MODERN BYU 14

Physics Solution – Q14

Solution: Threshold Energy

1. Reaction Analysis

The reaction is: $^4\text{He} + ^7\text{Li} \to ^{10}\text{B} + \text{n}$

The Q-value is $E = -2.87 \text{ MeV}$. Since $Q$ is negative, the reaction is endothermic. We need to supply this energy.

However, we cannot just supply $2.87 \text{ MeV}$. Momentum must be conserved. If the alpha particle hits the stationary lithium, the resulting system must still have some forward momentum (and thus kinetic energy). The energy available for the reaction is only the kinetic energy in the Center of Mass (COM) frame.

Alpha K_th Li (Rest) Compound System Must keep moving to conserve momentum

2. Formula for Threshold Energy

The minimum Kinetic Energy ($K_{th}$) of the projectile required to initiate an endothermic reaction is given by:

$$K_{th} = |Q| \left( 1 + \frac{m_{\text{projectile}}}{m_{\text{target}}} \right)$$

Here:

  • $|Q| = 2.87 \text{ MeV}$
  • $m_{\text{projectile}} = m_{\alpha} \approx 4 \text{ u}$
  • $m_{\text{target}} = m_{\text{Li}} \approx 7 \text{ u}$

3. Calculation

$$K_{th} = 2.87 \left( 1 + \frac{4}{7} \right)$$ $$K_{th} = 2.87 \left( \frac{11}{7} \right)$$ $$K_{th} = 0.41 \times 11$$ $$K_{th} = 4.51 \text{ MeV}$$
Answer: The minimum kinetic energy required is $4.51 \text{ MeV}$.