1) If the kinetic energy of a free electron doubles, it's de-Broglie wavelength changes by the factor A) 2 B) $\frac{1}{2}$ C) $\sqrt{2}$ D) $\frac{1}{\sqrt{2}}$ Answer: Option DExplanation:de-Broglie wavelength, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2.m.(K.E)}}$ $\lambda \propto \frac{1}{\sqrt{K.E}}$ If K.E is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$
