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. 50
B. 25
C. 75
D. 60

Answer

3

Explanation

`P^(2) = 2mK, P to "momentum", Kto "Kinetic energy" K_(2) = 16 K`
`P_(2)^(2) = 16 P_(2)^(2)`
`P_(2) = 4P_(1)`
Now `lambda = (h)/(p)`
`(lambda_(2))/(lambda_(1)) = (P_(1))/(P_(2)) = (P_(1))/(4P_(1)) = (1)/(4)`
`lambda_(2) = lambda_(1) (1 - (3)/(4)`
`lambda_(2) = lambda_(1) - 75 %`.