Solveeit Logo
Login

Question

Physics Question on Semiconductor electronics: materials, devices and simple circuits

The density of an electron-hole pair in a pure germanium is 3×1016 m−33 \times 10^{16} \, m^{-3} at room temperature. On doping with aluminium, the hole density increases to 4.5×1022 m−34.5 \times 10^{22} \, m^{-3}. Now the electron density (in m−3m^{-3}) in doped germanium will be

A

1×10101 \times 10^{10}

B

2×10102 \times 10^{10}

C

0.5×10100.5 \times 10^{10}

D

4×10104 \times 10^{10}

Answer

2×10102 \times 10^{10}

Explanation

Solution

n12=nhnen^2_1 = n_hn_e
⇒ne=(3×1016)24.5×1022=9×10324.5×1022\Rightarrow n_e = \frac{(3\times 10^{16})^2}{4.5 \times 10^{22}}= \frac{9 \times 10^{32}}{4.5\times 10^{22}}
=2×1010 m−3= 2 \times 10^{10} \,m ^{-3}