Q. The ratio of de-Broglie wavelength of a $\alpha$-particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector $\overrightarrow{B}$ is perpendicular to the velocity vectors of the $\alpha$-particle and the proton is
A) 1
B) 1/4
C) 1/2
D) 2
Explanation
(c) When a charged particle (charge $q,$ mass $m$) enters perpendicularly in a magnetic field $(B)$ then, radius of the path described by it $r = \frac{mv}{qB} \Rightarrow mv = qBr$
Also de-Broglie wavelength $\lambda = \frac{h}{mv}$
$\Rightarrow \lambda = \frac{h}{qBr} \Rightarrow \frac{\lambda_{\alpha}}{\lambda_{p}} = \frac{q_{p}r_{p}}{q_{\alpha}r_{\alpha}} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.