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. `lamda_e, lamda_p , lamda_n, lamda_alpha`
B. `lamda_e, lamda_n , lamda_p, lamda_alpha`
C. `lamda_alpha, lamda_n, lamda_p, lamda_e`
D. `lamda_p, lamda_e, lamda_alpha, lamda_n`

Answer

C

Explanation

`lamda = (h)/(sqrt(2mE) ) ` so `h prop (1)/(sqrtm)`
since `m_alpha gt m_n gt m_p gt m_c`
so de-broglie wavelength in increasing order will be
`lamda_alpha , lamda_n , lamda_p , lamda_e`