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

Electronic Devices PYQ — JEE Main 2005 · Shift 0 · Q139 with Solution

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Correct answer: D

For a semiconductor, conductivity increases when incident photons have enough energy to excite electrons across the band gap.

So, at the threshold wavelength $\lambda = 2480\,nm$ , the photon energy equals the band gap:

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

When $\lambda$ is in $nm$ ,

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

Thus,

$E_g = \frac{1240}{2480} = 0.5\,eV$

Therefore, the band gap is:

$\boxed{0.5\,eV}$