Solveeit Logo
Login

Question

Physics Question on Semiconductor electronics: materials, devices and simple circuits

A p -type semiconductor has acceptor level 66 meV66 \,meV above the valence band. The maximum wavelength of light required to create a hole is very near to :

A

2×10−5m2\times10^{-5}m

B

2×10−6m2\times10^{-6}m

C

2×10−8m2\times10^{-8}m

D

2×10−10m2\times10^{-10}m

Answer

2×10−5m2\times10^{-5}m

Explanation

Solution

E=66meV=66×10−3 eVE=66 meV =66 \times 10^{-3}\, eV E=hcλE=\frac{h c}{\lambda} ⇒λ=hcE \Rightarrow \lambda=\frac{h c}{E} λ=6.625×10−34×3×10866×10−3×1.6×10−19\lambda=\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{66 \times 10^{-3} \times 1.6 \times 10^{-19}} λ=0.188×10−4 m\lambda=0.188 \times 10^{-4} \,m λ=1.88×10−5 m\lambda=1.88 \times 10^{-5}\, m ≈2×10−5 m \approx 2 \times 10^{-5} \,m