{V}), so with ideal diodes, the capacitor charges to

V_{\max} = 34\ \text{V}

Time constant (from part (c)(ii)):

\tau = RC = 0.672\ \text{s}

After discharge time (t = 0.14\ \text{s}), the voltage is

V_{\min} = V_{\max} e^{-t/\tau} = 34 e^{-0.14/0.672}

Evaluate the exponent:

\frac{0.14}{0.672} = 0.208\quad\Rightarrow\quad e^{-0.208} \approx 0.812

Hence

V_{\min} \approx 34\times 0.812 = 27.6\ \text{V} \approx 28\ \text{V}

Key Takeaways

Smoothing is analysed as capacitor charge to near peak, then exponential discharge.
Use (V = V_0 e^{-t/RC}) to find the minimum at the end of the discharge interval.

Common Mistakes

Using a linear drop instead of exponential.
Using (t/\tau) the wrong way up (writing (e^{-\tau/t})).
Forgetting to use the peak voltage as the starting voltage for discharge.

Things to Be Careful About

This question typically assumes ideal diodes (no (0.7\ \text{V}) drop) unless stated.
Keep consistent units (seconds for (t) and (\tau)).
Quote the minimum voltage to a sensible number of significant figures.

Techniques used

model the capacitor discharge using an exponential decay equationsubstitute the discharge time and time constant into the decay expressioncalculate a minimum value after a decay interval

Q7MediumMagnetic FieldsMotion in a Circle

(a)

State Lenz’s law of electromagnetic induction.

DifficultyEasy

Worked solution

Answer

Lenz’s law: the induced e.m.f. (and hence induced current) is in a direction such that the magnetic effect produced opposes the change in magnetic flux (flux linkage) that causes it.

Final answer

The induced e.m.f./current acts so that its magnetic effect opposes the change in magnetic flux (flux linkage) producing it.

Detailed explanation

Background Concept

Electromagnetic induction occurs when the magnetic flux (or flux linkage) through a circuit changes. Faraday’s law relates the size of the induced e.m.f. to the rate of change of flux linkage:

\mathcal{E} = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

The negative sign is not “just a minus sign”: it encodes Lenz’s law, which gives the direction of the induced e.m.f./current.

Understanding the Question

You are asked to state Lenz’s law. For 2 marks, Cambridge typically expects both:

the induced e.m.f./current produces a magnetic effect, and
that magnetic effect opposes the change (in flux/flux linkage) that produced it.

Approach

Write a single clear sentence including:

“induced e.m.f./current is in such a direction that …”
“… it opposes the change in magnetic flux/flux linkage causing it.”

Step-by-Step Reasoning

When the flux through a circuit increases, the induced current must create a magnetic field that tends to reduce that increase.
When the flux decreases, the induced current must create a magnetic field that tends to increase it back.
This “opposition to the change” is the key idea.

Key Takeaways

Lenz’s law is about direction.
The induced current always acts to oppose the change in flux, not necessarily oppose the flux itself.

Common Mistakes

Saying “opposes the magnetic field” (it opposes the change in flux).
Omitting mention of induced current/e.m.f. direction.

Things to Be Careful About

Use the phrase “opposes the change” explicitly.
“Flux” or “flux linkage” are both acceptable provided the meaning is correct.

Techniques used

state Lenz’s law in terms of opposition to change in magnetic fluxrelate induced e.m.f. direction to the magnetic effect of the induced current

(b)

A helicopter hovering in stationary equilibrium has four rotors, each of length $12\ \text{m}$ , as shown in the view from above in Fig. 7.1.

Fig. 7.1

The vertical component of the Earth’s magnetic field at the helicopter is downwards with a flux density of $0.047\ \text{mT}$ .

The rotors each rotate in a horizontal plane in the direction shown with a frequency of $85\ \text{Hz}$ .

7M

(i)

Calculate the magnetic flux $\Phi$ cut by rotor OX during one complete rotation. Give a unit with your answer.

$\Phi$ = ______ unit ______

DifficultyMedium-Easy

Worked solution

Working

$B = 0.047\ \text{mT} = 4.7 \times 10^{-5}\ \text{T}$

Area swept in one rotation:

A = \pi r^2 = \pi (12)^2 = 452\ \text{m}^2

\Phi = BA = (4.7 \times 10^{-5})(452) = 2.13 \times 10^{-2}\ \text{Wb}

Answer

$\Phi = 2.1 \times 10^{-2}\ \text{Wb}$

Final answer

2.1 × 10^-2 Wb

Detailed explanation

Background Concept

Magnetic flux through an area is

\Phi = BA\cos\theta

where:

$B$ is the magnetic flux density (in $\text{T}$ ),
$A$ is the area (in $\text{m}^2$ ),
$\theta$ is the angle between the field direction and the normal to the area.

If the field is perpendicular to the area (i.e. parallel to the normal), then $\cos\theta = 1$ and $\Phi = BA$ .

