Understanding Alpha Decay and Q-Value

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle (\(\alpha\)) and thereby transforms into a different atomic nucleus, known as the daughter nucleus. An alpha particle is essentially a helium-4 nucleus, consisting of two protons and two neutrons. The general form of alpha decay is:

\[_Z^A X \rightarrow _{Z-2}^{A-4} Y + _2^4 \alpha\]

In this case, Thorium-232 (\(_{90}^{232}\text{Th}\)) decays into Radium-228 (\(_{88}^{228}\text{Ra}\)) by emitting an alpha particle (\(_{2}^{4}\alpha\)). The reaction is:

\[_{90}^{232}\text{Th} \rightarrow _{88}^{228}\text{Ra} + _{2}^{4}\alpha\]

The Q-value of a nuclear reaction or decay is the amount of energy released during the process. This energy comes from the mass difference between the reactants and the products according to Einstein's mass-energy equivalence principle \(E=mc^2\). For alpha decay, the Q-value is given by:

\[Q = (M_P - M_D - M_\alpha)c^2\]

where \(M_P\), \(M_D\), and \(M_\alpha\) are the masses of the parent nucleus, daughter nucleus, and alpha particle, respectively. The Q-value is the total kinetic energy shared by the daughter nucleus and the alpha particle, assuming the parent nucleus was initially at rest.

\[Q = K_D + K_\alpha\]

where \(K_D\) is the kinetic energy of the daughter nucleus and \(K_\alpha\) is the kinetic energy of the alpha particle.

Calculating Alpha Particle Kinetic Energy

In alpha decay, both energy and momentum are conserved. Assuming the parent nucleus is initially at rest, the total initial momentum is zero. Therefore, the total momentum of the products must also be zero.

\[p_D + p_\alpha = 0\]

This means the momenta of the daughter nucleus and the alpha particle are equal in magnitude and opposite in direction: \(|p_D| = |p_\alpha|\). Let's call this magnitude \(p\).

The kinetic energy of a particle is related to its momentum and mass by \(K = \frac{p^2}{2M}\). So, we have:

\[K_D = \frac{p_D^2}{2M_D} \quad \text{and} \quad K_\alpha = \frac{p_\alpha^2}{2M_\alpha}\]

Since \(|p_D| = |p_\alpha| = p\), we can write:

\[K_D = \frac{p^2}{2M_D} \quad \text{and} \quad K_\alpha = \frac{p^2}{2M_\alpha}\]

From these equations, we can relate \(K_D\) and \(K_\alpha\):

\[p^2 = 2M_D K_D = 2M_\alpha K_\alpha\]

\[M_D K_D = M_\alpha K_\alpha\]

\[K_D = \frac{M_\alpha}{M_D} K_\alpha\]

Now, substitute this expression for \(K_D\) into the energy conservation equation \(Q = K_D + K_\alpha\):

\[Q = \frac{M_\alpha}{M_D} K_\alpha + K_\alpha\]

\[Q = K_\alpha \left( \frac{M_\alpha}{M_D} + 1 \right)\]

\[Q = K_\alpha \left( \frac{M_\alpha + M_D}{M_D} \right)\]

Solving for \(K_\alpha\):

\[K_\alpha = Q \left( \frac{M_D}{M_\alpha + M_D} \right)\]

The masses of the nuclei and the alpha particle are approximately proportional to their mass numbers (A). So, \(M_D \approx A_D\) and \(M_\alpha \approx A_\alpha\).

For the decay of \(_{90}^{232}\text{Th} \rightarrow _{88}^{228}\text{Ra} + _{2}^{4}\alpha\):

• Mass number of daughter nucleus \(^{228}\text{Ra}\) is \(A_D = 228\).
• Mass number of alpha particle \(^{4}\alpha\) is \(A_\alpha = 4\).
• Mass number of parent nucleus \(^{232}\text{Th}\) is \(A_P = 232\).

Note that \(A_D + A_\alpha = 228 + 4 = 232 = A_P\). So, we can approximate \(M_D + M_\alpha \approx A_D + A_\alpha = A_P\).

Using mass numbers as approximation for masses:

\[K_\alpha \approx Q \left( \frac{A_D}{A_D + A_\alpha} \right) = Q \left( \frac{A_D}{A_P} \right)\]

Applying Values to Find Kinetic Energy

Given the Q-value \(Q = 4082 \text{ keV}\), the mass number of Radium-228 \(A_D = 228\), and the mass number of Thorium-232 \(A_P = 232\).

The maximum possible kinetic energy of the alpha particle is:

\[K_\alpha \approx 4082 \text{ keV} \times \left( \frac{228}{232} \right)\]

Let's calculate the fraction \(\frac{228}{232}\):

\[\frac{228}{232} = \frac{114}{116} = \frac{57}{58}\]

Now, calculate \(K_\alpha\):

\[K_\alpha \approx 4082 \times \frac{57}{58} \text{ keV}\]

\[K_\alpha \approx 70.3793... \times 57 \text{ keV}\]

\[K_\alpha \approx 4011.61 \text{ keV}\]

This calculated value, approximately 4011.61 keV, represents the maximum kinetic energy the alpha particle can have when the daughter nucleus recoils. This is the energy corresponding to the ground state decay, as the Q-value given is for the decay to the ground state of \(^{228}\text{Ra}\).

Comparing with Options

Let's compare our calculated value \(K_\alpha \approx 4011.61 \text{ keV}\) with the given options:

The calculated value \(4011.61 \text{ keV}\) is closest to \(4012 \text{ keV}\).

Conclusion

The maximum possible kinetic energy of the alpha particle in the alpha decay of Thorium-232 to the ground state of Radium-228, given the Q-value of 4082 keV, is approximately 4011.61 keV. Among the given options, the value closest to this is 4012 keV. The Q-value itself (4082 keV) is the total energy released, which is shared between the alpha particle and the recoiling daughter nucleus. The alpha particle carries the larger fraction of this energy because it has a much smaller mass than the daughter nucleus.