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)//

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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)//

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`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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`

4 views Answered 2023-02-05 15:29