E.2 Quantum physics (HL) – IB Physics exam questions with mark scheme

1HardHL only5 marks

Light of wavelength 400 nm is incident on a sodium surface of work function 2.28 eV.

(a) Calculate the energy of one photon in eV. [1]

(b) Calculate the maximum kinetic energy of the emitted photoelectrons in eV. [2]

(c) The emitted photoelectrons travel towards a collecting electrode. A potential difference is applied that just prevents the fastest photoelectrons from reaching the collecting electrode, so the photocurrent falls to zero. State the value of this stopping potential difference. [1]

(d) The intensity of the light is doubled at constant wavelength. State the effect on the maximum kinetic energy and on the number of photoelectrons emitted per second. [1]

(a) E = hc/λ = 4.97 × 10⁻¹⁹ J = 3.11 eV (allow 3.10 eV)(1)
(b) Einstein's equation: hf = Φ + KEmax(1)
(b) KEmax = 3.11 − 2.28 = 0.83 eV (allow 0.82–0.83 eV)(1)
(c) Stopping potential = 0.83 V (numerically equal to KEmax in eV; allow ECF from (b); unit V required)(1)
(d) KEmax unchanged; the number of photoelectrons emitted per second doubles (accept “increases”) (both needed)(1)
Answer: (a) 3.11 eV; (b) 0.83 eV; (c) 0.83 V; (d) KEmax same, rate doubles
Examiner tip: Intensity changes the number of photons per second, not the energy of each photon.

2CoreHL only3 marks

The work function of zinc is 4.3 eV.

(a) Calculate the threshold frequency for zinc. [2]

(b) Violet light of frequency $7.3\times10^{14}$ Hz shines on a clean zinc plate. State and explain whether photoelectrons are emitted. [1]

(a) Converts work function to joules: $\Phi=4.3\times1.60\times10^{-19}=6.9\times10^{-19}$ J and uses $hf_0=\Phi$(1)
(a) $f_0=\dfrac{6.88\times10^{-19}}{6.63\times10^{-34}}=1.0\times10^{15}$ Hz (allow $1.04\times10^{15}$ Hz; award [2] for a bare correct answer; no conversion from eV, giving $6.5\times10^{33}$ Hz, scores [0] for this mark)(1)
(b) No: $7.3\times10^{14}$ Hz is below the threshold frequency / each photon's energy (3.0 eV) is less than the work function, so no electron can gain enough energy to escape (allow ECF from (a); answer of 'no' without a reason scores [0])(1)
Answer: (a) $1.0\times10^{15}$ Hz (b) No, $f$ is below $f_0$ (photon energy 3.0 eV < 4.3 eV)
Examiner tip: Mark (b) needs the comparison with the threshold frequency or the work function; 'the light is not bright enough' scores nothing.

3StandardHL only5 marks

In an experiment on the photoelectric effect, a metal surface is illuminated with light of different frequencies $f$. The maximum kinetic energy $E_{\text{max}}$ of the emitted photoelectrons is found for each frequency. The graph shows the results.

(a) Use the graph to determine a value for the Planck constant. [2]

(b) Determine the work function of the metal, in eV. [2]

(c) The experiment is repeated with a second metal that has a larger work function. State how the line for the second metal compares with the line on the graph. [1]

(a) Gradient from a large triangle on the line, e.g. $\dfrac{2.9\text{ eV}}{(12-5)\times10^{14}\text{ Hz}}=4.1\times10^{-15}$ eV s (accept any correct pair of points on the line)(1)
(a) Converts to J s: $h=4.14\times10^{-15}\times1.60\times10^{-19}=6.6\times10^{-34}$ J s (accept $6.5$ to $6.8\times10^{-34}$ J s; unit not required if value correct; gradient left in eV s scores [1] max)(1)
(b) Threshold frequency read from the $f$-intercept: $f_0=5.0\times10^{14}$ Hz (or extrapolates the line to the $E_{\text{max}}$ axis, intercept $-\Phi$)(1)
(b) $\Phi=hf_0=4.14\times10^{-15}\times5.0\times10^{14}=2.1$ eV (accept 2.0 to 2.2 eV; allow ECF from (a))(1)
(c) A parallel line / same gradient, displaced to the right (larger threshold frequency, cuts the $f$-axis at a higher value) (both ideas needed)(1)
Answer: (a) $h\approx6.6\times10^{-34}$ J s (b) $\Phi\approx2.1$ eV (c) parallel line, shifted to higher $f$
Examiner tip: The gradient is in eV per Hz, so it must be multiplied by $e$ to give $h$ in J s; the gradient is the same for every metal because it is $h$ itself.

