The electrical conductivity of the semiconductor increases when it is illuminated with electromagnetic radiation whose photon energy is equal to or greater than its band gap, which, for the given data (wavelength shorter than 2480 nm), corresponds to a band gap of about 0.5 eV.

Quick Scoop: What’s Going On?

When light of sufficiently short wavelength (high enough energy) falls on a semiconductor, its photons can promote electrons from the valence band to the conduction band.

This creates extra electron–hole pairs, increasing the number of charge carriers and therefore increasing electrical conductivity.

For wavelength λ=2480 nm\lambda =2480,\text{nm}λ=2480nm, the photon energy is found using E≈1240λ(nm) eVE\approx \frac{1240}{\lambda(\text{nm})},\text{eV}E≈λ(nm)1240​eV, giving E≈12402480≈0.5 eVE\approx \frac{1240}{2480}\approx 0.5,\text{eV}E≈24801240​≈0.5eV.

So, if conductivity starts increasing for radiation shorter than 2480 nm, it means photons of energy ≳ 0.5 eV are just enough to cross the band gap, implying a band gap of about 0.5 eV.

In simple words: shine light whose photons have energy at least equal to the band gap, and the semiconductor “wakes up” with more mobile charges, so its conductivity rises.

Answer used in exams/forums: For the standard question
“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 …”
the correct option given in solutions is 0.5 eV.

TL;DR: Conductivity increases because photons with energy ≥ band gap create extra charge carriers; for 2480 nm, that band gap is about 0.5 eV.

Information gathered from public forums or data available on the internet and portrayed here.