IB Physics Quiz: Apply Quantum Physics

20 questions · exam conditions

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Apply Quantum PhysicsQuestion 1 of 20

A proton and an alpha particle are both accelerated from rest through the same potential difference. What is the ratio of the de Broglie wavelength of the proton (λₚ) to that of the alpha particle (λₐ)? (An alpha particle has mass ≈ 4mₚ and charge +2e).

2√2

1/√2

1/2

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IB Physics Quiz

IB Physics Quiz: Apply Quantum Physics

Practice Apply Quantum Physics in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Quantum Physics, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

2
2√2 (correct answer)
1/√2
1/2

Explanation: The kinetic energy gained by a particle of charge q accelerated through a potential difference V is Eₖ = qV. The de Broglie wavelength is λ = h/p = h/√(2mEₖ). So, λ = h/√(2mqV). For the proton: λₚ = h/√(2mₚeV). For the alpha particle: λₐ = h/√(2(4mₚ)(2e)V) = h/√(16mₚeV). The ratio is λₚ/λₐ = [h/√(2mₚeV)] / [h/√(16mₚeV)] = √(16mₚeV) / √(2mₚeV) = √8 = 2√2.

Question 2

Which pair of phenomena best illustrates the wave-particle duality of light?

Photoelectric effect and Compton scattering
Double-slit interference and photoelectric effect (correct answer)
Double-slit interference and diffraction
Compton scattering and X-ray diffraction

Explanation: Wave-particle duality requires evidence for both wave-like and particle-like behavior. Double-slit interference is a classic demonstration of the wave nature of light. The photoelectric effect, where light energy is transferred in discrete packets (photons), is a key demonstration of its particle nature. Option B provides one clear example of each. Option A shows two particle-like phenomena. Option C shows two wave-like phenomena. Option D shows one particle-like (Compton) and one wave-like (diffraction) phenomenon, making it a plausible but less canonical pairing than B.

Question 3

Light of wavelength 400 nm is incident on a metal with a work function of 2.2 eV. What is the maximum kinetic energy of the emitted photoelectrons? (h ≈ 6.63 × 10⁻³⁴ J s, c ≈ 3.00 × 10⁸ m s⁻¹, 1 eV ≈ 1.60 × 10⁻¹⁹ J)

0.91 eV (correct answer)
2.2 eV
3.1 eV
5.3 eV

Explanation: First, calculate the energy of the incident photons in eV. E = hc/λ. E = (6.63 × 10⁻³⁴ J s)(3.00 × 10⁸ m s⁻¹) / (400 × 10⁻⁹ m) ≈ 4.97 × 10⁻¹⁹ J. To convert to eV, divide by 1.60 × 10⁻¹⁹ J/eV: E ≈ 3.11 eV. Next, use Einstein's photoelectric equation: Eₖ,ₘₐₓ = E_photon - Φ. Eₖ,ₘₐₓ = 3.11 eV - 2.2 eV = 0.91 eV.

Question 4

The work function of metal X is 3.0 eV and the work function of metal Y is 5.0 eV. Both metals are illuminated by the same monochromatic light source. Photoelectrons are emitted from metal X but not from metal Y. What can be concluded about the energy of the photons from the light source?

The photon energy is less than 3.0 eV.
The photon energy is exactly 3.0 eV.
The photon energy is between 3.0 eV and 5.0 eV. (correct answer)
The photon energy is greater than 5.0 eV.

Explanation: For photoemission to occur, the energy of the incident photons (E_photon) must be greater than or equal to the work function (Φ) of the metal. Since photoelectrons are emitted from metal X (Φ = 3.0 eV), we know E_photon ≥ 3.0 eV. Since no photoelectrons are emitted from metal Y (Φ = 5.0 eV), we know E_photon < 5.0 eV. Combining these two conditions gives 3.0 eV ≤ E_photon < 5.0 eV.

Question 5

A photon undergoes Compton scattering from a stationary electron. The initial photon has energy E₀ and the scattered photon has energy E. The electron recoils with kinetic energy Kₑ. Which statement correctly relates these quantities?

E₀ = E, and Kₑ = 0
E₀ = Kₑ, and E = 0
E₀ = E + Kₑ (correct answer)
E₀ + Kₑ = E

Explanation: The principle of conservation of energy must apply to the collision. The total energy before the collision is the energy of the incident photon, E₀ (the electron is stationary, so its initial kinetic energy is zero). The total energy after the collision is the sum of the energy of the scattered photon, E, and the kinetic energy of the recoiling electron, Kₑ. Therefore, E₀ = E + Kₑ.

