1.

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

Answer»

`lamda_(p)=lamda_(N) GT lamda_(e)gt lamda_(alpha)`
`lamda_(alpha)gtlamda_(p)=lamda_(n)ltlamda_(e)`
`lamda_(e) lt lamda_(p)=lamda_(n) gt lamda_(alpha)`
`lamda_(e)=lamda_(p)lamda_(n)=lamda_(alpha)`.

Solution :From the relation `lamda=(h)/(mv)=(h)/(p)=(h)/(sqrt(2mK))`, it is clear that for amount of energy K, the de-Broglie WAVELENGTH is inversely PROPORTIONAL to square root of the mass of given particle i.e., `lamda prop(1)/(sqrt(m))`.
As `m_(alpha) gt m_(n)=m_(p)gtm_(e)`. HENCE `lamda_(alpha)ltlamda_(p)=lamda_(n) lt lamda_(e)`.


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