VCE Physics 2022

Every question from the 2022 VCE Physics examination. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle. This paper was set on the previous study design, so each question also says whether it still matches the course you are studying now.

How to use this page

• Questions are from the 2022 VCE Physics examination, copyright Victorian Curriculum and Assessment Authority (VCAA). Each is summarised briefly here; open the official examination PDF for the full wording, diagrams and the 2022 formula sheet (bound in at the back).
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2022 Physics external assessment report (Word document), and every Section B calculation was redone and compared with the report. Both files are listed on the VCAA Physics examinations page.
• Percentages after multiple-choice answers are the share of students who chose the correct option. In Section B, "From the report" notes summarise what the examiners said students got wrong.

Structure and timing

The paper is 130 marks in 150 minutes (plus 15 minutes reading time), with a scientific calculator, the formula sheet and one sheet of pre-written notes (one folded A3 sheet or two A4 sheets). That is about 1.15 minutes per mark. (Papers from 2024 on are 120 marks.)

• Section A (20 marks): 20 multiple-choice questions. Aim for about 20 minutes.
• Section B (110 marks): 17 questions, including a 16-mark practical investigation (Question 10). Show working, and give answers in the unit printed in the answer box. Take $g = 9.8$ m s⁻² unless told otherwise.

Study design check for 2026

The 2022 exam was written for the 2017 to 2023 study design. The current Units 3 and 4 study design (from 2024) keeps motion, fields, electricity generation and transmission, light and matter, and special relativity, but the old "waves and light" topics have gone: Snell's law, total internal reflection, dispersion and wave basics moved to Unit 1, and the Doppler effect and polarisation are now only in Unit 2 options. When you practise with this paper:

• Skip (no longer in Units 3 and 4): Section A Q10 (Doppler effect), Q12 (total internal reflection in a prism) and Q16 (polarisation); Section B Q13 (Snell's law and dispersion in a tank of water).
• Adapt: Section A Q11 (comparing transverse and longitudinal waves is no longer listed, though light as a transverse wave is) and Section B Q12d (it relies on $n_1v_1 = n_2v_2$, now Unit 1 content, although $\Delta x = \tfrac{\lambda L}{d}$ is still in).
• Still fully examinable: everything else, including the motor, satellite, mass spectrometer, induction, transmission, collision, projectile investigation (Q10), relativity, photoelectric and de Broglie questions.

Section A: Multiple choice

Q1

Direction of the magnetic field at the centre of a current loop. Answer: C - into the page, by the right-hand grip rule applied to the current around the loop. (80%)

Q2

Frequency of the flux variation in an AC generator (one cycle takes 0.04 s). Answer: D - $f = \tfrac{1}{T} = \tfrac{1}{0.04} = 25$ Hz. (73%)

Q3

Charges on particles K, L and M curving in a magnetic field directed into the page. Answer: A - L goes straight (no charge); by the right-hand slap rule, K curves the way a positive charge would and M the opposite way (negative). (83%)

Q4

Charges $Q$ and $4Q$ are 12 cm apart. Where between them is the field zero? Answer: D - $\tfrac{kQ}{a^2} = \tfrac{k(4Q)}{b^2}$ gives $b = 2a$, and $a + b = 12$ cm, so $a = 4$ cm from $Q$. (18%; 31% chose B.)

Q5

A generator's rotation frequency is doubled. New output? Answer: C - the flux changes twice as fast, so the peak EMF doubles and the period halves. (62%; 33% chose B.)

Q6

A 10 t truck at 3.0 m s⁻¹ joins a stationary truck and they move at 2.0 m s⁻¹. Mass of truck Y? Answer: B - $10\,000 \times 3.0 = (10\,000 + m_Y) \times 2.0$ gives $m_Y = 5000$ kg $= 5$ t. (78%)

Q7

Forces between the trucks at the instant of collision. Answer: C - $F_{X \text{ on } Y} = -F_{Y \text{ on } X}$: Newton's third law pairs are equal in size and opposite in direction. B leaves out the direction. (47%; 29% chose D.)

Q8

Compression needed to store 0.9 J in a spring (40 N at 0.08 m). Answer: B - $k = \tfrac{40}{0.08} = 500$ N m⁻¹, and $0.9 = \tfrac12 (500) x^2$ gives $x = 0.06$ m. (72%)

Q9

Two students each pull a rope with 400 N. Force of the rope on each student? Answer: B - 400 N: the rope pulls back on each student with a force equal and opposite to the student's pull. (52%; 33% chose A.)

