Light of frequency `10^(15)` Hz falls on a metal surface of work function 2.5 eV. The stopping potential of photoelectrons in volts is

Options

A. 1.6
B. 2.5
C. 4.1
D. 6.6

Explanation

To solve the problem step by step, we will use the photoelectric effect equation and the given data.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Frequency of light, \( \nu = 10^{15} \, \text{Hz} \)
- Work function of the metal, \( \phi_0 = 2.5 \, \text{eV} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

2. **Calculate the Energy of the Incident Photons:**
The energy of the photons can be calculated using the formula:
\[ E = h \nu \]
Substituting the values:
\[ E = (6.63 \times 10^{-34} \, \text{Js}) \times (10^{15} \, \text{Hz}) = 6.63 \times 10^{-19} \, \text{J} \]

3. **Convert the Energy from Joules to Electron Volts:**
To convert the energy from Joules to electron volts, we use the relation \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{6.63 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 4.14 \, \text{eV} \]

4. **Calculate the Maximum Kinetic Energy of the Photoelectrons:**
The maximum kinetic energy (\( K_{\text{max}} \)) of the emitted photoelectrons is given by:
\[ K_{\text{max}} = E - \phi_0 \]
Substituting the values:
\[ K_{\text{max}} = 4.14 \, \text{eV} - 2.5 \, \text{eV} = 1.64 \, \text{eV} \]

5. **Determine the Stopping Potential (\( V_0 \)):**
The stopping potential is related to the maximum kinetic energy by:
\[ K_{\text{max}} = eV_0 \]
Therefore, we can find \( V_0 \):
\[ V_0 = \frac{K_{\text{max}}}{e} = 1.64 \, \text{V} \]

### Final Answer:
The stopping potential of the photoelectrons is approximately \( 1.64 \, \text{V} \).