Question

If electron, proton, $\alpha$ and $L{i}^{++}$ particles are accelerated with same kinetic energy, then the smallest de broglie wavelength is for

• A. electron
• B. proton
• C. $\alpha$ particle
• D. $L{i}^{++}$

Answer 1

The correct option is D $L{i}^{++}$

$K=\frac{p^2}{2m},p=\sqrt{2mK}$
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}$
$\lambda \propto \frac{1}{\sqrt{m}}$
Since $L{i}^{++}$ is heaviest, de broglie wavelength will be smallest for it.