IB Physics HL E.2 Quantum physics Questions

SLSL HLHL

Practise IB Physics HL E.2 by applying photon, matter-wave, uncertainty and quantum-state models to evidence from spectra and microscopic experiments.

114 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Define photoelectric emission as electrons released from a metal by incident electromagnetic radiation.
Use E = hf to explain threshold emission: each photon must supply the work function; more insufficient photons do not supply a larger energy per photon.
Explain prompt emission by a single photon transferring its energy to one electron; weak classical waves would require slow energy accumulation.
When comparing classical and photon models, explain each observation fully; a rate increasing with intensity alone does not distinguish the models.
At fixed frequency, increasing intensity increases photon arrivals and emitted electron rate while leaving individual photon energy and Kmax unchanged.
At fixed intensity and beam area, higher photon energy means fewer photons per second and a lower ideal saturation current; state the reason.
For an intensity-only I–V change, keep the stopping intercept unchanged and change the current level with a similar curve shape.
For the same metal, equal stopping intercepts indicate equal photon frequencies; current levels distinguish intensities.
Explain zero collected current at large negative collector voltage using repulsion and a potential barrier greater than Kmax; emission may still occur.
Explain saturation with one complete argument: attraction collects all emitted electrons, or the maximum emission rate is limited by light intensity.
For a stopping-potential circuit, make the detector negative relative to the emitting surface, use a series ammeter and connect a voltmeter in parallel.
Monochromatic light can emit electrons with different KE because they are not equally bound; distinguish their KE from the maximum value.
In silicon pixels, photons liberate electrons internally and charge moves to an electrode; do not describe vacuum emission or import a capacitance formula.
For filtered sunlight reaching a receding spacecraft, fewer photons reduce charge-loss rate while the fixed photon frequency leaves electron KE unchanged.
A shorter wavelength may cross the emission threshold; intensity changes alone do not raise photon energy. Do not turn “may” into “always”.
For an emission table, state that photon energy is below the work function; unchanged yes/no outcomes after doubling intensity do not mean unchanged current.
Read f₀ from the frequency-axis intercept of Kmax against f; use λ₀ = c/f₀ = hc/Φ and the positive work-function magnitude.
With photon arrival rate fixed, use zero emission below threshold and constant ideal current above it; this condition differs from fixed intensity.
Apply Kmax = hf − Φ = hc/λ − Φ in one energy unit, converting joules and eV explicitly when required.
Define Φ as the minimum energy to remove a surface electron or the energy to remove the least tightly bound electron.
Near threshold, explain that photon energy is spent overcoming the surface escape potential energy, leaving almost no electron KE.
Calculate maximum speed from ½mₑv² = hf − Φ, subtracting escape energy before taking the square root.
Use Kmax = eVs with the stopping-voltage magnitude; an I–V intercept below zero does not imply negative kinetic energy.
Distinguish emitted KE from KE at a collector: a retarding 2 V removes 2 eV, so 6 eV photons and Φ = 3 eV leave 1 eV at that collector.
Recover incident photon energy by adding Φ and Kmax before finding λ; include the required unit in the final wavelength.
From Kmax against f, identify h as the gradient and Φ as the magnitude of the negative energy intercept; use experimental readings rather than quoting h.
