Question:
An $\alpha$ particle and a proton are accelerated from rest by a potential difference of $200 \mathrm{~V}$. After this, their de Broglie wavelengths are $\lambda_{\alpha}$ and
$\lambda_{\mathrm{p}}$ respectively. The ratio $\frac{\lambda_{p}}{\lambda_{\alpha}}$ is :
Correct Option: , 4
Solution:
$\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$
$\frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\frac{m_{a} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\frac{4 m_{p} \times 2 \mathrm{e}}{m_{p} \times e}}=\sqrt{8}=2 \sqrt{2}$
$=2 \sqrt{2}$
$\frac{\lambda_{p}}{\lambda_{\alpha}}=2 \times 1.4=2.8$