Q10

An observation of the Doppler effect for sound. Answer: C - moving towards a stationary source, you receive a higher frequency. (85%) Study design: outside Units 3 and 4 (Doppler is now only in a Unit 2 option).

Q11

Transverse and longitudinal waves. Answer: C - transverse waves travel perpendicular to their vibrations; longitudinal waves travel parallel to them. (85%) Study design: adapt (this comparison is no longer listed in Units 3 and 4).

Q12

Minimum refractive index of a 45° glass prism in water ($n = 1.33$) for total internal reflection. Answer: D - the critical angle must be at most 45°: $n \sin 45° = 1.33 \sin 90°$, so $n = 1.88$. (51%) Study design: outside Units 3 and 4 (total internal reflection is now Unit 1 content).

Q13

A standing wave of four loops between A and B (2.4 m apart) with period 1.6 s. Wave speed? Answer: A - each loop is half a wavelength, so $\lambda = \tfrac{2.4}{2} = 1.2$ m and $v = \tfrac{\lambda}{T} = \tfrac{1.2}{1.6} = 0.75$ m s⁻¹. (63%; 27% chose C.)

Q14

Best evidence of electrons behaving as waves. Answer: D - diffraction of electrons through a crystal. (68%)

Q15

Best evidence of light behaving as a particle. Answer: A - the photoelectric effect. (81%)

Q16

Which phenomenon shows light waves are transverse? Answer: A - polarisation: only transverse waves can be polarised, while longitudinal waves such as sound also interfere and diffract. (80%) Study design: outside Units 3 and 4 (polarisation is now only in a Unit 2 option).

Q17

Frequency of a 1.33 MeV gamma photon. Answer: B - $f = \tfrac{E}{h} = \tfrac{1.33 \times 10^6}{4.14 \times 10^{-15}} = 3.21 \times 10^{20}$ Hz. (73%)

Q18

An example of an inertial frame of reference. Answer: A - a bus travelling at constant velocity. The others are all accelerating (speeding up or changing direction). (82%)

Q19

Lorentz factor at $2.99 \times 10^8$ m s⁻¹ (with $c = 3.00 \times 10^8$ m s⁻¹). Answer: C - $\gamma = \tfrac{1}{\sqrt{1 - (2.99/3.00)^2}} = 12.3$. (85%)

Q20

Experimental uncertainty is best understood as... Answer: D - an estimate of the maximum likely difference between the measurement and the true value. (84%)

Section B

Question 1 (5 marks)

A single-turn DC motor coil turns in a uniform magnetic field (pointing to the right) that is parallel to the plane of the coil in Position 1. It rotates about an axis through the middles of sides LM and NK, with a commutator.

a

In Position 1, which way does the current flow in side KL, K to L or L to K? Justify. (2 marks)

b

In Position 3 (half a turn later), which way does the current flow in KL? (1 mark)

c

Side KL (0.10 m) feels a force of 0.15 N in a 0.5 T field. Calculate the current. (2 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]

From K to L. For the coil to turn in the direction shown, the force on side KL must be upwards. Using the right-hand slap rule with the fingers along the field (to the right) and the palm facing the force (up), the thumb gives the current: from K to L. (Name the rule you use; "a right-hand rule" alone is not enough.)

b. [1 mark]

From L to K. After half a turn the commutator has reversed the current in the coil, so the force on KL (now on the other side) still keeps the coil turning the same way.

c. [2 marks]

$$F = nIlB \quad\Rightarrow\quad I = \frac{F}{nlB} = \frac{0.15}{1 \times 0.10 \times 0.5} = 3.0 \text{ A}.$$

From the report. In a (49% full marks), the common reason for losing a mark was referring to "a right-hand rule" without saying which one or how it was applied; the grip rule does not apply here. Part c was well done (82% full marks).