Check graph units: energy in eV gives h in eV s, while stopping voltage against f has gradient h/e; convert to J s for SI h.
For a changed metal, keep the Kmax–f slope fixed: larger Φ shifts the line right and smaller Φ shifts it left with a positive threshold.
Retain the work-function offset: doubling f does not simply double Kmax, and changing 2f₀ to 4f₀ changes Kmax by a factor of three.
At fixed photon arrival rate, red-to-blue light keeps the ideal saturation current unchanged and makes the stopping voltage more negative.
For a specified emission probability, divide incident power by photon energy, apply that probability and convert the area units before reporting the rate.
Use preceding supplied results at their stated precision; a show-that value of about 10 eV is a rounded photon energy, not an exact atomic gap.
Identify electron diffraction as matter-wave evidence, distinct from photoelectric emission, atomic line spectra or Rutherford alpha scattering.
Explain bright and dark electron rings or angular intensity minima through interference or diffraction, then link the effect explicitly to wave behaviour.
Recognize crystal probing and electron double-slit fringes as electron-wave behaviour; specify the particle or photon property an experiment tests.
Use the supplied particle wavelength in sin θ₀ = λ/D to locate the first minimum; keep degrees and radians distinct.
State the de Broglie hypothesis for all particles: associated wavelength λ = h/p, with h and momentum p defined.
Compare a spatially extended electron wave with a localized electron on a definite classical Bohr trajectory; keep the model distinction qualitative.
Distinguish diffraction from particle descriptions of emitted or oscillating electrons without treating classical radiation as exclusive proof against waves.
For non-relativistic motion combine λ = h/p, p = mv and p² = 2mK; at fixed mass λ is proportional to K⁻¹ᐟ².
For acceleration from rest use K = qV and λ = h/√(2mqV); include alpha charge 2e and use inverse-square-root voltage ratios.
At equal KE, heavier non-relativistic particles have shorter wavelengths; equal charge and accelerating voltage give equal KE but different wavelengths.
If an electron and a photon have equal wavelengths, they share momentum; for photon energy E, that momentum is E/c.
Explain high-energy probing through larger momentum and shorter wavelength matching small target scales; lower energy may fail to resolve nuclear structure.
Use λ ≈ hc/E when explicitly supplied for a high-energy electron; do not replace it with non-relativistic p = mv in that regime.
Higher electron accelerating voltage shortens λ and brings diffraction maxima closer together; nuclear probing needs wavelengths around 10⁻¹⁵ m.
Evaluate supplied orbit or model expressions directly: the stated n = 3 circumference-to-wavelength ratio is about 3.
Use dimensional analysis to check an energy expression: among the listed options h²/(md²) has energy units, without deriving a confinement law.
Explain Compton’s additional particle evidence through photon momentum transfer in a collision, compared with photoelectric evidence for photon energy.
For an electron initially at rest, recoil KE equals photon energy lost: hc/λᵢ − hc/λf, with λf = λᵢ + Δλ.
A photon transferring energy to a recoiling electron has lower energy and longer wavelength; explain the change using E = hc/λ.
For the scattered photon, λ increases and f decreases while λf remains c, so the final-to-initial λf ratio is one.
Calculate Δλ = λf − λᵢ and use Δλ = h/(mₑc)(1 − cos θ), with θ measured for the photon relative to its incident direction.
A shift h/(2mₑc) gives cos θ = 1/2 and θ = 60°; the photon frequency decreases.
In the Compton energy balance, calculate the scattered photon energy using its final wavelength; 7.47 pm gives 0.166 MeV before rounding.

