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A proton and an alpha - particle are acc...

A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is

A

`2sqrt(2) : 1`

B

`1 : sqrt(2)`

C

`sqrt(2) : 1`

D

`1 : 2sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`lambda=h/sqrt(2mq V)`
`lambda prop 1/sqrt(mq)`
`lambda_(P)/lambda_(prop)=(m_(prop) q_(prop))/(m_(P)q_(P))=sqrt((4m_(P)xx2q_(p))/(m_(p)xxq_(P)))=(2sqrt(2))/1`
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