VCE Physics 2024

Every question from the 2024 VCE Physics examination, the first under the current study design. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle.

How to use this page

• Questions are from the 2024 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 formula sheet.
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2024 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 120 marks in 150 minutes (plus 15 minutes reading time), with a scientific calculator, the formula sheet and pre-written notes. That is 1.25 minutes per mark.

• Section A (20 marks): 20 multiple-choice questions. Aim for about 25 minutes.
• Section B (100 marks): 16 questions, ending with a 20-mark photoelectric effect investigation (Question 16). Show working on every question worth more than one mark.

Section A: Multiple choice

Q1

Jo (75 kg) stands in a lift moving up at a constant 2.0 m s⁻¹. Force from the floor? Answer: C - no acceleration, so the normal force equals her weight: $75 \times 9.8 \approx 740$ N. (54%)

Q2

A space observatory's mass doubles but its orbital speed stays the same. New orbital radius? Answer: B - $v = \sqrt{\tfrac{GM_E}{r}}$ does not depend on the satellite's mass, so the same speed means the same radius $R$. (39%; another 39% chose C.)

Q3

Forces on a ball moving in a horizontal circle on a string (a conical pendulum), seen front-on. Answer: A - only two forces: weight straight down and tension along the string. There is no separate centripetal force; it is the net of these two. (51%; 37% chose C.)

Q4

A spring (force rising linearly to 40 N at 10.0 cm compression) launches a 50 g ball. Launch speed? Answer: C - stored energy $=$ area $= \tfrac12 \times 0.100 \times 40 = 2.0$ J, so $\tfrac12(0.050)v^2 = 2.0$ gives $v \approx 8.9$ m s⁻¹. (63%)

Q5

Total energy of a rock falling from a cliff (no air resistance). Answer: A - a horizontal line: mechanical energy is conserved. (73%)

Q6

Field strength midway between the S poles of two identical facing magnets. Answer: A - very close to zero: the two fields are equal and opposite there. (79%)

Q7

A current loop in a uniform field, then rotated through 90°. Compare the force on side PQ and the torque. Answer: D - PQ carries the same current across the same field in both positions, so the forces are equal, but in Diagram B the force on PQ is parallel to the plane of the loop and gives no torque: $F_A = F_B$, $\tau_A > \tau_B$. (39%)

Q8

Force on a 4.0 cm wire carrying 5.0 A perpendicular to a $2.0 \times 10^{-3}$ T field. Answer: C - $F = BIl = 2.0 \times 10^{-3} \times 5.0 \times 0.040 = 4.0 \times 10^{-4}$ N. (83%)

Q9

The force between two charges becomes $6F$. New separation? Answer: B - $F \propto \tfrac{1}{d^2}$, so $d \to \tfrac{d}{\sqrt6} \approx 0.41d$. (54%)

Q10

An induced current is constant, steps up to a higher constant value, then returns. Matching flux graph? Answer: C - a constant current needs a constant rate of change of flux, so the flux rises in straight segments, steeper in the middle. (34%; 35% chose D.)

Q11

Why transmit at high voltage? Answer: B - for the same power, a higher voltage means a smaller current, and line losses are $I^2R$. (80%)

Q12

A coil moves upward away from the north pole of a magnet below it. Induced current (viewed from above) and induced field inside the coil? Answer: C - the upward flux through the coil is decreasing, so the induced field points up to oppose the decrease, which needs an anticlockwise current viewed from above. (42%; 30% chose A.)

Q13

Output of a generator with a split-ring commutator. Answer: A - DC with a varying value (a rectified sine wave): the commutator keeps the output in one direction. (51%; 32% chose D.)

Q14

A rocket passes a circular window at $0.85c$. What does its passenger measure? Answer: D - length contraction only along the motion: the horizontal width is shorter, the vertical height unchanged. (77%)

Q15

Synchrotron electrons go from $\gamma = 200$ to $\gamma = 2000$. Answer: B - speed barely increases (already almost $c$) while energy, proportional to $\gamma$, increases a lot. (47%; 31% chose A.)

