Nuclear physics: neutral atomic mass vs. atomic mass

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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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Answers and Replies

  • #2
Andrew Mason
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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
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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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Andrew Mason
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