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An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e, R_p, R_d \,$ and $\, R_\alpha$. It follows that
${R_e} = {R_p}$
${R_p} = {R_d}$
${R_d} = {R_\alpha }$
${R_p} = {R_\alpha }$
Solution
(c) By using $r = \frac{{mv}}{{qB}} = \frac{v}{{\left( {\frac{q}{m}} \right)B}} \Rightarrow r \propto \;\frac{1}{{\left( {q/m} \right)}}$
$\because \;{\left( {\frac{q}{m}} \right)_{{e^ – }}} > {\left( {\frac{q}{m}} \right)_{{p^ + }}} > \left\{ {{{\left( {\frac{q}{m}} \right)}_d} = {{\left( {\frac{q}{m}} \right)}_\alpha }} \right\}$
$\therefore \;{R_d} = {R_\alpha }$
