Work function of potassium metal is 2.30 eV . When light of frequency `8xx 10^(14)` Hz is incident on the metal surface, photoemission of electrons occurs. The stopping potential of the electrons will be equal to

Options

A. 0.1 V
B. 1.0 V
C. 2.3 V
D. 3.3 V

Explanation

To solve the problem, we need to calculate the stopping potential of the electrons emitted from potassium metal when light of a certain frequency is incident on it. The stopping potential can be determined using the photoelectric effect equation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Work function of potassium, \( \phi = 2.30 \, \text{eV} \)
- Frequency of incident light, \( \nu = 8 \times 10^{14} \, \text{Hz} \)

2. **Calculate the Energy of the Incident Photons:**
The energy of the photons can be calculated using the formula:
\[ E = h \nu \]
where \( h \) (Planck's constant) is approximately \( 6.63 \times 10^{-34} \, \text{Js} \).

Substituting the values:
\[ E = (6.63 \times 10^{-34} \, \text{Js}) \times (8 \times 10^{14} \, \text{Hz}) \]

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

4. **Calculate the Maximum Kinetic Energy of the Emitted Electrons:**
The maximum kinetic energy (Kmax) of the emitted electrons can be calculated using the photoelectric equation:
\[ K_{\text{max}} = E - \phi \]
where \( \phi \) is the work function.

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

6. **Substitute the Values and Solve:**
After calculating \( E \) and \( K_{\text{max}} \), we can find \( V_0 \).

- From the calculations, we find:
- \( E \approx 3.3 \, \text{eV} \)
- \( K_{\text{max}} = 3.3 \, \text{eV} - 2.3 \, \text{eV} = 1.0 \, \text{eV} \)
- Thus, \( V_0 = 1.0 \, \text{V} \)

### Final Answer:
The stopping potential of the electrons will be equal to **1 V**.

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