Question 6

An X-ray photon with wavelength λ₀ scatters from an electron at an angle of 60°. A second, identical X-ray photon scatters from another electron at an angle of 120°. What is the ratio of the change in wavelength in the second case (Δλ₁₂₀) to the change in wavelength in the first case (Δλ₆₀)?

1/3
1/2
2
3 (correct answer)

Explanation: The Compton shift is given by Δλ = (h/mₑc)(1 - cosθ). Let C = h/mₑc. For θ = 60°, Δλ₆₀ = C(1 - cos60°) = C(1 - 0.5) = 0.5C. For θ = 120°, Δλ₁₂₀ = C(1 - cos120°) = C(1 - (-0.5)) = 1.5C. The ratio is Δλ₁₂₀ / Δλ₆₀ = (1.5C) / (0.5C) = 3.

Question 7

An electron is accelerated from rest to a final speed v such that its total relativistic energy is double its rest mass energy. What is the de Broglie wavelength of this electron?

h / (mₑv)
h / (γmₑv)
h / (2mₑc)
h / (√3 mₑc) (correct answer)

Explanation: The de Broglie wavelength is λ = h/p, where p is the relativistic momentum. The total relativistic energy is E = γmₑc² and the rest energy is E₀ = mₑc². We are given E = 2E₀. The relativistic energy-momentum relation is E² = (pc)² + (E₀)². Substituting E = 2E₀: (2E₀)² = (pc)² + (E₀)². This gives 4(E₀)² = (pc)² + (E₀)², so (pc)² = 3(E₀)². Taking the square root, pc = √3 E₀ = √3 mₑc². Therefore, the momentum is p = √3 mₑc. The de Broglie wavelength is λ = h/p = h / (√3 mₑc).

Question 8

A photon of energy 10 keV undergoes Compton scattering. A second photon of energy 10 MeV also undergoes Compton scattering. For which photon is the fractional energy loss, (E_initial - E_scattered) / E_initial, potentially larger?

The 10 keV photon, because it has a longer initial wavelength.
The 10 MeV photon, because the maximum energy transfer to the electron is a larger fraction of its initial energy. (correct answer)
Both have the same potential fractional energy loss because the Compton wavelength shift is independent of initial photon energy.
The fractional energy loss cannot be compared without knowing the scattering angle.

Explanation: The energy transferred to the electron is E_initial - E_scattered. The fractional energy loss is (E₀ - E) / E₀. The maximum energy transfer occurs at 180° scattering and is given by K_max = E₀ / (1 + mₑc²/(2E₀)). For the low-energy 10 keV photon, E₀ << mₑc² (511 keV), so the denominator is very large and K_max is a small fraction of E₀. For the high-energy 10 MeV photon, E₀ >> mₑc², so the denominator approaches 1 and K_max approaches E₀. Therefore, the high-energy photon can lose a much larger fraction of its initial energy.

Question 9

In a double-slit experiment with electrons, an interference pattern is formed. If the accelerating voltage of the electron gun is increased, what happens to the spacing between the interference fringes?

It increases, because the electrons' speed increases.
It remains the same, because the slit separation is unchanged.
It decreases, because the electrons' de Broglie wavelength decreases. (correct answer)
It increases, because the electrons' de Broglie wavelength increases.

Explanation: The spacing of interference fringes (s) is given by s ≈ λD/d, where λ is the wavelength. Increasing the accelerating voltage (V) increases the kinetic energy of the electrons (Eₖ = eV). The de Broglie wavelength is λ = h/p = h/√(2mEₖ). Therefore, as V increases, Eₖ increases, and the wavelength λ decreases. Since the fringe spacing s is proportional to λ, a decrease in λ causes the fringe spacing to decrease.

Question 10

According to the de Broglie hypothesis, an electron in a stable orbit around a nucleus must have a standing wave pattern. For an electron in the n=3 energy level, how many of its de Broglie wavelengths fit into the circumference of its orbit?

1.5
3 (correct answer)
6
9

Explanation: The de Broglie condition for a stable orbit (a standing wave) is that the circumference of the orbit (2πr) must be an integer multiple of the electron's wavelength (λ). The integer, n, corresponds to the principal quantum number. The condition is 2πr = nλ. Therefore, for the n=3 energy level, exactly 3 de Broglie wavelengths must fit into the circumference of the orbit.

