Question

An alpha particle collides elastically with a stationary nucleus and continues moving at an angle of ${60}^{\circ }$ with respect to the original direction of motion. The nucleus recoils at an angle of ${30}^{\circ }$ with respect to initial direction of motion of alpha particle. Mass number of the nucleus is:

Answer 1

Let $v_1,v_2$ be velocity of alpha paticle and nucleus after collision.
Applying law of conservation of momentum,
$P_i=P_f$
$\Rightarrow 4mu=4m{v}_1\cos {60}^{\circ }+m^{\prime }{v}_2\cos {30}^{\circ }$ (along $x-$ axis)
$\Rightarrow 4mu=4m{v}_1\times \frac{1}{2}+m^{\prime }{v}_2\times \frac{\sqrt{3}}{2}$
$\Rightarrow 4mu=2m{v}_1+\frac{\sqrt{3}}{2}{m}^{\prime }{v}_2$ ... (1)
Along $y-$axis,
$4m{v}_1\sin {60}^{\circ }=m^{\prime }{v}_2\sin {30}^{\circ }$
(along $y-$ axis)
$\therefore 4m{v}_1\sqrt{3}=m^{\prime }{v}_2$ ... (2)

$\therefore$ From Eqs. (1) and (2),
$4mu=2m{v}_1+\frac{\sqrt{3}}{2}4m{v}_1\sqrt{3}$
$4mu=8m{v}_1$
$\therefore v_1=\frac{u}{2}$
As collision is elastic,
${\text{(K.E.)}}_i={\text{(K.E.)}}_f$
$\frac{1}{2}\left(4m\right)u^2=\frac{1}{2}\left(4m\right)v_1^2+\frac{1}{2}{m}^{\prime }{v}_2^2$
$\therefore 4m{u}^2=4m{v}_1^2+m^{\prime }{v}_2^2$ ... (3)

From (3), $4m{u}^2=4m\frac{u^2}{4}+2\sqrt{3}mu\times v_2$
$\Rightarrow 3u=2\sqrt{3}{v}_2$
$\therefore v_2=\frac{\sqrt{3}}{2}u$
From (2),
$4m\times \frac{u}{2}\times \sqrt{3}=\frac{m^{\prime }\sqrt{3}}{2}\times u$
$\therefore m^{\prime }=4m.$
So, mass number $=4$