Binding energy and radiactive decay?

[kraphysics]

We are studying nuclear physics right now and I don't quite understand the concept of binding energy. So, as I understand it binding energy is the work done to remove the nucleons which are bound together by the strong nuclear force. But then during radioactive decay, when you do the equations, how come you calculate energy using difference in mass of parent and daughter atoms and not nucleons/nucleus? I thought the electrons didn't have anything to do with binding energy? I am confused.

 

[eXorikos]

Because the mass of bare nuclei are nearly impossible to measure. Since you in real life situations you will always encounter atoms or ions it is more convenient to work with the atomic masses.

 

[jtbell]

Having to work with atomic masses can be confusing when the initial and final isotopes in a decay have different numbers of atomic electrons, because you have to account for the mass of an "extra" or "missing" electron somewhere. I like to handle this by writing my mass/energy balance equation using nuclear masses first, then substituting the atomic masses. For example, in a typical \beta^- decay:

X \rightarrow Y + e^- + \bar{\nu_e}

where X has atomic number Z and Y has atomic number Z+1:

m_{X,nuc} = m_{Y,nuc} + m_e + Q

where Q is the mass-equivalent of the energy released in the decay. The masses that we find in tables are atomic masses which include atomic electrons:

m_{X,nuc} = m_{X,atom} - Zm_e

m_{Y,nuc} = m_{Y,atom} - (Z+1)m_e

Substitute these into the mass-balance equation, and simplify, and you get

m_{X,atom} = m_{Y,atom} + Q

which doesn't look like it accounts for the mass of the ejected electron, but it really does because of the electron masses included in the atomic masses.

 

[kraphysics]

Having to work with atomic masses can be confusing when the initial and final isotopes in a decay have different numbers of atomic electrons, because you have to account for the mass of an "extra" or "missing" electron somewhere. I like to handle this by writing my mass/energy balance equation using nuclear masses first, then substituting the atomic masses. For example, in a typical \beta^- decay:

X \rightarrow Y + e^- + \bar{\nu_e}

where X has atomic number Z and Y has atomic number Z+1:

m_{X,nuc} = m_{Y,nuc} + m_e + Q

where Q is the mass-equivalent of the energy released in the decay. The masses that we find in tables are atomic masses which include atomic electrons:

m_{X,nuc} = m_{X,atom} - Zm_e

m_{Y,nuc} = m_{Y,atom} - (Z+1)m_e

Substitute these into the mass-balance equation, and simplify, and you get

m_{X,atom} = m_{Y,atom} + Q

which doesn't look like it accounts for the mass of the ejected electron, but it really does because of the electron masses included in the atomic masses.

Thank you for that. I think I understand that part but my confusion lies somewhere else. I am trying to conceptually make sense of what is happening in radioactive decay. I don't understand how energy is released? Binding energy is put into separate the nucleons so where does the energy that is released come from? Is it that during decay, more energy is available than the binding energy? I am confused!
Another question, does the mass-energy equivalence to ONLY binding energy or does a nucleus have potential energy as well?
I think I am thinking about it completely wrong.

 

[jtbell]

Binding energy is put into separate the nucleons so where does the energy that is released come from?

The total energy of a nucleus is the sum of the mass-energies of its nucleons (protons and neutrons), minus the binding energy.

In my example, nucleus X has more energy to begin with, than nucleus Y plus the mass-energy of the ejected electron. The two nuclei have different amounts of binding energy, even though they have the same number of nucleons. Some of the difference comes from the fact that one nucleus has more protons (more electrical repulsion), some comes from the different nuclear structures affecting the nuclear forces among all the nucleons. The difference in binding energy is enough to produce the electron and give it some kinetic energy.

If the difference in binding energy were smaller than the mass of the electron, or even in the other direction, beta decay would be impossible because it would require a net input of energy.

(You also have to include the small mass-energy difference between a proton and a neutron, which I omitted above to keep things simple.)

 

[mquirce]

From curiosity i will to know : what is the physics role of neutrino inside the nukleus and exist it as an entity there or "created after"?

 

[jtbell]

Neither the neutrino nor the emitted electron exists in the nucleus before the decay. They are created when a neutron in the original nucleus (X) converts into a proton.

Going a bit deeper, a down-quark in the neutron converts into an up-quark, thereby converting the neutron into a proton:

d \rightarrow u + e^- + \bar{\nu_e}

so

udd \rightarrow uud + e^- + \bar{\nu_e}

n \rightarrow p + e^- + \bar{\nu_e}

 

[kraphysics]

I donno.. I feel confused about this. Here is a formula that is in my textbook

mc^2(parent) = mc^2(daughter) + mc^2(alpha) + delta E

what does this really mean? I mean I don't see any delta m there so I assume there is some innate internal energy that each nucleus has. What does the formula really mean?

 

[kraphysics]

I think I get it. thanks