Q. A potential difference of $2V$ is applied between the opposite faces of a Ge crystal plate of area $1\ cm^{2}$ and thickness $0.5\ mm$. If the concentration of electrons in $Ge$ is $2 \times 10^{19}/m^{3}$ and mobilities of electrons and holes are $0.36\frac{m^{2}}{volt - s}$ and $0.14\frac{m^{2}}{volt - s}$ respectively, then the current flowing through the plate will be
A) 0.25 $A$
B) 0.45 $A$
C) 0.56 $A$
D) 0.64 $A$
Explanation
(d) $\sigma = ne\left( \mu_{e} + \mu_{h} \right) = 2 \times 10^{19} \times 1.6 \times 10^{- 19}(0.36 + 0.14)$
$= 1.6\ (\Omega - m)^{- 1}$
$R = \rho\frac{l}{A} = \frac{l}{\sigma A} = \frac{0.5 \times 10^{- 3}}{1.6 \times 10^{- 4}} = \frac{25}{8}\Omega$
$\therefore i = \frac{V}{R} = \frac{2}{25/8} = \frac{16}{25}A = 0.64\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.