Q153 / 180 · KCET 2018 · Physics
The density of an electron -hole pair in a pure germanium is \( 3 \times 10^{16} \mathrm{~m}^{-3} \) at room temperature
. On doping with aluminium, the hole density increases to \( 4.5 \times 10^{(22)} \mathrm{m}^{-3} \). Now the electron
density \( \left(\right. \) in \( \left.m^{-3}\right) \) in doped germanium is
. On doping with aluminium, the hole density increases to \( 4.5 \times 10^{(22)} \mathrm{m}^{-3} \). Now the electron
density \( \left(\right. \) in \( \left.m^{-3}\right) \) in doped germanium is
- A\( 1 \times 10^{10} \)
- Correct answerB\( 2 \times 10^{10} \)
- C\( 0.5 \times 10^{10} \)
- D\( 4 \times 10^{10} \)
Solution
Given, density of electron-hole pair in pure germanium $=3 \times 10^{16} \mathrm{~m}^{-3}$
On doping, hole density $=4.5 \times 10^{22} \mathrm{~m}^{-3}$
Electron density $=?$
We know $n_{1}^{2}=n_{h} n_{e}$
Where $n_{i}$ is electron-hole density in pure semiconductor; $n_{h}$ is hole density; $n_{e}$ is electron density
$\Rightarrow n_{e}=\frac{n_{i}^{2}}{n_{h}}=\frac{\left(3 \times 10^{16}\right)}{4.5 \times 10^{22}}=2 \times 10^{10} \mathrm{~m}^{-3}$
Therefore, electron density in doped germanium will be $2 \times 10^{10} \mathrm{~m}^{-3}$
Hence, the correct answer is option: 'B'.