Question 2 (9 marks)

Geostationary satellites orbit once a day (86 400 s) and stay above a fixed point on the equator.

a

Explain why they must be above the equator. (2 marks)

b

Using $G = 6.67 \times 10^{-11}$ N m² kg⁻², $M_E = 5.98 \times 10^{24}$ kg and $R_E = 6.37 \times 10^6$ m, show that the altitude is $3.59 \times 10^7$ m. (4 marks)

c

Calculate the orbital speed. (3 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]. The only force on the satellite is gravity, which points towards the centre of the Earth, so every orbit must be centred on the Earth's centre. To stay over one point on the surface, the satellite must also orbit in the same plane as the Earth's rotation (about the same axis). The only such orbit is in the equatorial plane, directly above the equator.

b. [4 marks]. Gravity provides the centripetal force:

$$\frac{GMm}{r^2} = \frac{4\pi^2 mr}{T^2} \quad\Rightarrow\quad r = \sqrt[3]{\frac{GMT^2}{4\pi^2}} = \sqrt[3]{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 86\,400^2}{4\pi^2}} = 4.225 \times 10^7 \text{ m}.$$

This is the orbital radius, measured from the Earth's centre. The altitude is

$$h = r - R_E = 4.225 \times 10^7 - 6.37 \times 10^6 = 3.59 \times 10^7 \text{ m}.$$

c. [3 marks].

$$v = \frac{2\pi r}{T} = \frac{2\pi \times 4.225 \times 10^7}{86\,400} = 3.07 \times 10^3 \text{ m s}^{-1}.$$

(Using $v = \sqrt{\tfrac{GM}{r}}$ gives the same answer.)

From the report. Part a was weak (2% full marks): most mentioned the plane of rotation but not the direction of the gravitational force. In b, errors were rearranging the equations and forgetting to subtract Earth's radius. In c, the common error was using the Earth's radius or the altitude instead of the orbital radius.

Question 3 (7 marks)

In a mass spectrometer, singly charged positive ions (+1.6 × 10⁻¹⁹ C, mass 4.80 × 10⁻²⁷ kg) are accelerated through 1500 V, then move in a semicircle of diameter $D$ in a 0.10 T magnetic field.

a

Calculate the increase in kinetic energy of each ion, in joules. (2 marks)

b

Show that each ion leaves at 3.16 × 10⁵ m s⁻¹, starting from rest. (2 marks)

c

Calculate $D$. (3 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]. $\Delta E_k = qV = 1.6 \times 10^{-19} \times 1500 = 2.4 \times 10^{-16}$ J.

b. [2 marks].

$$\tfrac12 mv^2 = qV \quad\Rightarrow\quad v = \sqrt{\frac{2 \times 2.4 \times 10^{-16}}{4.80 \times 10^{-27}}} = 3.16 \times 10^5 \text{ m s}^{-1}.$$

c. [3 marks].

$$r = \frac{mv}{qB} = \frac{4.80 \times 10^{-27} \times 3.16 \times 10^5}{1.6 \times 10^{-19} \times 0.10} = 9.5 \times 10^{-2} \text{ m}, \qquad D = 2r = 0.19 \text{ m}.$$

From the report. In a, some used $\tfrac12 mv^2$ without the speed. In c (52% full marks), the most common error was not doubling the radius.

Question 4 (6 marks)

A square loop with a resistor R sits to the right of a long vertical wire carrying a constant current upwards. It is moved three times: A, one full rotation at steady speed about the x-axis (a horizontal axis through the loop); B, at constant speed parallel to the wire (Position 1 to 2); C, at constant speed directly away from the wire (Position 2 to 3). For each, sketch any induced EMF against time and state whether the current is alternating, clockwise, anticlockwise or zero.

Study design: in the current course (field around a current-carrying wire, flux and induced EMF).

Show worked solution

[6 marks, 2 per movement]. To the right of a wire carrying current upwards, the magnetic field points into the page (right-hand grip rule), and it gets weaker further from the wire.

Movement	Induced EMF against time	Induced current
A (rotation)	One full cycle of an alternating, roughly sinusoidal EMF: the flux through the loop goes from maximum to zero to maximum (reversed) and back.	Alternating
B (parallel to wire)	Zero throughout: the loop stays the same distance from the wire, so the flux does not change.	No current
C (away from wire)	An EMF of one sign that gets smaller with time, since the field falls off more slowly further out (a range of shapes was accepted).	Clockwise

For C, the flux into the page decreases as the loop moves into a weaker field. By Lenz's law the induced current opposes the decrease by making its own field into the page inside the loop, which needs a clockwise current.

From the report. Only 7% scored full marks; many answers showed only a superficial understanding of induction. A missing graph for B was scored as zero, because a blank could not show "zero EMF".

