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Derive sigma = (n e^(2) tau)/(m) where the symbols have their usual meaning. |
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Answer» Solution :It is the time interval between two successive collisions of electron with the vibrating ATOMS of a CURRENT carrying conductor. We know that the relation between electric current (I) and drift velocity `(V_(d))` `I = N e A V_(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 = (n e A e E tau)/(m)` `rArr I = (n A e^(2) E tau)/(m)` But `E = (V)/(I)`, Where `V rarr` Potential across the conductor `I rarr` length of the conductor `I = (n A e^(2) V tau)/(ml)` `(I)/(V) = (n A e^(2) tau)/(ml)` `(V)/(I) = (ml)/(n A e^(2) tau) rArr R = ((m)/(n A e^(2) tau)) I` `R = ((m)/(n e^(2) tau))(I)/(A)` `R = (rho I)/(A)` Here `rho = (m)/(n e^(2) tau)` The conductivity of the material `sigma = (1)/(rho) = (1)/((m)/(n e^(2) tau))` `sigma = (n e^(2) tau)/(m)` |
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