Nuclear Instability and Q value in Alpha decay

Question

Alpha decay: $(Z,A) \rightarrow (Z-2,A-4)+ ^4_2He $

According to the book: "Nuclear and particle physics" by Williams, $Q_\alpha$ is the measure of available energy to permit an alpha decay.

It is defined as:

$$ Q_\alpha = M(Z,A)-M(Z-2,A-4)-M(2,4)$$

(in natural units)

where M is the nuclear mass, defined as: $M(Z,A) = Zm_p+Am_n-BE(Z,A) $

where BE is the binding energy.

Since the number of protons and neutrons doesn't change (in this case), $Q_\alpha$ can be written as:

$$ Q_\alpha = BE_{f}-BE_{i}=\Delta BE$$

If $Q_\alpha >0$ then the reaction is possible.

I am having difficulties to interpret the $Q_\alpha$-value in alpha decays. I will explain my reasoning hoping that someone can point out my mistake.

My reasoning:

If $Q_\alpha = \Delta BE > 0$ then it means that the energy of the system increased, and therefore energy from somewhere would be needed to make the reaction possible. Then $Q_\alpha $ wouldn't be the energy available for the reaction to happen, but the extra energy needed for the reaction to happen.

Answer 1

..... it means that the energy of the system increased

is not a correct statement as in a decay the (potential) energy of the system decreases.
You can think of it as a reduction in the potential energy of the system resulting in an increase in the kinetic energy of the system.

When the constituent parts of the parent nucleus $(A,Z)$ come together a certain amount of energy is released $Q_{\rm parent}$ and so the parent nucleus is at a lower energy state than its constituent parts.

When the constituent parts of the daughter nucleus $(A-4,Z-2)$ come together a certain amount of energy is released $Q_{\rm daughter}$ and so the daughter nucleus is at a lower energy state than its constituent parts.

When the constituent parts of the alpha particle $(4,2)$ come together a certain amount of energy is released $Q_{\rm alpha}$ and so the alpha particle is at a lower energy state than its constituent parts.

Now $Q_{\rm alpha}+ Q_{\rm daughter} > Q_{\rm parent}$ ie the alpha particle and daughter nucleus together are in a lower energy state than the parent nucleus and the Q-value of the decay is given by

$$Q_{\rm decay}=Q_{\rm alpha}+ Q_{\rm daughter} - Q_{\rm parent} $$.

which is a positive quantity, ie energy is released as a result of the decay and manifests itself as the kinetic energy of the alpha particle and the kinetic energy of the daughter nucleus.

Put another way, it takes more energy to break up a daughter nucleus and an alpha particle into their constituent parts than the energy required to break up the parent nucleus into its constituent parts and the difference in energy is the Q-value of the decay.