Stuck on E.2 Quantum physics (HL)?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4CoreHL only3 marks

An electron, initially at rest, is accelerated through a potential difference of 150 V.

(a) Show that the momentum of the electron is about $6.6\times10^{-24}$ kg m s$^{-1}$. [2]

(b) Calculate the de Broglie wavelength of the electron. [1]

(a) $E_k=eV=1.60\times10^{-19}\times150=2.4\times10^{-17}$ J(1)
(a) $p=\sqrt{2m_eE_k}=\sqrt{2\times9.11\times10^{-31}\times2.4\times10^{-17}}=6.61\times10^{-24}$ kg m s$^{-1}$ (a 'show that' needs the answer to at least 3 s.f. or the full substitution seen)(1)
(b) $\lambda=\dfrac{h}{p}=\dfrac{6.63\times10^{-34}}{6.61\times10^{-24}}=1.0\times10^{-10}$ m (accept 0.10 nm; use of $6.6\times10^{-24}$ gives the same answer)(1)
Answer: (a) $6.61\times10^{-24}$ kg m s$^{-1}$ (b) $1.0\times10^{-10}$ m
Examiner tip: In a 'show that' question the final value must be given to more significant figures than the value quoted in the question.

5HardHL only6 marks

The diagram shows an electron diffraction tube. Electrons emitted from the hot cathode are accelerated through a potential difference $V$ and pass through a thin film of graphite. Bright concentric rings are seen on the fluorescent screen.

(a) Explain how the pattern on the screen provides evidence for the wave nature of electrons. [2]

(b) $V$ is 3.0 kV. Show that the de Broglie wavelength of the electrons is about $2\times10^{-11}$ m. [2]

(c) With $V=3.0$ kV the radius $r$ of the inner ring is 24 mm. $V$ is increased to 4.5 kV. Assuming the angles of diffraction are small, predict the new radius of the inner ring. [2]

(a) Rings are a diffraction (interference) pattern: maxima and minima of intensity / electrons only arrive at certain angles(1)
(a) Diffraction is a wave property and occurs here because the electron wavelength is comparable to the spacing of the atoms (planes) in graphite(1)
(b) $p=\sqrt{2m_eeV}=\sqrt{2\times9.11\times10^{-31}\times1.60\times10^{-19}\times3.0\times10^{3}}=2.96\times10^{-23}$ kg m s$^{-1}$(1)
(b) $\lambda=h/p=2.24\times10^{-11}$ m (answer to at least 2 s.f. better than the 'about' value, i.e. 2.2 or 2.24)(1)
(c) $\lambda\propto 1/\sqrt{V}$ and $r\propto\sin\theta\propto\lambda$ (small angles), so $r\propto1/\sqrt{V}$(1)
(c) $r=24\times\sqrt{3.0/4.5}=20$ mm (accept 19.6 mm; an answer of 16 mm, from $r\propto1/V$, scores [1] max for the idea that $r$ decreases)(1)
Answer: (a) rings = diffraction, a wave effect, since $\lambda$ is comparable to atomic spacing (b) $2.2\times10^{-11}$ m (c) 20 mm
Examiner tip: Ring radius scales with wavelength, and wavelength scales with $1/\sqrt{V}$, not $1/V$: many candidates forget the square root.

6StandardHL only5 marks

X-ray photons of wavelength 0.0710 nm are scattered by loosely bound electrons in a carbon block. Photons scattered through an angle of 90° are detected.

(a) Calculate the wavelength of the scattered photons. [2]

(b) Calculate the kinetic energy given to an electron that scatters a photon through 90°. Give your answer in eV. [3]

(a) $\Delta\lambda=\dfrac{h}{m_ec}(1-\cos90°)=\dfrac{6.63\times10^{-34}}{9.11\times10^{-31}\times3.00\times10^{8}}=2.43\times10^{-12}$ m(1)
(a) $\lambda'=0.0710+0.00243=0.0734$ nm / $7.34\times10^{-11}$ m (award [2] for a bare correct answer; giving only $\Delta\lambda$ scores [1])(1)
(b) Energy of incident photon $hc/\lambda=2.80\times10^{-15}$ J and of scattered photon $hc/\lambda'=2.71\times10^{-15}$ J(1)
(b) Electron's kinetic energy = difference in photon energies (energy conserved) $=9.3\times10^{-17}$ J (allow ECF from (a))(1)
(b) $=\dfrac{9.26\times10^{-17}}{1.60\times10^{-19}}=580$ eV (accept 570 to 590 eV; premature rounding of $\lambda'$ to 2 s.f. that gives a very different value scores [2] max)(1)
Answer: (a) 0.0734 nm (b) about 580 eV
Examiner tip: Keep at least 3 or 4 significant figures in the two photon energies: their difference is small, so early rounding wrecks the answer.

