Q 12-14-172JEE MainJEE Main 2020 (3 Sep, Shift 2)Easy
If a semiconductor photo diode can detect a photon with a maximum wavelength of $400\ \text{nm}$, then its band gap energy is: (Planck's constant $h = 6.63\times10^{-34}\ \text{J s}$, speed of light $c = 3\times10^{8}\ \text{m s}^{-1}$)
Answer: (D) $3.1\ \text{eV}$
$$E_g = \frac{hc}{\lambda_{max}} = \frac{6.63\times10^{-34}\times3\times10^{8}}{400\times10^{-9}\times1.6\times10^{-19}} \approx 3.1\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics