Question

# An electron having mass m with an initial velocity v=v0^i in an electric field E=E0^i. If λ0=hmv0, its de-Broglie wave length at time t is given by-

A
λ0[1+eE0mv0t]
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B
λ01+e2E2t2m2v20
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C
λ0 1+e2E20t2m2v20
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D
λ0[1+eE0mv0t]
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Solution

## The correct option is A λ0[1+eE0mv0t]Given, E=E0^i ; initial velocity and de-Broglie wavelength as v=v0^i & λ0=hmv0 When, the electron subjected to an electric field E=E0^i, Fe=ma qE0^i=ma ⇒ a=eE0m^i At any time t, net velocity of the electron, v=v0+at v=v0^i+eE0mt^i=(v0+eE0mt)^i ∴ v=v0(1+eE0mv0t)^i Now, de-Broglie wavelength at time t is, λ=hmv=hm[v0(1+eE0mv0t)] λ=hmv=hmv0[1+eE0mv0t]=λ0[1+eE0mv0t] <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.

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