The electrical conductivity of semicondutor increases when electromagnetic radiation of wavelength shorter than `24800Å` is incident on it. The band gap for the semiconductor is

Options

A. 0.9ev
B. 0.7 Ev
C. 0.5 eV
D. 1.1 eV

Answer

To find the band gap of the semiconductor given that its electrical conductivity increases when electromagnetic radiation of wavelength shorter than 24800 Å is incident on it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between wavelength and energy**: The energy of a photon is related to its wavelength by the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3 \times 10^{8} \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

2. **Convert the wavelength from Ångströms to meters**:
\[ 24800 \, \text{Å} = 24800 \times 10^{-10} \, \text{m} = 2.48 \times 10^{-6} \, \text{m} \]

3. **Substitute the values into the energy formula**:
\[ E_g = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^{8} \, \text{m/s})}{2.48 \times 10^{-7} \, \text{m}} \]

4. **Calculate the energy**:
- First, calculate \(hc\):
\[ hc = (6.626 \times 10^{-34})(3 \times 10^{8}) = 1.9878 \times 10^{-25} \, \text{J m} \]
- Now divide by the wavelength:
\[ E_g = \frac{1.9878 \times 10^{-25}}{2.48 \times 10^{-7}} \approx 8.01 \times 10^{-19} \, \text{J} \]

5. **Convert energy from joules to electron volts**:
- Use the conversion \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ E_g \approx \frac{8.01 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 5.00625 \, \text{eV} \]

6. **Final result**: The band gap for the semiconductor is approximately:
\[ E_g \approx 0.5 \, \text{eV} \]

### Conclusion:
The band gap for the semiconductor is **0.5 eV**.