A photodiode is fabricated with band gap of 3 eV. It can detect

The correct answer is Option (1) → Visible light

A photodiode is an optical detector: a photon can create an electron-hole pair at the junction only if its energy is at least the band gap, \( h\nu \ge E_g \). The longest wavelength it can detect is

    \( \displaystyle \lambda_{\max} = \frac{hc}{E_g} = \frac{1240\,\mathrm{eV\,nm}}{3\,\mathrm{eV}} \approx 413\,\mathrm{nm} \)

This lies at the violet end of the visible spectrum (about 400 to 700 nm), so the photodiode responds to visible light of the shortest wavelengths.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Photodiode – “… When the photodiode is illuminated with light (photons) with energy (hν) greater than the energy gap (E_g) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons.”

Light Emitting Diode – “… The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV (spectral range of visible light is from about 0.4 μm to 0.7 μm, i.e., from about 3 eV to 1.8 eV).”