InterviewSolution
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InterviewSolution
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Calculate the relaxation time and mean free path at room temperature (i.e. `27^(@)C`). If the number of free electrons per unit volume is `8.5 xx 10^(28)//m^(3)` and resistivity `rho = 1.7 xx 10^(8) Omega-m`. Given that mass of electron `= 9.1 xx 10^(-31) kg` `e = 1.6 xx 10^(-19)C and k = 1 .36 xx 10^(-23) JK^(-1)` |
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Answer» Correct Answer - `2.46 xx 10^(-14)s ;28.7Å` Relaxation time, ` tau =(m)/(n e^(2)rho)` `= (9.1 xx 10^(-31))/( (8.5 xx 10^(28)) (1.6 xx 10^(-19))^(2) xx (1.7 xx 10^(-8)))` `= 2.46 xx 10^(-14)s` From kinetic theory `upsilon_(rms) = sqrt((3kT)/(m)) = sqrt((3 xx (1.38 xx 10^(-23)) xx 300)/(9.1 xx 10^(-31)))` ` = 1.17 xx 10^(5)ms^(-1)` Therefore mean free path `lambda = upsilon_(rms) xx tau` `= 1.17 xx 10^(5) xx 2.46 xx 10^(-14) = 2.87 xx 10^(-9)m = 28.7Å` |
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