Solution: Inelastic Scattering Analysis
Key Concept: We analyze the collision using conservation of momentum and the energy condition. The particles scatter in “mutually perpendicular directions,” which imposes a specific geometric constraint on the momentum vectors.
Step 1: Momentum Constraint
Since the scattered alpha particle and the recoil nucleus move perpendicularly: $$ \vec{p}_{\alpha} = \vec{p}’_{\alpha} + \vec{p}_{N} $$ From the geometry (Pythagoras theorem applied to momentum vectors): $$ p_{\alpha}^2 = (p’_{\alpha})^2 + p_{N}^2 $$
Since the scattered alpha particle and the recoil nucleus move perpendicularly: $$ \vec{p}_{\alpha} = \vec{p}’_{\alpha} + \vec{p}_{N} $$ From the geometry (Pythagoras theorem applied to momentum vectors): $$ p_{\alpha}^2 = (p’_{\alpha})^2 + p_{N}^2 $$
Step 2: Energy Analysis
Initial Kinetic Energy: $K_i = \frac{p_{\alpha}^2}{2m_{\alpha}}$
Final Kinetic Energy: $K_f = \frac{(p’_{\alpha})^2}{2m_{\alpha}} + \frac{p_{N}^2}{2M_N}$
Substitute $(p’_{\alpha})^2 = p_{\alpha}^2 – p_{N}^2$ into the $K_f$ equation: $$ K_f = \frac{p_{\alpha}^2 – p_{N}^2}{2m_{\alpha}} + \frac{p_{N}^2}{2M_N} $$ $$ K_f = K_i – \frac{p_{N}^2}{2m_{\alpha}} + \frac{p_{N}^2}{2M_N} $$
Initial Kinetic Energy: $K_i = \frac{p_{\alpha}^2}{2m_{\alpha}}$
Final Kinetic Energy: $K_f = \frac{(p’_{\alpha})^2}{2m_{\alpha}} + \frac{p_{N}^2}{2M_N}$
Substitute $(p’_{\alpha})^2 = p_{\alpha}^2 – p_{N}^2$ into the $K_f$ equation: $$ K_f = \frac{p_{\alpha}^2 – p_{N}^2}{2m_{\alpha}} + \frac{p_{N}^2}{2M_N} $$ $$ K_f = K_i – \frac{p_{N}^2}{2m_{\alpha}} + \frac{p_{N}^2}{2M_N} $$
Step 3: Condition for Energy Loss
The problem states kinetic energy is lost, so $K_f < K_i$. $$ K_i - \frac{p_{N}^2}{2} \left( \frac{1}{m_{\alpha}} - \frac{1}{M_N} \right) < K_i $$ $$ -\frac{p_{N}^2}{2} \left( \frac{1}{m_{\alpha}} - \frac{1}{M_N} \right) < 0 $$ $$ \frac{1}{m_{\alpha}} - \frac{1}{M_N} > 0 \implies \frac{1}{m_{\alpha}} > \frac{1}{M_N} $$ $$ M_N > m_{\alpha} $$
The problem states kinetic energy is lost, so $K_f < K_i$. $$ K_i - \frac{p_{N}^2}{2} \left( \frac{1}{m_{\alpha}} - \frac{1}{M_N} \right) < K_i $$ $$ -\frac{p_{N}^2}{2} \left( \frac{1}{m_{\alpha}} - \frac{1}{M_N} \right) < 0 $$ $$ \frac{1}{m_{\alpha}} - \frac{1}{M_N} > 0 \implies \frac{1}{m_{\alpha}} > \frac{1}{M_N} $$ $$ M_N > m_{\alpha} $$
Correct Option: (d) a nucleus heavier than an $\alpha$-particle