Question

Pure $Si$ at $500\, K$ has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5 \times 10^{16}\, m^{-3}$ Doping by indium increases $n_h$ to 4.5 $\times 10^{22}\, m^{-3}$ The doped semiconductor is of

• A. p-type having electron concentration $n_e=5\times 10^9\, m^{-3}$
• 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

The Correct Option is A

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