Question

If the kinetic energy of the particle is increased to $16$ times previous the percentage change in the deBrogille wavelength of the particle is

Answer 1

$\lambda =\frac{h}{\sqrt{2mh}}$

${\lambda }^1=\frac{h}{\sqrt{2m\times 16h}}$ [Increase by $16$ times]

$\lambda =\frac{\lambda }{4}$

Percentage change in de-Broglie wavelength

$\frac{\lambda -{\lambda }^1}{\lambda }\times 100=(1-\frac{{\lambda }^1}{\lambda })\times 100$

$=(1-\frac{1}{4})\times 100$

$=75$%