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. `400P_0`
B. `P_0`
C. `4P_0`
D. `100P_0`

Answer

A

Explanation

`lamda = (h)/(mv) = h/P i.e. lamda prop 1/P " or " P prop 1/lamda`
`therefore |(DeltaP)/(P)| = |(Deltalamda)/(lamda)| therefore (P_0)/(P) = (0.25)/(100) = 1/400`
`therefore P= 400 P_0`