Covered syllabus points

E.2.1 (HL)—Photoelectric effect E.2.2 (HL)—Threshold frequency E.2.3 (HL)—Photoelectric equation E.2.4 (HL)—Particle diffraction E.2.5 (HL)—Matter wave-particle duality E.2.6 (HL)—de Broglie wavelength E.2.7 (HL)—Compton scattering E.2.8 (HL)—Photon wavelength increase E.2.9 (HL)—Compton wavelength shift

Exam points

Define photoelectric emission as electrons released from a metal by incident electromagnetic radiation; distinguish emitted electrons from emitted photons.
Use E = hf: below threshold, each photon lacks the work function energy; increasing intensity cannot make an insufficient photon eject an electron.
Explain prompt emission by one electron absorbing a single photon’s energy; classical weak waves would require time to accumulate escape energy.
At fixed frequency above threshold, increasing intensity raises photon and electron arrival rates; it does not raise individual photon energy or Kmax.
At fixed intensity and beam area, higher photon energy means fewer photons per second; explain the reduced ideal saturation current, not just its sign.
For an intensity-only change, draw a similar I–V curve with a changed current level and the same stopping-voltage intercept.
Read equal stopping-voltage intercepts as equal photon frequencies for the same metal; different current levels indicate different intensities.
Connect a negative collector potential to an opposing energy barrier: current vanishes when even the fastest emitted electrons cannot reach the collector.
Explain saturation with one complete argument: positive collector attraction collects all emitted electrons, so a larger voltage adds no further current.
For a stopping-potential circuit, make the detector negative relative to the emitting surface; use a series ammeter and a parallel voltmeter.
Distinguish maximum and individual photoelectron energies: monochromatic light can eject electrons with different KE because they are not equally bound.
Identify threshold behaviour, frequency-dependent Kmax and prompt emission as photon evidence; an emission rate rising with intensity alone is insufficient.
For a receding spacecraft receiving filtered sunlight, link fewer incident photons per second to a lower rate of photoelectric charge loss.
In a CCD context, describe photons liberating electrons within silicon and charge moving to an electrode; do not describe electrons escaping into vacuum.

[Maximum number: 4]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

(a)

The diagram shows part of an experimental arrangement used to investigate the photoelectric effect.

[ 2 ]

(i)

On the diagram, draw the power supply and other necessary components needed in order to carry out the experiment in (b)(i).

[ 2 ]

(b)

In an experiment, light at a particular frequency is incident on a surface and electrons are emitted. Explain what happens to the number of electrons emitted per second when the intensity of this light is increased.

[ 2 ]

Exam points

Use E = hf and the work function to explain threshold emission: a single photon must supply the minimum energy needed to release an electron.
When no electrons are emitted, a shorter wavelength may cross the threshold; intensity changes alone do not increase an individual photon’s energy.
A longer wavelength lowers photon energy and may stop emission; explain a violet-to-red change by comparing photon energy with the work function.
For non-emission comparisons, explicitly state that photon energy is below the work function; doubling intensity does not change that energy.
Read threshold frequency from the frequency-axis intercept of Kmax against f, extrapolating the line and retaining the axis power of ten.
For a higher work function, shift the Kmax–f line parallel to the right: the threshold frequency rises while the gradient remains h.
Find the longest ejecting wavelength using λ₀ = hc/Φ = c/f₀; convert eV to joules when using SI constants.
With the photon arrival rate held constant, use zero emission below threshold and constant ideal emitted current above it; intensity is not held fixed.

[Maximum number: 3]

This question is in two parts. Part 1 is about the oscillation of a mass. Part 2 is about the photoelectric effect.
Part 1 Oscillation of a mass
A mass of 0.80 kg rests on a frictionless surface and is connected to two identical springs both of which are fixed at their other ends. A force of 0.030 N is required to extend or compress each spring by 1.0 mm . When the mass is at rest in the centre of the arrangement, the springs are not extended.

(a)

Part 2 Photoelectric effect
A student carries out a photoelectric experiment in which radiation is incident on a metal surface in a vacuum.

Explain why photoelectrons are not emitted from the metal surface unless the frequency of incident light exceeds a minimum value.

[ 2 ]

(b)

A graph of the results of the experiment show how the maximum kinetic energy $E_{\text {max }}$ of the emitted photoelectrons varies with the frequency f of the incident radiation.

Use the graph to

[ 1 ]

(i)

identify the minimum value of the frequency $f_{0}$ for photoelectrons to be emitted.

[ 1 ]

(value of intercept on f-axis $=) 9.3 \times 10^{14} \mathrm{~Hz}$ ;
Allow answers in the range of 9.0 to 9.5, but with correct power of ten.

(ii)

determine the Planck constant.

