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Pure Si at 500 K has equal number of electron ($n_e$) and hole ($n_h$) concentrations of $1.5\times10^{16}\ m^{-3}$. Doping by indium increases $n_h$ to $4.5\times10^{22}\ m^{-3}$. The doped semiconductor is of
A
n-type with electron concentration $n_e = 2.5\times10^{23}\ m^{-3}$
B
p-type having electron concentration $n_e = 5\times10^9\ m^{-3}$
C
n-type with electron concentration $n_e = 2.5\times10^{22}\ m^{-3}$
D
p-type with electron concentration $n_e = 2.5\times10^{10}\ m^{-3}$
Detailed Solution
Indium is a trivalent impurity (acceptor); it creates holes, so the doped semiconductor is p-type.
For a semiconductor in thermal equilibrium, $n_en_h = n_i^2$ (mass action law).
$n_i = 1.5\times10^{16}\ m^{-3}$, so $n_i^2 = 2.25\times10^{32}\ m^{-6}$
$n_e = \frac{n_i^2}{n_h} = \frac{2.25\times10^{32}}{4.5\times10^{22}}$
$n_e = 0.5\times10^{10} = 5\times10^9\ m^{-3}$
So it is p-type with electron concentration $n_e = 5\times10^9\ m^{-3}$.
For a semiconductor in thermal equilibrium, $n_en_h = n_i^2$ (mass action law).
$n_i = 1.5\times10^{16}\ m^{-3}$, so $n_i^2 = 2.25\times10^{32}\ m^{-6}$
$n_e = \frac{n_i^2}{n_h} = \frac{2.25\times10^{32}}{4.5\times10^{22}}$
$n_e = 0.5\times10^{10} = 5\times10^9\ m^{-3}$
So it is p-type with electron concentration $n_e = 5\times10^9\ m^{-3}$.