In a p-n juction diode, an electric field of magnitude `2xx10^(5)V` / m exists in the depletion region. A particle with charge -3e can diffuse from n-side to p-side, if it has minimum kinetic energy 0.6 eV.
The width of the depletion region of the p-n juction is

Options

A. 300 nm
B. 600 nm
C. 1000 nm
D. 1200 nm

Answer

C

Explanation

For depletion width d, developed potential is
`V=E.d = 2xx10^(5)xxd`
For charged particle, experoenced potential is
`V= ("Energy")/("Charge") = (0.6 eV)/(3e)= 0.2 V`
So, `2xx10^(5)xxd=0.2`
`rArr d=10^(-6)m` or 1000 nm