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. 60
B. 50
C. 25
D. 75

Answer

D

Explanation

`lamda = (h)/(mv) = (h)/(sqrt(2mE)) i.e. lamda prop 1/sqrtE`
`therefore lamda_2/lamda_1 = sqrt(E_1/E_2) = sqrt((E_1)/(16E_1)) = 1/4`
`therefore lamda_2 = (lamda_1)/(4) therefore lamda_1 -lamda_2 = 3/4 lamda_1`
Thus `(lamda_1 - lamda_2)/(lamda_1) = 3/4 " or " 75%`
Thus `lamda` decreases by 75%.