The work function of a metallic surface is `5.01 eV`. The photo - electrons are emitted when light of wavelength `2000 Å` falls on it . The potential difference applied to stop the fastest photo - electrons is `[ h = 4.14 xx 10^(-15) eV sec]`

Options

A. 1.2 volt
B. 2.24 volt
C. 3.6 volt
D. 4.8 volt

Answer

To solve the problem, we will use Einstein's photoelectric equation, which relates the energy of the incident photons to the work function of the metal and the kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Identify the given values:**
- Work function \( W_0 = 5.01 \, \text{eV} \)
- Wavelength of light \( \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} \)
- Planck's constant \( h = 4.14 \times 10^{-15} \, \text{eV sec} \)
- Speed of light \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Calculate the energy of the incident photons:**
The energy \( E \) of the photons can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV sec})(3 \times 10^8 \, \text{m/s})}{2000 \times 10^{-10} \, \text{m}} \]

3. **Perform the calculation:**
\[ E = \frac{(4.14 \times 3) \times 10^{-7}}{2000} \, \text{eV} \]
\[ E = \frac{12.42 \times 10^{-7}}{2000} \, \text{eV} \]
\[ E = 6.21 \times 10^{-10} \, \text{eV} \]

4. **Use Einstein's photoelectric equation:**
According to the photoelectric equation:
\[ E = W_0 + K_{\text{max}} \]
where \( K_{\text{max}} = eV_0 \) (the maximum kinetic energy of the emitted electrons). Rearranging gives:
\[ eV_0 = E - W_0 \]

5. **Substituting the values:**
\[ V_0 = \frac{E - W_0}{e} \]
Since we are working in eV, we can directly substitute:
\[ V_0 = E - W_0 \]
\[ V_0 = 6.21 \, \text{eV} - 5.01 \, \text{eV} \]

6. **Calculate the stopping potential \( V_0 \):**
\[ V_0 = 1.20 \, \text{V} \]

### Final Answer:
The potential difference applied to stop the fastest photo-electrons is \( V_0 = 1.20 \, \text{V} \).

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