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Question

The ratio of de-Broglie wavelengths of proton and α-particle having same kinetic energy is

A
2:1
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B
22:1
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C
2 :1
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D
4 : 1
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Solution

The correct option is C 2 :1
De Broglie wavelength is given by:

λ=hp

Writing momentum as a a function of kinetic energy and mass

λ=h2Em

i.e. λprotonλalpha=malphamproton

malpha=4mproton
So,

λprotonλalpha=malphamproton=41=2

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