QJEEM21Phf7a9: JEE Main 2021 Physics - Class 12 JEEM21Phf7a9 Solution: A deuteron and an alpha particle having equal kinetic energy…

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.]

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