When the mkomentum of a proton is changed by an amount `p_(0)`, the corresponding change in the de-Broglie wavelength is found to be `0.25%`. Then, the original momentum of the proton was

Options

A. `p_(0)`
B. `100 p_(0)`
C. `400 p_(0)`
D. `4 p_(0)`

Answer

C

Explanation

`lambda prop (1)/(p)` …(i)
`lambda(1- (0.25)/(100))prop(1)/((p+p_(0))` …(ii)
(i)/(ii)
`(1)/(1- (0.25)/(100)) =(p+p_(0))/(p)`
`(1- (0.25)/(100))^(-1) =1 +(p_(0))/(p)`
`1+ (0.25)/(100) =1+(p_(0))/(p)`
`p= 400 p_(0)`