Q16

Match five transitions (three to $E_1$, and $E_5$ to $E_2$ and to $E_3$) to a line spectrum. Answer: D - three high-frequency lines (the drops to $E_1$) and two low-frequency lines. (43%)

Q17

A 50 Hz standing wave: shape at $t = 0.010$ s. Answer: B - that is half a period, so the pattern is inverted. (59%)

Q18

Double-slit spacing when $L$ doubles and $d$ halves. Answer: D - $\Delta x = \tfrac{\lambda L}{d}$ becomes $\tfrac{\lambda (2L)}{0.5d} = 4\Delta x$. (63%)

Q19

Which linear graph fits $v = \sqrt{\tfrac{GM}{r}}$? Answer: D - $v^2 = GM \times \tfrac1r$, so $v^2$ against $\tfrac1r$ is a straight line through the origin. (69%)

Q20

Data are precise when? Answer: B - repeated results show little variation. (Closeness to the accepted value is accuracy.) (85%)

Section B

Question 1 (3 marks)

A tug pulls Boat 2 ($5.0 \times 10^5$ kg) and Boat 3 ($6.0 \times 10^5$ kg) in a line with Rope 1 (tension $9.2 \times 10^4$ N) to Boat 2 and Rope 2 from Boat 2 to Boat 3. Resistances are $1.2 \times 10^4$ N on Boat 2 and $1.4 \times 10^4$ N on Boat 3.

a. Show that the towed boats accelerate at $6.0 \times 10^{-2}$ m s⁻². (1 mark)

b. Calculate the tension in Rope 2. (2 marks)

Show worked solution

a. [1 mark]. Treat Boats 2 and 3 as one system:

$$a = \frac{\sum F}{m} = \frac{9.2 \times 10^4 - 1.2 \times 10^4 - 1.4 \times 10^4}{5.0 \times 10^5 + 6.0 \times 10^5} = \frac{6.6 \times 10^4}{1.1 \times 10^6} = 6.0 \times 10^{-2} \text{ m s}^{-2}.$$

b. [2 marks]. Apply Newton's second law to Boat 3 alone:

$$T - 1.4 \times 10^4 = 6.0 \times 10^5 \times 6.0 \times 10^{-2} \quad\Rightarrow\quad T = 3.6 \times 10^4 + 1.4 \times 10^4 = 5.0 \times 10^4 \text{ N}.$$

From the report. In a, common errors were ignoring friction or using the mass of all three boats, and some "fudged" working to reach the given answer. In b, many used both boats' mass, or subtracted the resistance from $ma$ instead of adding it.

Question 2 (4 marks)

A test vehicle travels around a circular track of radius 1.2 km banked at 33°.

a. Calculate the speed at which no sideways friction is needed. (3 marks)

b. Explain how the normal force provides the centripetal force. (1 mark)

Show worked solution

a. [3 marks]. With no friction, the normal force's vertical component balances the weight and its horizontal component provides the centripetal force: $N\cos\theta = mg$ and $N\sin\theta = \tfrac{mv^2}{r}$. Dividing, $\tan\theta = \tfrac{v^2}{rg}$:

$$v = \sqrt{rg\tan\theta} = \sqrt{1200 \times 9.81 \times \tan 33°} \approx 87.4 \text{ m s}^{-1}.$$

b. [1 mark]. The normal force acts perpendicular to the tilted track, so it has a horizontal component pointing towards the centre of the turn. This unbalanced horizontal component is the centripetal force.

From the report. In a, the common errors were not converting km to m and using the wrong trigonometric function. In b, it is wrong to say the normal force points to the centre of the turn; only its horizontal component does.

