For wave concerned with proton ,de-Broglie wavelength changes by 0.25 %.If its momentum changes by `p_(0)`, initial momentum

Options

A. `100 p_(0)`
B. `(p_(0))/(400)`
C. `401 p_(0)`
D. `(p_(0))/(100)`

Answer

C

Explanation

`lambda =(h)/(p)`
Equation shows that as p decreases `lambda` increases.
`therefore +(0.25)/(100)lambda=(h)/(p-p_(0))`
`therefore (100.25 lambda)/(100)=(h)/(p-p_(0))` .........(2)
Taking ratio of equation (2) and (1),
`(100.25)/(100)=(h)/(p-p_(0))`
`therefore 100.25 p-100.25 p_(0)=100p`
`therefore 0.25 p=100.25 p_(0)`
`therefore p=(100.25 p_(0))/(0.25) therefore p=401 p_(0)`