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Question

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

A
22
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B
122
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C
2
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D
2
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Solution

The correct option is A 22
Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity v is given by

λ=hmv (p=mv)

de-Broglie wavelength of a proton of mass m1 and kinetic energy k is given by

λ1=h2m1k (p=2mk)

λ1=h2m1qV....(i) [k=qV]

For an alpha particle mass m2 carrying charge q0 is accelerated through potential V, then

λ2=h2m2q0V

For αparticle (42He) : q0=2q and m2=4m1

λ2=h2×4m1×2q×V....(ii)

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get
λ1λ2=h2m1qV×2×m1×4×2qVh=42×22

We get λ1λ2=22

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