Understanding the Question

The rotor OX is a radius of length $12\ \text{m}$ rotating in a horizontal plane. The Earth’s magnetic field component given is vertical downward, so it is perpendicular to the rotor plane.

During one complete revolution, the radius sweeps out a full circle of radius $r=12\ \text{m}$ . The question asks for the total flux cut in one rotation, which corresponds to the flux through that swept circular area.

Approach

Convert $B$ from mT to T.
Find the area of the circle swept out: $A = \pi r^2$ .
Use $\Phi = BA$ (since the field is normal to the area).

Step-by-Step Reasoning

Convert units:

0.047\ \text{mT} = 0.047 \times 10^{-3}\ \text{T} = 4.7 \times 10^{-5}\ \text{T}

Swept area in one full rotation:

A = \pi r^2 = \pi (12\ \text{m})^2 = \pi \times 144 = 452\ \text{m}^2

Flux:

\Phi = BA = (4.7 \times 10^{-5})(452) = 2.13 \times 10^{-2}\ \text{Wb}

The unit is the weber (Wb).

Key Takeaways

For a uniform field perpendicular to a surface, flux is simply $\Phi = BA$ .
A rotating radius sweeps an area $\pi r^2$ in one revolution.

Common Mistakes

Forgetting to convert mT to T (a factor of $10^3$ error).
Using $A = 2\pi r$ (circumference) instead of $A = \pi r^2$ (area).
Including a $\cos\theta$ factor incorrectly (here $\theta=0$ ).

Things to Be Careful About

Always state the unit of flux: $\text{Wb}$ .
Use a sensible number of significant figures (here 2 s.f. matches the given $B$ ).

Techniques used

convert magnetic flux density to teslacalculate area swept out by a rotating radiususe $\Phi = BA$ for a uniform field normal to the area

(ii)

Determine the magnitude of the electromotive force (e.m.f.) induced across the length of rotor OX.

e.m.f. = ______ $\text{V}$

DifficultyMedium-Easy

Worked solution

Working

Time for one rotation:

T = \frac{1}{f} = \frac{1}{85}\ \text{s}

\mathcal{E} = \frac{\Delta\Phi}{\Delta t} = \frac{\Phi}{T} = \Phi f

\mathcal{E} = (2.13 \times 10^{-2})(85) = 1.81\ \text{V}

Answer

$\mathcal{E} = 1.8\ \text{V}$

Final answer

1.8 V

Detailed explanation

Background Concept

Faraday’s law gives the magnitude of induced e.m.f. as the rate of change of flux linkage. For a single conductor sweeping out area so that the flux changes by $\Delta\Phi$ in time $\Delta t$ :

\mathcal{E} = \left|\frac{\Delta\Phi}{\Delta t}\right|

For steady rotation, the flux change per revolution is constant, and the time per revolution is the period $T = 1/f$ .

Understanding the Question

You already found the flux cut in one complete rotation (part (i)). The rotor completes $f=85$ rotations per second, so the flux cut per second is $\Phi f$ . That equals the induced e.m.f. magnitude.

Approach

Use either of these equivalent routes:

$\mathcal{E} = \Phi/T$ with $T=1/f$ .
Directly $\mathcal{E} = \Phi f$ .

Step-by-Step Reasoning

Period of rotation:

T = \frac{1}{f} = \frac{1}{85}\ \text{s}

Use Faraday’s law with one-rotation flux change $\Phi$ :

\mathcal{E} = \frac{\Phi}{T} = \Phi f

Substitute:

\mathcal{E} = (2.13 \times 10^{-2}\ \text{Wb})(85\ \text{s}^{-1}) = 1.81\ \text{V}

So the magnitude is about $1.8\ \text{V}$ .

(You may also see the rotating-rod formula $\mathcal{E} = \tfrac{1}{2}B\omega r^2$ ; it is consistent because $\omega = 2\pi f$ and $\Phi = B\pi r^2$ .)

Key Takeaways

Induced e.m.f. is a rate: flux change per unit time.
If you know flux change per revolution and revolutions per second, multiply them.

Common Mistakes

Using $T=f$ instead of $T=1/f$ .
Forgetting that $85\ \text{Hz} = 85\ \text{s}^{-1}$ .
Carrying forward the wrong flux from part (i) due to mT-to-T conversion.

Things to Be Careful About

The question asks for magnitude, so no minus sign is needed.
Check units: $\text{Wb s}^{-1} = \text{V}$ .

Techniques used

use $\mathcal{E} = \Delta\Phi/\Delta t$ with one rotation periodcalculate period from frequencycombine steps to obtain induced e.m.f. magnitude

(iii)

Use Lenz’s law to explain whether end O or end X of the rotor is at the higher potential.

DifficultyMedium

Worked solution

Answer

The induced current must produce a magnetic force that opposes the rotation (Lenz’s law). With $\vec{B}$ downward and rotation clockwise (viewed from above), this requires conventional current along the blade from $O \to X$ so that $I\vec{L} \times \vec{B}$ gives a force opposing the motion.

Hence positive charge accumulates at $X$ , so end $X$ is at the higher potential.