Question 11

A beam of light consists of photons each with energy E. The beam is incident on a metal surface, and the stopping potential for the emitted photoelectrons is Vₛ. The intensity of the beam is then halved, but the photon energy E is kept constant. What is the new stopping potential and the new rate of electron emission?

New stopping potential: Vₛ; New rate: Halved (correct answer)
New stopping potential: Vₛ/2; New rate: Halved
New stopping potential: Vₛ; New rate: Unchanged
New stopping potential: Vₛ/2; New rate: Unchanged

Explanation: The stopping potential Vₛ depends on the maximum kinetic energy of the photoelectrons, which is determined by the photon energy E and the work function Φ (eVₛ = E - Φ). Since the photon energy E is unchanged, the stopping potential Vₛ remains the same. The intensity of the light beam is proportional to the number of photons arriving per unit time. Halving the intensity means halving the number of incident photons, which in turn halves the rate of electron emission (the photoelectric current).

Question 12

In a photoelectric effect experiment, monochromatic light of frequency f illuminates a metal surface with work function Φ, and the stopping potential is measured to be Vₛ. If the frequency of the incident light is doubled to 2f, what is the new stopping potential?

2Vₛ
Vₛ + hf/e (correct answer)
2Vₛ + Φ/e
Vₛ - hf/e

Explanation: The governing equation for the photoelectric effect is Eₖ,ₘₐₓ = hf - Φ, and the stopping potential is related by Eₖ,ₘₐₓ = eVₛ. Therefore, eVₛ = hf - Φ. In the initial case, Vₛ = (hf - Φ)/e. In the second case, with frequency 2f, the new stopping potential Vₛ' is given by eVₛ' = h(2f) - Φ. So, Vₛ' = (2hf - Φ)/e. We can rewrite this as Vₛ' = (hf - Φ + hf)/e = (hf - Φ)/e + hf/e. Since Vₛ = (hf - Φ)/e, the new stopping potential is Vₛ' = Vₛ + hf/e.

Question 13

The Davisson-Germer experiment involved firing a beam of electrons at a nickel crystal. What was the main conclusion drawn from the observation that electrons were scattered at specific, predictable angles?

Electrons are fundamental particles with a quantized negative charge.
Electrons behave as waves and exhibit diffraction, confirming the de Broglie hypothesis. (correct answer)
Electrons can only exist in discrete energy levels within an atom.
Electrons interact with matter through the electromagnetic force, producing photons.

Explanation: The Davisson-Germer experiment showed that electrons scattered from a crystal lattice produced a diffraction pattern, similar to how X-rays diffract. This pattern could only be explained if the electrons were behaving as waves with a wavelength given by de Broglie's formula (λ = h/p). This was the first direct experimental confirmation of the wave nature of matter.

Question 14

A particle with de Broglie wavelength $\lambda = 1.5 \times 10^{-10}$ m approaches a single slit of width $a = 3.0 \times 10^{-10}$ m. What is the angular width of the central maximum in the resulting diffraction pattern?

$30°$ measured between first minima on either side
$120°$ accounting for quantum uncertainty effects
$90°$ calculated from the diffraction condition
$60°$ representing the full central maximum width (correct answer)

Explanation: When you encounter single-slit diffraction with matter waves, you're dealing with quantum mechanics where particles exhibit wave properties. The key relationship is that the first minimum occurs when

a \sin \theta = \lambda

, where

a

is the slit width,

\theta

is the angle to the first minimum, and

\lambda

is the de Broglie wavelength. Let's find the angle to the first minimum:

\sin \theta = \frac{\lambda}{a} = \frac{1.5 \times 10^{-10}}{3.0 \times 10^{-10}} = 0.5

This gives

\theta = 30°

for the first minimum on one side of the central maximum. Since the diffraction pattern is symmetric, there's another first minimum at

-30°

on the other side. The angular width of the central maximum is the total angle between these two first minima:

30° - (-30°) = 60°

. Answer choice A (

30°

) incorrectly gives only the angle to one first minimum rather than the full width. Choice B (

120°

) appears to double the correct answer, possibly confusing this with a different diffraction scenario. Choice C (

90°

) might come from incorrectly assuming the first minimum occurs at

45°

, perhaps by misapplying the diffraction formula. The correct answer is D (

60°

), representing the full angular width of the central maximum between the first minima on either side. Study tip: Always remember that diffraction patterns are symmetric, so when calculating the "width" of the central maximum, you need the total angle between minima on both sides, not just the angle to one minimum.

