If the kinetic energy of a particle is increased by 16 times, 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)/(sqrt(2mK)) , lambda. = (h)/(sqrt(2m xx 16 K)) = (lambda)/(4)`
% change in de-Broglie wavelength, `(lambda - lambda.)/(lambda) = (1 - (lambda.)/(lambda)) xx 100 = (1-(1)/(4)) xx 100 = 75%`