Final answer

End X is at the higher potential.

Detailed explanation

Background Concept

Lenz’s law says the induced current flows in the direction that opposes the change that produces it. In a moving conductor, the induced e.m.f. can be understood via:

charges in the conductor moving with the metal and experiencing a magnetic force $q\vec{v} \times \vec{B}$ , causing charge separation, and/or
the induced current producing a magnetic force on the conductor $\vec{F} = I\vec{L} \times \vec{B}$ that tends to oppose the motion (a resisting torque).

Both descriptions give the same polarity.

Understanding the Question

The rotor blade is a conductor rotating in a horizontal plane.
The Earth’s field component is vertical downwards (into the rotor plane when viewed from above).
The rotor rotates clockwise when viewed from above.

You must decide which end becomes more positive (higher potential): the centre $O$ or the tip $X$ .

Approach

Use Lenz’s law in its “opposes the motion” form:

The induced e.m.f. would drive a current (if a circuit exists).
That current in the downward magnetic field experiences a magnetic force.
For Lenz’s law, that force must oppose the rotation (provide a braking torque).
From the required current direction, infer which end is at higher potential.

A helpful visual is to look at one instant when the blade is horizontal to the right.

diagram-7biii-1

Step-by-Step Reasoning

Take the instant when the blade points to the right from the centre (so $X$ is to the right of $O$ ).

Direction of motion: since the blade rotates clockwise, the velocity of the tip at that instant is down the page (tangential).
The magnetic field is downward (into the page).
Suppose conventional current flowed outward ( $O \to X$ ) along the blade (to the right). The force on a current-carrying conductor is

\vec{F} = I\vec{L} \times \vec{B}

Here $\vec{L}$ is to the right and $\vec{B}$ is into the page. Using the right-hand rule, $\vec{L} \times \vec{B}$ points up the page.

At this instant the blade’s motion is down the page, so a force up the page opposes the motion (braking). This is exactly what Lenz’s law requires.

So the induced e.m.f. must drive conventional current from $O$ to $X$ along the blade.

If conventional current is driven from $O$ to $X$ , positive charge is driven toward and accumulates at the tip $X$ . Therefore $X$ is at a higher electric potential than $O$ .

Key Takeaways

Lenz’s law often appears as “the induced effects oppose the motion/change producing them”.
Determine polarity by demanding a braking magnetic force/torque on the moving conductor.

Common Mistakes

Reversing clockwise/anticlockwise when viewed from above.
Treating “opposes the flux” rather than “opposes the change / opposes the motion”.
Stating a current direction without linking it to a force that opposes motion (missing the Lenz’s law reasoning).

Things to Be Careful About

Be consistent with the viewing direction: the question specifies the view from above.
The Earth’s field is downwards, i.e. into the page in that view.
The question asks for higher potential, so you must conclude which end is positive (not just current direction).

Techniques used

apply Lenz’s law to determine the direction of the induced current effectuse the motor-effect rule $\vec{F} = I\vec{L} \times \vec{B}$ to link current direction to a resisting torquededuce charge separation and potential difference direction along the rotor

Q8MediumMedical PhysicsNuclear PhysicsIdeal Gases

Oxygen-15 ( $^{15}_{8}\text{O}$ ) is radioactive and has a half-life of $2.04$ minutes.

The decay of oxygen-15 produces positrons. For this reason, oxygen-15 is sometimes used as a tracer in positron emission tomography (PET scanning).

(a)

State what is meant by a tracer.

DifficultyEasy

Worked solution

Answer

A tracer is a (small amount of) radioactive substance introduced into the body/system so that its movement or location can be followed by detecting the radiation it emits outside the body.

Final answer

A tracer is a radioactive substance introduced into the body/system so its location/movement can be followed by detecting the radiation it emits.

Detailed explanation

Background Concept

In nuclear medicine, a tracer is a substance used to follow what happens inside the body without opening it up. The key idea is that the substance emits radiation that can escape the body and be detected externally.

Understanding the Question

You are asked for what “tracer” means in the context of PET scanning, where a radioactive isotope is administered and its emissions are used to locate where it is in the body.

Approach

Give two clear points that typically earn the marks:

it is a (small quantity of) radioactive material introduced into a system/body,
its distribution/movement is determined by detecting the radiation it emits.

Step-by-Step Reasoning

A tracer must be detectable from outside, so it must emit radiation.
In medical use it is given to the patient (injected/inhale/ingest) in small quantity so it does not significantly change the process being studied.
The emitted radiation is detected outside the body to infer where the tracer is, so we can track flow or uptake.

Key Takeaways

A tracer is defined by radioactivity and detectability.
It is used to determine location/movement inside a system.

Common Mistakes

Saying only “a radioactive substance” (misses the purpose of tracking via detection).
Saying only “a dye used to trace” without mentioning radiation and external detection.

Things to Be Careful About

Mention introduced into the body/system and detected externally to secure both marks.
Avoid vague statements like “used to see organs” without stating how (radiation detection).