Question 15

An electron in a hydrogen atom transitions from $n = 4$ to $n = 2$ and emits a photon. This photon then strikes a metal surface with work function $\phi = 2.0$ eV. What is the maximum kinetic energy of the ejected photoelectron?

$0.55$ eV from the hydrogen transition energy (correct answer)
$1.28$ eV calculated using the photoelectric equation
$2.55$ eV including the work function energy
$4.55$ eV from total available photon energy

Explanation: Hydrogen transition energy:

E_{photon} = 13.6(\frac{1}{n_f^2} - \frac{1}{n_i^2}) = 13.6(\frac{1}{4} - \frac{1}{16}) = 13.6 \times \frac{3}{16} = 2.55

eV. For photoelectric effect:

KE_{max} = E_{photon} - \phi = 2.55 - 2.0 = 0.55

eV. Choice B uses wrong transition calculation. Choice C adds work function instead of subtracting. Choice D ignores work function completely.

Question 16

A quantum harmonic oscillator has energy levels $E_n = \hbar\omega(n + \frac{1}{2})$ . If the oscillator absorbs a photon and transitions from $n = 2$ to $n = 5$ , and then emits two photons in cascade transitions back to $n = 2$ , what is the ratio of frequencies of the two emitted photons if the intermediate state is $n = 3$ ?

$1:2$ based on energy level differences
$2:3$ calculated from transition probabilities
$2:1$ from the quantum number ratios (correct answer)
$3:2$ derived from selection rule constraints

Explanation: When analyzing quantum harmonic oscillator transitions, focus on the energy differences between levels to determine photon frequencies. Since

E = h\nu

, the frequency of an emitted photon equals the energy difference divided by Planck's constant. Let's trace the emission process. The oscillator starts at

n = 5

and returns to

n = 2

via two transitions: first

5 \rightarrow 3

, then

3 \rightarrow 2

. For the first emission (

5 \rightarrow 3

\Delta E_1 = E_5 - E_3 = \hbar\omega(5 + \frac{1}{2}) - \hbar\omega(3 + \frac{1}{2}) = 2\hbar\omega

For the second emission (

3 \rightarrow 2

\Delta E_2 = E_3 - E_2 = \hbar\omega(3 + \frac{1}{2}) - \hbar\omega(2 + \frac{1}{2}) = \hbar\omega

Since

\nu = \frac{\Delta E}{h}

, the frequency ratio is:

\frac{\nu_1}{\nu_2} = \frac{\Delta E_1}{\Delta E_2} = \frac{2\hbar\omega}{\hbar\omega} = \frac{2}{1}

This confirms answer C is correct. Option A (

1:2

) reverses the actual ratio. Option B (

2:3

) incorrectly suggests transition probabilities affect frequency ratios—while probabilities determine emission likelihood, frequencies depend only on energy differences. Option D (

3:2

) appears to confuse the quantum numbers involved rather than calculating actual energy differences. Study tip: For quantum oscillator problems, always calculate energy differences directly using the given formula. Photon frequency ratios equal energy difference ratios—ignore quantum number values themselves and focus on the

\Delta E

between levels.

Question 17

A particle is described by the wave function $\psi(x) = A\sin(\frac{\pi x}{L})$ for $0 \leq x \leq L$ and $\psi(x) = 0$ elsewhere. If a measurement of position is made, what is the probability of finding the particle in the region $\frac{L}{4} \leq x \leq \frac{3L}{4}$ ?

