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Three materials A, B and C have electrical conductivities sigma , 2 sigma and 2 sigma respectively. Their numbers densities of free electrons are 2 n, n and 2n respectively. For which material is a average collision time of free electrons maximum? |
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Answer» Solution :CONDUCTIVITY `SIGMA=1/rho = (n E^(2) tau)/m` Relaxation TIME, `tau = (m sigma)/(n e^(2)`. i.e., `tau prop sigma/n` `:. tau_(A) : tau_(B):tau_(C) = sigma/2n : 2sigma/n : 2sigma/2n = sigma/2n : 2sigma/n : sigma/n` Thus, `tau_(B) gt tau_(C ) gt tau_(A)`. So average collision time for material B is maximum. |
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