Techniques used

recall the definition of a radioactive tracerlink detectability of emitted radiation to tracking within the body

(b)

The equation for the decay of oxygen-15 is

^{15}_{8}\text{O} \rightarrow ^{P}_{Q}\text{X} + ^{R}_{S}\beta^+ + Z

where X is the nucleus formed during the decay and Z is another particle.

3M

(i)

State the values of the integers $P$ , $Q$ , $R$ and $S$ .

$P$ = ______ $R$ = ______
$Q$ = ______ $S$ = ______

DifficultyMedium-Easy

Worked solution

Working

Conservation of nucleon number:

15 = P + R

For a positron, $R = 0$ so $P = 15$ .

Conservation of proton number:

8 = Q + S

For $\beta^+$ , $S = +1$ so $Q = 7$ .

Answer

$P = 15$ , $Q = 7$ , $R = 0$ , $S = +1$ .

Final answer

P = 15, Q = 7, R = 0, S = +1

Detailed explanation

Background Concept

A nuclear decay equation must conserve:

nucleon (mass) number $A$ (total number of protons + neutrons),
proton (atomic) number $Z$ (number of protons).

For beta-plus decay ( $\beta^+$ emission), a proton changes into a neutron, so the nucleus keeps the same nucleon number but its proton number decreases by 1.

A positron has:

^{0}_{+1}\beta^+

So it contributes $A=0$ and $Z=+1$ .

Understanding the Question

You are given:

^{15}_{8}\text{O} \rightarrow ^{P}_{Q}\text{X} + ^{R}_{S}\beta^+ + Z

and asked to state the integers $P, Q, R, S$ .

Approach

Use conservation of nucleon number to relate $15$ to $P$ and $R$ .
Use conservation of proton number to relate $8$ to $Q$ and $S$ .
Use the known identity of $\beta^+$ to set $R$ and $S$ .

Step-by-Step Reasoning

For a positron, $R = 0$ and $S = +1$ .
Nucleon number conservation:

15 = P + 0 \Rightarrow P = 15.

Proton number conservation:

8 = Q + (+1) \Rightarrow Q = 7.

So the daughter nucleus has $A=15$ and $Z=7$ (nitrogen-15).

Key Takeaways

In $\beta^+$ decay: $A$ unchanged, $Z$ decreases by 1.
A positron is $^{0}_{+1}\beta^+$ .

Common Mistakes

Using $S=-1$ (that would be $\beta^-$ / electron emission).
Changing $P$ to $14$ (nucleon number must be conserved, so it stays $15$ ).

Things to Be Careful About

The symbol $\beta^+$ is the positron.
Always check both conserved numbers independently to avoid sign errors.

Techniques used

apply conservation of nucleon numberapply conservation of proton (atomic) numberidentify a positron as a beta-plus particle

(ii)

State the name of particle Z.

(c)

7M

(i)

Define the activity of a sample.

(ii)

Calculate the decay constant of oxygen-15. Give a unit with your answer.

decay constant = ______ unit ______

DifficultyMedium-Easy

Worked solution

Working

t_{1/2} = 2.04\ \text{min} = 2.04 \times 60 = 122.4\ \text{s}

\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{122.4} = 5.66 \times 10^{-3}\ \text{s}^{-1}

Answer

$\lambda = 5.66 \times 10^{-3}\ \text{s}^{-1}$ .

Final answer

5.66 × 10^-3 s^-1

Detailed explanation

Background Concept

The decay constant $\lambda$ is the probability per unit time that a nucleus will decay.

Half-life $t_{1/2}$ is related to $\lambda$ by:

\lambda = \frac{\ln 2}{t_{1/2}}.

This comes from the exponential decay law $N = N_0 e^{-\lambda t}$ .

Understanding the Question

You are given the half-life of oxygen-15 as $2.04\ \text{min}$ and asked to calculate $\lambda$ with a unit.

Approach

Convert the half-life to seconds (so the unit of $\lambda$ becomes $\text{s}^{-1}$ ).
Substitute into $\lambda = \ln 2 / t_{1/2}$ .

Step-by-Step Reasoning

Convert minutes to seconds:

2.04\ \text{min} = 2.04 \times 60 = 122.4\ \text{s}.

Apply the formula:

\lambda = \frac{0.693}{122.4} = 5.66 \times 10^{-3}.

Unit: since we used seconds, $\lambda$ is in $\text{s}^{-1}$ .

Key Takeaways

Always convert time to SI units before calculating $\lambda$ .
$\lambda$ has dimension $\text{time}^{-1}$ .

Common Mistakes

Leaving half-life in minutes but writing unit $\text{s}^{-1}$ .
Using $\lambda = 1/t_{1/2}$ (missing the $\ln 2$ factor).

Things to Be Careful About

Keep sufficient significant figures: half-life is 3 s.f., so $\lambda$ to 3 s.f. is appropriate.
State the unit explicitly ( $\text{s}^{-1}$ or Bq per nucleus is not appropriate here).

