Questions & explanations
1. A metal surface is illuminated by light of intensity I and frequency ν. The photocurrent is i. If the intensity is doubled and frequency is halved, what is the new photocurrent?
- i
- 2i
- 0
- i/2
Answer: 0
The photocurrent is proportional to the number of photons per second, which is intensity divided by photon energy (hν). Doubling intensity doubles the number of photons, but halving frequency halves the photon energy, so the number of photons becomes (2I)/(h(ν/2)) = 4(I/hν) = 4 times. However, if the frequency is halved, the photon energy may fall below the work function, resulting in zero photocurrent. Since the problem does not specify the work function, but the frequency is halved, it is likely below threshold, so photocurrent becomes zero.
2. A metal with work function 2.0 eV is illuminated by light containing 400 nm and 600 nm wavelengths. The intensity of 600 nm light is doubled. How does the saturation current and stopping potential change?
- Saturation current increases, stopping potential unchanged
- Saturation current unchanged, stopping potential increases
- Both saturation current and stopping potential increase
- Both saturation current and stopping potential unchanged
Answer: Saturation current increases, stopping potential unchanged
Only photons with energy ≥ work function cause emission. 400 nm (3.1 eV) is above threshold; 600 nm (2.07 eV) is also above threshold. Doubling intensity of 600 nm increases number of above-threshold photons, so saturation current increases. Stopping potential depends only on maximum kinetic energy, set by the highest photon energy (400 nm), which is unchanged. Hence V₀ remains same.
3. A student increases the intensity of monochromatic light incident on a metal surface but keeps the frequency constant. What happens to the stopping potential?
- It remains the same
- It decreases
- It increases
- It becomes zero
Answer: It remains the same
According to Einstein's photoelectric equation, K_max = hν - φ, and stopping potential V₀ = K_max/e = (h/e)ν - (φ/e). This depends only on frequency ν and work function φ, not on intensity. Increasing intensity increases the number of photons but does not change the maximum kinetic energy of emitted electrons, so V₀ remains unchanged.
4. A metal has work function 2.0 eV. Light of frequency 4.0 × 10¹⁴ Hz is incident on it. What happens if the intensity is increased tenfold? (h = 4.14 × 10⁻¹⁵ eV·s)
- Photoelectrons are emitted with higher kinetic energy
- Photoelectrons are emitted with lower kinetic energy
- More photoelectrons are emitted with same kinetic energy
- No photoelectrons are emitted
Answer: No photoelectrons are emitted
Einstein's photoelectric equation: Kmax = hν - φ. Here hν = (4.14×10⁻¹⁵ eV·s)(4.0×10¹⁴ Hz) = 1.656 eV, which is less than φ = 2.0 eV. Hence no photoelectrons are emitted regardless of intensity. Increasing intensity only increases the number of sub-threshold photons, but each photon still lacks the energy to liberate an electron.
5. In a photoelectric effect experiment, the intensity of incident light is fixed above threshold. If the frequency is increased, which of the following remains approximately constant?
- Stopping potential
- Maximum kinetic energy of photoelectrons
- Saturation current
- Threshold frequency
Answer: Saturation current
With fixed intensity, the number of photons per second is constant, so the number of emitted photoelectrons per second (saturation current) remains approximately constant. Increasing frequency increases the energy per photon, raising the stopping potential and maximum kinetic energy, but does not change the emission rate.
6. According to wave theory, if light intensity is increased, the maximum kinetic energy of photoelectrons should:
- increase as square of intensity
- decrease linearly with intensity
- remain constant
- increase linearly with intensity
Answer: increase linearly with intensity
Wave theory treats light as a continuous wave. Intensity is proportional to amplitude squared, so a more intense wave delivers more energy per unit area. An electron absorbs energy continuously; higher intensity means more energy absorbed, thus higher kinetic energy. Hence K_max increases linearly with intensity.
7. In a photoelectric experiment, the I-V curves for two different frequencies f1 and f2 (f1 > f2) at same intensity are shown. Which statement correctly explains the curves using the photon model?
- Saturation current is higher for f1 because each photon has more energy, ejecting more electrons per photon.