[ 3 ]

attempted use of gradient $\left(\frac{h}{e}\right)$ to evaluate h;
eg $\frac{2.4}{6 \times 10^{14}}=4 \times 10^{-15} \mathrm{eVs}$ ; (allow answers in the range of 3.9 to $4.2 \times 10^{-15}$ ) $6.4 \times 10^{-34} \mathrm{Js}$ ; (allow ECF, eg answers in the range of 6.2 to $6.7 \times 10^{-34}$ )

Marking guidance:

Do not allow a bald answer of $6.6 \times 10^{-34}$ as value is known.
or
use of y-intercept $=\frac{-h f_{0}}{e}$ ;
$h=\frac{3.77 \times 1.6 \times 10^{-19}}{0.93 \times 10^{15}} ;($ allow intercept between -3.6 and -3.8 eV)
$6.4 \times 10^{-34} \mathrm{Js}$ ; (allow ECF, eg answers in the range of 6.0 to $6.8 \times 10^{-34}$ )
Do not allow a bald answer of $6.6 \times 10^{-34}$ as value is known.

(iii)

calculate the work function, in eV , for the metal surface.

[ 2 ]

$E=6.62 \times 10^{-34} \times 9.3 \times 10^{14}\left(=6.2 \times 10^{-19} \mathrm{~J}\right)$ ;
an answer in the range of 3.4 to 4.0 eV ;
(allow ECF from (e)(i) and (e)(ii))
or
identifies y intercept as work function;
an answer in the range of 3.6 to 3.8 eV ;

(c)

The student repeats the experiment with a different metal surface that has a smaller value for the work function. On the graph in (e), draw a line to show how $E_{\text {max }}$ varies with f.

[ 2 ]

Exam points

Apply Kmax = hf − Φ = hc/λ − Φ using a single energy unit throughout; convert joules and eV explicitly when the question specifies a unit.
Define the work function as the minimum energy needed to remove an electron from the metal surface, or to remove its least tightly bound electron.
Near threshold, photon energy is spent overcoming the surface escape energy, leaving little KE; state where the energy goes, not merely that it is conserved.
Find maximum speed from ½mₑv² = hf − Φ; subtract the work function before taking the square root.
Determine Kmax experimentally from the stopping-voltage magnitude: Kmax = eVs. Use the negative I–V intercept as a retarding voltage, not a negative KE.
To recover the incident wavelength, add the work function to the maximum electron KE before using λ = hc/Ephoton; include the requested unit.
Distinguish KE at emission from KE at the collector: a retarding potential removes eV from the emitted electron’s kinetic energy.
For Kmax against frequency, the slope is h, the frequency intercept is f₀, and the negative energy intercept has magnitude Φ.
Measure h from the experimental gradient and show the readings and calculation; quoting its known value alone does not meet the experimental task.
Check gradient units: an energy graph in eV gives h in eV s; a stopping-voltage graph gives h/e. Convert to J s when reporting SI h.
When changing metals, keep the Kmax–f gradient unchanged: larger Φ shifts the line right, while smaller Φ shifts it left with a positive threshold.
Higher frequency or lower work function raises Kmax; increasing intensity at fixed frequency changes photon number rather than the maximum electron energy.
Retain the work-function offset in frequency comparisons: doubling f does not simply double Kmax, and 2f₀ to 4f₀ changes Kmax by a factor of three.
At fixed photon arrival rate, changing red light to blue leaves the ideal saturation current unchanged and makes the stopping voltage more negative.
For equal intensity, higher-frequency photons mean fewer arrivals per second; compare this condition separately from a fixed photon arrival rate.
For a stated emission probability, divide incident power by photon energy, then multiply by the probability; convert surface area units before counting.
Use explicitly supplied preceding results in linked tasks, keeping their stated precision and units; do not treat a rounded photon energy as exact.

[Maximum number: 3]

(a)

Part 2 Photoelectric effect
A student carries out a photoelectric experiment in which radiation is incident on a metal surface in a vacuum.

Explain why photoelectrons are not emitted from the metal surface unless the frequency of incident light exceeds a minimum value.

[ 2 ]

(b)

A graph of the results of the experiment show how the maximum kinetic energy $E_{\text {max }}$ of the emitted photoelectrons varies with the frequency f of the incident radiation.

