Solution: Derivation of $E_{CM}$ and Threshold Calculation
Key Concept: In endothermic nuclear reactions, the “Available Energy” to drive the reaction is the Kinetic Energy in the Center of Mass (CM) frame. The kinetic energy associated with the motion of the center of mass itself cannot be used for the reaction (conservation of momentum).
Derivation of Available Energy ($E_{CM}$)
Consider projectile $m_1$ moving with velocity $v$ hitting stationary target $m_2$.
Consider projectile $m_1$ moving with velocity $v$ hitting stationary target $m_2$.
- Total Momentum: $P = m_1 v$.
- Velocity of CM: $V_{CM} = \frac{m_1 v}{m_1 + m_2}$.
- Total Kinetic Energy (Lab): $K_{lab} = \frac{1}{2} m_1 v^2$.
- Kinetic Energy of CM motion: $K_{CM\_motion} = \frac{1}{2} (m_1+m_2) V_{CM}^2$.
- Available Energy ($K_{rel}$ or $E_{CM}$): $K_{avail} = K_{lab} – K_{CM\_motion}$.
Application to Case 1 (Proton hits Deuteron)
$m_1 = m$ (proton), $m_2 = 2m$ (deuteron). $K_{lab} = 1.4$ MeV. $$ E_{required} = 1.4 \times \frac{2m}{m + 2m} = 1.4 \times \frac{2}{3} \text{ MeV} $$
$m_1 = m$ (proton), $m_2 = 2m$ (deuteron). $K_{lab} = 1.4$ MeV. $$ E_{required} = 1.4 \times \frac{2m}{m + 2m} = 1.4 \times \frac{2}{3} \text{ MeV} $$
Application to Case 2 (Deuteron hits Proton)
$m_1 = 2m$ (deuteron), $m_2 = m$ (proton). Let required lab energy be $K’$. $$ E_{required} = K’ \times \frac{m}{2m + m} = K’ \times \frac{1}{3} $$
$m_1 = 2m$ (deuteron), $m_2 = m$ (proton). Let required lab energy be $K’$. $$ E_{required} = K’ \times \frac{m}{2m + m} = K’ \times \frac{1}{3} $$
Solve for $K’$
Since the reaction physics ($E_{required}$) is invariant: $$ 1.4 \times \frac{2}{3} = K’ \times \frac{1}{3} $$ $$ K’ = 2.8 \text{ MeV} $$
Since the reaction physics ($E_{required}$) is invariant: $$ 1.4 \times \frac{2}{3} = K’ \times \frac{1}{3} $$ $$ K’ = 2.8 \text{ MeV} $$
Correct Option: (d) 2.8 MeV