An electron, a doubly ionized helium ion
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 :

1. (1) $\lambda_{\mathrm{e}}>\lambda_{\mathrm{He}++}>\lambda_{\mathrm{p}}$

2. (2) $\lambda_{\mathrm{e}}<\lambda_{\mathrm{He}++}=\lambda_{\mathrm{p}}$

3. (3) $\lambda_{e}>\lambda_{p}>\lambda_{\mathrm{He}++}$

4. (4) $\lambda_{\mathrm{e}}<\lambda_{\mathrm{p}}<\lambda_{\mathrm{He}++}$

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 $\lambda_{e}>\lambda_{P}>\lambda_{\mathrm{He}^{++}}$