# Nuclear bound states

"The $3/2$$^-$ ground state of gallium-61, Ga-61, is bound by only $190$ keV relative to the system: Zn-60 + p, where p is a proton. There are excited Ga-61 states at $271$ keV and $1000$ keV"

How do I interpret the first sentence of that statement?

My attempt was to think that, if I were to draw an energy level diagram for this situation, I would put the ground state of Ga-61 at the zero energy, then the two excited states above it at $271$ keV and $1000$ keV. Then on this same diagram, I would put the system: (Zn-60 + p) at $-190$ keV, that is, $190$ keV below the zero energy ground state of Ga-61. Is this the correct way to interpret this situation?

Here's an ASCII energy-level diagram based on your description:

. ------- 1000 keV
.
.
.
.
.
.
. --------  271 keV
.           190 keV -----------
.
. --------    0 keV
Ga-61            Zn-60 + p


The key here (and the difference between my diagram and the way I read your question) is that gallium-61 is bound, which means that decay to zinc+proton is energetically forbidden. However the two gallium excited states you mention could decay by proton emission, releasing either 80 keV or 800 keV kinetic energy.

It means that if $E_0(Ga-61)$ is the energy of the Ga-61 in rest, and $E_0(p)+E_0(Zn-60)$ the energy of the proton and Zn-60 in rest when both are separated by an infinite distance (so they don't interact), that: $E_0(Ga-61)-E_0(p)-E_0(Zn-60) = -190 keV$. The Galiumsystem is bound with respect to the proton and Zinksystem, so Galium is more stable than a zink+proton. When you combine the zink and the proton it will release $190 keV$, for instance by emitting a photon.

• The way I understand your response is that if I were to illustrate the above-described energy relationship, say, on a nuclear energy level diagram or a decay scheme, the Zn + p will have an energy above the ground state energy of Gallium, making it less stable relative to the Ga-61 ground state. The energy of Zn + p on my illustration would then be between the Ga-61 ground state and the first excited Ga-61 state (at 271 keV). Is this what you are saying? Feb 5 '16 at 0:05
• Yes, I think the sentence in the question should be interpreted this way. Galium ground state is then more stable than Zn+p, but any excitation would be unstable and could decay -> Zn+p Feb 5 '16 at 0:09