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
Concept. $\lambda = \dfrac{h}{\sqrt{2mK}} \propto \dfrac{1}{\sqrt K}$.
Step 1. Double $K$. $\lambda$ becomes $\dfrac{1}{\sqrt2}$ times.
Answer. $\dfrac{1}{\sqrt2}$ times (Option A).
Solution by Sreeraj P, M.Sc Physics