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In pure silicon at 300 K the electron and hole concentration is each equal to $1.5 \times 10^{16} m^{- 3}$ . When doped with indium, the hole concentration increases to $4.5 \times 10^{22} m^{- 3}$ . What is the electron concentration in doped silicon?

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

$4 \times 10^{9} m^{- 3}$

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

$6 \times 10^{9} m^{- 3}$

answer is C.

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Detailed Solution

$Given n_{i} = 1.5 \times 10^{16} m^{- 3} and n_{h} = 4.5 \times 10^{22} m^{- 3} Now, we know that n_{e} n_{h} = n_{i}^{2} or n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{{(1.5 \times 10^{16})}^{2}}{4.5 \times 10^{22}} = 5.0 \times 10^{9} m^{- 3}$

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