Question

The work function for a photosensitive surface is 3.315 eV. The cut-off wavelength for photoemission of electrons from this surface is:

• A. 150 nm
• B. 200 nm
• C. 375 nm
• D. 500 nm

Hint:

The cut-off wavelength is inversely proportional to the work function. Higher work functions result in shorter cut-off wavelengths.

Answer 1

The Correct Option is C

The cut-off wavelength \( \lambda_{\text{cut}} \) can be found using the photoelectric equation: \[ E_{\text{photon}} = \phi = \frac{hc}{\lambda_{\text{cut}}}, \] where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \phi \) is the work function. Rearranging to solve for \( \lambda_{\text{cut}} \): \[ \lambda_{\text{cut}} = \frac{hc}{\phi}. \] Substitute \( h = 6.626 \times 10^{-34} \, \text{J}\cdot\text{s} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \phi = 3.315 \, \text{eV} = 3.315 \times 1.602 \times 10^{-19} \, \text{J} \): \[ \lambda_{\text{cut}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{3.315 \times 1.602 \times 10^{-19}} \approx 375 \, \text{nm}. \]