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Derive the expression from drift velocity of free electron in terms of relaxation time and electric field applied across a conductor. |
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Answer» Solution :In the absence of an external electric field E, the conduction electron electron in a conductor MOVE randomly and their net velocity `vecu=0`. In the presence of an electric field `vecE`, electrons EXPERIENCE an acceleration `veca=-(evecE)/(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 `vec(v_(1))=vec(u_(1))+vecat_(1), vec(v_(2))=vec(u_(2))+veca(t_(2), ... vec(v_(n))=vec(u_(n))+vecat_(n))` `THEREFORE` Mean value of electron velocity in the presence of an electric field is called drift velocity `vec(v_(d))`. Then `vec(v_(d))=(vec(v_(1))+vec(v_(2))+vec(v_(3))+....+vec(v_(n)))/(n)=((vec(u_(1))+vec(u_(2))+vec(u_(3))+...+vec(u_(n)))/(n))+veca((t_(1)+t_(2)+t_(3)+....t_(n))/(n))` `=veca tau=-(evecE)/(m)tau`, where `tua=(t_(1)+t_(2)+t_(3)+...+t_(n))/(n)="relaxation time"` |
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