1. Use conservation of momentum and energy

When a nucleus of mass number $A=184$ initially at rest emits an $\alpha$-particle, the daughter nucleus has mass number

$A_d = 184-4 = 180.$

Let

• kinetic energy of $\alpha$ be $K_\alpha$
• kinetic energy of daughter nucleus be $K_d$

Since initial momentum is zero, after decay the two particles have equal and opposite momenta:

$p_\alpha = p_d = p.$

2. Relate kinetic energies using same momentum

For non-relativistic recoil,

$K = \frac{p^2}{2m}.$

So,

$\frac{K_\alpha}{K_d} = \frac{m_d}{m_\alpha} = \frac{180}{4} = 45.$

Hence,

$K_\alpha = 45K_d.$

3. Use Q-value

The total kinetic energy released is the Q-value:

$K_\alpha + K_d = Q = 5.5\ \text{MeV}.$

Substitute $K_\alpha = 45K_d$:

$45K_d + K_d = 5.5$ $46K_d = 5.5$ $K_d = \frac{5.5}{46} \approx 0.1196\ \text{MeV}.$

Therefore,

$K_\alpha = 5.5 - 0.1196 = 5.3804\ \text{MeV}.$

Equivalently,

$K_\alpha = Q\cdot \frac{m_d}{m_d+m_\alpha} = 5.5\cdot \frac{180}{184} \approx 5.38\ \text{MeV}.$

4. Match with options

$K_\alpha \approx 5.38\ \text{MeV}$

So the correct option is C.