Techniques used

convert half-life into secondsuse the relation \lambda = \ln 2 / t_{1/2}quote decay constant with correct unit

(iii)

Determine the rate at which positrons are produced in a sample of oxygen-15 that has a mass of $2.85 \times 10^{-6}\ \text{kg}$ .

rate = ______ $\text{s}^{-1}$

4M Similar

DifficultyMedium

Worked solution

Working

Molar mass of $^{15}\text{O}$ : $M = 15\ \text{g mol}^{-1} = 1.5 \times 10^{-2}\ \text{kg mol}^{-1}$ .

n = \frac{m}{M} = \frac{2.85 \times 10^{-6}}{1.5 \times 10^{-2}} = 1.90 \times 10^{-4}\ \text{mol}

N = n N_A = (1.90 \times 10^{-4})(6.02 \times 10^{23}) = 1.14 \times 10^{20}

Using $\lambda = 5.66 \times 10^{-3}\ \text{s}^{-1}$ ,

\text{rate} = A = \lambda N = (5.66 \times 10^{-3})(1.14 \times 10^{20}) = 6.46 \times 10^{17}\ \text{s}^{-1}

Answer

$6.46 \times 10^{17}\ \text{s}^{-1}$ .

Final answer

6.46 × 10^17 s^-1

Detailed explanation

Background Concept

For a radioactive sample:

The activity $A$ (decays per second) is

A = \lambda N

where $\lambda$ is the decay constant and $N$ is the number of undecayed nuclei.

If each decay produces one positron (as in $\beta^+$ decay), then:

rate of positron production = activity.

To find $N$ from a mass, use the mole concept:

n = \frac{m}{M}, \qquad N = n N_A

where $M$ is molar mass and $N_A$ is the Avogadro constant.

Understanding the Question

You are given a mass of oxygen-15: $2.85 \times 10^{-6}\ \text{kg}$ . You must find how many $^{15}\text{O}$ nuclei this corresponds to, then multiply by $\lambda$ to get the decay rate (and hence positron production rate) in $\text{s}^{-1}$ .

Approach

Convert the isotope’s molar mass to $\text{kg mol}^{-1}$ .
Find number of moles $n = m/M$ .
Convert moles to number of nuclei: $N = nN_A$ .
Use $A = \lambda N$ .

Step-by-Step Reasoning

Molar mass of $^{15}\text{O}$ is approximately $15\ \text{g mol}^{-1}$ (mass number in grams per mole):

M = 15\ \text{g mol}^{-1} = 15 \times 10^{-3}\ \text{kg mol}^{-1} = 1.5 \times 10^{-2}\ \text{kg mol}^{-1}.

Moles in the sample:

n = \frac{2.85 \times 10^{-6}}{1.5 \times 10^{-2}} = 1.90 \times 10^{-4}\ \text{mol}.

Number of nuclei:

N = nN_A = (1.90 \times 10^{-4})(6.02 \times 10^{23}) = 1.14 \times 10^{20}.

Use $\lambda$ from part (c)(ii): $\lambda = 5.66 \times 10^{-3}\ \text{s}^{-1}$ .
Then:

A = \lambda N = (5.66 \times 10^{-3})(1.14 \times 10^{20}) = 6.46 \times 10^{17}\ \text{s}^{-1}.

Because one $\beta^+$ is produced per decay, this is also the positron production rate.

Key Takeaways

Convert mass to number of nuclei using $n = m/M$ and $N = nN_A$ .
Activity gives decays per second: $A = \lambda N$ .
For $\beta^+$ emitters, positron rate equals decay rate.

Common Mistakes

Using $M = 15\ \text{kg mol}^{-1}$ (wrong by a factor $10^3$ ).
Forgetting to convert grams to kilograms.
Using $A = N/\lambda$ or $A = \lambda/t$ (incorrect formula).
Treating activity as “per minute” because the half-life was in minutes.

Things to Be Careful About

Keep track of powers of ten carefully when multiplying $10^{-3}$ by $10^{20}$ type factors.
Ensure the final unit is $\text{s}^{-1}$ .
Assume one positron per decay only because the stem explicitly states oxygen-15 decay produces positrons.

Techniques used

convert sample mass to number of nuclei using molar mass and Avogadro constantuse A = \lambda N to find the decay rateinterpret activity as positron production rate for one positron per decayhandle powers of ten consistently

(d)

The particles that are emitted from the body and detected outside it during PET scanning are not positrons but another type of particle.

3M

(i)

State the name of the particles that are detected.

(ii)

Explain how these particles are formed inside the body.

DifficultyMedium-Easy

Worked solution

Answer

The emitted positron travels a short distance in tissue and annihilates with an electron. Their mass is converted to energy, producing two gamma-ray photons emitted in opposite directions.

Final answer

Positron annihilates with an electron producing two oppositely directed gamma photons.

