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

Options

A. `lamda_(p)=lamda_(n) gt lamda_(e)gt lamda_(alpha)`
B. `lamda_(alpha)gtlamda_(p)=lamda_(n)ltlamda_(e)`
C. `lamda_(e) lt lamda_(p)=lamda_(n) gt lamda_(alpha)`
D. `lamda_(e)=lamda_(p)lamda_(n)=lamda_(alpha)`.

Answer

B

Explanation

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)`.