7HardHL only5 marks

Gamma-ray photons of energy 0.662 MeV from a caesium-137 source are Compton scattered by free electrons. Some photons are scattered straight back, through an angle of 180°.

(a) Calculate the change in wavelength of a photon scattered through 180°. [1]

(b) Determine the energy, in MeV, of a photon scattered through 180°. [3]

(c) Visible light of wavelength 500 nm is shone onto the same target. Suggest why the Compton shift is very hard to observe for visible light. [1]

(a) $\Delta\lambda=\dfrac{h}{m_ec}(1-\cos180°)=2\times2.43\times10^{-12}=4.85\times10^{-12}$ m(1)
(b) Incident wavelength $\lambda=\dfrac{hc}{E}=\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{0.662\times10^{6}\times1.60\times10^{-19}}=1.88\times10^{-12}$ m(1)
(b) $\lambda'=1.88\times10^{-12}+4.85\times10^{-12}=6.73\times10^{-12}$ m (allow ECF from (a))(1)
(b) $E'=\dfrac{hc}{\lambda'}=2.96\times10^{-14}$ J $=0.18$ MeV (accept 0.18 to 0.19 MeV)(1)
(c) The shift (about $5\times10^{-12}$ m at most) is the same for any wavelength, so for 500 nm it is a tiny fraction (about $10^{-5}$) of the wavelength / too small to detect (accept: visible photon energy is comparable to binding energies so electrons are not free) (1)
Answer: (a) $4.85\times10^{-12}$ m (b) 0.18 MeV (c) $\Delta\lambda$ is a negligible fraction of 500 nm
Examiner tip: The Compton shift $\Delta\lambda$ does not depend on the incident wavelength, which is why the effect is only noticeable for short-wavelength X-rays and gamma rays.

8HardHL only6 marks

When ultraviolet radiation falls on a clean metal surface, electrons are emitted. Experiments show that:

no electrons are emitted below a certain frequency, however intense the radiation
above this frequency, emission begins with no measurable delay, even at very low intensity
the maximum kinetic energy of the electrons depends on frequency but not on intensity.

Discuss how these observations support the photon model of light and why they cannot be explained by a classical wave model. [6]

Photon model: light consists of photons, each with energy $E=hf$(1)
Each electron absorbs the energy of one photon only (one-to-one interaction)(1)
An electron escapes only if $hf\ge\Phi$ (work function = minimum energy to remove an electron), giving a threshold frequency $f_0=\Phi/h$(1)
Intensity is the number of photons per second (per unit area), so a more intense beam below $f_0$ still has no photon with enough energy(1)
Energy is delivered in a single photon, so emission is instantaneous; a single photon can release an electron even at low intensity(1)
$E_{\text{max}}=hf-\Phi$: depends on $f$ only; greater intensity gives more electrons per second but the same $E_{\text{max}}$(1)
Wave model: energy delivered continuously and spread over the wavefront, with energy depending on amplitude/intensity(1)
Wave model predicts emission at any frequency if intense enough / given enough time(1)
Wave model predicts a time delay at low intensity while an electron accumulates enough energy(1)
Wave model predicts that higher intensity gives more energetic electrons(1)

Level	Marks	What it looks like
Level 3	5–6	Explains all three observations using photons of energy $hf$ absorbed one at a time by single electrons, with the work function and $E_{\text{max}}=hf-\Phi$ used correctly, and contrasts each with a specific failed prediction of the wave model. Reasoning is logical and terms are used accurately.
Level 2	3–4	Explains at least two observations using the photon model, with some contrast with the wave model, but one observation or the wave-model comparison is missing or unclear.
Level 1	1–2	Relevant statements such as '$E=hf$' or 'light is made of photons' with little explanation of how they account for the observations; little or no reference to the wave model.
Level 0	0	No relevant content.

Answer: Photons of energy $hf$ interact one-to-one with electrons: this explains the threshold frequency, instant emission and $E_{\text{max}}=hf-\Phi$ independent of intensity. A wave model, with energy spread continuously and set by intensity, predicts emission at any frequency, delays at low intensity and $E_{\text{max}}$ increasing with intensity.
Examiner tip: Top-band answers deal with each observation twice: what the photon model says, and what the wave model would wrongly predict.

9StretchHL only7 marks

A student uses a vacuum photocell and a mercury lamp with filters to measure the stopping potential $V_s$ for four wavelengths $\lambda$. The table shows her results.