Question 3 (7 marks)

A golf ball is hit from a cliff top at 48 m s⁻¹ at 35° above the horizontal and lands on the ground below 6.2 s later (no air resistance).

a

Find the height of the cliff. (3 marks)

b

Find the horizontal distance travelled. (2 marks)

c

Describe how air resistance would change the path. (2 marks)

Show worked solution

a. [3 marks]. Take up as positive: $u_y = 48\sin 35° = 27.53$ m s⁻¹ and $a = -9.81$ m s⁻².

$$s_y = u_y t + \tfrac12 at^2 = 27.53 \times 6.2 - \tfrac12 \times 9.81 \times 6.2^2 = -17.9 \text{ m},$$

so the ball lands about 18 m below the launch point: $h \approx 18$ m.

b. [2 marks]. $u_x = 48\cos 35° = 39.32$ m s⁻¹, constant, so $x = 39.32 \times 6.2 \approx 244$ m.

c. [2 marks]. Air resistance would reduce both the maximum height the ball reaches and the horizontal distance it travels (the path would be shorter and lower, and less symmetrical).

From the report. Common errors in a were an inconsistent sign convention (gravity acting upwards), splitting the flight into phases and losing track of times, and misusing pre-derived formulas such as the range equation. In c, most mentioned the shorter distance but not the lower height.

Question 4 (4 marks)

Without airbags, the force on a crash-test dummy rises linearly from 0 to 24 000 N at 120 ms and falls back to 0 at 240 ms.

a. Determine the impulse on the dummy. (2 marks)

b. With airbags (same car and speed), the collision lasts twice as long. Sketch the force-time graph. (2 marks)

Show worked solution

a. [2 marks]. Impulse is the area under the force-time graph:

$$I = \tfrac12 \times 0.240 \times 24\,000 = 2.88 \times 10^3 \text{ N s}.$$

b. [2 marks]. The change in momentum (the area) must stay the same while the duration doubles to 480 ms, so the peak force halves to 12 000 N (at 240 ms):

From the report. In a, not converting 240 ms to seconds was the common error. In b, most had the duration right but the wrong peak force.

Question 5 (4 marks)

A 0.63 kg basketball hits the floor moving down at 12 m s⁻¹ and rebounds up at 10 m s⁻¹.

a. Find the magnitude and direction of its change in momentum. (2 marks)

b. Is the collision elastic or inelastic? Justify. (2 marks)

Show worked solution

a. [2 marks]. Take up as positive: $\Delta p = p_f - p_i = 0.63 \times 10 - 0.63 \times (-12) = 13.9$ kg m s⁻¹. The change is 14 kg m s⁻¹ upwards.

b. [2 marks]. Inelastic. The ball's speed after the bounce is lower, so its kinetic energy ($\tfrac12 mv^2$, same mass) is lower: kinetic energy is not conserved.

From the report. In a, the common error was $0.63 \times 2 = 1.26$ kg m s⁻¹ with a guessed direction. In b, saying "momentum is not conserved" or "energy is not conserved" (instead of kinetic energy) was wrong.

Question 6 (3 marks)

Kwan suggests a supply craft ($1.50 \times 10^4$ kg) could orbit the ISS ($4.50 \times 10^5$ kg) at a radius of 500 m and a speed of 0.436 m s⁻¹, held by their mutual gravitational attraction. Will this work? Justify with calculations. (3 marks)

Show worked solution

[3 marks]. Compare the gravitational force available with the centripetal force needed.

$$F_g = \frac{GMm}{r^2} = \frac{6.67 \times 10^{-11} \times 4.50 \times 10^5 \times 1.50 \times 10^4}{500^2} = 1.80 \times 10^{-6} \text{ N}$$

$$F_c = \frac{mv^2}{r} = \frac{1.50 \times 10^4 \times 0.436^2}{500} = 5.70 \text{ N}$$

The force needed is millions of times larger than gravity can provide, so no, Kwan's suggestion will not work. (Comparing the required and possible orbital speeds or radii works too.)

