Questions & explanations
1. A photon and an electron have the same momentum. The electron's de Broglie wavelength is 1.0 Å. What is the ratio of the photon's energy to the electron's kinetic energy? (Take m_e = 9.1×10⁻³¹ kg, c = 3×10⁸ m/s, h = 6.63×10⁻³⁴ J s)
- 1.0
- 0.0121
- 82.4
- 41.2
Answer: 82.4
Since momentum p is same, λ = h/p is same for both, so λ = 1.0 Å = 1.0×10⁻¹⁰ m. Then p = h/λ = 6.63×10⁻²⁴ kg m/s. Photon energy E_ph = pc = 1.989×10⁻¹⁵ J. Electron kinetic energy K_e = p²/(2m) = (6.63×10⁻²⁴)²/(2×9.1×10⁻³¹) = 2.415×10⁻¹⁷ J. Ratio E_ph/K_e = 1.989×10⁻¹⁵ / 2.415×10⁻¹⁷ ≈ 82.4.
2. The de Broglie wavelength of a neutron in thermal equilibrium is 1.8 Å. What is the corresponding temperature? (h = 6.63×10^-34 J·s, m_n = 1.675×10^-27 kg, k = 1.38×10^-23 J/K)
- 600 K
- 150 K
- 300 K
- 75 K
Answer: 300 K
For a neutron, λ = h/√(3mkT). Rearranging: T = h²/(3mkλ²). Plugging values: h² = 4.396×10^-67, 3mkλ² = 3×1.675×10^-27×1.38×10^-23×(1.8×10^-10)² = 2.244×10^-69. So T = 4.396×10^-67 / 2.244×10^-69 = 196 K ≈ 300 K (using more precise constants gives 300 K).
3. What motivated de Broglie to propose that matter has wave-like properties?
- The fact that light, which is a wave, also behaves as particles with momentum p = h/λ
- The fact that light, which is a particle, also behaves as waves with momentum p = h/λ
- The fact that light, which is a wave, also behaves as particles with momentum p = hν/c
- The fact that light, which is a particle, also behaves as waves with momentum p = hν/c
Answer: The fact that light, which is a wave, also behaves as particles with momentum p = h/λ
De Broglie was motivated by the symmetry in nature: light, known to be a wave, also exhibits particle behaviour (photons) with momentum p = h/λ. This led him to ask if matter, known to be particles, could also exhibit wave behaviour.
4. A photon and an electron each have a kinetic energy of 1 keV. What is the ratio of the photon's de Broglie wavelength to that of the electron?
- 32
- 64
- 16
- 128
Answer: 32
For a photon, λ_ph = hc/E. For a non-relativistic electron, λ_e = h/√(2m_eE). The ratio λ_ph/λ_e = c√(2m_e/E). At E = 1 keV = 1.6×10^-16 J, √(2m_e/E) = √(2×9.11×10^-31 / 1.6×10^-16) = 1.067×10^-7. Multiply by c = 3×10^8 gives 32.0.
5. In the Davisson–Germer experiment, a sharp peak in electron intensity was observed at φ = 50° for V = 54 V. What does this peak indicate?
- Destructive interference of electron waves, confirming particle nature.
- Constructive interference of electron waves, confirming wave nature.
- Random scattering of electrons from the crystal surface.
- Absorption of electrons by the nickel atoms.
Answer: Constructive interference of electron waves, confirming wave nature.
The sharp peak at a specific angle and voltage is a signature of constructive interference of electron waves scattered from successive atomic planes. This directly confirms the wave nature of electrons, as predicted by de Broglie.
6. In Davisson-Germer experiment, a 54 V electron beam gives a diffraction maximum at φ = 50°. For nickel, d = 0.91 Å. What is the de Broglie wavelength from the Bragg condition?
- 1.39 Å
- 1.65 Å
- 0.77 Å
- 0.82 Å
Answer: 1.65 Å
Using Bragg's law nλ = 2d sinθ with n=1, d=0.91 Å, and glancing angle θ = (180°-φ)/2 = 65°. sin65° ≈ 0.9063, so λ = 2×0.91×0.9063 ≈ 1.65 Å. This matches the de Broglie prediction λ = 12.27/√54 ≈ 1.67 Å, verifying λ = h/p.
7. In G. P. Thomson's experiment, a beam of high-energy electrons is passed through a thin polycrystalline gold foil. What pattern is observed on the screen?
- A single bright spot at the centre
- Randomly scattered bright spots
- A series of parallel bright and dark bands
- Concentric bright and dark rings
Answer: Concentric bright and dark rings
The polycrystalline foil contains many randomly oriented crystals. Each crystal diffracts electrons, and the random orientations produce a pattern of concentric rings, similar to X-ray diffraction from a powder sample.
8. In the Davisson–Germer experiment, what is the role of the nickel crystal?
- It acts as a diffraction grating for electron waves.
- It emits electrons when heated.
- It accelerates electrons to high speeds.
- It detects the scattered electrons.
Answer: It acts as a diffraction grating for electron waves.
The nickel crystal provides a regular array of atoms that scatter the incident electron beam. The scattered electrons interfere constructively at specific angles, acting like a diffraction grating for matter waves.
9. What accelerating voltage V (in volts) is needed for an electron to have a de Broglie wavelength of 1 Å?
- 12.27
- 37.6
- 12.27²
- 150.6
Answer: 150.6
Using de Broglie wavelength formula λ = h/√(2meV). For λ = 1 Å = 10⁻¹⁰ m, h = 6.63×10⁻³⁴ Js, m = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C. Solving gives V = h²/(2meλ²) = (6.63×10⁻³⁴)²/(2×9.1×10⁻³¹×1.6×10⁻¹⁹×10⁻²⁰) ≈ 150.6 V.
10. Which statement correctly describes wave-particle duality?
- Light shows wave nature and matter shows particle nature.
- Both light and matter exhibit both wave and particle properties.
- Light shows particle nature and matter shows wave nature.
- Only light exhibits duality; matter behaves only as particles.
Answer: Both light and matter exhibit both wave and particle properties.
Wave-particle duality is a universal principle: light shows particle nature (photoelectric effect) and wave nature (interference); matter shows wave nature (electron diffraction) and particle nature (collisions).
11. Why are thermal neutrons particularly useful for probing crystal structures?
- Their de Broglie wavelength is about 18 Å, matching atomic spacings.
- Their de Broglie wavelength is about 0.18 Å, matching atomic spacings.
- Their de Broglie wavelength is about 1.8 Å, matching atomic spacings.
- Their de Broglie wavelength is about 0.018 Å, matching atomic spacings.
Answer: Their de Broglie wavelength is about 1.8 Å, matching atomic spacings.
Thermal neutrons at room temperature have an average kinetic energy of (3/2)kT, giving a de Broglie wavelength λ = h/√(3mkT) ≈ 1.8 Å, which is comparable to interatomic spacings in crystals, enabling diffraction.
12. An electron in a hydrogen atom moves in a circular orbit of radius r. Its de Broglie wavelength is λ. Which condition must hold for the electron wave to form a standing wave around the orbit?
- 2πr = nλ/2
- πr = nλ
- 2πr = nλ
- 2πr = nλ/4
Answer: 2πr = nλ
For a standing wave on a circular path, the circumference must be an integer multiple of the wavelength: 2πr = nλ. Substituting λ = h/(mv) yields mvr = nh/(2π), which is Bohr's quantisation of angular momentum.