Question

Pure Si at $500 \,K$ has equal number of electron $\left( n _{ e }\right)$ and hole $\left( n _{ h }\right)$ concentrations to $1.5 \times 10^{16} \,m ^{-3}$. Doping by indium increases $n_{ h }$ to $4.5 \times 10^{22} \,m ^{-3}$. The doped semiconductor is of :-

• A. $p$-type having elecron concentrations $n _{ e }=5 \times 10^{9} \,m ^{-3}$
• B. $n$-type with electron concentration $n _{ e }=5 \times 10^{22} \,m ^{-3}$
• C. $P$-type with electron concentration $n _{ e }=2.5 \times 10^{10} \,m ^{-3}$
• D. $n$-type with electron concentration $n _{ e }=2.5 \times 10^{23}\, m ^{-3}$

Answer 1

Answer: (A)

$n _{ e }=\frac{ n _{ i }^{2}}{ n _{ h }}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}$
$=5 \times 10^{9} \,m ^{-3}$