Nuclear physics: neutral atomic mass vs. atomic mass

In summary: But all the sources I have show a binding energy of 1783 MeV for the first four significant digits. That's an error of 0.72%. Less than one, sure, but I would expect, using reliable sources, to get at least those first four digits correct. What's wrong?You are probably using the wrong atomic mass for the uranium-235 nucleus.
  • #1
rhombus
6
0

Homework Statement



I am preparing a presentation on nuclear reactor technology.

To demonstrate the mass-energy equivalence, I am trying to calculate the binding energy of some heavier isotopes. The problem is that, when I substitute the values that I have into the equation, I get a binding energy that is substantially below what I have found in a range of sources.

Homework Equations



[tex]E_{B} = (Zm_{p}+Nm_{n}-^{A}_{Z}m)c^{2}[/tex]

The Attempt at a Solution



Substitution, using uranium 235 as an example:

[tex]E_{B} = ((143*1.007974)+(92*1.0086649156)-235.04393005)931.5*10^{6}\frac{eV}{c^{2}} [/tex]

[tex]=1763.81777851379 MeV[/tex]

But the binding energy reported in the same source that the atomic mass came from was 1783.890991 MeV, which is substantially more than my result.

I understand that I need to be using the neutral atomic mass, but quite honestly, I don't know what that means exactly, and I don't know where to find it. I tried subtracting the electron masses from the atomic mass, but then I ended up with a binding energy that was too large.

A textbook I have used an abridged table that was lifted from a nuclear physics textbook (which I don't have).

What am I doing wrong here?
 
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  • #2
rhombus said:
What am I doing wrong here?
It would help to explain the concept. The binding energy is the difference in mass between the nucleus and the separated nucleons x c^2.

So you have the right concept. But you have to use the atomic mass of just the U235 nucleus. You are using the atomic mass of the whole U235 atom.

If you use the neutral atomic mass, you have to include the mass of 92 electrons (one for each proton). Then you can use the atomic mass of the whole U235 atom (which includes 92 electrons).

AM
 
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  • #3
Andrew Mason said:
But you have to use the atomic mass of just the U235 nucleus. You are using the atomic mass of the whole U235 atom.

If you use the neutral atomic mass, you have to include the mass of 92 electrons (one for each proton). Then you can use the atomic mass of the whole U235 atom (which includes 92 electrons).

AM

Yike. I must have been tired when I did this.

My first mistake was switching the neutron and proton numbers.

Secondly, my notation was not correct. The first mass is in fact the mass of the complete hydrogen atom, including electrons, so it does balance out the mass of the neutral U-235 atom. Perhaps the notation should have looked like this:


[tex]
E_{B} = (Zm_{H}+Nm_{n}-^{A}_{Z}m)c^{2}
[/tex]

Now, the number looks a bit better, though still off:

[tex]
E_{B} = ((92*1.007974)+(143*1.0086649156)-235.04393005)931.5*10^{6}\frac{eV}{u}
[/tex]

[tex]
=1796.64076046521 MeV
[/tex]

But all the sources I have show a binding energy of 1783 MeV for the first four significant digits. That's an error of 0.72%. Less than one, sure, but I would expect, using reliable sources, to get at least those first four digits correct. What's wrong?

Here are my sources:

For the hydrogen mass: http://www.iupac.org/publications/pac/2006/pdf/7811x2051.pdf (IUPAC)

For the neutron mass: http://pdg.lbl.gov/2006/tables/bxxx.pdf (Lawrence Berkeley Lab)

For the uranium-235 mass: http://t2.lanl.gov/cgi-bin/quecalc?192,335 (that's from Los Alamos).
 
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  • #5


Thank you for sharing your progress on your presentation. It seems like you are on the right track with your calculations, but there are a few important concepts to understand when working with nuclear physics.

Firstly, when we talk about atomic mass in the context of nuclear physics, we are usually referring to the mass of the nucleus, which is composed of protons and neutrons. This is also known as the mass number (A) of the isotope. However, in most sources, the atomic mass listed is actually the average atomic mass, which takes into account the relative abundance of different isotopes of an element. This is why the atomic mass you used for uranium 235 is slightly different from the actual mass of the uranium 235 nucleus.

Secondly, the equation you used for binding energy is correct, but it is important to use the neutral atomic mass (M) rather than the atomic mass (A). The neutral atomic mass takes into account the mass of the electrons surrounding the nucleus, which is important for calculating the mass defect (difference between the mass of the nucleus and the sum of its component particles). The neutral atomic mass can be found in a table of atomic masses, which can be found in textbooks or online sources.

Finally, it is important to note that the binding energy value you calculated is an approximation, as it assumes that all the protons and neutrons are bound in a perfect nucleus. In reality, there are small variations in the mass of the nucleus due to the strong nuclear force and other factors. This may explain the slight difference in your calculated binding energy compared to the value reported in your source.

I hope this helps clarify some of the concepts and helps you to accurately calculate the binding energy for your presentation. Good luck!
 

1. What is the difference between neutral atomic mass and atomic mass in nuclear physics?

In nuclear physics, the neutral atomic mass refers to the mass of an atom when it is in its most stable, neutral state. This includes the mass of the nucleus as well as the electrons orbiting around it. On the other hand, atomic mass refers to the mass of the nucleus alone, without taking into account the electrons. This is because the mass of electrons is much smaller compared to the mass of the nucleus.

2. How is neutral atomic mass calculated?

The neutral atomic mass is calculated by adding the masses of protons and neutrons in the nucleus. This can be determined by using a mass spectrometer, which separates and measures different isotopes of an element based on their mass-to-charge ratio. The average mass of all the isotopes, weighted by their abundance, gives the neutral atomic mass.

3. Why do we use atomic mass instead of neutral atomic mass in most calculations?

In most calculations, we use atomic mass instead of neutral atomic mass because it is a more convenient and precise value to work with. Atomic mass is a property of the nucleus alone and does not change with the addition or removal of electrons. Therefore, it is a more fundamental and consistent value to use in nuclear physics calculations.

4. How does the concept of neutral atomic mass relate to nuclear stability?

The neutral atomic mass is closely related to nuclear stability. The more stable an atom is, the closer its neutral atomic mass will be to its atomic mass. This is because a stable atom has a balanced number of protons and neutrons, resulting in a more compact and tightly bound nucleus. Any deviation from this balance can result in an unstable nucleus, which may undergo radioactive decay to achieve a more stable state.

5. Can neutral atomic mass be changed?

No, the neutral atomic mass of an element cannot be changed under normal circumstances. It is a fundamental property of an element's nucleus and is determined by the number of protons and neutrons it contains. However, in certain nuclear reactions such as fission or fusion, the number of protons and neutrons in the nucleus can change, resulting in a different neutral atomic mass for the resulting nuclei.

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