Question

The threshold frequency for a given metal is \( 3.6 \times 10^{14} \) Hz. If monochromatic radiations of frequency \( 6.8 \times 10^{14} \) Hz are incident on this metal, find the cut-off potential for the photoelectrons.
Given: - Threshold frequency, \( \nu_0 = 3.6 \times 10^{14} \) Hz
- Frequency of incident radiation, \( \nu = 6.8 \times 10^{14} \) Hz
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \)
- Charge of the electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

Hint:

The cut-off potential is obtained from the difference between the frequency of the incident light and the threshold frequency. The greater the frequency difference, the higher the energy of the emitted photoelectrons.

Answer 1

Einstein’s Photoelectric Equation and Calculation of Cut-off Potential 

According to Einstein’s photoelectric equation:

\[ K_{\text{max}} = h (\nu - \nu_0) \]

Where \( K_{\text{max}} \) is the maximum kinetic energy of the photoelectrons.

The kinetic energy of the photoelectrons is related to the cut-off potential \( V_{\text{cut-off}} \) by:

\[ K_{\text{max}} = eV_{\text{cut-off}} \]

Therefore, we can write:

\[ eV_{\text{cut-off}} = h (\nu - \nu_0) \]

Substituting the given values:

\[ V_{\text{cut-off}} = \frac{h (\nu - \nu_0)}{e} \]

Substitute \( h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), \( \nu = 6.8 \times 10^{14} \, \text{Hz} \), and \( \nu_0 = 3.6 \times 10^{14} \, \text{Hz} \):

\[ V_{\text{cut-off}} = \frac{(6.63 \times 10^{-34}) \times (6.8 \times 10^{14} - 3.6 \times 10^{14})}{1.6 \times 10^{-19}} \]

\[ V_{\text{cut-off}} = \frac{6.63 \times 10^{-34} \times 3.2 \times 10^{14}}{1.6 \times 10^{-19}} \]

\[ V_{\text{cut-off}} = \frac{2.12 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.33 \, \text{V} \]

Thus, the cut-off potential for the photoelectrons is \( 1.33 \, \text{V} \).