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 $\frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}}$ is :
Correct Option:
Solution:
(2)
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q v}}$
$\frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\frac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\frac{4 \times 2}{1 \times 1}}$
$=2 \sqrt{2}=2.8$