The semiconductor used for fabrication of visible LEDs must at least have a band gap of:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.8 eV
An LED emits a photon of energy about equal to the band gap, \( E_g = h\nu = hc/\lambda \). The visible spectrum runs from about 0.4 μm to 0.7 μm, and the lowest photon energy in it is for the red end.
\( \displaystyle E_g = \frac{1240\,\mathrm{eV\,nm}}{700\,\mathrm{nm}} \approx 1.8\,\mathrm{eV} \)
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):
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).”