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

Options

A. `lambda_(p)=lambda_(n)gtlambda_(e)gtlambda_(alpha)`
B. `lambda_(alpha) lt lambda_(p)=lambda_(n)gtlambda_(e)`
C. `lambda_(e)ltlambda_(p)=lambda_(n)gtlambda_(alpha)`
D. `lambda_(e)=lambda_(p)=lambda_(n)=lambda_(alpha)`

Answer

B

Explanation

We know that the relation between `lambda` and `K` is given by `lambda=h/sqrt(2mk)`
Here, for the given value of energy `K,h/sqrt(2k)` is a constant.
Thus, `lambda prop 1sqrtm`
`therefore lambda_(p):lambda_(n):lambda_(e):lambda_(alpha)`
`1/sqrtm_(p):1/sqrtm_(n):1/sqrtm_(e):1/sqrtm_(alpha)`
Since, `m_(p)=m_(n)`, hence `lambda_(p)=lambda_(alpha)`
As,`m_(alpha) gt m_(p)`, therefore `lambda_(alpha) lt lambda_(p)`
As,`m_(e) lt m_(n)`, therefore `lambda_(e) gt lambda_(n)`
Hence,`lambda_(alpha) lt lambda_(p)=lambda_(n) lt lambda_(e)`