If the kinetic energy of the particle is increased to `16` times its previous value , the percentage change in the de - Broglie wavelength of the particle is

Options

A. `25%`
B. `75%`
C. `60%`
D. `50%`

Answer

b

Explanation

`lambda=h/(mv)=h/(sqrt(2mE))` or `lambda prop 1/(sqrt(E))`
`:. (lambda')/(lambda)=sqrt(E/(E'))=sqrt(1/16)=1/4`
`%` change in wavelength `=((lambda-lambda')/lambda)xx100`
`=(1-(lambda')/lambda)100=(1-1/4)xx100=75%`