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Question

Physics Question on Photoelectric Effect

In a pp - type semiconductor the donor level is at 50meV50 \,meV above the valence band. To produce one electron, the maximum wavelength of light photon required is (Planck's constant, h=6.6×1034Jsh=6.6 \times 10^{-34} \,Js and speed of light in vacuum, c=3×108ms1c=3 \times 10^{8} \,ms ^{-1} )

A

0.0248μm0.0248\, \mu m

B

0.248μm0.248 \, \mu m

C

2.48μm2.48 \, \mu m

D

24.8μm24.8\, \mu m

Answer

24.8μm24.8\, \mu m

Explanation

Solution

Given,
pp -type semiconductor donor energy level,
E=50meV=50×103×1.6×1019VE=50\, meV =50 \times 10^{-3} \times 1.6 \times 10^{-19} \,V
Planck's constant, h=6.6×1034Jsh=6.6 \times 10^{-34} \,Js
speed of light in vacuum, c=3×108m/sc=3 \times 10^{8} \,m / s
Now, for the maximum wavelength of light photon's required (p)(p).
According to the Planck's quantum theory,
E=hv\therefore \, E=h v
E=hcλ\Rightarrow E=\frac{h c}{\lambda}
[v=cλ]\left[\because v=\frac{c}{\lambda}\right]
Putting the given values, we get
50×103×1.6×1019=6.6×1034×3×108λ50 \times 10^{-3} \times 1.6 \times 10^{-19}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{\lambda}
λ=6.6×1034×3×10850×103×1.6×1019\lambda= \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{50 \times 10^{-3} \times 1.6 \times 10^{-19}}
=6.6×3×1034×1085×16×1022=6.6×3×1045×16= \frac{6.6 \times 3 \times 10^{-34} \times 10^{8}}{5 \times 16 \times 10^{-22}}=\frac{6.6 \times 3 \times 10^{-4}}{5 \times 16}
=2.475×105m=2.475 \times 10^{-5} \,m
or λ=24.75×106=24.75μm\lambda=24.75 \times 10^{-6}=24.75 \,\mu m
Hence. the maximum wavelength of light photon required is 24.8μm24.8\, \mu \,m