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  4. A particular semiconductor in equilibrium has 1× 1016 cm- 3 donor atoms, 1.1 × 1017 cm- 3 acceptor atoms. If the intrinsic carrier density (ni) of the semiconductor is 1012 cm- 3 , then the electron density in it will be

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Q. A particular semiconductor in equilibrium has $1\times 10^{16} \, cm^{- 3}$ donor atoms, $1.1 \, \times \, 10^{17} \, cm^{- 3}$ acceptor atoms. If the intrinsic carrier density $\left(n_{i}\right)$ of the semiconductor is $10^{12} \, cm^{- 3 \, }$ , then the electron density in it will be

NTA AbhyasNTA Abhyas 2020Semiconductor Electronics: Materials Devices and Simple Circuits

Solution:

From mass action law.
$n_{e} n_{h}=n_{i}^{2}$
and $n_{h}=\left(1.1 \times 10^{17}-10^{16}\right) cm ^{-3}=10^{17} cm ^{-3}$
$\therefore \quad n_{e}=\frac{\left(10^{12}\right)^{2}}{10^{17}}=10^{7} cm ^{-3}$