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

Options

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

Answer

B

Explanation

We know that the relation between `lamda` and K is given by
`" "lamda(h)/(sqrt(2mk))`
Here, for the given value of energy `K,(h)/(sqrt(2mk))` is a constant.
Thus, `" "lamda prop(1)/(sqrtm)`
`therefore" "lamda_(p):lamda_(n):lamda_(e):lamda_(alpha)`
`implies" "=(1)/(sqrt(m_(p)))`
Since, `" "m_(p)=m_(n)"hence"lamda_(p)=lamda_(n)`
As, `" "m_(alpha)gtm_(p),` therefore `lamda_(alpha)lt lamda_(p)`
As, `m_(e)ltm_(n),` therefore `lamda_(e)ltlamda_(n)`
Hence, `" "lamda_(alpha)ltlamda_(p)=lamda_(n)ltlamda_(e)`