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Physics Question on Electronic devices

In a p-type semiconductor, the acceptor level is situated 50 meV above the valence band. The maximum wavelength of light required to produce a hole will be:

A

24.8 \times 10^{-5} \, \text{m} \\\

B

0.248 \times 10^{-5} \, \text{m} \\\

C

2.48 \times 10^{-5} \, \text{m} \\\

D

248×10−5 m248 \times 10^{-5} \, \text{m}

Answer

2.48 \times 10^{-5} \, \text{m} \\\

Explanation

Solution

 The energy required to produce a hole is 50 meV=50×10−3 eV=8.0×10−21 J.\text{ The energy required to produce a hole is } 50 \, \text{meV} = 50 \times 10^{-3} \, \text{eV} = 8.0 \times 10^{-21} \, \text{J}.\\\\
The wavelength corresponding to this energy is given by:\text{The wavelength corresponding to this energy is given by:}
E=hcλ  ⟹  λ=hcEE = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}

Substitute h=6.626×10−34 J.s, c=3×108 m/s, and E=8.0×10−21 J:\text{Substitute } h = 6.626 \times 10^{-34} \, \text{J.s}, \, c = 3 \times 10^8 \, \text{m/s}, \text{ and } E = 8.0 \times 10^{-21} \, \text{J}:

λ=6.626×10−34×3×1088.0×10−21=2.48×10−5 m\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{8.0 \times 10^{-21}} = 2.48 \times 10^{-5} \, \text{m}