Question

What is the number density of donor atoms which must be added to a pure germanium semiconductor to produce an N-type semiconductor of conductivity $6.4\times {10}^2{\text{ohm}}^{–1}{\text{m}}^{–1}$ ? The mobility of electrons in N-type germanium is $4\times {10}^{–1}{\text{m}}^2{\text{V}}^{–1}{\text{s}}^{–1}$. Neglect contribution of holes to conductivity.

• A. ${10}^{20}{\text{m}}^{-3}$
• B. ${10}^{21}{\text{m}}^{-3}$
• C. ${10}^{22}{\text{m}}^{-3}$
• D. ${10}^{23}{\text{m}}^{-3}$

Answer 1

The correct option is C ${10}^{22}{\text{m}}^{-3}$

Mobility of the electron is
$\mu =\frac{V}{E}\Rightarrow V=\mu E...\left(1\right)$
current density is
$J=\sigma E$
$neV=\sigma E$
Putting value of $V$ from equation $\left(1\right)$,
$ne\mu E=\sigma E$
$\sigma =ne\mu$
$n=\frac{\sigma }{e\mu }=\frac{6.4\times {10}^2}{1.6\times {10}^{-19}\times 4\times {10}^{-1}}$
$={10}^{22}{\text{/m}}^3$
$={10}^{22}{\text{m}}^{-3}$
Hence, option (C) is correct.