Solveeit Logo
Login

Question

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

The acceptor level of a p-type semiconductor is 6eV. The maximum wavelength of light which can create a hole would be : Given hc = 1242 eV nm.

A

407 nm

B

414 nm

C

207 nm

D

103.5 nm

Answer

207 nm

Explanation

Solution

The energy EE of a photon is given by:
E=hcλE = \frac{hc}{\lambda}
3535
Substitute E=6 eVE = 6 \, \text{eV} and hc=1240 eVnmhc = 1240 \, \text{eV} \text{nm}:
6=1240λ (nm)6 = \frac{1240}{\lambda \, (\text{nm})}
Rearrange to find λ\lambda:
λ=12406=207 nm\lambda = \frac{1240}{6} = 207 \, \text{nm}