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Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time. |
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Answer» Solution : We know that in the absence of an external ELECTRIC field E, the conduction electrons in aconductor move randomly with VELOCITIES `vecu_1 , vecu_2, vecu_3,.......vecu_n`such that their mean VALUE`(vecu_1 + vecu_2 + vecu_3 + ..........+vecu_n )/(N) = 0.` However, in the presence of an external electric field `vecE`, electrons experience an acceleration `veca = - (e vecE)/(m) `. If `t_1, t_2, t_3`..... be the times before two successive collisions for different electrons, then the final velocities acquired by different electrons are `vecv_1 = vecu_1 + veca t_1 , vecv_2 = vecu_2 + veca t_2 ,.... vecv_n = vecu_n + veca t_n` ` therefore `Mean value of electron velocity in the presence of an electric field = Drift velocity `vecv_d = (vecv_1 + vecv_2 + vec v_3 +........+vecv_n)/(n)` ` = ( (vecu_1+ vecu_2 + vecu_3 + .........+vecu_n)/(n) ) + veca ( (t_1 + t_2 + t_3+.......+t_n)/(n))` ` = veca tau = (e vecE)/(m) tau` Where `tau = (t_1 + t_2 + t_3 +......+t_n)/(n)` = RELAXATION time . |
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