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. `lamda_(0)`B. `lamda_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))`C. `lamda_(0)/(sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2))))`D. `lamda_(0)/((1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2))))` |
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Answer» Correct Answer - C (c ) : Initial de Broglie wavelength of electron, `lamda_(0)=(h)/(mv_(0))` Force on electron in electric field, `vec(F)=-evec(E)=-eE_(0)hat(j)` Acceleration of electron, `vec(a)=(vec(F))/(m)=-(eE_(0))/(m)hat(j)` It is acting along negative along x-axis `vec(v_(x0))=v_(0)hat(i)`. Initial velocity of electron along y-axis `vec(v_(y0))=v_(0)hat(i)` `( :.` there is no acceleration of electron along x-axis) Velocity of electron after time t y- axis, `vec(v_(y))=0+(-(eE_(0))/(m)hat(j))t=-(eE_(0))/(m)that(j)` Magnitude of velocity of elecron after time t is `|vec(v)|=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(v_(0)^(2)+((-eE_(0))/(m)t)^(2))=v_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/(m^(2)v_(0)^(2)))` de Broglie wavelength associated with electron at time t is `lamda=(h)/(mv)=(h)/(mv_(0)sqrt(1+(e^(2)E_(0)^(2)t^(2))/((m^(2)v_(0)^(2)))))=(lamda_(0))/sqrt(1+(e^(2)E_(0)^(2)t^(2))/((m^(2)v_(0)^(2))))` |
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