GATE 2026 EC Q.54: solution and answer

Official answer: A, B, DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. An LED based on a direct-bandgap semiconductor emits photons whose energy is essentially equal to the material's bandgap energy: $E_{\gamma }\approx E_g$. So the emission is concentrated at a single wavelength ${\lambda }_g=\frac{hc}{E_g}$, not spread across an arbitrary range.

2. Compute $hc$: $hc=6.63\times 10^{-34}\times 3\times 10^8=1.989\times 10^{-25}\text{J m}$.

3. Convert to eV·nm using $q=1.6\times 10^{-19}$ C: $\frac{hc}{q}=\frac{1.989\times 10^{-25}}{1.6\times 10^{-19}}=1.24313\times 10^{-6}\text{eV m}=1243.13\text{eV nm}$.

4. Find the emission wavelength: ${\lambda }_g=\frac{1243.13\text{eV nm}}{1.3\text{eV}}\approx 956\text{nm}$.

5. Check each option's window against ${\lambda }_g\approx 956$ nm: (A) $1390$–$1430$ nm does not contain 956 nm. (B) $1070$–$1110$ nm does not contain 956 nm. (C) $930$–$970$ nm DOES contain 956 nm. (D) $490$–$530$ nm does not contain 956 nm.

6. Since the LED emits only near ${\lambda }_g\approx 956$ nm, it will NOT emit in the ranges that exclude this value: (A), (B) and (D). It WILL emit within range (C), so C is not part of the answer.