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The mean free path of electrons in a metal is `4 xx 10^(-8)m` The electric field which can give on an average `2eV` energy to an electron in the metal will be in the units `V//m`A. `5 xx 10^(-11)`B. `8 xx 10^(-11)`C. `5 xx 10^(7)`D. `8 xx 10^(7)` |
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Answer» Correct Answer - c Average energy = workdone `= q E lambda` `:. 2eV = eE xx 4 xx 10^(-8)` or `E = (2)/(4 xx 10^(-8)m) = 8 xx 10^(7)Vm^(-1)` |
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