Question 5 (7 marks)

A wind generator produces 415 V RMS and 100 A RMS. The factory, 2.00 km away, needs 40 kW; the transmission wires have a total resistance of 2.00 Ω.

a

Calculate the power produced, in kW. (1 mark)

b

Can the system supply the factory? Justify with calculations. (3 marks)

c

Ideal transformers (1 : 10 step-up at the generator, 10 : 1 step-down at the factory) are added. Find the power now supplied, in kW. (3 marks)

Study design: in the current course.

Show worked solution

a. [1 mark]. $P = VI = 415 \times 100 = 41\,500$ W $= 41.5$ kW.

b. [3 marks].

$$P_{\text{loss}} = I^2R = 100^2 \times 2.00 = 20\,000 \text{ W}, \qquad P_{\text{supplied}} = 41\,500 - 20\,000 = 21\,500 \text{ W} = 21.5 \text{ kW}.$$

No: 21.5 kW is less than the 40 kW the factory needs.

c. [3 marks]. Stepping the voltage up by 10 steps the line current down by 10:

$$I_{\text{line}} = \frac{100}{10} = 10 \text{ A}, \qquad P_{\text{loss}} = 10^2 \times 2.00 = 200 \text{ W}, \qquad P_{\text{supplied}} = 41\,500 - 200 = 41.3 \text{ kW}.$$

From the report. The common errors were with powers of ten, for example $100^2 \times 2 = 2000$ W.

Question 6 (5 marks)

An alternator coil of area 0.060 m² and 200 turns has a flux of 1.2 × 10⁻³ Wb through it in the vertical position shown (perpendicular to the field).

a. Calculate the magnetic field strength. (2 marks)

b. The coil turns at 2.5 Hz. Find the average EMF in the first quarter turn from that position. (3 marks)

Study design: in the current course.

Show worked solution

a. [2 marks]. $\Phi = BA$, so $B = \dfrac{1.2 \times 10^{-3}}{0.060} = 2.0 \times 10^{-2}$ T. (The number of turns is not part of $\Phi = BA$.)

b. [3 marks]. The period is $\tfrac{1}{2.5} = 0.40$ s, so a quarter turn takes $\Delta t = 0.10$ s. In that time the flux falls from $1.2 \times 10^{-3}$ Wb to zero:

$$\varepsilon_{\text{av}} = \frac{N\Delta\Phi}{\Delta t} = \frac{200 \times 1.2 \times 10^{-3}}{0.10} = 2.4 \text{ V}.$$

From the report. In a, many wrongly included the number of turns. In b, the common error was using the full period, 0.4 s, as $\Delta t$.

Question 7 (10 marks)

A 2.0 kg trolley is released on a 25° ramp and accelerates at 3.2 m s⁻². At the bottom it moves at 4.0 m s⁻¹ on a frictionless surface and sticks to an identical stationary trolley.

a. i. Show that the component of gravity down the slope is 8.3 N. (2 marks) ii. Calculate the constant friction force on the ramp. (2 marks)

b. i. Calculate the speed after the collision, stating the principle used. (3 marks) ii. Is the collision elastic or inelastic? Show working. (3 marks)

Study design: in the current course.

Show worked solution

a. i. [2 marks]

$F = mg\sin\theta = 2.0 \times 9.8 \times \sin 25° = 8.28 \approx 8.3$ N.

ii. [2 marks]

$F_{\text{net}} = ma = 2.0 \times 3.2 = 6.4$ N $= 8.3 - F_f$, so $F_f = 1.9$ N.

b. i. [3 marks]

Conservation of momentum (the system is isolated on the frictionless surface):

$$2.0 \times 4.0 + 2.0 \times 0 = (2.0 + 2.0)v \quad\Rightarrow\quad v = 2.0 \text{ m s}^{-1}.$$

ii. [3 marks].

$$E_{k,\text{before}} = \tfrac12 (2.0)(4.0)^2 = 16 \text{ J}, \qquad E_{k,\text{after}} = \tfrac12 (4.0)(2.0)^2 = 8.0 \text{ J}.$$

Kinetic energy is not conserved (8 J is transformed into heat, sound and deformation), so the collision is inelastic.

From the report. In b.i, most named conservation of momentum but could not express it mathematically. In b.ii, the common slip was forgetting to square the speed.

Question 8 (5 marks)

A Formula 1 car and driver (800 kg) take a horizontal corner of radius 80.0 m at a constant 40 m s⁻¹.

a

Calculate the net force. (2 marks)

b

Draw its direction on the top view. (1 mark)

c

Explain why a net horizontal force is needed, and state what exerts it. (2 marks)

Study design: in the current course.

