Questions & explanations
1. Which of the following experimental observations of the photoelectric effect cannot be explained by classical wave theory?
- Existence of threshold frequency
- Increase in photoelectric current with increase in intensity
- Dependence of stopping potential on frequency of light
- Both (A) and (C)
Answer: Both (A) and (C)
Classical wave theory predicts that photoelectric emission should occur at all frequencies if the intensity is sufficient, failing to explain the existence of a threshold frequency. It also predicts that stopping potential (related to electron kinetic energy) should increase with intensity, not with frequency. The increase in current with intensity is actually consistent with classical theory, so only the threshold frequency and the frequency-dependence of stopping potential are inexplicable.
2. Which of the following is an experimental observation of the photoelectric effect?
- Photoelectric emission occurs at all frequencies of incident light.
- Photoelectrons are emitted instantaneously when light of suitable frequency falls on the metal surface.
- The kinetic energy of emitted electrons increases with increase in intensity of light.
- The number of photoelectrons emitted per second depends only on the frequency of light.
Answer: Photoelectrons are emitted instantaneously when light of suitable frequency falls on the metal surface.
The correct answer is that emission of photoelectrons occurs with no measurable time lag, i.e., instantaneously, which is one of the key experimental observations. Other options are false: emission only occurs above threshold frequency, kinetic energy depends on frequency (not intensity), and the number of photoelectrons depends on intensity (not frequency).
3. Which of the following experiments provides direct evidence for the wave nature of electrons?
- Photoelectric effect
- Davisson‑Germer experiment
- Millikan oil drop experiment
- Rutherford's alpha‑particle scattering experiment
Answer: Davisson‑Germer experiment
The Davisson‑Germer experiment observed diffraction of electrons from a nickel crystal, confirming the wave nature of electrons as predicted by de Broglie. The photoelectric effect demonstrates the particle nature of light, Millikan's experiment measures the electron charge, and Rutherford's experiment reveals the nuclear model of the atom.
4. According to de Broglie's hypothesis, which of the following statements is correct about any moving particle?
- It exhibits wave-like properties and its associated wavelength is inversely proportional to its momentum.
- It behaves only as a particle and never as a wave.
- It has a wavelength that is directly proportional to its kinetic energy.
- It exhibits wave-like properties only when it is at rest.
Answer: It exhibits wave-like properties and its associated wavelength is inversely proportional to its momentum.
De Broglie proposed that all moving particles have an associated matter wave with wavelength λ = h/p, where h is Planck's constant and p is the momentum. Hence the wavelength is inversely proportional to momentum. The other options contradict the wave-particle duality or the relationship between wavelength and kinetic energy (λ ∝ 1/√K).
5. According to Einstein's photoelectric equation, the maximum kinetic energy (Kmax) of an emitted photoelectron is given by:
- Kmax = hf - φ
- Kmax = hf + φ
- Kmax = φ - hf
- Kmax = hf / φ
Answer: Kmax = hf - φ
Einstein's photoelectric equation is Kmax = hf - φ, where h is Planck's constant, f is frequency of incident light, and φ is the work function. This equation correctly accounts for the conservation of energy: the photon energy (hf) is partly used to overcome the work function and the remainder appears as kinetic energy of the electron.
6. The minimum energy required to eject an electron from a metal surface is called:
- Threshold frequency
- Stopping potential
- Work function
- Kinetic energy
Answer: Work function
The work function (φ) is defined as the minimum energy needed to remove an electron from the metal surface. Threshold frequency is the minimum frequency of light required to cause emission, stopping potential is the potential that stops the most energetic photoelectrons, and kinetic energy is the energy of the ejected electron.
7. When the intensity of incident light is increased while keeping its frequency constant (above the threshold frequency), the photoelectric current:
- Increases
- Decreases
- Remains the same
- Becomes zero
Answer: Increases
Increasing intensity means more number of photons per second falling on the metal surface. Since each photon can eject one photoelectron (provided frequency is above threshold), the number of photoelectrons emitted per second increases, thereby increasing the photoelectric current.
8. An electron (mass m = 9.1 × 10⁻³¹ kg, charge e = 1.6 × 10⁻¹⁹ C) is accelerated through a potential difference of 100 V. What is its de Broglie wavelength? (Take h = 6.63 × 10⁻³⁴ J s)
- 1.23 Å
- 12.3 Å
- 0.123 Å
- 0.0123 Å
Answer: 1.23 Å
Kinetic energy acquired = eV = 1.6×10⁻¹⁹ × 100 = 1.6×10⁻¹⁷ J. Momentum p = √(2mE) = √(2×9.1×10⁻³¹×1.6×10⁻¹⁷) = √(2.912×10⁻⁴⁷) ≈ 5.396×10⁻²⁴ kg m/s. λ = h/p = 6.63×10⁻³⁴ / 5.396×10⁻²⁴ ≈ 1.229×10⁻¹⁰ m = 1.23 Å. The other options result from factor‑10 errors in the calculation.
9. In a graph of stopping potential (V0) versus frequency (f) of incident light, the slope of the straight line obtained is equal to:
- Planck's constant (h)
- h/e
- Work function (φ)
- e/h
Answer: h/e
From Einstein's equation eV0 = hf - φ, rearranging gives V0 = (h/e)f - φ/e. Hence a graph of V0 vs f is a straight line with slope h/e (where e is the elementary charge). Planck's constant h can be found by multiplying the slope by e.
10. Two particles A and B have masses m and 2m respectively. They are moving with the same kinetic energy. What is the ratio of their de Broglie wavelengths λ_A : λ_B?
- 1 : 1
- √2 : 1
- 1 : 2
- 2 : 1
Answer: √2 : 1
For equal kinetic energy K, de Broglie wavelength λ = h/√(2mK). Thus λ ∝ 1/√m. So λ_A / λ_B = √(m_B / m_A) = √(2m / m) = √2. Hence the ratio is √2 : 1. The other ratios arise from misunderstanding the inverse‑square‑root dependence.
11. The de Broglie wavelength λ of a non‑relativistic particle is related to its kinetic energy K as:
- λ ∝ 1/√K
- λ ∝ √K
- λ ∝ K
- λ ∝ 1/K
Answer: λ ∝ 1/√K
For a non‑relativistic particle, momentum p = √(2mK), so λ = h/p = h/√(2mK). Therefore λ is inversely proportional to the square root of kinetic energy. The other proportionalities are incorrect because λ varies with K as 1/√K.
12. A photon has a frequency of 5.0 × 10^14 Hz. Its momentum is: (h = 6.63 × 10^{-34} J·s, c = 3.0 × 10^8 m/s)
- 1.105 × 10^{-27} kg·m/s
- 3.315 × 10^{-28} kg·m/s
- 1.105 × 10^{-33} kg·m/s
- 3.315 × 10^{-27} kg·m/s
Answer: 1.105 × 10^{-27} kg·m/s
Photon momentum p = hf / c = (6.63×10^{-34} × 5.0×10^{14}) / (3.0×10^8) = (3.315×10^{-19}) / (3.0×10^8) = 1.105×10^{-27} kg·m/s. The other options result from incorrect calculation or unit errors.