GATE 2022 IN Q.27: solution and answer

Official answer: 870.0 to 880.0Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. A photodiode responds only to photons whose energy is at least the bandgap, so the cut-off wavelength is ${\lambda }_c=\frac{hc}{E_g}$.

2. Convert the bandgap to joules: $E_g=1.42\times 1.6\times 10^{-19}=2.272\times 10^{-19}\text{J}$.

3. Compute the photon constant: $hc=6.626\times 10^{-34}\times 3\times 10^8=1.9878\times 10^{-25}\text{J m}$.

4. $${\lambda }_c=\frac{1.9878\times 10^{-25}\text{J m}}{2.272\times 10^{-19}\text{J}}=8.749\times 10^{-7}\text{m}$$

5. Since $1\text{nm}=10^{-9}\text{m}$, ${\lambda }_c\approx 874.9\text{nm}$.