Given, kinetic energy of $$\alpha$$-particle (K_$$\alpha$$) = kinetic energy of deuteron (K_d)

Since, kinetic energy, K = $${1 \over 2}$$ mv²

$$\Rightarrow$$ mv² = 2K $$\Rightarrow$$ v² = $${{2K} \over m}$$

$$\Rightarrow$$ v = $$\sqrt {{{2K} \over m}} $$ .... (i)

We know that,

r = $${{mv} \over {Bq}}$$ .... (ii)

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.

From Eqs. (i) and (ii), we get

$$r = {{m\sqrt {{{2K} \over m}} } \over {Bq}}$$

$$ \Rightarrow r = {{\sqrt {2Km} } \over {Bq}}$$ .... (iii)

Since, m, K and B are same for both deuteron and $$\alpha$$-particle.

From Eq. (iii), we get

$$\gamma \propto {{\sqrt m } \over q}$$

$$\therefore$$ $${{{r_d}} \over {{r_\alpha }}} = \sqrt {{{{m_d}} \over {{m_\alpha }}}} .{{{q_\alpha }} \over {{q_d}}} = \sqrt {{2 \over 4}} \left( {{2 \over 1}} \right)$$

[$$\because$$ m_d = 2m_p and m_$$\alpha$$ = 4m_p

q_$$\alpha$$ = 2e and q_d = e.]

$${{{r_d}} \over {{r_\alpha }}} = \sqrt 2 $$