Question

If the kinetic energy of a particle is reduced to half, de Broglie wavelength becomes:

• A. $2$ times
• B. $\frac{1}{\sqrt{2}}$ times
• C. $4$ times
• D. $\sqrt{2}$ times

Answer 1

The correct option is D $\sqrt{2}$ times

$P=mv$ momentum

$KE=\frac{1}{2}m{v}^2$ kinetic energy

$⟹p=\sqrt{2mKE}$

De Broglie wavelength $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mKE}}$

When $KE$ is reduced to half, $p$ becomes $\frac{1}{\sqrt{2}}$ times hence $\lambda$ becomes $\sqrt{2}$ times.