Q. A proton accelerated through a potential $V$ has de-Broglie wavelength $\lambda.$ Then the de-Broglie wavelength of an $\alpha$-particle, when accelerated through the same potential $V$ is
A) $\frac{\lambda}{2}$
B) $\frac{\lambda}{\sqrt{2}}$
C) $\frac{\lambda}{2\sqrt{2}}$
D) $\frac{\lambda}{8}$
Explanation
(c) $Using\ \lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2mqV}} \Rightarrow \lambda \propto \frac{1}{\sqrt{mq}}$
$\Rightarrow \frac{\lambda_{\alpha}}{\lambda_{p}} = \sqrt{\frac{m_{p}q_{p}}{m_{\alpha}q_{\alpha}}}\because q_{\alpha} = 2q_{p},m_{\alpha} = 4m_{p},\lambda_{p} = \lambda$ [Given]
$\Rightarrow \lambda_{\alpha} = \frac{\lambda}{2\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.