InterviewSolution
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InterviewSolution
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An electron of mass `m` with an initial velocity `vec(v) = v_(0) hat`(i) `(v_(0) gt 0)` enters an electric field `vec(E ) =- E_(0) hat (i)` `(E_(0) = constant gt 0)` at `t = 0` . If `lambda_(0)` is its de - Broglie wavelength initially, then its de - Broglie wavelength at time `t` isA. `(lambda_(0))/((1+(eE_(0))/(m)(t)/(v_(0))))`B. `lambda_(0)(1+(eE_(0)t)/(mv_(0)))`C. `lambda_(0)`D. `lambda_(0)t` |
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Answer» Correct Answer - a Initial de-Broglie wavelength of electron, `lambda_(0)=h/(mv_(0)).......(i)` Force on electron in electric field, `vecF=-evecE=-e[-E_(0)hati]=eE_(0)hati` Acceleration of electron, `veca=(vecF)/m=(eE_(0)hati)/m` velocity of electron after time t, `vecv=v_(0)hati+((eE_(0)hati)/m)t=(v_(0)+(eE_(0))/mt)hati=v_(0)(1+(eE_(0))/(mv_(0))t)hati` De-Broglie wavelength associated with electron at time t is `lambda=h/(mv)=h/(m[v_(0)(1+(eE_(0))/(mv_(0))t)])=(lambda_(0))/([1+(eE_(0))/(mv_(0))t])` [from (i)] |
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