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A proton, a netron, an electron and an a...

A proton, a netron, an electron and an `alpha`-particle have same energy. Then their de-Broglie wavelengths compare as

A

`lambda_P gt lambda_e gt lambda_alpha`

B

`lambda_alpha gt lambda_P gt lambda_e`

C

`lambda_alpha lt lambda_P lt lambda_e`

D

`lambda_e = lambda_P lt lambda_alpha`

Text Solution

Verified by Experts

The correct Answer is:
C
`lambda=h/sqrt(2mKE)` if KE same `lambda prop 1/sqrtm` `m uarr` `lambdadarr`
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