The width of the depletion region in a p-n junction diode is 400 nm and an intense electric field of `8 xx 10^5` V/m exists in it. What is the kinetic energy which a conduction electron must have in order to diffuse from then region to p region?

Options

A. 0.16eV
B. 0.24eV
C. 0. 8 eV
D. 0.32 eV

Answer

B

Explanation

`E="dV"/"dx" therefore dV= 8xx10^5 xx 400 xx 10^(-9)`
`=32xx10^(-2)` = 0.32 V
The minimum K.E. of the electron to cross the potential barrier U is e x V = 0.32 eV