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

Correct Answer:

75

Explanation:

For de-Broglie wavelength, λ₁ = h / p = h / √(2mK) —–(i)
λ₂ = h / √(2m 16K) = h / 4√(2mK) = λ₁ / 4 ——-(ii)
λ₂ = 25% of λ₁
There is 75% change in the wavelength.