Q. The length of germanium rod is 0.928 cm and its area of cross-section is $1\ mm^{2}$. If for germanium $n_{i} = 2.5 \times 10^{19}\ m^{- 3},\ \ \mu_{h} = 0.19\ m^{2}V^{- 1}s^{- 1},\ \mu_{e} = 0.39\ m^{2}V^{- 1}s^{- 1}$
A) $2.5\ k\Omega$
B) $4.0\ k\Omega$
C) $5.0\ k\Omega$
D) $10.0\ k\Omega$
Explanation
(b) $R = \frac{\rho l}{A} = \frac{L}{n_{i}e(\mu_{e} + \mu_{h})A}$
$= \frac{0.928 \times 10^{- 2}}{2.5 \times 10^{19} \times 1.6 \times 10^{- 19}(0.39 + 0.19) \times 10^{- 6}}$
$= 4000\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.