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

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

Einstein’s photoelectric equation is:

\[ K_{\text{max}} = h (u - u_0) \]

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

The relationship between the kinetic energy of the photoelectrons and the cut-off potential \( V_{\text{cut-off}} \) is:

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

Equating these two expressions yields:

\[ eV_{\text{cut-off}} = h (u - u_0) \]

Rearranging to solve for the cut-off potential:

\[ V_{\text{cut-off}} = \frac{h (u - u_0)}{e} \]

Using the given constants and frequencies: \( h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), \( u = 6.8 \times 10^{14} \, \text{Hz} \), and \( u_0 = 3.6 \times 10^{14} \, \text{Hz} \):

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

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

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

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