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I Energy Released in Stellar Nucleosynthesis

  1. Jul 19, 2016 #1
    This is taken from page 226 in Essential Astrophysics by Lang:

    "The mass defect, ##ΔM##, for a nucleus containing ##A## nucleons, ##Z## protons, and ##A-Z## neutrons is
    $$ΔM = Z m_p + (A - Z) m_n - m_{nuc}$$ where ##A## is the mass number of the nucleus, ##Z## is the atomic number, ##m_p## is the mass of the proton, ##m_n## is the mass of the neutron, and ##m_{nuc}## is the mass of the nucleus.

    The binding energy, ##E_B##, used to assemble the nucleus from its constituent nucleons is:
    $$E_B = ΔM c^2$$ The binding energy measures how tightly bound a nucleus is."

    So mass defect represents both the difference between the mass of a composite particle and the sum of the masses of its parts and the binding energy released during nuclear fusion? If I wanted to calculate the binding energy released during the proton-proton chain reaction, would I simply plug in 938.272 MeV for ##m_p##, 939.5654 MeV for ##m_n##, 2 for ##Z##, 2 for ##A-Z## (4 total nucleons minus 2 protons for a helium atom), and 3727.379 MeV for ##m_{nuc}##? This gives me 28.2958 MeV. Multiplying that by (3 x 10^8 m/s)^2 gives me 2.54662 x 10^18. Is this correct? Is that the energy released during a nuclear fusion reaction in a main sequence star?

    If it helps to know, I took both calculus-based E&M (level of Purcell) and multi-variable calculus/vector calculus last semester. I have not taken university-level chemistry yet.

    Thanks, guys!
     
  2. jcsd
  3. Jul 19, 2016 #2

    Bandersnatch

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    The first part is (roughly that many MeV), but you shouldn't multiply that by c^2 - MeV is already a unit of energy. I.e., the values you used are not in MeV, but in the convenient mass units of MeV/c^2. If you multiply a mass of 1 MeV/c^2 by c^2 you get 1 MeV of energy.

    (Or you can just forget about mass altogether and use units of energy from start to finish)
     
  4. Jul 19, 2016 #3

    D H

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    @Bandersnatch already caught your big error, what was not using units correctly. The answer to the highlighted question is no. Some of that binding energy represents two protons becoming two neutrons. An easy way to get the correct answer is to look at what the system starts and ends with. The proton-proton chain changes four protons and two electrons into an alpha particle. Add the numbers up and you get 26.731 MeV rather than 28.2958 MeV.
     
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