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Question

The de-Broglie wavelength and kinetic energy of a particle is 2000˚A and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de-Broglie wavelength is

A
2˚A
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B
1˚A
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C
4˚A
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D
10˚A
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E
5˚A
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Solution

The correct option is D 2˚A
The de-Broglie wavelength, λ=hp
where h= Plank's constant and p= momentum of the particle
Kinetic energy, E=p22m=h22mλ2
As mass m and h are constants in both case, so E1λ2
Thus, E1E2=λ22λ21
or λ22=(E1/E2)λ21=1eV1×106eV(2000Ao)2
or λ2=2Ao

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