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Question

An electron (mass m) with an initial velocity ¯v=v0^i is in an electric field ¯E=E0^j. If λ0=hmv0, its de Broglie wavelength at time t is given by

A
λ0
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B
λ0 1+e2E20t2m2v20
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C
λ0 1+e2E20t2m2v20
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D
λ0(1+e2E20t2m2v20)
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Solution

The correct option is C λ0 1+e2E20t2m2v20
Initial de Broglie wavelength of electron, λ0=hmv0
Force on electron in electric field, ¯F=e¯E=eE0^j

Acceleration of electron, ¯a=¯Fm=eE0m^j

It is acting along negative y-axis.
The initial velocity of electron along x-axis ¯vx0=v0^i. Initial velocity of electron alon y-axis ¯vyo=0. Velocity of electron after time t along x-axis ,¯vx=v0^i
( there is no acceleration of electron along x-axis.)
Velocity of electron after time t along y-axis, ¯vy=0+(eE0m^j)t=eE0mt^j

Magnitude of velocity of electron after time t is |¯v|=v2x+v2y=v20+(eE0mt)2=v0 1+e2E20t2m2v20

de Broglie wavelength associated with electron at time t is λ=hmv=hmv0 1+e2E20t2m2v20=λ0 1+e2E20t2m2v20

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