Question

In a $p-n$ junction diode, an electric field of magnitude $2 \times 10^{5} \,V / m$ exists in the depletion region. A particle with charge $-3 e$ 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 $p-n$ junction is

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

Answer 1

Answer: (C)

For depletion width $d$, developed potential is
$V=E \cdot d=2 \times 10^{5} \times d$
For charged particle, experienced potential is
$V=\frac{\text { Energy }}{\text { Charge }}=\frac{0 \cdot 6 eV }{3 e}=0 \cdot 2 V$
So, $2 \times 10^{5} \times d=0 \cdot 2$
$ \Rightarrow \,d =10^{-6} \,m$ or $1000 \,nm$