$\lambda$ / nm	365	405	436	546
$V_s$ / V	1.31	0.97	0.75	0.18

(a) Explain what is meant by the stopping potential. [1]

(b) Show that $V_s=\dfrac{hc}{e}\cdot\dfrac{1}{\lambda}-\dfrac{\Phi}{e}$, where $\Phi$ is the work function of the cathode. [1]

(c) The student plots a graph of $V_s$ against $1/\lambda$. Using the data, determine the gradient of this graph and hence a value for the Planck constant. [3]

(d) Determine the work function of the cathode in eV. [1]

(e) Another student says the values of $V_s$ would all increase if the lamp were moved closer to the photocell. Comment on this claim. [1]

(a) The minimum (reverse) potential difference between collector and cathode that stops the most energetic photoelectrons reaching the collector, so the photocurrent just becomes zero(1)
(b) $eV_s=E_{\text{max}}$ and $E_{\text{max}}=hf-\Phi=hc/\lambda-\Phi$, so dividing by $e$ gives the result(1)
(c) Calculates $1/\lambda$ values, e.g. $2.74\times10^{6}$ m$^{-1}$ (365 nm) and $1.83\times10^{6}$ m$^{-1}$ (546 nm)(1)
(c) Gradient $=\dfrac{1.31-0.18}{(2.74-1.83)\times10^{6}}=1.24\times10^{-6}$ V m (accept 1.2 to 1.3 $\times10^{-6}$ V m; best-fit or any two widely spaced points accepted)(1)
(c) $h=\dfrac{\text{gradient}\times e}{c}=\dfrac{1.24\times10^{-6}\times1.60\times10^{-19}}{3.00\times10^{8}}=6.6\times10^{-34}$ J s (accept 6.4 to 6.9 $\times10^{-34}$ J s; allow ECF from gradient)(1)
(d) Uses one data pair, e.g. $\Phi/e=1.24\times10^{-6}\times2.74\times10^{6}-1.31$, so $\Phi=2.1$ eV (accept 2.0 to 2.2 eV; or from the magnitude of the $V_s$-intercept)(1)
(e) Claim is wrong: moving the lamp closer increases intensity, i.e. more photons per second (larger photocurrent), but photon energy and so $E_{\text{max}}$ and $V_s$ are unchanged(1)
Answer: (a) see mark scheme (c) gradient $1.24\times10^{-6}$ V m, $h\approx6.6\times10^{-34}$ J s (d) $\Phi\approx2.1$ eV (e) incorrect: $V_s$ depends on $f$, not intensity
Examiner tip: Convert nm to m before finding $1/\lambda$; a gradient in V nm leads to an answer for $h$ that is out by $10^{9}$.

10StretchHL only5 marks

A beam of electrons and a beam of protons each have a de Broglie wavelength of 0.15 nm, similar to the spacing of atoms in a crystal.

(a) Calculate the speed of the electrons. [2]

(b) Calculate the kinetic energy of the electrons in eV. [1]

(c) Without calculating the proton's speed, determine the ratio $\dfrac{\text{kinetic energy of an electron}}{\text{kinetic energy of a proton}}$, and hence the kinetic energy of a proton in the beam, in eV. [2]

(a) $p=\dfrac{h}{\lambda}=\dfrac{6.63\times10^{-34}}{1.5\times10^{-10}}=4.42\times10^{-24}$ kg m s$^{-1}$(1)
(a) $v=\dfrac{p}{m_e}=\dfrac{4.42\times10^{-24}}{9.11\times10^{-31}}=4.9\times10^{6}$ m s$^{-1}$ (accept $4.85\times10^{6}$)(1)
(b) $E_k=\tfrac12m_ev^2$ or $\dfrac{p^2}{2m_e}=1.07\times10^{-17}$ J $=67$ eV (accept 66 to 68 eV; allow ECF from (a))(1)
(c) Same wavelength so same momentum; $E_k=\dfrac{p^2}{2m}$ so ratio $=\dfrac{m_p}{m_e}=\dfrac{1.673\times10^{-27}}{9.11\times10^{-31}}=1840$ (accept 1836 to 1840)(1)
(c) Proton kinetic energy $=\dfrac{67}{1840}=0.036$ eV (accept 0.036 to 0.037 eV; allow ECF from (b); ratio inverted scores [1] max)(1)
Answer: (a) $4.9\times10^{6}$ m s$^{-1}$ (b) 67 eV (c) ratio 1840; proton $E_k\approx0.036$ eV
Examiner tip: Equal wavelengths mean equal momenta, not equal speeds or energies: use $E_k=p^2/2m$ to compare particles of different mass.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session