Question

Monochromatic light of frequency \( 8 \times 10^{14} \, \text{Hz} \) is incident on a metal surface whose threshold frequency is \( 5 \times 10^{14} \, \text{Hz} \). The stopping potential is approximately:
(Given \( h = 6.6 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

• A. 0.62 V
• B. 1.24 V
• C. 2.48 V
• D. 3.72 V

Hint:

To perform quick calculations in photoelectric effect questions, convert energies into electron-volts (eV). Since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \), dividing the energy in Joules by \( e \) directly gives the stopping potential in Volts.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the stopping potential of photoelectrons emitted from a metal surface when monochromatic light of a given frequency is incident on it.

Step 2: Key Formula or Approach:
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by:
\[ K_{\max} = h\nu - h\nu_0 \] The stopping potential \( V_0 \) is related to the maximum kinetic energy by:
\[ eV_0 = K_{\max} \implies V_0 = \frac{h(\nu - \nu_0)}{e} \] where:
- \( h \) is Planck's constant
- \( \nu \) is the frequency of incident light
- \( \nu_0 \) is the threshold frequency
- \( e \) is the elementary charge

Step 3: Detailed Explanation:
Given:
- \( \nu = 8 \times 10^{14} \, \text{Hz} \)
- \( \nu_0 = 5 \times 10^{14} \, \text{Hz} \)
- \( h = 6.6 \times 10^{-34} \, \text{J s} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
Substitute the given values into the formula:
\[ V_0 = \frac{6.6 \times 10^{-34} \times (8 \times 10^{14} - 5 \times 10^{14})}{1.6 \times 10^{-19}} \] \[ V_0 = \frac{6.6 \times 10^{-34} \times 3 \times 10^{14}}{1.6 \times 10^{-19}} \] \[ V_0 = \frac{19.8 \times 10^{-20}}{1.6 \times 10^{-19}} \] \[ V_0 = \frac{1.98}{1.6} \approx 1.2375 \, \text{V} \] Thus, the stopping potential is approximately \( 1.24 \, \text{V} \).

Step 4: Final Answer:
(B) 1.24 V