Question

The density of electrons and holes in a pure germanium sample at room temperature are equal and its value is $3\times 10^{16}$ per $m^{3}$ . On doping with Aluminium, increases to $4.5\times 10^{22}$ per $m^{3}$ . Then the electron density in doped germanium is $k\times 10^{10}m^{- 3}$ . The value of $k$ is:

Answer 1

The electron and hole concentration in a semiconductor in thermal equilibrium is given by $n_{e} n_{h}=n_{i}^{2}$
$n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{\left(3 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}=2 \times 10^{10} m ^{-3}$