2023 H2 Physics Paper 2 Q7 — A Photocell and Its Stopping Potential

The model, live

Current at this intensityOriginal intensityStopping potentialEnergy taken by the p.d.

Part (a)(i)

1 mark

Name a phenomenon that is evidence for the particulate nature of electromagnetic radiation.

The photoelectric effect.

Part (a)(ii)

1 mark

Name a phenomenon that is evidence for the wave nature of electromagnetic radiation.

Diffraction (or interference).

Watch

Reflection and refraction are not accepted: particles can be pictured doing both. Diffraction and interference are uniquely wave behaviour.

Part (b)(i)

1 mark

The supply is turned up until the current just falls to zero. Mark the polarity of the supply.

Approach

To stop the photoelectrons, the supply must push back on them: the collector must be negative relative to the metal electrode, so that electrons arriving at it are repelled.

The terminal connected (through the ammeter) to the metal electrode is $+$ ; the terminal connected to the collector electrode is $-$ .

Part (b)(ii)

2 marks

The photon energy is $12.4\,\text{eV}$ and the work function is $4.2\,\text{eV}$ . Find the stopping potential.

Approach

Einstein's photoelectric equation gives the maximum kinetic energy. The stopping potential is the p.d. that removes exactly that much energy from an electron, $eV_s = E_{k,\max}$ . Working in electronvolts makes the last step a relabelling.

E_{k,\max} = hf - \Phi = 12.4 - 4.2 = 8.2\,\text{eV}

eV_s = E_{k,\max} \quad\Longrightarrow\quad V_s = 8.2\,\text{V}

stopping potential $= 8.2\,\text{V}$

Part (b)(iii)

3 marks

Explain why the stopping potential does not depend on the intensity of the radiation.

Approach

Build the chain in three links, one per mark: the stopping potential is set by the maximum kinetic energy; that is set by the energy of one photon, because one electron absorbs one photon; and intensity changes the number of photons, not the energy of each.

The stopping potential depends only on the maximum kinetic energy of the photoelectrons, $eV_s = hf - \Phi$ . Each electron gains its energy by absorbing a single photon, whose energy $hf$ depends only on the frequency. Increasing the intensity increases the number of photons arriving per second (and so the number of photoelectrons), but not the energy of each photon, so $E_{k,\max}$ and $V_s$ are unchanged.

Part (c)(i)

1 mark

Label the stopping potential $V_S$ on the I–V graph.

$V_S$ is marked on the negative $V$ axis where the curve first reaches zero current.

Part (c)(ii)

2 marks

Sketch the I–V graph when the intensity is increased.

A curve of the same shape that starts from zero at the same stopping potential $-V_S$ and levels off at a higher saturation current.

Note

The two marks are separate: one for the same intercept on the $V$ axis, one for the higher plateau. The plateau is higher because more photoelectrons are emitted each second, and at saturation all of them reach the collector.

What this question was really testing

Particle and wave evidence. Photoelectric effect; diffraction or interference.
What "stopping" requires. The collector negative, pushing the electrons back.
$eV_s = hf - \Phi$ . In electronvolts the arithmetic is one subtraction.
One photon, one electron. Intensity is the number of photons; the energy of each is set by frequency alone.

Turn up the light, and the stopping potential doesn't move

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

Part (c)(i)

Part (c)(ii)

What this question was really testing

This is how we mark, and how we teach.