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

A we know
`lamda = h/p = (h)/(sqrt(2mK)) (because p = sqrt(2mKE))`
`lamda_1/lamda_2 = sqrt(K_2/K_1) = sqrt((16K)/(K)) = 4/1`
therefore the percentage change in de-broglie
wavelength ` = (1-4)/(4) xx 100 = -75%`