1.

Obtain the expression for the conductivity of a conductor in terms of its relaxation time. Or Deduce sigma = ("ne"^2 tau)/(m)where the symbols have their usual meaning.

Answer»

Solution : We know that the relation between electric CURRENT (I) and drift velocity `(V_d)`
`I = "neAV"_d`......(1)
EXPRESSION for drift velocity in TERMS of electric field and relaxation TIME is given by
` V_d = (e E TAU)/(m)`.....(2)
Substituting (2) in (1) ` rArr I = ("ne" A e E tau)/(m)`
` rArr I = (n A e^2 E tau)/(m)`
But `E = V/l`
`I = (nAe^2 V tau)/(ml)`
`I/V = (nA e^2 tau)/(ml)`
`V/I = (ml)/(nAe^2 tau) rArr R = ((m)/(nAe^2 tau)) I`
`R = ((m)/("ne"^2 tau) ) l/A`
` R =(rhol)/(A)`
Here `rho = (m)/("ne"^2 tau)`
The conductivity of the material
`sigma = 1/rho = ( 1)/((m)/("ne"^2 tau))`
`sigma = ("ne"^2 tau)/(m)`


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