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:
λ√2
λ2
√2λ
2λ
A
λ2
B
2λ
C
λ√2
D
√2λ
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Solution
Verified by Toppr
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=h√2mE
Given E2=2E1
⟹λ2=λ1√2
Given λ1=λ,∴λ2=λ√2
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