1.

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

Answer»

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

SOLUTION :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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