A nucleus with mass number 184 initially at rest emits an $$\alpha$$-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the $$\alpha$$-particle.

The nucleus has mass number $$A = 184$$ and is initially at rest. It emits an alpha particle (mass number 4), leaving a daughter nucleus with mass number $$A - 4 = 180$$.

By conservation of momentum (initial momentum = 0): $$m_\alpha v_\alpha = M_D v_D$$ $$v_D = \frac{m_\alpha v_\alpha}{M_D} = \frac{4}{180}v_\alpha$$

The total kinetic energy released equals the Q-value: $$K_\alpha + K_D = Q$$ $$\frac{1}{2}m_\alpha v_\alpha^2 + \frac{1}{2}M_D v_D^2 = Q$$

Substituting $$v_D = \frac{4}{180}v_\alpha$$: $$K_D = \frac{1}{2}M_D\left(\frac{4}{180}\right)^2 v_\alpha^2 = \frac{4}{180} \cdot \frac{1}{2}m_\alpha v_\alpha^2 = \frac{4}{180}K_\alpha$$

Therefore: $$K_\alpha + \frac{4}{180}K_\alpha = Q$$ $$K_\alpha\left(1 + \frac{4}{180}\right) = Q$$ $$K_\alpha \cdot \frac{184}{180} = 5.5 \text{ MeV}$$ $$K_\alpha = 5.5 \times \frac{180}{184} = 5.5 \times 0.9783 = 5.38 \text{ MeV}$$

The kinetic energy of the $$\alpha$$-particle is approximately $$5.38 \text{ MeV}$$.