Q 12-14-066JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
The acceptor level of a p-type semiconductor is $6\ \text{eV}$. The maximum wavelength of light which can create a hole would be (Given $hc = 1242\ \text{eV nm}$)
Answer: (C) $207\ \text{nm}$
The photon energy must be at least $6\ \text{eV}$: $\lambda_{max} = \dfrac{1242}{6} = 207\ \text{nm}$.
Solution by Sreeraj P, M.Sc Physics