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. `5xx10^(7)`B. `8xx 10^(7)`C. `5xx10^(-11)`D. `8xx10^(-11)`

The mean free path of electrons in a metal is `4 xx 10^(-8)m` The elec

1.	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. `5xx10^(7)`B. `8xx 10^(7)`C. `5xx10^(-11)`D. `8xx10^(-11)`
Answer» Correct Answer - A `qV=2eV` `implies 1.6xx10^(-19)V=2xx 1.6xx10^(-19)V ` `implies V=2V` `E=(V)/(d)` `implies E=(2V)/(4 xx 10^(-8))=5xx10^(7)`

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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. `5xx10^(7)`B. `8xx 10^(7)`C. `5xx10^(-11)`D. `8xx10^(-11)`

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