A2 · Paper 4 practice · Quantum physics

Photons, electrons, and energy gaps.

Ten original structured questions in the style of Paper 4, covering the whole of Quantum physics: photon energy and momentum, the photoelectric effect, de Broglie wavelengths, and atomic energy levels. The later questions send a photon from a transition straight into a metal, linking the lessons. Each is tagged with its lessons; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Data: h = 6.63 × 10⁻³⁴ J s · c = 3.00 × 10⁸ m s⁻¹ · e = 1.60 × 10⁻¹⁹ C · m₋ = 9.11 × 10⁻³¹ kg · 1 eV = 1.60 × 10⁻¹⁹ J · hc ≈ 1240 eV nm

Keep these straight

Energy, threshold, wavelength.

Photon: E = hf = hc/λ, momentum p = E/c = h/λ.
Photoelectric: hf = Φ + KE_max; threshold f₀ = Φ/h; intensity sets rate, frequency sets KE_max.
de Broglie: λ = h/p; for an electron through voltage V, p = √(2m₋eV).
Energy levels: hf = E₁ − E₂ gives the wavelength of each spectral line.

Paper 4

[8 marks]

Photon energy & momentum →

Green light has a wavelength of 520 nm in a vacuum.

(a) Calculate the frequency of this light. [2]

(b) Calculate the energy of one photon, in joules. [2]

(c) Express this energy in electronvolts. [1]

(d) Calculate the momentum of one photon. [2]

(e) State how the photon energy would change if the wavelength were doubled. [1]

(a)f = c/λ = 3.00 × 10⁸ / 520 × 10⁻⁹ ✓ = 5.77 × 10¹⁴ Hz ✓
(b)E = hf = 6.63 × 10⁻³⁴ × 5.77 × 10¹⁴ ✓ = 3.83 × 10⁻¹⁹ J ✓
(c)E = 3.83 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.39 eV ✓
(d)p = h/λ = 6.63 × 10⁻³⁴ / 520 × 10⁻⁹ ✓ = 1.28 × 10⁻²⁷ N s ✓
(e)E = hc/λ, so doubling λ halves the energy ✓

Paper 4

[7 marks]

Photon energy & momentum →

A helium-neon laser emits light of wavelength 633 nm with an output power of 2.0 mW.

(a) Calculate the energy of a single photon. [2]

(b) Calculate the number of photons emitted per second. [3]

(c) Calculate the momentum of one photon. [2]

(a)E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 633 × 10⁻⁹ ✓ = 3.14 × 10⁻¹⁹ J (≈ 1.96 eV) ✓
(b)number per second = P/E = 2.0 × 10⁻³ / 3.14 × 10⁻¹⁹ ✓✓ = 6.4 × 10¹⁵ s⁻¹ ✓
(c)p = h/λ = 6.63 × 10⁻³⁴ / 633 × 10⁻⁹ ✓ = 1.05 × 10⁻²⁷ N s ✓

Paper 4

[8 marks]

Photoelectric effect →

The work function of sodium is 2.3 eV.

(a) State what is meant by the work function. [1]

(b) Express the work function in joules. [1]

(c) Calculate the threshold frequency for sodium. [2]

(d) Calculate the corresponding threshold wavelength. [2]

(e) State and explain what happens when light of wavelength 600 nm shines on sodium. [2]

(a)The minimum energy needed to remove an electron from the surface of the metal ✓
(b)Φ = 2.3 × 1.60 × 10⁻¹⁹ = 3.7 × 10⁻¹⁹ J ✓
(c)f₀ = Φ/h = 3.7 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ ✓ = 5.6 × 10¹⁴ Hz ✓
(d)λ₀ = c/f₀ = 3.00 × 10⁸ / 5.6 × 10¹⁴ ✓ = 540 nm ✓
(e)600 nm is longer than 540 nm, so f < f₀ ✓; the photon energy is below Φ, so no electrons are emitted ✓

Paper 4

[9 marks]

Photoelectric effect →

Ultraviolet light of wavelength 400 nm is shone on a metal of work function 2.0 eV.

