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Question

# An electron (mass m) with initial velocity →v=v0^i+v0^j is in an electric field →E=−E0^k. If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by

A
λ02 1+e2E20t2m2v20
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B
λ0 1+e2E20t2m2v20
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C
λ0 1+e2E20t22m2v20
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D
λ0 2+e2E20t2m2v20
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Solution

## The correct option is C λ0 ⎷1+e2E20t22m2v20Given, Inital velocity u=v0^i+v0^j Acceleration a=qE0m=eE0m Using v = u + at v=v0^i+v0^j+eE0mt^k ∴|→v|=√2v20+(eE0tm)2 de -Broglie wavelength, λ=hP ⇒λ=hmv (∵ p=mv) Initial wavelength, λ0=hmv0√2 Final wavelength, λ=hm√2v20+(eE0tm)2 λλ0=1 ⎷1+(eE0t√2mv0)2 ⇒λ=λ0 ⎷1+e2E20t22m2 v20

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