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To show my current understanding, I'll use alpha decay as an example and list my questions at the end. Could you please correct me if I'm wrong.

An alpha particle forms in the parent nucleus. It's formation releases energy (the binding energy of the alpha particle) causing it to gain kinetic energy. The alpha particle tunnels through the Coulomb barrier and is emitted out of the nucleus.

By conservation of momentum, the daughter nucleus must also gain kinetic energy.

The mass energy of $D$ (daughter nucleus) and of the $\alpha$-particle is less than that of $P$ (the parent nucleus) by conservation of energy with the new kinetic energy.

The $D$ binding energy increases (1), releasing energy, and the $\alpha$-particle's binding energy has also caused energy to be released.

The energy released is given by: binding energy of $D$ $+$ binding energy of $\alpha$-particle $-$ binding energy of $P$. This energy is used as: kinetic energy $D$ $+$ kinetic energy $\alpha$-particle.

This energy released is from the loss of mass energy:

$$(M_p - (M_d + M_{\alpha}))\times c^2 = (B_d + B_{\alpha}) - B_p$$ where $M$ is mass and $B$ is binding energy.

  1. The D binding energy per nucleon always increases, but does the binding energy of $D$ always increase? If not, does this mean that the kinetic energy of the alpha particle and daughter nucleus is all from the release of the alpha particle's binding energy (as the binding energy of $D$ does not increase)?

  2. Daughter nuclei are usually produced with excess energy after a decay. Is this excess energy the kinetic energy of the daughter nucleus, or something else?

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This example of calculating an alpha lifetime using a tunneling nuclear model may help.

alphadecay

Please note that the alpha particle after decay has the energy it had in the bound energy level.

the incredible range of alpha decay half-lives can be modeled with quantum mechanical tunneling. The illustration represents the barrier faced by an alpha particle in polonium-212, which emits an 8.78 MeV alpha particle with a half-life of 0.3 microseconds.

Momentum conservation will give the daughter nucleus a momentum .

The daughter nucleus might be in an excited state and then there would be a gamma decay to a lower stable level.

The binding energy curve rises up to iron and then decreases, so your statement " D binding energy per nucleon always increases" is true only if the daughter nucleus has a higher atomic number than iron.

This handbook goes into the energy details in page 4.

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An alpha particle forms in the parent nucleus. Its formation releases energy (the binding energy of the alpha particle) causing it to gain kinetic energy. The alpha particle tunnels through the Coulomb barrier, and is emitted out of the nucleus.

The repulsive Coulomb force between the two fragments has an infinite range, so after being split apart the parts gain some speed to opposite directions.

The mass energy of $D$ (daughter nucleus) and of the $\alpha$-particle is less than that of $P$ (the parent nucleus) by conservation of energy with the new kinetic energy.

We are not interested if the binding energy of $D$ increases; we just care about total binding energy.

The energy released is given by: binding energy of $D$ $+$ binding energy of $\alpha$-particle $-$ binding energy of $P$. This energy is used as: kinetic energy $D$ $+$ kinetic energy $\alpha$-particle.

This energy released is from the loss of mass energy:

$$(M_p - (M_d + M_{\alpha}))\times c^2 = (B_d + B_{\alpha}) - B_p$$ where $M$ is mass and $B$ is binding energy.

The questions in the next paragraph are not very meaningful, I mean, what are we comparing the $D$ to? Or what are we comparing the binding energy of the Alpha particle to? The binding energy of $\alpha$-particle is large. It did not increase, or rather, it makes no sense to say that it increased.

The binding energy per nucleon of $D$ always increases, but does the binding energy of $D$ always increase? If not, does this mean that the kinetic energy of the $\alpha$-particle and daughter nucleus is all from the release of the alpha particle's binding energy (as the binding energy of $D$ does not increase)?

Daughter nuclei are usually produced with excess energy after a decay. Is this excess energy the kinetic energy of the daughter nucleus, or something else?

I guess the daughter nucleus is in excited state, oh yes, it says so here: How can the nucleus of an atom be in an excited state?

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  • $\begingroup$ Please use mathjax to format mathematical expressions. To learn more about mathjax, please read MathJax basic tutorial and quick reference. $\endgroup$ – Yashas Feb 26 '17 at 8:53
  • $\begingroup$ Thanks for the answer! 1) I understand that the BE of the alpha particle did not increase - it just came about during formation of the alpha particle. I was just wondering if in some cases the daughter nucleus could have less binding energy than the parent nucleus (despite greater binding energy per nucleon) in which case all energy released would be due to the alpha particle's binding energy. 2) So is the extra energy it has in its excited state equal to its kinetic energy? In other words, is its excess energy = extra energy in excited state = kinetic energy? $\endgroup$ – John Feb 26 '17 at 9:47
  • $\begingroup$ 1) I don't know. 2) I don't understand. Nucleus has more internal energy when it is in exited state, like atom has more internal energy when it is in exited state. What different types of internal energies different nuclei may have, that I don't know. $\endgroup$ – stuffu Feb 27 '17 at 9:03

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