Question

A moving $\alpha$ particle encounters a stationary unknown nucleus. After collision the two nuclei are scattered in mutually perpendicular directions. If some kinetic energy is lost in collision then unknown nucleus must be

• A. A proton
• B. A nucleus lighter than $\alpha$ - particle
• C. An $\alpha$ - particle
• D. A nucleus heavier than $\alpha$ - particle

Answer 1

The correct option is D A nucleus heavier than $\alpha$ - particle

For inelastic collision mass of nucleus must be more than incident particle so that both go perpendicular after collision.

Let $v$ be the initial velocity of $\alpha$ particle and $v_1$ and $v_2$ be the final velocities.

Using conservation of momentum and conservation of energy for inelastic collision,

$m_1.\overrightarrow{v}=m_1.\overrightarrow{v_1}+m_2.\overrightarrow{v_2}\left[1\right]\overrightarrow{v_1}.\overrightarrow{v_2}=0\left[2\right]\frac{1}{2}{m}_1{v}^2>\frac{1}{2}(m_1{v_1}^2+m_2{v_2}^2)\left[3\right]$Squaring [1] on both sides and using [2],
${m_1}^2{v}^2={m_1}^2{v_1}^2+{m_2}^2{v_2}^2$
Divide by $m_1$ on both side and use [3] to obtain
$\frac{m_2{v_2}^2}{m_1}>m_2{v_2}^2\Rightarrow m_2>m_1$