Q 12-04-239JEE MainJEE Main 2018 (8 Apr)Easy
An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e$, $r_p$, $r_\alpha$ respectively in a uniform magnetic field $B$. The relation between $r_e$, $r_p$, $r_\alpha$ is:
Answer: (C) $r_e < r_p = r_\alpha$
$$r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}\propto\frac{\sqrt m}{q}$$
- proton: $\dfrac{\sqrt{m_p}}{e}$
- alpha: $\dfrac{\sqrt{4m_p}}{2e} = \dfrac{\sqrt{m_p}}{e}$, the same as the proton
- electron: much smaller mass, so smallest radius
Hence $r_e < r_p = r_\alpha$.
Solution by Sreeraj P, M.Sc Physics