Question-Proper form for nuclear addition

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Homework Help Overview

The discussion revolves around the proper form for calculating the sum of masses in binding energy calculations, specifically for Carbon-12 and other isotopes. Participants explore different equations and methods for determining the total mass of nucleons involved in nuclear reactions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to identify the correct equation for calculating the total mass of nucleons, comparing several forms and expressing uncertainty about the implications of including or excluding electrons. Some participants question the validity of different methods and their impact on binding energy results.

Discussion Status

Participants are actively discussing various methods and their outcomes, with some suggesting that certain equations yield differing results. There is an ongoing exploration of the assumptions behind the calculations, particularly regarding the treatment of protons and neutrons versus hydrogen isotopes.

Contextual Notes

There is mention of potential discrepancies between calculated and published mass values, as well as the importance of distinguishing between atomic mass and nuclear mass in binding energy calculations. The original poster expresses confusion regarding the proper technique for mass addition, leading to a deeper inquiry into the subject.

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Question--Proper form for nuclear addition

Homework Statement



Say I am doing Binding Energy calculations, and I want to know the sum of mass for all the parts for, say, Carbon12 (C12). Which of the following is the correct form:

Homework Equations



1.) 6p + 6n = total

2.) 6(H1) + 6n = total

3.) 6(H2) = total

The Attempt at a Solution



I am betting on #2. For #1, it seems to me that neglecting electrons could change the result in huge atoms. In #3 the measured mass of H2 is NOT the same as H1 + n, because H2 has already lost mass in fusion.

OK, I'm assuming #2 was the winner, would the following be accurate:

total mass of parts for Pt78 = 78(H1) + 78n
or
total mass of parts for Bx = x(H1) + x(n)

How'd I do?
 
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I'd use equation 1 since electrons are not nucleons and they do not lose any mass to contribute to the binding energy. Secondly, be careful with elements of a higher atomic number as they generally have more neutrons than protons and you can't assume that hey are the same. Although for the most common occurrence of Boron they happen to be the same.
 
Really?? I have found using #1 SOMETIMES give results that are different from other's results, once you get out past 4 decimals or so.
 
I don't know why your results differ, but only the nucleons contribute to the binding energy of the nucleus which makes sense.
 
I meant the masses listed in published charts, tables, etc are a little different than the masses I get using Method #1. Doubtful if these differences are large enough, however, to alter the Binding Energy results in any meaningful manner. I was just kinda curious, is all.
 
The masses listed are often the ATOMIC mass, so you mist first calculate the nuclear mass first, then calculate the binding energy. But that is unessisary, just consider:

B(A,Z) = [Z*M(H1) + (A-N)M(n) - M(A,Z)]*c^2 (povh, parcticles and nuclei 5th ed, p 14)

Where B(A,Z) is nuclear binding energy, M(H1) mass of hydrogen-1, M(n) neutron mass, M(A,Z) is ATOMIC mass of isotope with A and Z. The electron bindning energy is of the order keV, and the nuclear bindingenergy is of order MeV and can thus be neglected. This is usally the way you calculate bindning energy from tables.
 
I didn't mean atomic mass--it showed a few "sum of parts" masses in examples, from which you would then subtract the measured mass of whole atom to get the mass defect, right? I was really only interested in the proper form for addition--the BE calculation was just as an example of why; not part of what I wanted to know. However, I HAVE learned some things by doing an hours worth of math. Below are 8 different techniques for getting the "mass of all parts". I used all 8 techniques to find the BE for He4--(28.295993). Only one of these techniques provided the correct answer, Kurdt, and it was #2 from above, not #1 from above.

1. p+p+n+n=2p+2n==>BE 27.273213
2. H1+H1+n+n=2(H1)+2n==>BE 28.295993 **winner**
3. H2+H2=2(H2)==>BE 23.845314
4. H1+p+n+n=H1+p+2n==>BE 27.784603
5. H2+H1+n==>BE 26.070654
6. H2+p+n==>BE 25.559264
7. H3+p==>BE 19.302419
8. H3+H1==>BE 19.813809

It is the same #2 here as the #2 above (#1, #2, and #3 here are identical to those above), and it is the only one which results in the correct answer. It is the same as part of the equation malawi used:

H1+H1+n+n=2(H1)+2n=mass of parts for He4

or in general:

x(H1)+y(n)= Z*M(H1) + (A-N)M(n)=mass of parts, where x= # of p, and y= # of n.
For my example with He4, x=2 and y=2

I used Method #1 in the past--which is why my fractions were always off a little.

OK, now how'd I do??

PS---malawi, please define for me A, Z, and A-N. Do you have a A up there that you meant to be an A-N?
 
Last edited:
I still don't understand much of what you are trying to calculate by those things..
Give us/me an example of a specific BE you want to calculate. I just pointed you that in tables, very often, the atomic mass of an isotope is listed. Therefore we must take into account the electrons.

A is the mass number, i.e the number of neutrons + protons

Z is the atomic number, i.e the number of protons.

Do you want good reading tips for introductore nuclear physics, I can surely provide you. Also your library should have plenty good books, and also there is a lot on the internet to be found.
 
I'm sorry for all the confusion--I wasn't trying to calculate ANYTHING---I just wanted to know the proper technique for the addition. It began when I noticed ny fractions were never exactly like the answers. Then I noticed that H1 had a mass that WAS NOT the same as p (a proton). Well Hell's Bells, sometimes I had done my addition using the weight of a proton, times how many there were in the atom. Other times I had used the weight of H1 for how many protons there were in the atom. Now do you see the basis for my question? I would get different answers for the same equation! That is the WHOLE PURPOSE of my thread here--which form is the proper one. I guessed the proper technique way up there, in post #1, but Kurdt said I was wrong. So, I did all those calculations with 8 different techniques to see which one gave the EXACT answer that was easy to verify (the BE of He4 is listed everywhere). And I was right--#2 is the only correct form. Which is exactly the same as the form you gave, only mine uses x and y, while yours uses A and Z, and continues on to calculate BE. I do understand yours now--I found it in my book last night and studied it. I originally thought your "A-Z" meant "A through Z", not "A minus Z"! lol That is what threw me, and that is why I asked you what A, Z, and A-Z meant. I think I'm OK now.
I'd love some book titles, and thanks for the help.
 
  • #10
okay, now I understand.

Nuclear and Particle Physics: An Introduction ; Brian R. Martin (has solutiuons to problems)

Nuclear Physics: Principles and Applications; J. S. Lilley (has answers to almost all problems)

Introductory Nuclear Physics by Kenneth S. Krane (a classic, no aswers altough)

Fundamentals of Nuclear Science and Engineering by Richard E. Faw (many applications to energy use and so on)
 
  • #11
Excellent! Many thanx!
 

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