The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than `2480 nm` is incident on it. The band gap in `(eV)` for the semiconductor is.

Options

A. 0.9
B. 0.7
C. 0.5
D. 1.1

Explanation

To find the band gap energy (Eg) of the semiconductor in electron volts (eV), we can use the formula:

\[ E_g = \frac{hc}{\lambda} \]

where:
- \( E_g \) is the band gap energy in joules,
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.

Given that the wavelength is 2480 nm, we first need to convert this into meters:

\[ \lambda = 2480 \, \text{nm} = 2480 \times 10^{-9} \, \text{m} \]

Now, we can substitute the values into the formula:

1. **Calculate \( hc \)**:
\[ hc = (6.63 \times 10^{-34} \, \text{Js}) \times (3 \times 10^{8} \, \text{m/s}) = 1.989 \times 10^{-25} \, \text{Jm} \]

2. **Calculate \( E_g \)**:
\[ E_g = \frac{1.989 \times 10^{-25} \, \text{Jm}}{2480 \times 10^{-9} \, \text{m}} = 8.014 \times 10^{-19} \, \text{J} \]

3. **Convert \( E_g \) from joules to electron volts**:
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E_g = \frac{8.014 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} = 5.009 \, \text{eV} \]

However, since we are looking for the maximum wavelength that can cause an increase in conductivity, we need to consider the threshold wavelength. The band gap energy is calculated for the wavelength of 2480 nm, which means we need to find the energy corresponding to this wavelength.

4. **Final Calculation**:
\[ E_g = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \, \text{Js}) \times (3 \times 10^{8} \, \text{m/s})}{2480 \times 10^{-9} \, \text{m}} = 0.5 \, \text{eV} \]

Thus, the band gap energy \( E_g \) for the semiconductor is approximately **0.5 eV**.