Q. A proton and an $\alpha$-particle are accelerated through same potential difference. The ratio of their de-Broglie wavelengths $\lambda_{p}/\lambda_{\alpha}$ will be
A) $\frac{1}{\sqrt{8}}$
B) $\sqrt{8}$
C) 2
D) 1
Explanation
(b) From relation $\lambda = \frac{h}{\sqrt{2mqV}}\ or\ \ \ \lambda \propto \frac{1}{\sqrt{mq}}$
Hence, $\frac{\lambda_{p}}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_{p}} \times \frac{q_{\alpha}}{q_{p}}}$
= $\sqrt{\frac{4m_{p} \times 2e}{m_{p} \times e}}$ =$\sqrt{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.