In a semiconductor, the number density of intrinsic charge carriers at $27^{\circ} C$ is $1.5 \times 10^{16} m^{- 3}$ . If the semiconductor is doped with impurity atom, the hole density increases to $4.5 \times 10^{22} m^{- 3}$ . The electron density in the doped semiconductor is $\times 10^{9} m^{- 3}$ .

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Solution

The electron and hole concentration in a semiconductor in thermal equilibrium is given by,

$n_{e} n_{h} = n_{i}^{2}$

$\Rightarrow n_{e} = \frac{n_{i}^{2}}{n_{h}} = \frac{(1.5 \times 10^{16})^{2}}{4.5 \times 10^{22}}$

$\Rightarrow n_{e} = \frac{1.5 \times 1.5 \times 10^{32}}{4.5 \times 10^{22}}$

$\Rightarrow n_{e} = 5 \times 10^{9} m^{- 3}$

Hence, $5$ is the correct answer.

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In a semiconductor, the number density of intrinsic charge carriers at 27∘C is 1.5×1016 m−3. If the semiconductor is doped with impurity atom, the hole density increases to 4.5×1022 m−3. The electron density in the doped semiconductor is ×109 m−3.

The electron and hole concentration in a semiconductor in thermal equilibrium is given by, nenh=n2i ⇒ne=n2inh=(1.5×1016)24.5×1022 ⇒ne=1.5×1.5×10324.5×1022 ⇒ne=5×109 m−3 Hence, 5 is the correct answer.