(a) Calculate the energy of each photon, in electronvolts. [2]

(b) Calculate the maximum kinetic energy of the photoelectrons, in electronvolts and in joules. [2]

(c) Calculate the maximum speed of the photoelectrons. [3]

(d) State the stopping voltage that would just prevent these electrons from reaching a collector. [1]

(e) State the effect on the maximum kinetic energy of doubling the intensity. [1]

(a)E = 1240/400 ✓ = 3.10 eV ✓
(b)KE_max = hf − Φ = 3.10 − 2.0 = 1.10 eV ✓ = 1.10 × 1.60 × 10⁻¹⁹ = 1.76 × 10⁻¹⁹ J ✓
(c)½mv² = 1.76 × 10⁻¹⁹; v = √(2 × 1.76 × 10⁻¹⁹ / 9.11 × 10⁻³¹) ✓✓ = 6.2 × 10⁵ m s⁻¹ ✓
(d)Vₛ = KE_max/e = 1.10 V ✓
(e)No effect ✓ (intensity changes only the rate, not KE_max)

Paper 4

[8 marks]

Photoelectric effect →

Fig. 5.1 shows how the maximum kinetic energy of photoelectrons varies with the frequency of the incident light for a particular metal.

Fig. 5.1

(a) State what the gradient of the line represents. [1]

(b) State what is found at the point marked f₀. [1]

(c) State what the intercept marked −Φ represents. [1]

(d) The line crosses the frequency axis at f₀ = 5.0 × 10¹⁴ Hz. Calculate the work function of the metal, in joules and in electronvolts. [3]

(e) A second metal has a larger work function. State how its line would differ from the one shown. [2]

(a)The Planck constant h ✓ (KE_max = hf − Φ, gradient = h)
(b)The threshold frequency: below it, no photoelectrons are emitted ✓
(c)The (negative of the) work function Φ of the metal ✓
(d)Φ = hf₀ = 6.63 × 10⁻³⁴ × 5.0 × 10¹⁴ ✓ = 3.3 × 10⁻¹⁹ J ✓ = 2.1 eV ✓
(e)Same gradient (parallel line) ✓, but a larger threshold frequency and a more negative intercept ✓

Paper 4

[8 marks]

Wave-particle duality →

An electron moves with a speed of 2.0 × 10⁶ m s⁻¹.

(a) Calculate the momentum of the electron. [2]

(b) Calculate its de Broglie wavelength. [2]

(c) State why a beam of these electrons can be diffracted by a crystal but not by a wide slit. [2]

(d) A cricket ball of mass 0.16 kg moves at 30 m s⁻¹. State, with a reason, whether its wave nature could be observed. [2]

(a)p = mv = 9.11 × 10⁻³¹ × 2.0 × 10⁶ ✓ = 1.8 × 10⁻²⁴ N s ✓
(b)λ = h/p = 6.63 × 10⁻³⁴ / 1.8 × 10⁻²⁴ ✓ = 3.6 × 10⁻¹⁰ m (0.36 nm) ✓
(c)Diffraction is strong only when the gap is comparable to λ ✓; this λ matches atomic spacings in a crystal but is far smaller than a slit width ✓
(d)No ✓; p = 4.8 N s gives λ ≈ 1.4 × 10⁻³⁴ m, far too small to diffract through any gap ✓

Paper 4

[8 marks]

Wave-particle duality →

Electrons are accelerated from rest through a potential difference of 2500 V.

(a) Calculate the kinetic energy of an electron, in joules. [2]

(b) Show that the momentum of an electron is about 2.7 × 10⁻²³ N s. [2]

(c) Calculate the de Broglie wavelength of the electrons. [2]

(d) State and explain the effect on the wavelength of doubling the accelerating voltage. [2]

(a)KE = eV = 1.60 × 10⁻¹⁹ × 2500 ✓ = 4.0 × 10⁻¹⁶ J ✓
(b)p = √(2m₋KE) = √(2 × 9.11 × 10⁻³¹ × 4.0 × 10⁻¹⁶) ✓ = 2.7 × 10⁻²³ N s ✓
(c)λ = h/p = 6.63 × 10⁻³⁴ / 2.7 × 10⁻²³ ✓ = 2.5 × 10⁻¹¹ m (25 pm) ✓
(d)λ ∝ 1/√V ✓, so doubling V multiplies λ by 1/√2 ≈ 0.71 (the wavelength falls to about 18 pm) ✓

Paper 4

[9 marks]

Energy levels →Photon energy & momentum →

Fig. 8.1 shows some of the electron energy levels of a hydrogen atom.

