Quantum physics practice questions.

In the Bohr model of hydrogen the electron energies are given by Eₙ = −13.6/n² eV. Determine the energy of the photon emitted when the electron makes a transition from n = 3 to n = 2.

• 1.51 eV
• 2.27 eV
• 1.89 eV
• 4.91 eV

Answer: 1.89 eV

ΔE = E₃ − E₂ = (−13.6/9) − (−13.6/4) = −1.51 + 3.40 = 1.89 eV. A common error is using 1/n instead of 1/n², which gives 2.27 eV.

In a photoelectric experiment the frequency of the incident light is increased while the intensity is kept constant. The frequency remains above the threshold frequency. What happens to the stopping potential?

• It increases
• It decreases
• It stays the same
• It becomes zero

Answer: It increases

eVₛ = hf − Φ, so a higher frequency gives photoelectrons a larger maximum kinetic energy and a larger stopping potential is needed to halt them. Intensity affects only the photocurrent, not Vₛ.

An X-ray photon is scattered by a free electron initially at rest (the Compton effect). Compared with the incident photon, the scattered photon has

• a shorter wavelength
• a longer wavelength
• the same wavelength but lower speed
• a higher frequency

Answer: a longer wavelength

The photon transfers some energy and momentum to the electron, so its energy decreases and its wavelength increases by Δλ = (h/mₑc)(1 − cos θ). The photon speed is always c.

An electron and a proton have the same kinetic energy. Which statement about their de Broglie wavelengths is correct?

• The wavelengths are equal because the energies are equal
• The proton has the longer wavelength because it has more mass
• The wavelengths depend on charge, which is equal in magnitude, so they are equal
• The electron has the longer wavelength because it has less mass

Answer: The electron has the longer wavelength because it has less mass

For the same kinetic energy, p = √(2mE_k), so the lighter electron has the smaller momentum and hence, from λ = h/p, the longer de Broglie wavelength.

According to the Bohr model, the angular momentum of the electron in the hydrogen atom is quantized. The allowed values are given by

• nh/2π
• nh
• n²h/2π
• h/(2πn)

Answer: nh/2π

Bohr postulated mvr = nh/2π, where n is a positive integer. This quantization condition leads directly to the discrete energy levels of hydrogen.

An electron (mass 9.11e-31 kg) moves at 1.0e6 m/s. Find its de Broglie wavelength in nanometres (h = 6.63e-34). Give to 2 decimal places.

Answer: 0.73

λ = h/(mv) = 6.63e-34 / (9.11e-31 × 1.0e6) ≈ 7.3e-10 m = 0.73 nm.