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

• A. 25
• B. 75
• C. 60
• D. 50

Answer 1

The correct option is B 75

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}(\because p=\sqrt{2mE}){\lambda }^{\prime }=\frac{h}{\sqrt{2m\left(16E\right)}}=\frac{\lambda }{4}=0.25\lambda$
% change=-75% ie. De-Broglie wave length will decrease by 75%