Detailed explanation

Background Concept

A positron is the antimatter counterpart of an electron. When a positron meets an electron, they can undergo annihilation, in which their rest mass is converted into electromagnetic radiation (photons):

e^- + e^+ \rightarrow \gamma + \gamma

To conserve momentum, the photons are produced in opposite directions (approximately $180^{\circ}$ apart) when the annihilation occurs with negligible net momentum.

Understanding the Question

You are asked to explain how the detected particles (gamma photons) are formed inside the body during PET.

Approach

Explain the sequence:

$\beta^+$ decay emits a positron.
Positron slows down in tissue.
Positron annihilates with an electron.
Two gamma photons are produced and escape to detectors.

Step-by-Step Reasoning

Oxygen-15 decays by $\beta^+$ emission, producing a positron.
In the body, the positron interacts with matter and rapidly loses kinetic energy.
It then encounters an electron; because they are particle-antiparticle, they annihilate.
The combined rest mass energy becomes photon energy (gamma rays).
Two gamma photons are emitted in opposite directions to satisfy conservation of momentum, which is used in PET coincidence detection.

Key Takeaways

PET signal comes from annihilation radiation.
The detected photons originate from $e^+e^-$ annihilation, typically producing two gamma photons in opposite directions.

Common Mistakes

Saying the positron itself is detected outside the body (it is stopped quickly in tissue).
Mentioning only “gamma is emitted” without describing annihilation with an electron.
Stating one photon only (momentum conservation requires two in the simplest case).

Things to Be Careful About

Use the word annihilation and mention the electron.
Include the key detail: two gamma photons in opposite directions (this is central to PET).

Techniques used

describe positron-electron annihilationapply mass-energy conversion to photon productionstate the emission of two gamma photons in opposite directions

Q9MediumQuantum Physics

(a)

State what is meant by the photoelectric effect.

DifficultyEasy

Worked solution

Answer

Emission of electrons from a metal surface when electromagnetic radiation is incident on it, provided the radiation frequency is above a threshold value (for that metal).

Final answer

Emission of electrons from a metal surface when EM radiation of frequency above a threshold is incident.

Detailed explanation

Background Concept

The photoelectric effect is the emission of electrons (photoelectrons) from a metal surface when it absorbs electromagnetic radiation (photons). The key quantum idea is that the energy arrives in packets of energy

E = hf

where $h$ is the Planck constant and $f$ is the radiation frequency.

A metal has a work function $\Phi$ , which is the minimum energy needed to liberate one electron from the surface. If the incoming photon energy is too small, electrons cannot be emitted.

Understanding the Question

You are asked to state what is meant by the photoelectric effect. For full credit, the definition must include (i) what happens (electrons are emitted from a metal surface) and (ii) the condition for it (radiation must have frequency above a threshold / photon energy must exceed the work function).

Approach

Write a concise definition with two points:

emission of electrons from a metal surface due to incident EM radiation;
only occurs when the radiation frequency exceeds a threshold frequency for that metal.

Step-by-Step Reasoning

Mention the effect: electrons leave the metal surface.
Mention the cause: incident electromagnetic radiation (light).
Mention the condition: there is a minimum frequency (threshold frequency) below which no electrons are emitted.

Key Takeaways

Photoelectric effect = electron emission due to incident EM radiation.
There is a threshold frequency related to the work function.

Common Mistakes

Saying only “electrons are emitted by light” but not stating the existence of a threshold frequency.
Confusing threshold frequency with intensity (intensity affects number of electrons, not whether emission occurs).

Things to Be Careful About

The threshold is a property of the metal surface (and its cleanliness), not of the light source.
Use “electromagnetic radiation” or “light” and specify emission from the surface of a metal.

Techniques used

recall the definition of the photoelectric effectstate the condition for emission in terms of threshold frequency

(b)

The photoelectric effect is investigated using two clean metal plates. One plate is made from metal X and the other is made from metal Y.

Metal X has work function energy $\Phi$ . Metal Y has work function energy $2\Phi$ .
Metal X has threshold frequency $F$ .

State expressions, in terms of either or both of $\Phi$ and $F$ , for

2M

(i)

the threshold frequency of metal Y

threshold frequency = ______

DifficultyMedium-Easy

Worked solution

Working

At threshold for metal X:

\Phi = hF

For metal Y, work function $2\Phi$ :

2\Phi = hf_Y \Rightarrow f_Y = \frac{2\Phi}{h} = \frac{2\Phi}{\Phi/F} = 2F

Answer

$2F$

Final answer

2F

Detailed explanation

Background Concept

At the threshold frequency $f_0$ , emitted electrons just have zero maximum kinetic energy, so all the photon energy goes into overcoming the work function:

\Phi = hf_0

This means the threshold frequency is

f_0 = \frac{\Phi}{h}

So, for a given $h$ , the threshold frequency is directly proportional to the work function.

Understanding the Question

Metal X has work function $\Phi$ and threshold frequency $F$ , so it satisfies $\Phi = hF$ . Metal Y has work function $2\Phi$ . You must write the threshold frequency of Y in terms of $\Phi$ and/or $F$ .

