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Difficulty: MediumPhotoelectric Effect and Work Function

Monochromatic light of frequency 8.0×1014 Hz8.0 \times 10^{14}\text{ Hz} illuminates a metal surface with a work function of 2.2 eV2.2\text{ eV}. If the intensity of the incident light is doubled while keeping the frequency constant, what is the maximum kinetic energy of the emitted photoelectrons? (h=6.6×10−34 J⋅sh = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}, 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. 1.1 eV1.1\text{ eV}Answer
  2. B
    2.2 eV2.2\text{ eV}
  3. C
    3.3 eV3.3\text{ eV}
  4. D
    4.4 eV4.4\text{ eV}

Answer

The maximum kinetic energy of the emitted photoelectrons remains 1.1 eV1.1\text{ eV}.
According to Einstein's photoelectric equation, Kmax⁡=hf−W0K_{\max} = hf - W_0. For incident light of frequency 8.0×1014 Hz8.0 \times 10^{14}\text{ Hz}, the photon energy is 3.3 eV3.3\text{ eV}. Subtracting the work function of 2.2 eV2.2\text{ eV} yields Kmax⁡=1.1 eVK_{\max} = 1.1\text{ eV}. Because maximum kinetic energy depends solely on photon frequency and the metal's work function, changing the light intensity has no impact on Kmax⁡K_{\max}.

Step-by-Step Solution

1
Calculate the energy of the incident photon in Joules and convert to electron-volts (eV).
E=hf=(6.6×10−34 J⋅s)×(8.0×1014 Hz)=5.28×10−19 J=5.28×10−191.6×10−19 eV=3.3 eVE = hf = (6.6 \times 10^{-34}\text{ J}\cdot\text{s}) \times (8.0 \times 10^{14}\text{ Hz}) = 5.28 \times 10^{-19}\text{ J} = \frac{5.28 \times 10^{-19}}{1.6 \times 10^{-19}}\text{ eV} = 3.3\text{ eV}.
Einstein's photoelectric equation requires knowing the energy of each incident photon.
2
Apply Einstein's photoelectric equation to calculate the maximum kinetic energy (Kmax⁡K_{\max}).
Kmax⁡=E−W0=3.3 eV−2.2 eV=1.1 eVK_{\max} = E - W_0 = 3.3\text{ eV} - 2.2\text{ eV} = 1.1\text{ eV}.
The work function (W0W_0) is the minimum energy needed to liberate an electron from the metal surface.
3
Analyze the effect of doubling light intensity at constant frequency.
Kmax⁡K_{\max} remains 1.1 eV1.1\text{ eV}.
Light intensity is proportional to the number of photons striking the surface per second, affecting the rate of electron emission (photocurrent), but not the energy of individual photons or the maximum kinetic energy of the emitted electrons.

Key Concept

Independence of photoelectron maximum kinetic energy from light intensity
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Topics

Photoelectric Effect and Work Function

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