Question 41: Atoms & Nuclei

Question

A nucleus with mass number 190 initially at rest emits an alpha particle. If the $$\mathrm{Q}$$ value of the reaction is $$4.5 \mathrm{~MeV}$$, the kinetic energy of the alpha particle is

Accepted Answer

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 Ã¢ÂÂ 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.
HereÃ¢ÂÂ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:
$$ T_\alpha = \frac{p^2}{2m_\alpha} \quad 	ext{and} \quad T_d = \frac{p^2}{2m_d} $$
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:
$$ Q = T_\alpha + \frac{m_\alpha}{m_d} T_\alpha = T_\alpha \left(1 + \frac{m_\alpha}{m_d}ight) $$
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:
$$ m_\alpha = 4 $$
$$ m_d = 186 $$
$$ Q = 4.5~	ext{MeV} $$
Thus,
$$ T_\alpha = 4.5~	ext{MeV} 	imes \frac{186}{4 + 186} = 4.5~	ext{MeV} 	imes \frac{186}{190} $$
Evaluating the fraction:
$$ \frac{186}{190} \approx 0.9789 $$
Therefore,
$$ T_\alpha \approx 4.5~	ext{MeV} 	imes 0.9789 \approx 4.4~	ext{MeV} $$
Thus, the kinetic energy of the alpha particle is approximately 4.4 MeV.
Answer: Option D.

Answer

4 MeV

Answer

3.2 MeV

Answer

0.43 MeV

Answer

4.4 MeV

Atoms & Nuclei

A nucleus with mass number 190 initially at rest emits an alpha particle. If the $\mathrm{Q} value of the reaction is 4.5 \mathrm{~MeV}$, the kinetic energy of the alpha particle is

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