Approach

Use the threshold condition for each metal:

For X: $\Phi = hF$ .
For Y: $2\Phi = hf_Y$ .
Eliminate $h$ using the first equation to express $f_Y$ in terms of $F$ .

Step-by-Step Reasoning

For metal X:

\Phi = hF \Rightarrow h = \frac{\Phi}{F}

For metal Y:

2\Phi = hf_Y \Rightarrow f_Y = \frac{2\Phi}{h}

Substitute $h = \Phi/F$ :

f_Y = \frac{2\Phi}{\Phi/F} = 2F

So doubling the work function doubles the threshold frequency.

Key Takeaways

Threshold condition: $\Phi = hf_0$ .
If work function doubles (with $h$ fixed), threshold frequency doubles.

Common Mistakes

Writing $f_Y = F/2$ (inverting the proportionality).
Using $E_K = hf - \Phi$ at threshold but forgetting that $E_K = 0$ at threshold.

Things to Be Careful About

Use consistent symbols: $F$ here is a frequency, not a force.
Don’t confuse threshold frequency with stopping potential (a different experiment quantity).

Techniques used

apply the threshold condition $\Phi = hf_0$use proportional reasoning between work function and threshold frequency

(ii)

the Planck constant.

Planck constant = ______

(c)

The maximum kinetic energy $E_K$ of photoelectrons is determined for each of the plates in (b) for different frequencies $f$ of incident radiation.

On Fig. 9.1, sketch the variation of $E_K$ with $f$ for each plate. Label your lines X and Y to identify which line relates to which plate.

Fig. 9.1

4M Similar

DifficultyMedium

Worked solution

Working

E_K = hf - \Phi

So the $E_K$ – $f$ graph is a straight line of gradient $h$ .

For plate X: $E_K = 0$ at $f = F$ and $E_K = -\Phi$ at $f = 0$ .

For plate Y:

E_K = hf - 2\Phi

Same gradient $h$ , with threshold at $f = 2F$ and $E_K = -2\Phi$ at $f = 0$ .

diagram-9c

Answer

Two parallel straight lines of gradient $h$ : line X crosses the $f$ -axis at $F$ , line Y crosses at $2F$ ; label the lines X and Y accordingly.

Final answer

Two parallel straight lines (gradient h): X crosses at f = F, Y crosses at f = 2F.

Detailed explanation

Background Concept

Einstein’s photoelectric equation relates the maximum kinetic energy of emitted electrons to the photon frequency:

E_K = hf - \Phi

$E_K$ is the maximum kinetic energy of the photoelectrons.
$h$ is Planck’s constant.
$f$ is the frequency of the incident radiation.
$\Phi$ is the work function (minimum energy needed to release an electron).

This has the form of a straight line $y = mx + c$ if you plot $E_K$ (vertical axis) against $f$ (horizontal axis):

gradient $m = h$
intercept $c = -\Phi$
the x-intercept (where $E_K = 0$ ) is the threshold frequency $f_0 = \Phi/h$ .

Understanding the Question

You have two metals:

Metal X: work function $\Phi$ , threshold frequency $F$ .
Metal Y: work function $2\Phi$ .

You must sketch on the provided axes (which already mark $F, 2F, 3F$ on the frequency axis and values like $\pm\Phi$ , $\pm 2\Phi$ on the energy axis) the straight-line relationships for both metals, and label which line is X and which is Y.

Approach

Write the photoelectric equation for each metal.
Identify three key graph features for each line:
- gradient (both are $h$ , so the lines are parallel),
- y-intercept at $f=0$ (negative work function),
- x-intercept where $E_K=0$ (threshold frequency).
Use the given axis markings to place the lines accurately and label them.

Step-by-Step Reasoning

For metal X:

E_K = hf - \Phi

At $f = 0$ :

E_K = -\Phi

so the line crosses the $E_K$ axis at $-\Phi$ .

Threshold frequency is given as $F$ , meaning $E_K = 0$ when $f=F$ . So the line crosses the $f$ axis at $F$ .

So line X must go through $(F,0)$ and $(0,-\Phi)$ .

For metal Y:

E_K = hf - 2\Phi

Same gradient $h$ (so parallel to X).
At $f = 0$ , intercept is $E_K = -2\Phi$ .
Threshold frequency $f_{0Y}$ satisfies $0 = hf_{0Y} - 2\Phi$ , so:

f_{0Y} = \frac{2\Phi}{h} = 2\left(\frac{\Phi}{h}\right) = 2F

So line Y crosses the $f$ axis at $2F$ and the $E_K$ axis at $-2\Phi$ .

Using the markings on the given grid, convenient check points are:

For X at $f = 3F$ :

E_K = h(3F) - \Phi = 3(hF) - \Phi = 3\Phi - \Phi = 2\Phi

so X should pass through $(3F, 2\Phi)$ .

For Y at $f = 3F$ :

E_K = 3hF - 2\Phi = 3\Phi - 2\Phi = \Phi

so Y should pass through $(3F, \Phi)$ .