From the report. Most students calculated something but did not compare it with anything. Plan a strategy before starting longer questions.

Question 7 (6 marks)

An electron accelerated from rest enters a 5.00 mT magnetic field at right angles and moves in a circle of radius 1.50 cm (ignore relativity).

a

Show that its speed is $1.32 \times 10^7$ m s⁻¹. (1 mark)

b

Calculate the accelerating voltage. (2 marks)

c

A proton at the same speed enters the field. Describe its path compared with the electron's, with justification. (3 marks)

Show worked solution

a. [1 mark].

$$v = \frac{qBr}{m} = \frac{1.60 \times 10^{-19} \times 5.00 \times 10^{-3} \times 1.50 \times 10^{-2}}{9.11 \times 10^{-31}} = 1.32 \times 10^7 \text{ m s}^{-1}.$$

b. [2 marks].

$$qV = \tfrac12 mv^2 \quad\Rightarrow\quad V = \frac{9.11 \times 10^{-31} \times (1.32 \times 10^7)^2}{2 \times 1.60 \times 10^{-19}} \approx 496 \text{ V}.$$

c. [3 marks]. The proton curves the opposite way (anticlockwise instead of clockwise) because its charge is positive. Its path has a much larger radius, because $r = \tfrac{mv}{qB}$ and its mass is about 1800 times the electron's (same speed, same size of charge).

From the report. In a, converting the radius to metres was the common error. In c, answers needed both changes, each justified.

Question 8 (6 marks)

A negatively charged thundercloud base is 850 m above positively charged ground.

a

Sketch at least five electric field lines between the ground and the cloud. (2 marks)

b

The potential difference is 1.20 GV. Calculate the field strength. (2 marks)

c

A lightning bolt at 1.20 GV carries 30.0 kA for 60.0 µs. Calculate the energy transferred, to three significant figures. (2 marks)

Show worked solution

a. [2 marks]. At least five evenly spaced, parallel, straight lines between the ground and the cloud, with arrows pointing upwards (from the positive ground to the negative cloud).

b. [2 marks].

$$E = \frac{V}{d} = \frac{1.20 \times 10^9}{850} \approx 1.41 \times 10^6 \text{ V m}^{-1}.$$

c. [2 marks].

$$E = VIt = 1.20 \times 10^9 \times 30.0 \times 10^3 \times 60.0 \times 10^{-6} = 2.16 \times 10^9 \text{ J}.$$

From the report. In a, drawing the arrows downwards was the common error. In b, converting GV to volts caused errors.

Question 9 (5 marks)

A single-coil DC motor (sides EF, FG, GH) sits in a uniform field between N and S poles, powered by a 6.0 V battery.

a

Describe an orientation, if any, where the force on side FG is zero. Justify. (2 marks)

b

Describe a starting orientation, if any, from which the coil will not start to rotate. Justify. (2 marks)

c

Suggest one way to increase the torque with the same coil and battery. (1 mark)

Show worked solution

a. [2 marks]

When the coil is horizontal, side FG is parallel to the magnetic field, so no force acts on it (the force needs a component of current perpendicular to the field).

b. [2 marks]

When the coil is vertical (its plane perpendicular to the field), the forces on EF and GH are vertical, in line with the coil, so they have no turning effect and the coil will not start to rotate.

c. [1 mark]

Increase the strength of the magnetic field (for example, stronger magnets).

From the report. In a, restating the orientation was not a justification. In b, many confused motors with generators and discussed induction.

