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If n, e, tau, m, are representing electr...

If n, e, `tau`, m, are representing electron density charge, relaxation time and mass of an electron respectively then the resistance of wire of length 1 and cross sectional area A is given by

A

`(ml)/(n e^(2)tauA)`

B

`(2mA)/(n e^(2)tau)`

C

`n e^(2)tauA`

D

`(n e^(2)tauA)/(2m)`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • If n, e, tau and m represent electron density, charge, relaxation time and mass of an electron respectively, then the resistance of a wire of length 1 and cross-sectional area A is

    A
    `(ml)/("ne"^(2)tauA)`
    B
    `(m tauA)/("ne"^(2)l)`
    C
    `("ne"^(2)tauA)/(ml)`
    D
    `("ne"^(2)ml)/(tauA)`

  • The resistance will be least in a wire with length, cross-section area respectively,

    A
    L/2, 2A
    B
    2L, A
    C
    L, A
    D
    L, 2A

  • If n,e,m and tau are free electron density in conductor , charge of electron and relaxation time of free electrons, then resistivity rho of the conductor can be expressed as .

    A
    `rho = (m)/("ne"^(2) tau)`
    B
    `rho = ("ne"^(2) tau)/(m)`
    C
    `rho = ("ne"^(2) tau)/(m)`
    D
    `rho = (m)/("ne" tau)`

    • Similar Questions

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    If in a conductor number density of electrons is 8.5xx10^(28) average relaxation time 25 femtosecond mass of electron being 9.1xx10^(-31) kg, the resistivity would be of the order.

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