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

The correct option is 0.5 eV.

Step-by-Step Explanation:

The electrical conductivity of a semiconductor increases when the incident electromagnetic radiation has enough energy to excite electrons from the valence band to the conduction band. The minimum energy required for this transition is equal to the band gap energy (E_g) of the semiconductor.

The energy (E) of a photon is related to its wavelength (λ) by the Planck-Einstein relation:

$$E=\frac{hc}{\lambda }$$

Where:
• h is Planck's constant (≈ 6.63 × 10^-34 J·s)
• c is the speed of light (≈ 3 × 10⁸ m/s)
• λ is the wavelength of the incident radiation

To find the band gap energy in electron-volts (eV), we can use the convenient relation:

$$E\left(\text{in eV}\right)=\frac{1240}{\lambda \left(\text{in nm}\right)}$$

Given that the threshold wavelength is 2480 nm, we substitute λ = 2480 nm into the equation:

$$E_g=\frac{1240}{2480}\text{eV}$$

$$E_g=0.5\text{eV}$$

Therefore, the band gap for the semiconductor is 0.5 eV.