Show worked solution

a. [2 marks].

$$F_{\text{net}} = \frac{mv^2}{r} = \frac{800 \times 40^2}{80.0} = 1.6 \times 10^4 \text{ N}.$$

b. [1 mark]. Horizontally towards the centre of the circle (on the top view, not the front view).

c. [2 marks]. The car's velocity keeps changing direction, so it is accelerating towards the centre; by Newton's second law this needs a net (centripetal) force towards the centre. On a flat corner this horizontal force is the sideways friction of the road on the tyres.

From the report. In b, some drew the arrow on the wrong figure. In c (21% full marks), many named friction but did not link it to circular motion or the centripetal force.

Question 9 (2 marks)

A star transforms energy at 2.90 × 10²⁵ W. Explain the transformation and its effect, if any, on the star's mass. No calculations are required.

Study design: in the current course (mass-energy conversion in the Sun).

Show worked solution

[2 marks]. Nuclear fusion in the core converts mass into energy ($E = mc^2$), which is radiated away. Because energy leaves the star, its mass decreases (by $\tfrac{2.90 \times 10^{25}}{c^2}$ kg every second).

From the report. Almost 40% scored zero. Some explained fusion in detail without the mass change; others said the mass decreases without a reason; a number wrongly named fission.

Question 10 (16 marks)

Students launch 56 g tennis balls from ground level on a level oval. The launcher can be set to speeds $u$ from 8 to 30 m s⁻¹ and angles $\theta$ from 10° to 80°. At $u = 25$ m s⁻¹, average ranges (three trials each) are: 17 m at 10°, 30 m at 20°, 37 m at 30°, 40 m at 40° and 50°, 36 m at 60°, 29 m at 70° and 15 m at 80°. The range uncertainty is ±1 m.

a

Give one controlled, one dependent and one independent variable. (3 marks)

b

Plot the data with scales, uncertainty bars on at least three points and a smooth curve of best fit. (6 marks)

c

From your graph, estimate the maximum range and the angle that gives it. (2 marks)

d. i

Using $R = \tfrac{u^2 \sin 2\theta}{g}$, calculate the theoretical range at 25 m s⁻¹ and 30°. (2 marks) ii. Evaluate whether air resistance can be ignored. (3 marks)

Study design: in the current course (projectile motion, investigation skills and uncertainty).

Show worked solution

a. [3 marks].

• Controlled: the launch speed (25 m s⁻¹ for this set of trials) or the mass of the ball.
• Dependent: the range, $R$.
• Independent: the launch angle, $\theta$ (the launch speed is the other possible independent variable for the launcher).

b. [6 marks]. Use most of the grid, label both axes with units, draw ±1 m bars and a smooth curve (not straight segments joining the dots):

c. [2 marks]. From the curve: maximum range about 40.5 m at an angle of about 45° (the values the report reads from its curve).

d. i. [2 marks].

$$R = \frac{u^2 \sin 2\theta}{g} = \frac{25^2 \times \sin 60°}{9.8} = 55.2 \text{ m}.$$

ii. [3 marks]. No, air resistance cannot be ignored. The theoretical range (55.2 m) is much greater than the measured range at 30° (37 m), and the difference of about 18 m is far larger than the ±1 m uncertainty, so it cannot be put down to measurement uncertainty. Air resistance is the likely cause of the shorter ranges.

From the report. Part b was well done (51% full marks); uncertainty bars were needed on at least three points. In d.i, the common error was using 30° instead of $2\theta = 60°$. In d.ii only 4% scored full marks: most cited the difference but did not say it lies outside the uncertainty.

Question 11 (2 marks)

Explain why muons formed in the upper atmosphere reach the Earth's surface although their half-lives suggest they should decay first.

Study design: in the current course (named in the study design).

Show worked solution

[2 marks]. Either explanation, with the frame stated:

• Time dilation (Earth frame): the muons move close to $c$, so in the Earth's frame their half-life is dilated, $t = \gamma t_0$, much longer than their proper half-life. They survive long enough to travel the whole distance.
• Length contraction (muon frame): in the muons' frame the distance from the upper atmosphere to the ground is contracted, $L = \tfrac{L_0}{\gamma}$, so it is short enough to cover within their proper half-life.

From the report. 52% scored zero; students struggled to identify the two frames of reference and what each observer measures.

