Question

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

• 25

• 75

• 60

• 50

Answer 1

75

Explanation:

As we know

λ = `"h"/"P" = "h"/sqrt(2"mK")` ....`(∵ "P" = sqrt(2"mK"))`

or `λ_1/λ_2 = sqrt("K"_2/"K"_1) = sqrt((16"K")/"K") = 4/1`

Therefore the percentage chanbe in de-Broglie wavelength

= `(1 - 4)/4 xx 100`

= − 75%