Question

If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is ____________.

विकल्प

• 50

• 75

• 25

• 5

Answer 1

If the kinetic energy of a particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is 75.

Explanation:

`"Kinetic energy k" = "p"^2/(2"m")  "where p is momentum"`

`therefore "k"_2/"k"_1 = "p"_2^2/"p"_1^2`

`therefore "p"_2/"p"_1 = sqrt("k"^2/"k"^1)`

`= sqrt16 = 4`

`"de Broglie wavelength"  lambda = "h"/"p"`

`therefore lambda_2/lambda_1 = "p"_1/"p"_2 = 1/4`

`therefore lambda_2 = lambda_1/4`

`therefore lambda_1 - lambda_2 = 3/4 lambda_1`