Question 12 (8 marks)

Laser light (6.00 × 10¹⁴ Hz) passes through two slits 0.10 mm apart onto a screen 2.00 m away. C is the bright centre of the pattern.

a

Explain why C is bright and why there is a dark band to its left. (2 marks)

b

Explain how Young's experiment supported the wave model of light. (2 marks)

c

Calculate the spacing of the dark bands. (2 marks)

d

The whole apparatus is immersed in a liquid with refractive index greater than air. Describe and explain the change in the band spacing. (2 marks)

Study design: a to c are in the current course. Part d relies on $n_1v_1 = n_2v_2$, which is now Unit 1 content.

Show worked solution

a. [2 marks]

At C the path difference from the two slits is zero, so the waves arrive in phase and interfere constructively (bright). At the first dark band to the left, the path difference is half a wavelength, $\tfrac{\lambda}{2}$, so the waves arrive out of phase and interfere destructively.

b. [2 marks]

The bands are an interference pattern, and interference is a property of waves that a particle model could not explain, so the result supported the wave model.

c. [2 marks]

$\lambda = \tfrac{c}{f} = \tfrac{3.0 \times 10^8}{6.00 \times 10^{14}} = 5.0 \times 10^{-7}$ m.

$$\Delta x = \frac{\lambda L}{d} = \frac{5.0 \times 10^{-7} \times 2.00}{0.10 \times 10^{-3}} = 1.0 \times 10^{-2} \text{ m} \;(0.01 \text{ m}).$$

d. [2 marks]. The spacing decreases. In the liquid the light travels more slowly ($v = \tfrac{c}{n}$), and since the frequency is unchanged, its wavelength ($\lambda = \tfrac{v}{f}$) is shorter. A shorter wavelength gives a smaller $\Delta x = \tfrac{\lambda L}{d}$.

From the report. Part a was weak (20% full marks): answers had to explain the two specific places, not give generic notes on interference. In b, Young's experiment shows interference; answers that relied on diffraction did not score. In c, errors came from converting frequency to wavelength.

Question 13 (7 marks)

A green LED ray strikes the surface of 80.00 cm deep water ($n = 1.335$) at 40.00° to the surface, entering at a point directly above O, and reaches the bottom at X.

a. Calculate OX in centimetres. (4 marks)

b. The LED is replaced by a narrow beam of white light. Describe the colour arriving left of X, at X and right of X. (3 marks)

Study design: outside Units 3 and 4 (Snell's law and dispersion are now Unit 1 content).

Show worked solution

a. [4 marks]. The angle of incidence is measured from the normal: $90.00° - 40.00° = 50.00°$.

$$1.000 \sin 50.00° = 1.335 \sin\theta_2 \quad\Rightarrow\quad \theta_2 = 35.02°$$

$$\text{OX} = 80.00 \tan 35.02° = 56.05 \text{ cm} \approx 56 \text{ cm}.$$

b. [3 marks]. Water's refractive index is higher for shorter wavelengths, so blue light bends most (closer to the normal, landing nearer O) and red light bends least:

• left of X: blue (and violet);
• at X: green;
• right of X: red (with yellow and orange between green and red).

From the report. In a (25% full marks), the most common errors were using 40° or 60° as the angle of incidence.

Question 14 (9 marks)

Sam measures photocurrent against voltage with a green filter (Figure 13): the current is zero at a negative voltage P and levels off at a saturation current for positive voltages. A later graph of $E_{k\max}$ against $f$ (Figure 16) is a straight line with a vertical intercept of $-3.4$ eV and a horizontal intercept of $8.1 \times 10^{14}$ Hz.

a

What does point P represent? (1 mark)

b

Sketch the graph when the light intensity is significantly increased. (2 marks)

c

Sketch the graph with a violet filter, at the increased intensity. (2 marks)

d

Determine the work function in eV. (1 mark)

e

Calculate Planck's constant in J s from Figure 16. (2 marks)

f

State one limitation of the wave model in explaining the photoelectric effect. (1 mark)

Study design: in the current course.

