A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let r_d and r_$\alpha$ be their respective radii of circular path. The value of ${{{r_d}} \over {{r_\alpha }}}$ is equal to :

Given, kinetic energy of $\alpha$-particle (K_$\alpha$) = kinetic energy of deuteron (K_d)<br/><br/>Since, kinetic energy, K = ${1 \over 2}$ mv²<br/><br/>$\Rightarrow$ mv² = 2K $\Rightarrow$ v² = ${{2K} \over m}$<br/><br/>$\Rightarrow$ v = $\sqrt {{{2K} \over m}}$ .... (i)<br/><br/>We know that,<br/><br/>r = ${{mv} \over {Bq}}$ .... (ii)<br/><br/>where, r = radius of curvature of path of a charged particle, m = mass of the charged particle, q = charge of the particle, v = velocity of charged particle and B = magnetic field.<br/><br/>From Eqs. (i) and (ii), we get<br/><br/>$r = {{m\sqrt {{{2K} \over m}} } \over {Bq}}$<br/><br/>$\Rightarrow r = {{\sqrt {2Km} } \over {Bq}}$ .... (iii)<br/><br/>Since, m, K and B are same for both deuteron and $\alpha$-particle.<br/><br/>From Eq. (iii), we get<br/><br/>$\gamma \propto {{\sqrt m } \over q}$<br/><br/>$\therefore$ $${{{r_d}} \over {{r_\alpha }}} = \sqrt {{{{m_d}} \over {{m_\alpha }}}} .{{{q_\alpha }} \over {{q_d}}} = \sqrt {{2 \over 4}} \left( {{2 \over 1}} \right)$$ <br/><br/>[$\because$ m_d = 2m_p and m_$\alpha$ = 4m_p<br/><br/>q_$\alpha$ = 2e and q_d = e.]<br/><br/>${{{r_d}} \over {{r_\alpha }}} = \sqrt 2$