Fig. 8.1

(a) Explain why the energy values are negative. [1]

(b) Calculate the energy of the photon emitted in the n = 3 to n = 2 transition, in eV and in joules. [2]

(c) Calculate the wavelength of this photon. [2]

(d) State the region of the spectrum in which this line appears. [1]

(e) Explain why the spectrum of hydrogen consists of discrete lines rather than a continuous band. [3]

(a)Energy must be supplied to free the electron from the atom; the zero is taken at infinity, so bound states are negative ✓
(b)ΔE = −1.51 − (−3.40) = 1.89 eV ✓ = 1.89 × 1.60 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J ✓
(c)λ = hc/ΔE = 1240/1.89 ✓ = 656 nm ✓
(d)Visible (red) ✓
(e)The energy levels are discrete ✓; only certain transition energies are possible ✓; each gives one fixed photon energy and wavelength, so the spectrum is a set of lines ✓

Paper 4

[10 marks]

Energy levels →Photoelectric effect →

A hydrogen atom emits a photon when its electron falls from the n = 4 level (−0.85 eV) to the n = 1 level (−13.6 eV). The photon then strikes a metal of work function 4.5 eV.

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

(b) Calculate the wavelength of the photon and state the region of the spectrum. [3]

(c) Calculate the maximum kinetic energy of the photoelectron released from the metal, in eV. [2]

(d) A photon from the n = 3 to n = 2 transition (1.89 eV) instead strikes the same metal. Explain why no photoelectron is emitted. [2]

(e) State what determines the number of photoelectrons emitted per second by the metal. [1]

(a)ΔE = −0.85 − (−13.6) = 12.75 eV ✓
(b)λ = 1240/12.75 ✓ = 97 nm ✓; ultraviolet ✓
(c)KE_max = 12.75 − 4.5 = 8.25 eV ✓✓
(d)1.89 eV is less than the work function 4.5 eV ✓; a single photon cannot supply enough energy to free an electron, so none is emitted ✓
(e)The intensity (number of photons arriving per second) ✓

Paper 4

[11 marks]

Photon →Photoelectric effect →Duality →

Ultraviolet light of wavelength 254 nm falls on a zinc surface of work function 4.3 eV. This question follows the energy from photon to photoelectron.

(a) Calculate the energy of each photon, in eV. [2]

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

(c) Calculate the momentum of one of these photoelectrons. [2]

(d) Calculate the de Broglie wavelength of such a photoelectron. [2]

(e) Calculate the momentum of one incident photon, and comment on how it compares with the photoelectron's momentum. [3]

(a)E = 1240/254 ✓ = 4.88 eV ✓
(b)KE_max = 4.88 − 4.3 = 0.58 eV ✓ = 0.58 × 1.60 × 10⁻¹⁹ = 9.3 × 10⁻²⁰ J ✓
(c)p = √(2m₋KE) = √(2 × 9.11 × 10⁻³¹ × 9.3 × 10⁻²⁰) ✓ = 4.1 × 10⁻²⁵ N s ✓
(d)λ = h/p = 6.63 × 10⁻³⁴ / 4.1 × 10⁻²⁵ ✓ = 1.6 × 10⁻⁹ m (1.6 nm) ✓
(e)pₙₘₒₜₒₙ = h/λ = 6.63 × 10⁻³⁴ / 254 × 10⁻⁹ = 2.6 × 10⁻²⁷ N s ✓; this is far smaller than the photoelectron's momentum (4.1 × 10⁻²⁵ N s) ✓, about 160 times smaller ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.