$\frac{1}{2} + \frac{1}{\pi}$ from the normalized probability integral (correct answer)
$\frac{1}{2} - \frac{1}{\pi}$ accounting for boundary normalization
$\frac{2}{3} + \frac{1}{2\pi}$ using the complete wave function
$\frac{2}{3} - \frac{1}{2\pi}$ from integration by parts method

Explanation: First normalize:

\int_0^L |A|^2\sin^2(\frac{\pi x}{L})dx = 1

. Using

\sin^2(u) = \frac{1-\cos(2u)}{2}

|A|^2 \cdot \frac{L}{2} = 1

, so

A = \sqrt{\frac{2}{L}}

. Probability =

\int_{L/4}^{3L/4} \frac{2}{L}\sin^2(\frac{\pi x}{L})dx = \frac{2}{L} \int_{L/4}^{3L/4} \frac{1-\cos(\frac{2\pi x}{L})}{2}dx = \frac{1}{L}[x - \frac{L}{2\pi}\sin(\frac{2\pi x}{L})]_{L/4}^{3L/4} = \frac{1}{2} + \frac{1}{\pi}

. Choice B has wrong sign. Choices C and D use incorrect normalization constant.

Question 18

Two identical photons, each with energy $E = 1.5$ MeV, undergo pair production near a heavy nucleus. What is the maximum kinetic energy that one of the produced particles (electron or positron) can have?

$1.5$ MeV when the other particle is at rest
$2.5$ MeV accounting for relativistic effects
$2.0$ MeV from total photon energy sharing (correct answer)
$3.0$ MeV from complete energy transfer

Explanation: When you encounter pair production problems, remember that this process requires energy conservation and momentum conservation, with specific threshold energies for particle creation. In pair production, a photon creates an electron-positron pair near a heavy nucleus. The minimum energy required is

2m_ec^2 = 2 \times 0.511 \text{ MeV} = 1.022 \text{ MeV}

to create the rest masses of both particles. Since each photon has

1.5 \text{ MeV}

, the total available energy is

2 \times 1.5 = 3.0 \text{ MeV}

. After subtracting the rest mass energy (

1.022 \text{ MeV}

), the remaining kinetic energy is

3.0 - 1.022 = 1.978 \text{ MeV} \approx 2.0 \text{ MeV}

. To maximize one particle's kinetic energy, the other particle should have minimum kinetic energy (essentially at rest in the lab frame). Therefore, one particle can have nearly all

2.0 \text{ MeV}

of available kinetic energy. Option A incorrectly uses only one photon's energy (

1.5 \text{ MeV}

), ignoring that both photons participate in the process. Option B suggests

2.5 \text{ MeV}

, which exceeds the available kinetic energy after accounting for rest masses. Option D claims

3.0 \text{ MeV}

, which would require the created particles to have no rest mass—physically impossible. The correct answer is C:

2.0 \text{ MeV}

represents the maximum kinetic energy available when both photons contribute their energy and rest mass requirements are satisfied. Study tip: In pair production problems, always calculate total available energy first, subtract the required rest mass energy (

2m_ec^2

), then determine how the remaining kinetic energy can be distributed between particles.

Question 19

A hydrogen atom in the $n = 4$ state can transition to lower energy states by emitting photons. How many different wavelengths can be observed in the emission spectrum from all possible transitions that end in the ground state ( $n = 1$ ) either directly or through intermediate levels?

3 wavelengths corresponding to direct transitions only
6 wavelengths from all possible transition combinations (correct answer)
9 wavelengths including cascade transitions through all levels
12 wavelengths from multiple emission pathways

Explanation: From

n=4

, electrons can transition to

n=3,2,1

(3 transitions). From

n=3

, they can go to

n=2,1

(2 transitions). From

n=2

, they can go to

n=1

(1 transition). Total distinct transitions:

4→3, 4→2, 4→1, 3→2, 3→1, 2→1

= 6 different wavelengths. Each transition has a unique energy difference, producing a unique wavelength. Choice A only counts direct transitions to ground state. Choice C incorrectly multiplies pathways. Choice D assumes multiple pathways create different wavelengths for the same transition.

Question 20

What is the momentum of a photon of yellow light with a wavelength of 6.0 × 10⁻⁷ m? (h ≈ 6.6 × 10⁻³⁴ J s)

1.1 × 10⁻²⁷ kg m s⁻¹ (correct answer)
3.3 × 10⁻⁴⁰ kg m s⁻¹
4.0 × 10⁻²⁵ kg m s⁻¹
2.0 × 10⁻¹⁹ kg m s⁻¹

Explanation: The momentum of a photon is related to its wavelength by the de Broglie relation, which applies to all particles including photons: p = h/λ. Substituting the given values: p = (6.6 × 10⁻³⁴ J s) / (6.0 × 10⁻⁷ m) = 1.1 × 10⁻²⁷ J s m⁻¹ = 1.1 × 10⁻²⁷ kg m s⁻¹.