Show worked solution

a. [1 mark]. The stopping voltage (cut-off potential): the reverse voltage at which even the most energetic photoelectrons are stopped.

b. [2 marks]. Same stopping voltage P (the photon energy is unchanged), but a higher saturation current (more photons per second, so more photoelectrons).

c. [2 marks]. Violet photons have a higher frequency, so the photoelectrons have more kinetic energy and the stopping voltage is further from zero than P. The saturation current is still above the original.

d. [1 mark]. The vertical intercept of $E_{k\max} = hf - \phi$ is $-\phi$, so $\phi = 3.4$ eV.

e. [2 marks]. Planck's constant is the gradient:

$$h = \frac{\text{rise}}{\text{run}} = \frac{3.4}{8.1 \times 10^{14}} = 4.2 \times 10^{-15} \text{ eV s}.$$

Converting to joule-seconds: $h = 4.2 \times 10^{-15} \times 1.6 \times 10^{-19} = 6.7 \times 10^{-34}$ J s.

f. [1 mark]. For example, the wave model cannot explain the threshold frequency (below it no electrons are emitted, however bright the light), or why there is no time delay at low intensity, or why the electrons' maximum kinetic energy does not depend on intensity.

From the report. In d, calculating $\phi = hf_0$ with the accepted $h$ assumes the graph's gradient equals it, which may not be the case; read the intercept. In e (48% full marks), errors were run over rise and leaving the answer in eV s.

Question 15 (2 marks)

Some neon energy levels: 0, 13.50, 19.80, 20.02, 20.65 and 21.74 eV.

a. Show that a 640 nm photon corresponds to a 1.94 eV transition. (1 mark)

b. Draw the transition that emits it. (1 mark)

Study design: in the current course.

Show worked solution

a. [1 mark].

$$E = \frac{hc}{\lambda} = \frac{4.14 \times 10^{-15} \times 3.0 \times 10^8}{640 \times 10^{-9}} = 1.94 \text{ eV}.$$

b. [1 mark]. A downward arrow from 21.74 eV to 19.80 eV ($21.74 - 19.80 = 1.94$ eV).

From the report. In a, noting that $21.74 - 19.80 = 1.94$ only identifies the levels; the mark needed the photon energy calculated from the wavelength.

Question 16 (3 marks)

A distant sodium lamp (589 nm) is seen through a window covered with fine steel mesh and through an open window. Only the mesh gives a diffraction pattern.

a. What condition must the mesh satisfy for diffraction to form? (1 mark)

b. Why does this condition not apply to the open window? (2 marks)

Study design: in the current course (diffraction and the $\tfrac{\lambda}{w}$ ratio).

Show worked solution

a. [1 mark]. The gaps in the mesh must be very small, comparable with the wavelength of the light, so that $\tfrac{\lambda}{w}$ is large enough for noticeable spreading.

b. [2 marks]. The open window's width is enormous compared with 589 nm, so $\tfrac{\lambda}{w}$ is tiny and the light spreads by a negligible amount: no visible diffraction pattern forms.

From the report. Part b was answered fully by 32%; the key was comparing the gap widths through the $\tfrac{\lambda}{w}$ ratio.

Question 17 (7 marks)

Electrons of energy 10.0 keV diffract through a thin metal foil.

a. Calculate their de Broglie wavelength in nanometres. (4 marks)

b. The electron energy, and hence speed, is increased slightly. Explain the effect on the wavelength. (3 marks)

Study design: in the current course.

Show worked solution

a. [4 marks]. Convert the energy to joules and find the speed:

$$E_k = 10.0 \times 10^3 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-15} \text{ J}, \qquad v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-15}}{9.1 \times 10^{-31}}} = 5.93 \times 10^7 \text{ m s}^{-1}.$$

$$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^7} = 1.23 \times 10^{-11} \text{ m} = 0.0123 \text{ nm}.$$

(The electrons move at about $0.2c$, so a relativistic calculation gives a wavelength about 0.5% shorter; the classical method is what the report uses.)

b. [3 marks]. A higher speed means a larger momentum, $p = mv$. Since $\lambda = \tfrac{h}{p}$, the de Broglie wavelength decreases (so the diffraction rings get closer together).

From the report. In a (29% full marks), many could not work methodically through the steps or convert metres to nanometres. In b, the common error was claiming the wavelength increases with speed.

General advice from the 2022 report

• Show working for every calculation worth more than one mark, and lay out multi-step answers so the assessor can follow them. Write indices clearly, and write in pen.
• Annotate the right diagram and use the whole grid when drawing graphs; when a question says "from the graph", refer to your graph.
• Know the difference between diffraction and interference: Young's double-slit result is interference (Q12).
• Do not copy definitions from the notes sheet; they rarely answer the specific question asked.