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The density of electrons and holes in a pure germanium sample at room temperature are equal and its value is $3 \times 10^{16} p e r m^{3} .$ On doping with aluminium, the hole density increases to $4.5 \times 10^{22} p e r m^{3}$ . Then the electron density in dopped germanium is

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$2 \times 10^{10} m^{- 3}$

$5 \times 10^{9} m^{- 3}$

$4.5 \times 10^{- 10} m^{- 3}$

$3 \times 10^{9} m^{- 3}$

answer is A.

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Detailed Solution

The electron and hole concentration in a semi conductor in thermal equilibrium is given by $n_{e} n_{h} = n_{1}^{2}$

$n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(3 \times 10^{16})}^{2}}{4.5 \times 10^{22}} = 2 \times 10^{10} m^{- 3}$

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