Question:
An electron, a doubly ionized helium ion $\left(\mathrm{He}^{++}\right)$and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths $\lambda_{e}, \lambda_{\mathrm{He}+t}$ and $\lambda_{\mathrm{p}}$ is :
Correct Option: , 3
Solution:
(3) de-Broglie wavelength, $\lambda=\frac{h}{P}=\frac{h}{\sqrt{2 m(\mathrm{KE})}}$
$\therefore \lambda \propto \frac{1}{\sqrt{m}}$
As $m_{\mathrm{He}^{++}}>m_{P}>m_{e}$
$\lambda_{\mathrm{He}^{++}}>\lambda_{P}>\lambda_{e}$ or $\lambda_{e}>\lambda_{P}>\lambda_{\mathrm{He}^{++}}$