The work function of a surface of a photosensitive material is `6.2 eV`. The wavelength of the incident radiation for which the stopping potential is `5 V` lies in the

Options

A. ultraviolet region
B. visible region
C. infrared region
D. X - ray region

Explanation

To solve the problem, we need to find the wavelength of the incident radiation using the given work function and stopping potential. We will use the photoelectric equation derived from Einstein's photoelectric effect.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Work function (Φ) = 6.2 eV
- Stopping potential (V₀) = 5 V

2. **Use the Photoelectric Equation:**
The energy of the incident photons (E) can be expressed as:
\[ E = eV_0 + \Phi \]
where \( e \) is the charge of an electron (1 eV = 1.6 × 10⁻¹⁹ J).

3. **Calculate the Total Energy:**
Substitute the values into the equation:
\[ E = (5 \, \text{V}) + (6.2 \, \text{eV}) = 11.2 \, \text{eV} \]

4. **Convert Energy to Wavelength:**
The energy of a photon is also given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

Rearranging for wavelength (\( \lambda \)):
\[ \lambda = \frac{hc}{E} \]

5. **Substituting Values:**
Substitute \( h \), \( c \), and \( E \) into the equation:
\[ \lambda = \frac{(4.1357 \times 10^{-15} \, \text{eV s}) \times (3 \times 10^8 \, \text{m/s})}{11.2 \, \text{eV}} \]

6. **Calculate the Wavelength:**
\[ \lambda = \frac{1.24071 \times 10^{-6} \, \text{m eV}}{11.2 \, \text{eV}} \approx 1.107 \times 10^{-7} \, \text{m} = 110.7 \, \text{nm} \]

7. **Determine the Region of Wavelength:**
The calculated wavelength of approximately 110.7 nm falls within the ultraviolet (UV) region of the electromagnetic spectrum.

### Conclusion:
The wavelength of the incident radiation for which the stopping potential is 5 V lies in the ultraviolet region.

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