Question

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

• A. $1 \times 10^{10}$
• B. $2 \times 10^{10}$
• C. $0.5 \times 10^{10}$
• D. $4 \times 10^{10}$

Answer 1

The Correct Option is B

$n^2_1 = n_hn_e$
$ \Rightarrow n_e = \frac{(3\times 10^{16})^2}{4.5 \times 10^{22}}= \frac{9 \times 10^{32}}{4.5\times 10^{22}} $
$= 2 \times 10^{10} \,m ^{-3}$