The magnitude of the de-Broglie wavelength `(lambda)` of an electron `(e)`,proton`(p)`,neutron `(n)` and `alpha` particle `(a)` all having the same energy of `Mev`, in the increasing order will follow the sequence:

Options

A. `lambda_(e),lambda_(p), lambda_(n),lambda_(alpha)`
B. `lambda_(e),lambda_(n),lambda_(p),lambda_(alpha)`
C. `lambda_(alpha), lambda_(n),lambda_(p),lambda_(e)`
D. `lambda_(p), lambda_(e),lamda_(alpha), lambda_(n)`

Answer

C

Explanation

`lambda = (h)/(sqrt(2mE)) so h prop(1)/(sqrt(m))`
Since `m_(alpha) gt m_(n) gt m_(p) gt m_(e)`
so de-Broglie wavelegth in increasing order will be
`lambda_(alpha) , lambda_(n),lambda_(p), lambda_(e)`