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

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

A

`lambda_(p)=lambda_(n)gtlambda_(e)gtlambda_(alpha)`

B

`lambda_(alpha)ltlambda_(p)=lambda_(n)ltlambda_(e)`

C

`lambda_(e)ltlambda_(p)=lambda_(n)ltlambda_(alpha)`

D

`lambda_(e)=lambda_(p)=lambda_(n)=lambda_(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
B
de-Broglie wavelength,
`lambda=(h)/(p)=(h)/(sqrt(2mk))`…….(1)
(`therefore` Kinetic energy of a particle `K=(p^(2))/(2m)`)
Here ,K is same and so,
`lambdaprop(1)/(m)`…….(2)
Here `m_(alpha) gt m_(p)=m_(n)gtm_(e)` and so from equation (2),`lambda_(alpha)ltlambda_(p)=lambda_(n)ltlambda_(e)`
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