Quantum Mechanics

Problems on Photoelectric Effect

Quick Exam Notes
  • Maximum Kinetic Energy: \( KE_{max} = h\nu - \phi \)
  • Threshold Frequency: \( \nu_0 = \phi / h \)
  • Stopping Potential: \( eV_0 = K_{max} \)
  • Threshold Wavelength: \( \lambda_0 = hc / \phi \)

1. Light with a wavelength of 500 nm is incident on sodium, which has a work function of 2.28 eV. Determine the maximum kinetic energy of the ejected photoelectron. Given data:

  • Wavelength: \( \lambda = 500 \) nm = \( 500 \times 10^{-9} \) m
  • Work function of sodium: \( \phi = 2.28 \) eV
  • Speed of light: \( c = 3.00 \times 10^8 \) m/s
  • Planck’s constant: \( h = 6.626 \times 10^{-34} \) J·s
Frequency of incident light
\[ \nu = \frac{c}{\lambda} \] \[ \nu = \frac{3.00 \times 10^8}{500 \times 10^{-9}} \] \[ \nu = 6.00 \times 10^{14} \text{ Hz} \]
Energy of the incident photon
\[ E = h\nu \] \[ E = (6.626 \times 10^{-34} \times 6.00 \times 10^{14}) \] \[ E = 3.976 \times 10^{-19} \text{ J} \] Converting to electron volts: \[ E = \frac{3.976 \times 10^{-19}}{1.602 \times 10^{-19}} \] \[ E \approx 2.48 \text{ eV} \]
The maximum kinetic energy is given by
\[ KE_{\text{max}} = E - \phi \] \[ KE_{\text{max}} = 2.48 - 2.28 \] \[ KE_{\text{max}} = 0.20 \text{ eV} \]

2. Given a metal with a work function of 6.63 eV, calculate its threshold wavelength. Given data:

  • Work function: \( \phi = 6.63 \) eV
  • Planck’s constant: \( h = 6.626 \times 10^{-34} \) J·s
  • Speed of light: \( c = 3.00 \times 10^8 \) m/s
  • Conversion: \( 1 \) eV \( = 1.602 \times 10^{-19} \) J
Work Function to Joules
\[ \phi = 6.63 \times 1.602 \times 10^{-19} \] \[ \phi = 1.062 \times 10^{-18} \text{ J} \]
Threshold wavelength is given by
\[ \lambda_0 = \frac{hc}{\phi} \] \[ \lambda_0 = \frac{(6.626 \times 10^{-34}) (3.00 \times 10^8)}{1.062 \times 10^{-18}} \] \[ \lambda_0 = \frac{1.988 \times 10^{-25}}{1.062 \times 10^{-18}} \] \[ \lambda_0 = 1.87 \times 10^{-7} \text{ m} = 187 \text{ nm} \]

3. For a metal with a threshold wavelength of 12.4 nm, find its work function. Given data:

  • Threshold wavelength: \( \lambda_0 = 12.4 \) nm = \( 12.4 \times 10^{-9} \) m
  • Planck’s constant: \( h = 6.626 \times 10^{-34} \) J·s
  • Speed of light: \( c = 3.00 \times 10^8 \) m/s
  • Conversion: \( 1 \) eV \( = 1.602 \times 10^{-19} \) J
Work function in Joules
\[ \phi = \frac{hc}{\lambda} \] \[ \phi = \frac{(6.626 \times 10^{-34}) (3.00 \times 10^8)}{12.4 \times 10^{-9}} \] \[ \phi = \frac{1.988 \times 10^{-25}}{12.4 \times 10^{-9}} \] \[ \phi = 1.603 \times 10^{-17} \text{ J} \]
Convert to Electron Volts
\[ \phi = \frac{1.603 \times 10^{-17}}{1.602 \times 10^{-19}} \] \[ \phi \approx 100 \text{ eV} \]

4. A copper surface is irradiated with light of wavelength 1849 Å, resulting in a stopping potential of 2.7 V. Calculate the threshold frequency and work function for the copper surface. Given data:

  • Stopping potential: \( V_s = 2.7 \) V
  • Wavelength: \( \lambda = 1849 \) Å = \( 1849 \times 10^{-10} \) m
  • Speed of light: \( c = 3.00 \times 10^8 \) m/s
  • Planck’s constant: \( h = 6.626 \times 10^{-34} \) J·s
  • Charge of electron: \( e = 1.602 \times 10^{-19} \) C
Frequency of Incident Light
\[ \nu = \frac{c}{\lambda} \] \[ \nu = \frac{3.00 \times 10^8}{1849 \times 10^{-10}} \] \[ \nu = 1.62 \times 10^{15} \text{ Hz} \]
Work Function (\( \phi \)) is given by
\[ \phi = h\nu - eV_s \] \[ \phi = (6.626 \times 10^{-34} \times 1.62 \times 10^{15}) - (1.602 \times 10^{-19} \times 2.7) \] \[ \phi = (1.073 \times 10^{-18}) - (4.325 \times 10^{-19}) \] \[ \phi = 6.41 \times 10^{-19} \text{ J} \] Converting to electron volts: \[ \phi = \frac{6.41 \times 10^{-19}}{1.602 \times 10^{-19}} \] \[ \phi \approx 4.0 \text{ eV} \]
The threshold Frequency (\( \nu_0 \)) is
\[ \nu_0 = \frac{\phi}{h} \] \[ \nu_0 = \frac{6.41 \times 10^{-19}}{6.626 \times 10^{-34}} \] \[ \nu_0 \approx 9.67 \times 10^{14} \text{ Hz} \]

Practice Problems

  1. Light of frequency \( 8 \times 10^{14} \) Hz is incident on a metal with work function 2 eV. Calculate stopping potential.
  2. Find the threshold frequency of a metal with work function 5 eV.
  3. If threshold wavelength of sodium is 680 nm, what is its work function in eV?

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📌 Summary

Problems on the photoelectric effect test concepts of Einstein’s equation, threshold frequency, stopping potential, and work function. By applying formulas like \( K_{max} = h\nu - \phi \) and \( eV_0 = K_{max} \), students can calculate kinetic energy, photon energy, and emission conditions. These numericals strengthen understanding of quantum theory and its applications in devices such as photoelectric cells and solar panels.

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