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Question

The de-Broglie wavelength of a free electron with kinetic energy E is λ. If the kinetic energy of the electron is doubled, the de-Broglie wavelength is:

A
λ2
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B
2λ
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C
λ2
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D
2λ
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Solution

The correct option is A λ2
Kinetic energy initially is given by:
E=12mv2=p22m
p=2mE
where
m: Mass of the particle
v: Speed of the particle
p: Momentum of the particle

De-broglie wavelength is given by,
λ=hp
λ=h2mE

Given E2=2E1
λ2=λ12
Given λ1=λ,λ2=λ2

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