These points match the provided axis labels and make the sketch precise.

diagram-9c

Key Takeaways

$E_K$ vs $f$ is linear with gradient $h$ .
x-intercept gives threshold frequency.
Increasing work function shifts the line down and to the right (higher threshold frequency), without changing the gradient.

Common Mistakes

Drawing curves instead of straight lines.
Giving different gradients for X and Y (the gradient depends on $h$ , which is universal).
Placing metal Y’s threshold at $F$ instead of $2F$ .
Confusing y-intercept with threshold frequency.

Things to Be Careful About

Use the axis markings: ensure X crosses exactly at $F$ and Y at $2F$ .
Label both lines clearly (X and Y) to get the identification mark.
The y-intercepts are negative: $-\Phi$ for X and $-2\Phi$ for Y, as indicated on the vertical axis.

Techniques used

use Einstein’s photoelectric equation to identify gradient and interceptsdetermine threshold frequencies from work functionssketch straight-line graphs with correct relative positions

Q10MediumAstronomy and Cosmology

(a)

State what is meant by redshift.

DifficultyEasy

Worked solution

Answer

Redshift is the shift of spectral lines towards the red end of the spectrum, i.e. an increase in the observed wavelength (decrease in frequency) compared with the emitted value.

Final answer

Shift of spectral lines to longer wavelength (towards red end of spectrum).

Detailed explanation

Background Concept

Light from atoms produces spectral lines at specific wavelengths. When the relative motion between the source and observer changes, the observed wavelength can differ from the emitted wavelength. A shift of a spectral line to a longer wavelength corresponds to the red end of the visible spectrum and is called redshift.

Understanding the Question

You are being asked to state what “redshift” means in astronomy, based on what happens to the observed spectral lines compared with their known (laboratory) wavelengths.

Approach

Give a clear definition in terms of spectral lines and wavelength (or frequency). For full credit, specify the direction of the shift: towards red / longer wavelength.

Step-by-Step Reasoning

Spectral lines have known rest wavelengths (measured in the lab).
If the observed line is at a longer wavelength than the rest value, the line has moved towards the red end of the spectrum.
This is the definition of redshift.

Key Takeaways

Redshift means spectral lines are observed at longer wavelengths than their rest values.
Equivalent statement: observed frequency is lower.

Common Mistakes

Saying only “light turns red” without mentioning spectral lines or wavelength shift.
Confusing redshift with “the object is red in colour”; it is specifically about a change in spectral line position.

Things to Be Careful About

Use a precise physics statement: “increase in wavelength” (or “decrease in frequency”), not vague wording.
Redshift is a comparison between emitted (rest) and observed wavelengths.

Techniques used

define redshift in terms of change in wavelength of spectral linesrelate redshift to a shift toward the red end of the spectrum

(b)

Explain how observations of redshift lead to the idea that the universe is expanding.

DifficultyMedium-Easy

Worked solution

Answer

Light from distant galaxies shows redshifted spectral lines, which indicates (Doppler effect) that the galaxies are moving away from us. Since most distant galaxies in all directions show redshift, space must be expanding so that galaxies are receding from each other.

Final answer

Most galaxies show redshift → they are receding (Doppler) in all directions → the universe is expanding.

Detailed explanation

Background Concept

For waves, relative motion between source and observer causes a Doppler shift. For light, if the source is receding, the observed wavelength is increased (redshift). Astronomers measure redshift by comparing the observed wavelengths of known spectral lines with their laboratory values.

Understanding the Question

The question asks how the observation that galaxies’ spectra are redshifted leads to the conclusion that the universe is expanding. You must connect: (1) redshift measurement, to (2) recession speeds, to (3) the idea of expansion on a cosmic scale.

Approach

State what redshift implies about motion (receding).
Use the observational fact: redshift is seen for most distant galaxies and in all directions.
Conclude that this is consistent with an overall expansion of the universe (not just a local motion).

Step-by-Step Reasoning

Astronomers observe spectral lines from galaxies at wavelengths longer than the same lines measured in the lab.
A longer observed wavelength is a redshift.
For the Doppler effect, redshift means the source is moving away from the observer.
When nearly all distant galaxies show redshift (whichever direction we look), it suggests recession is a general property of the universe.
The simplest explanation is that distances between galaxies are increasing with time: the universe is expanding.

Key Takeaways

Redshift is evidence of recession.
Widespread recession of galaxies in all directions implies expansion of the universe.

Common Mistakes

Claiming “redshift proves the Big Bang” directly (that is the next part; here it’s just expansion).
Forgetting to mention that redshift corresponds to galaxies moving away.

Things to Be Careful About

Avoid wording that suggests we are at the centre; expansion means all galaxies separate from each other, and any observer would see others receding.
Make the logical chain explicit: observation (shift) → interpretation (receding) → conclusion (expanding universe).

Techniques used

interpret redshift as Doppler effect due to recessioninfer global expansion from widespread recession of galaxies

(c)

Explain how Hubble’s law leads to the Big Bang theory of the origin of the universe.