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Step-by-Step Solution
Step 1: Identify the total energy released (Q value)
The total energy released in the decay process is given as:
$ Q = 5.5 \,\text{MeV}.$
This energy is shared between the kinetic energy of the emitted $ \alpha $-particle $(k_\alpha)$ and the recoiling daughter nucleus $(k_N)$. Thus,
$$
k_\alpha + k_N = 5.5 \,\text{MeV}.
$$
Step 2: Relate momenta of the $ \alpha $-particle and daughter nucleus
Since the nucleus is initially at rest, conservation of momentum ensures that the $ \alpha $-particle and the daughter nucleus have equal and opposite momenta. Let the momentum of the $ \alpha $-particle be $ p $. Hence, the momentum of the daughter nucleus is also $ p $ (in magnitude).
Step 3: Express the kinetic energy in terms of momentum
The kinetic energy $ k $ of a particle of mass $ m $ and momentum $ p $ is
$$
k = \frac{p^2}{2m}.
$$
Therefore, for the $ \alpha $-particle of mass $ m_\alpha $ and the daughter nucleus of mass $ m_N $, we have:
$$
k_\alpha = \frac{p^2}{2m_\alpha}, \quad
k_N = \frac{p^2}{2m_N}.
$$
Step 4: Find the ratio of kinetic energies
The ratio of their kinetic energies is given by
$$
\frac{k_\alpha}{k_N} = \frac{\frac{p^2}{2m_\alpha}}{\frac{p^2}{2m_N}} = \frac{m_N}{m_\alpha}.
$$
According to the problem, the mass number of the parent nucleus is 184. After emitting an $ \alpha $-particle (mass number 4), the daughter nucleus has a mass number of 180. Thus,
$$
\frac{k_\alpha}{k_N} = \frac{180}{4} = 45.
$$
Step 5: Relate kinetic energies to Q value
We have:
$$
k_\alpha + k_N = 5.5 \,\text{MeV},
$$
along with
$$
\frac{k_\alpha}{k_N} = 45.
$$
From the second equation:
$$
k_\alpha = 45 \, k_N.
$$
Substitute in the first equation:
$$
45 \, k_N + k_N = 5.5 \,\text{MeV} \\
46 \, k_N = 5.5 \,\text{MeV} \\
k_N = \frac{5.5}{46} \,\text{MeV}.
$$
Therefore,
$$
k_\alpha = 45 \times \frac{5.5}{46} \,\text{MeV} = \frac{45}{46} \times 5.5\ \text{MeV}.
$$
Step 6: Calculate the kinetic energy of the $ \alpha $-particle
$$
k_\alpha = \frac{45}{46} \times 5.5 \,\text{MeV} \approx 5.38 \,\text{MeV}.
$$
This is the kinetic energy of the $ \alpha $-particle.
Answer: 5.38 MeV