- Saturation current is same for both frequencies because intensity is same; stopping potential is more positive for f1.
- Stopping potential is more negative for f1 because higher frequency photons impart more kinetic energy to electrons.
- Stopping potential is more positive for f1 because higher frequency photons require less reverse voltage to stop electrons.
Answer: Saturation current is same for both frequencies because intensity is same; stopping potential is more positive for f1.
According to Einstein's photoelectric equation, eV₀ = hf - φ. For higher frequency f1, the stopping potential V₀ is larger (more positive relative to the emitter). Saturation current depends on number of photons, which is same for same intensity, so it is same for both frequencies. Option c correctly states both.
8. The photoelectric emission from a metal surface begins within what approximate time after the light strikes?
- 10⁻⁶ s
- 10⁻¹² s
- 10⁻³ s
- 10⁻⁹ s
Answer: 10⁻¹² s
Photoelectric emission is nearly instantaneous. According to Einstein's photoelectric effect, an electron absorbs a photon and is emitted without any measurable time lag. Modern experiments show the emission occurs within about 10⁻¹² s (femtoseconds), which is the time scale for electron-photon interaction.
9. In a photoelectric effect experiment, the frequency of incident light is fixed above threshold. If the intensity is doubled, what happens to the stopping potential V₀?
- V₀ becomes half
- V₀ remains unchanged
- V₀ becomes double
- V₀ increases by a factor of √2
Answer: V₀ remains unchanged
According to Einstein's photoelectric equation, eV₀ = hν - φ, the stopping potential V₀ depends only on frequency ν and work function φ, not on intensity. Doubling intensity increases the number of photons but does not change the maximum kinetic energy of emitted electrons, so V₀ remains unchanged.
10. Photoelectrons emitted from a metal surface have kinetic energies ranging from 0 to K_max. Which statement correctly explains this?
- Electrons have different binding energies; those more tightly bound require more than work function to escape, leaving less kinetic energy.
- Electrons are emitted with the same kinetic energy K_max, but some lose energy to the metal after emission.
- Electrons lose energy due to collisions inside the metal, so they emerge with varying energies.
- The photon energy is shared among multiple electrons, so each gets a fraction of hν.
Answer: Electrons have different binding energies; those more tightly bound require more than work function to escape, leaving less kinetic energy.
Electrons in a metal have different binding energies. The work function φ is the minimum energy to escape from the surface. Deeper electrons need more than φ to escape, so after absorbing a photon of energy hν, their kinetic energy is less than hν - φ, giving a range from 0 to K_max.
11. A monochromatic light source of power 10 W emits light of wavelength 400 nm. If the power is doubled at fixed wavelength, what happens to the saturation photocurrent and stopping potential? (Work function = 2.0 eV)
- Current doubles; stopping potential unchanged
- Current doubles; stopping potential also doubles
- Current unchanged; stopping potential doubles
- Current becomes half; stopping potential unchanged
Answer: Current doubles; stopping potential unchanged
Doubling power at fixed wavelength doubles the number of photons per second. Each photon ejects at most one electron, so the saturation current doubles. Stopping potential depends only on photon energy (hν) and work function, which are unchanged, so it remains the same.
12. A light source of power 100 W emits photons of wavelength 400 nm. The quantum efficiency of the metal is 0.5%. What is the saturation current? (h = 6.63×10⁻³⁴ J s, c = 3×10⁸ m/s, e = 1.6×10⁻¹⁹ C)
- 0.64 A
- 0.32 A
- 0.08 A
- 0.16 A
Answer: 0.16 A
Photon energy E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(400×10⁻⁹) = 4.97×10⁻¹⁹ J. Photon rate n = P/E = 100 / 4.97×10⁻¹⁹ = 2.01×10²⁰ s⁻¹. Electron rate n_e = η n = 0.005 × 2.01×10²⁰ = 1.005×10¹⁸ s⁻¹. Saturation current i_sat = n_e e = 1.005×10¹⁸ × 1.6×10⁻¹⁹ = 0.1608 A ≈ 0.16 A.