Question 10 (7 marks)

A single coil rotates in a uniform field, connected through slip rings to an oscilloscope. The output is a sine wave with peaks of +4 V and -4 V and a period of 0.2 s. After a change, the peaks are +8 V and -8 V with the same period.

a

Explain the purpose of the slip rings. (2 marks)

b

Find the peak-to-peak voltage and the frequency of rotation. (2 marks)

c

Explain two possible modifications that would give the new trace. (3 marks)

Show worked solution

a. [2 marks]

The slip rings keep a continuous electrical connection between the rotating coil and the external circuit (through brushes), so the generated AC output reaches the oscilloscope without the wires twisting.

b. [2 marks]

Peak-to-peak voltage $= 4 - (-4) =$ 8 V. Frequency $= \tfrac{1}{0.2} =$ 5 Hz.

c. [3 marks]

The peak EMF has doubled while the frequency is unchanged, so the rotation rate is the same. By Faraday's law, $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$, the EMF doubles if the change in flux doubles. Any two of: double the number of turns (from 1 to 2); double the magnetic field strength; double the area of the coil.

From the report. In a, many described only "a connection to the loop" (a commutator also does that) or the role of slip rings in a motor. In b, the common errors were giving the peak (4 V) or the period (0.2 s). In c, answers had to say "double", not just "increase".

Question 11 (5 marks)

A charger's ideal transformer steps 230 V RMS down to 36 V RMS.

a

Find the number of secondary turns if the primary has 460. (1 mark)

b

The RMS output current is 0.80 A. Find the RMS input current. (2 marks)

c

Explain why the transformer needs AC input and would not work with constant DC. (2 marks)

Show worked solution

a. [1 mark]. $N_s = N_p \times \dfrac{V_s}{V_p} = 460 \times \dfrac{36}{230} = 72$ turns.

b. [2 marks]. For an ideal transformer, $V_pI_p = V_sI_s$:

$$I_p = \frac{36 \times 0.80}{230} \approx 0.13 \text{ A}.$$

c. [2 marks]. A transformer works by induction: an alternating current in the primary produces a changing magnetic flux in the core, which induces an EMF in the secondary. A constant DC input produces a constant flux, so no EMF is induced and there is no output (except momentarily when switched on or off).

From the report. In b, converting to peak values was the common error. In c, "because it needs alternating current" repeats the question; the answer must explain the changing flux.

Question 12 (6 marks)

A store installs 50 solar panels (600 W each, maximum 20 V each) as five parallel strings of 10 panels in series, feeding an inverter.

a

Find the maximum total output power in kW. (1 mark)

b

Find the maximum voltage and current from one string. (3 marks)

c

Find the maximum voltage and current from the whole array. (1 mark)

d

What is the inverter's function? (1 mark)

Show worked solution

a. [1 mark]

$50 \times 600 = 30\,000$ W $=$ 30 kW.

b. [3 marks]

In series, voltages add: $10 \times 20 =$ 200 V. The current is the same through every panel in series: $I = \tfrac{P}{V} = \tfrac{600}{20} =$ 30 A (or $\tfrac{6000}{200}$).

c. [1 mark]

In parallel, the voltage stays the same and the currents add: 200 V and $5 \times 30 =$ 150 A.

d. [1 mark]

It converts the panels' DC output into AC for the mains supply.

From the report. In a, some wrote 30 000 kW; give the answer in the unit printed in the box. In c, 1000 V and 150 A was the common error, showing weak understanding of parallel circuits.

Question 13 (9 marks)

Muons (mean lifetime 2.20 µs) are produced 9700 m above the ground and travel down at $0.985c$ to a detector on a mountain top at 6500 m.

a

Using classical mechanics, show that a muon takes 10.8 µs to reach the detector. (1 mark)

b

Classically, is a muon likely to reach the detector? Justify. (2 marks)

c

Calculate the Lorentz factor at $0.985c$. (2 marks)

d

State the mean lifetime in the physicists' frame of reference. (1 mark)

e

Explain how special relativity accounts for many muons reaching the detector. (3 marks)

Show worked solution

a. [1 mark].

