Question

If the kinetic energy of a particle is doubled, de Broglie wavelength becomes:

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

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$ times

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

$p=mv$ (P = momentum)

$⟹p=\sqrt{2mKE}$

$\lambda =\frac{h}{p}$ (de Broglie)

If KE is doubled, p will become $\sqrt{2}$ times.

$\lambda$ will become $\frac{1}{\sqrt{2}}$ times.

Hence, option $D$ is correct.