1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question-Proper form for nuclear addition

  1. Aug 29, 2007 #1
    Question--Proper form for nuclear addition

    1. The problem statement, all variables and given/known data

    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:

    2. Relevant equations

    1.) 6p + 6n = total

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

    3.) 6(H2) = total

    3. 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
    total mass of parts for Bx = x(H1) + x(n)

    How'd I do?
  2. jcsd
  3. Aug 30, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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 occurence of Boron they happen to be the same.
  4. Aug 30, 2007 #3
    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.
  5. Aug 30, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't know why your results differ, but only the nucleons contribute to the binding energy of the nucleus which makes sense.
  6. Aug 30, 2007 #5
    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.
  7. Aug 30, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    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.
  8. Aug 30, 2007 #7
    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: Aug 30, 2007
  9. Aug 31, 2007 #8


    User Avatar
    Science Advisor
    Homework Helper

    I still dont 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.
  10. Aug 31, 2007 #9
    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 thanx for the help.
  11. Aug 31, 2007 #10


    User Avatar
    Science Advisor
    Homework Helper

    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)
  12. Aug 31, 2007 #11
    Excellent! Many thanx!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Question-Proper form for nuclear addition
  1. Vector addition question (Replies: 12)