$$t = \frac{d}{v} = \frac{9700 - 6500}{0.985 \times 3.00 \times 10^8} = 1.08 \times 10^{-5} \text{ s} = 10.8 \text{ µs}.$$

b. [2 marks]. No. The trip takes 10.8 µs, nearly five times the 2.20 µs mean lifetime, so most muons would decay before reaching the detector.

c. [2 marks].

$$\gamma = \frac{1}{\sqrt{1 - 0.985^2}} = \frac{1}{\sqrt{0.0298}} \approx 5.8.$$

d. [1 mark]. VCAA awarded full marks to all students for this part after review. (In the physicists' frame the muon's mean lifetime is dilated to $5.8 \times 2.20 \approx 12.8$ µs, the figure the report uses in part e.)

e. [3 marks]. Time dilation: in the physicists' (Earth) frame, the muon's clock runs slow, so its mean lifetime is dilated from 2.20 µs to about 12.8 µs. That is longer than the 10.8 µs trip, so many muons reach the detector. Equivalently, using length contraction in the muon's frame, the 3200 m distance is contracted to about $\tfrac{3200}{5.8} \approx 552$ m, which it covers in about 1.9 µs, within its 2.20 µs lifetime. Use one frame consistently.

From the report. Many answers confused frames of reference, for example saying "the muon experiences time dilation" (it does not, in its own frame), or mixed time dilation and length contraction arguments.

Question 14 (3 marks)

In a PET scanner, an electron and a positron at negligible speed annihilate to give two gamma rays travelling in opposite directions.

a. Calculate the total energy released. (2 marks)

b. Explain why the gamma rays travel in opposite directions. (1 mark)

Show worked solution

a. [2 marks]. Both particles' masses are converted:

$$E = mc^2 = (2 \times 9.11 \times 10^{-31}) \times (3.00 \times 10^8)^2 \approx 1.64 \times 10^{-13} \text{ J}.$$

b. [1 mark]. Momentum is conserved. The initial momentum is (about) zero, so the two photons must carry equal and opposite momenta.

From the report. In a, using only one electron's mass was the common error. Part b was poorly answered (15%): most wrongly attributed it to the opposite charges.

Question 15 (8 marks)

An electron microscope beam has a de Broglie wavelength of $2.00 \times 10^{-10}$ m.

a

Calculate the momentum of an electron. (2 marks)

b

The beam passes through a crystal with atomic spacing $3.00 \times 10^{-10}$ m. Sally says no detectable diffraction pattern will form because the wavelength is smaller than the spacing; Max says a useful pattern will form. Who is correct? Justify. (2 marks)

c

600 eV electrons ($\lambda = 5.01 \times 10^{-11}$ m) and X-rays give identical diffraction patterns through the same foil. Explain. (2 marks)

d

Calculate the energy of the X-ray photons in eV. (2 marks)

Show worked solution

a. [2 marks].

$$p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{2.00 \times 10^{-10}} = 3.32 \times 10^{-24} \text{ kg m s}^{-1}.$$

b. [2 marks]

Max. Significant diffraction occurs when the wavelength and the gap (atomic spacing) are of similar size; here they are the same order of magnitude ($\tfrac{\lambda}{w} \approx 0.7$), so a clear pattern forms. A wavelength slightly smaller than the spacing still diffracts noticeably.

c. [2 marks]

The spread of a diffraction pattern depends on the wavelength (relative to the spacing). Identical patterns through the same foil mean the electrons and X-rays have the same wavelength, $5.01 \times 10^{-11}$ m.

d. [2 marks]

$$E = \frac{hc}{\lambda} = \frac{4.14 \times 10^{-15} \times 3.00 \times 10^8}{5.01 \times 10^{-11}} \approx 2.48 \times 10^4 \text{ eV}.$$

From the report. In b, many wrongly believed diffraction does not occur when the wavelength is less than the spacing; a name without a justification scored nothing. In c, few linked the pattern spacing to wavelength. In a and d, using the wrong Planck's constant was the common error.