Use the graph to

[ 1 ]

(i)

identify the minimum value of the frequency $f_{0}$ for photoelectrons to be emitted.

[ 1 ]

(value of intercept on f-axis $=) 9.3 \times 10^{14} \mathrm{~Hz}$ ;
Allow answers in the range of 9.0 to 9.5, but with correct power of ten.

(ii)

determine the Planck constant.

[ 3 ]

(iii)

calculate the work function, in eV , for the metal surface.

[ 2 ]

(c)

The student repeats the experiment with a different metal surface that has a smaller value for the work function. On the graph in (e), draw a line to show how $E_{\text {max }}$ varies with f.

[ 2 ]

Exam points

Identify electron diffraction from a crystal as evidence for matter waves; distinguish it from photoelectric emission, energy spectra and Rutherford scattering.
Explain electron diffraction rings through bright and dark maxima and minima, constructive and destructive interference, and interference as a wave property.
Link maxima and minima in scattered-electron intensity to diffraction by nuclei, then state that diffraction demonstrates wave behaviour.
Recognize both crystal probing and electron double-slit fringes as wave behaviour; photoelectric emission is a different kind of evidence.
Estimate the first minimum using sin θ₀ = λ/D in the stated scattering model; use the particle wavelength and keep radians and degrees distinct.

[Maximum number: 3]

This question is in two parts. Part 1 is about thermal properties of matter. Part 2 is about quantum physics.

Part 1 Thermal properties of matter

Part 2 Quantum physics

The diagram shows the end of an electron diffraction tube.

A pattern forms when diffracted electrons are incident on a fluorescent layer at the end of the tube.

Explain how the pattern demonstrates that electrons have wave properties.

Exam points

State de Broglie’s hypothesis for all particles: each has an associated wavelength λ = h/p; define h as Planck’s constant and p as momentum.
Compare the spatially extended electron-wave description with a localized electron following a definite Bohr orbit; state both sides of the model comparison.
Match electron diffraction to the wave nature of electrons and the photoelectric effect to the particle nature of light; keep the object being tested explicit.
Distinguish electron diffraction from particle descriptions of emitted electrons or oscillating charges; the same electron can require different models.

[Maximum number: 2]

This question is in two parts. Part 1 is about momentum and energy. Part 2 is about the de Broglie hypothesis and radioactive decay.

Describe the de Broglie hypothesis.

Exam points

Use λ = h/p and define p as momentum; when momentum is supplied, no confinement or uncertainty model is needed.
For non-relativistic particles, combine p = mv and K = p²/(2m) with λ = h/p; at fixed mass, λ varies as K⁻¹ᐟ².
For acceleration from rest, use K = qV and λ = h/√(2mqV); doubling V divides λ by √2, while quadrupling V halves it.
At the same kinetic energy, a heavier non-relativistic particle has a shorter wavelength: λₑ/λₚ = √(mₚ/mₑ).
Equal electron and photon wavelengths imply equal momenta; for a photon of energy E, that shared momentum is E/c.
Explain probe resolution with a complete chain: larger accelerating voltage gives higher energy and momentum, shorter wavelength and access to smaller scales.
When λ ≈ hc/E is explicitly supplied for a high-energy electron, use it with consistent energy units; do not substitute p = mv in that regime.
Read the first diffraction minimum, then use D = λ/sin θ in the stated nuclear-scattering model; preserve reading tolerances and error carried forward.
Higher accelerating voltage shortens the electron wavelength and brings diffraction maxima closer together.
Calculate orbit circumference divided by a supplied electron wavelength using the given radius; for the stated n = 3 orbit, the ratio is about 3.
Use dimensional analysis to reject proposed energy expressions: h²/(md²) has energy units, while the other listed expressions do not.
Rearrange an explicitly supplied model equation with its stated lowest integer value to estimate a length scale; keep that estimate tied to the model.

