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. `400 P_(0)`
B. `P_(0)`
C. `4P_(0)`
D. `100P_(0)`

Answer

A

Explanation

`lambda = (h)/(mv)= h/p " "i.e lambda prop(1)/(p) or p prop (1)/(lambda)`
`therefore |(Deltap)/(p)|=|(Delta lambda)/(lambda)| therefore (p_0)/(p) = (0.25)/(100) = 1/400`
`therefore p=400p0`