Question

A proton, a deuteron and an $\alpha$-particle enter a region of perpendicular magnetic field (to their velocities) with same kinetic energy. If $r_{p}, r_{d}$ and $r_{\alpha}$ are the radii of circular paths of these particles, then

• A. $r_{\alpha}=r_{d} < r_{p}$
• B. $r_{\alpha}=r_{d}=r_{p}$
• C. $r_{\alpha} < r_{d} < r_{p}$
• D. $r_{\alpha}>r_{d}>r_{p}$

Answer 1

Answer: (A)

As we know that radius of circle,
$r=\frac{m v}{B q}=\frac{1}{2}\left(\frac{2 m v^{2}}{B q v}\right) $
$\Rightarrow r=\frac{2 E}{B q v}$
Energy is same. So, $r \propto 1 / q$ but in case of deuteron and $\alpha$-particle $q$ is same and $q$ for proton is least
$\therefore r_{p} >r_{\alpha}=r_{d}$