Question

What is the conductivity of a semiconductor if electron density $=5\times {10}^{12}/c{m}^3$ and hole density $=8\times {10}^{13}/c{m}^3({\mu }_e=2.3m^2{V}^{-1}{s}^{-1},{\mu }_h=0.01m^2{V}^{-1}{s}^{-1}$)

• A. $5.634$
• B. $1.968$
• C. $3.421$
• D. $8.964$

Answer 1

The correct option is B $1.968$

Electron density $=n_e=5\times {10}^{12}/c{m}^3=5\times {10}^{18}/m^3$

Hole density $=n_h=8\times {10}^{13}/c{m}^3=8\times {10}^{19}/m^3$

The conductivity of the semiconductor

$=e({\mu }_e{n}_e+{\mu }_h{n}_h)$

$=1.6\times {10}^{-19}\left(\right(2.3\left)\right(5\times {10}^{18})+(0.01\left)\right(8\times {10}^{19}\left)\right)$

$=1.968$