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The length of a germanium rod is 0.58 cm...

The length of a germanium rod is `0.58 cm` and its area of cross-section is `10cm^(2)`.
If for germanium `n_(i)=2.5xx10^(19)m^(-3),mu_(h)=0.19 m^(2)//V-s,mu_(e)=0.39 m^(2)//V-s`, then the resistance of the rod will be-

A

`2.5k Omega`

B

`4.0kOmega`

C

`5.0kOmega`

D

`10.0kOmega`

Text Solution

Verified by Experts

The correct Answer is:
B
We have, `R=(rhol)/(A)=(L)/(n_(i)e(mu_(e)+mu_(h))A)`
`=(0.928xx10^(-2))/(2.5xx10^(19)xx1.6xx10^(-19)(0.19)xx10^(-6))`
`=4000Omega=4.0kOmega`
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