In an alpha decay, the parent nucleus decays at rest into an alpha particle (mass number 4) and a daughter nucleus (mass number equal to 190 – 4 = 186). The Q value represents the total kinetic energy released. Because the momentum of the products is conserved, the kinetic energy is divided between the alpha particle and the daughter nucleus.

Here’s a step-by-step explanation:

Conservation of momentum (initial momentum is zero) ensures that the momentum of the alpha particle, $$p_\alpha$$, and that of the daughter nucleus, $$p_d$$, are equal in magnitude:

$$ p_\alpha = p_d $$

The kinetic energies are given by:

$$ T_\alpha = \frac{p^2}{2m_\alpha} \quad \text{and} \quad T_d = \frac{p^2}{2m_d} $$

where $$m_\alpha$$ is the mass of the alpha particle (approximately 4 atomic mass units) and $$m_d$$ is the mass of the daughter nucleus (approximately 186 atomic mass units).

With momentum conservation, the ratio of the kinetic energies becomes:

$$ \frac{T_\alpha}{T_d} = \frac{m_d}{m_\alpha} $$

This shows that the lighter particle (the alpha) gets most of the kinetic energy.

Since the total energy released (the Q value) is shared between the two products:

$$ Q = T_\alpha + T_d $$

By expressing $$T_d$$ in terms of $$T_\alpha$$:

$$ T_d = \frac{m_\alpha}{m_d} T_\alpha $$

so that:

$$ Q = T_\alpha + \frac{m_\alpha}{m_d} T_\alpha = T_\alpha \left(1 + \frac{m_\alpha}{m_d}\right) $$

Solving for $$T_\alpha$$:

$$ T_\alpha = \frac{Q}{1 + \frac{m_\alpha}{m_d}} = Q \cdot \frac{m_d}{m_\alpha + m_d} $$

Substitute the values:

$$ m_\alpha = 4 $$

$$ m_d = 186 $$

$$ Q = 4.5~\text{MeV} $$

Thus,

$$ T_\alpha = 4.5~\text{MeV} \times \frac{186}{4 + 186} = 4.5~\text{MeV} \times \frac{186}{190} $$

Evaluating the fraction:

$$ \frac{186}{190} \approx 0.9789 $$

Therefore,

$$ T_\alpha \approx 4.5~\text{MeV} \times 0.9789 \approx 4.4~\text{MeV} $$

Thus, the kinetic energy of the alpha particle is approximately 4.4 MeV.

Answer: Option D.

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