Q 12-11-029NEETJEE MainMedium
If the kinetic energy of a free electron is doubled, its de Broglie wavelength becomes
Answer: (A) $\dfrac{1}{\sqrt{2}}$ times
$\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt{K}}$, so doubling $K$ makes $\lambda$ smaller by $\sqrt{2}$.
Solution by Sreeraj P, M.Sc Physics