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

As we know `lambda = (h)/(p) = (h)/(sqrt(2 mK))` `(because P = sqrt( 2mKE))`
or `(lambda_(1))/(lambda_(2)) = sqrt((K_(2))/(K_(1))) = sqrt((16 K)/( K)) = (4)/(1)`
Therefore the percentage change in de - Broglie wavelength ` = ( 1 - 4)/(4) xx 100 = - 75%`