[Maximum number: 5]

(a)

The existence of atomic energy levels can be understood by applying the de Broglie hypothesis to an electron confined to move in one dimension in a box of length L.

[ 4 ]

(i)

State the de Broglie hypothesis as it applies to an electron.

[ 1 ]

(ii)

Show that the energy $E_{\mathrm{n}}$ of the electron in the box is given by

E_{\mathrm{n}}=\frac{n^{2} h^{2}}{8 m_{\mathrm{e}} L^{2}}

where $n=1,2,3 \ldots, h$ is the Planck constant and $m_{\mathrm{e}}$ is the mass of the electron.

[ 3 ]

(b)

The magnitude of the least energy that an electron in the hydrogen atom can have is $2.2 \times 10^{-18} \mathrm{~J}$ . Estimate, using the equation in (b)(ii), the radius of the hydrogen atom.

[ 1 ]

$1.7 \times 10^{-10} \mathrm{~m}$ ;

[Maximum number: 3]

(a)

Outline why Compton scattering offers better evidence for the particle nature of light than the photoelectric effect.

A photon of wavelength $6.40 \times 10^{-12} \mathrm{~m}$ and energy 0.194 MeV is incident on an electron at rest. The photon scatters at an angle $\theta$ with wavelength $7.47 \times 10^{-12} \mathrm{~m}$ . The electron recoils.

[ 2 ]

(b)

Calculate the kinetic energy of the recoiling electron.

[ 1 ]

$E=\lessdot 0.194-0.166=>0.028 \mathrm{MeV} \quad$

Accept 0.024 from the value given as show that.

Allow ECF from c) i)

Accept $4.5 \times 10^{-15} \mathrm{~J}$ .

(c)

Show that the energy of the scattered photon is about 0.17 MeV .

[ 1 ]

$E=<\frac{h c}{\lambda}=>\frac{1.24 \times 10^{-6}}{7.47 \times 10^{-12}}$ OR 0.166 MeV

(d)

For an electron, $\frac{h}{m_{\mathrm{e}} c}=2.43 \times 10^{-12} \mathrm{~m}$ . Determine $\theta$ .

[ 2 ]

$7.47 \times 10^{-12}-6.40 \times 10^{-12}=2.43 \times 10^{-12}(1-\cos \theta)^{}$

« $\cos \theta=0.5597$ so » $\theta=56.0^{\underline{\mathrm{o}}}$

Accept angle in radians ( 0.98 « rad »).

Award [2] if 56 « ${ }^{\circ}$ » or 0.98 « rad» are seen as the answer without working

[Maximum number: 2]

Outline why the wavelength of the scattered photon is longer than that of the incident photon.

The photon loses energy / transfers energy to the electron

$E=\frac{h c}{\lambda} /$ lower energy implies longer wavelength

[Maximum number: 3]

(a)

[ 2 ]

(b)

Calculate the kinetic energy of the recoiling electron.

[ 1 ]

$E=\lessdot 0.194-0.166=>0.028 \mathrm{MeV} \quad$

Accept 0.024 from the value given as show that.

Allow ECF from c) i)

Accept $4.5 \times 10^{-15} \mathrm{~J}$ .

(c)

Show that the energy of the scattered photon is about 0.17 MeV .

[ 1 ]

$E=<\frac{h c}{\lambda}=>\frac{1.24 \times 10^{-6}}{7.47 \times 10^{-12}}$ OR 0.166 MeV

(d)

For an electron, $\frac{h}{m_{\mathrm{e}} c}=2.43 \times 10^{-12} \mathrm{~m}$ . Determine $\theta$ .

[ 2 ]

$7.47 \times 10^{-12}-6.40 \times 10^{-12}=2.43 \times 10^{-12}(1-\cos \theta)^{}$

« $\cos \theta=0.5597$ so » $\theta=56.0^{\underline{\mathrm{o}}}$

Accept angle in radians ( 0.98 « rad »).

Award [2] if 56 « ${ }^{\circ}$ » or 0.98 « rad» are seen as the answer without working

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