Question 16 (20 marks)

Nicola shines light through filters onto a photocell and increases the voltage $V$ until the current reaches zero. Results: $(6.0 \times 10^{14}$ Hz, 0.1 V), $(7.0$, 0.5), $(8.0$, 0.9), $(9.0$, 1.2), $(10.0 \times 10^{14}$ Hz, 1.4 V), each ±0.1 V.

a

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

b

Plot the data with uncertainty bars and a straight line of best fit. (3 marks)

c

From the graph find i. Planck's constant (2 marks)

ii

the threshold frequency (1 mark)

iii

the work function. (1 mark)

d

What do the voltage measurements tell you about the photoelectrons? (2 marks)

e

How would a plate of a different metal change the graph? (1 mark)

f

A current-voltage graph with green light shows the current falling to zero at a negative voltage P and levelling off at a maximum N-M. With a blue filter at the same light power, does P move left, right or stay the same? Justify. (2 marks)

g

Is the level N-M for blue light higher, the same or lower? Justify. (2 marks)

h

State one feature of photoelectric results that supports the particle model, explaining why the wave model fails and how the particle model succeeds. (3 marks)

Show worked solution

a. [3 marks]. Controlled: the intensity of the light (or the metal plate, or the light source). Dependent: the stopping voltage. Independent: the frequency of the light (set by the filter).

b. [3 marks]. Plot the five points with ±0.1 V bars and rule a line that balances all the points, not one forced through the first and last points:

c. i. [2 marks]. Planck's constant is the gradient, using two points on the line, for example $(5.5 \times 10^{14}, 0)$ and $(10 \times 10^{14}, 1.5)$:

$$h = \frac{1.5 - 0}{(10 - 5.5) \times 10^{14}} \approx 3.3 \times 10^{-15} \text{ eV s}.$$

ii. [1 mark]

The frequency-axis intercept: about $5.5 \times 10^{14}$ Hz.

iii. [1 mark]

The magnitude of the voltage-axis intercept (extend the line): about 1.8 eV.

d. [2 marks]

The stopping voltage measures the maximum kinetic energy of the emitted photoelectrons: $E_{k\max} = qV_{\text{stop}}$, so a stopping voltage of 1.4 V means the fastest electrons have 1.4 eV.

e. [1 mark]

The line would shift (left or right, with different intercepts) but keep the same gradient, since the gradient is Planck's constant and does not depend on the metal.

f. [2 marks]

P moves to the left. Blue light has a higher frequency, so each photon has more energy ($E = hf$). The photoelectrons have a higher maximum kinetic energy, so a larger reverse (stopping) voltage is needed.

g. [2 marks]

Lower. The light power is the same but each blue photon carries more energy, so fewer photons arrive per second. Fewer photons release fewer photoelectrons, so the maximum photocurrent is lower.

h. [3 marks]

For example, the threshold frequency. The wave model predicts that light of any frequency should eventually eject electrons if it is intense enough (energy builds up over time). In fact, below a threshold frequency no electrons are emitted, however intense the light. The particle model explains this: each electron absorbs one photon of energy $hf$, and if $hf$ is less than the work function, no electron can escape. (The lack of a time delay, or the maximum kinetic energy depending on frequency not intensity, can be argued the same way.)

From the report. In b, ruling the line through the first and last points was the most common error. In c.i, the gradient calculation had to be clearly shown using points on the line, not forced towards $4.14 \times 10^{-15}$. Part g (3% full marks) was the hardest: many repeated an old answer that the line stays the same, ignoring the equal-power condition. In h, fewer than 30% explained both the wave prediction and the particle explanation.

General advice from the 2024 report

• Show working on every multi-mark question; the working in the report is the minimum expected.
• Justify choices. When you pick between two options (as in Question 15b), a name without an explanation earns nothing.
• Lines of best fit use all the points, not just the first and last.
• Plan longer questions first. Decide what you will calculate and what you will compare it with before you start (Question 6).