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Q value of 3H(p,n)3He nuclear reaction

  1. May 30, 2009 #1
    Problem:

    I'm trying to calculate the Q-value for the nuclear reaction H-3 + p > He-3 + n, where H-3 is Deuterium, p is a proton, He-3 is Helium 3 and n is of course a neutron. Doing the calculation, I get -1.2715 MeV for the Q, but my textbook and an online 'Q-calculator' both give -763.15 KeV - I cannot work out where I'm going wrong - can anybody help?!

    My method:

    The Q-value for a reaction is given by the difference in the initial and final masses of the system (times the speed of light squared), and is equal to the energy 'liberated' by the reaction. I.e. Q = (Mi - Mf)*c^2.

    Now Mi = M(3H) + M(p) = 3.016049u + 1.00728u = 4.023329u

    and Mf = M(3He) + M(n) = 3.016029u + 1.008665u = 4.024694u

    Mi - Mf = -0.001365

    Q = -0.001365 * 931.5 MeV/u = -1.2715 MeV

    But all other sources that I can find listing the Q for this reaction give -0.76375 MeV.

    Where am I going wrong and how do I make it better?!?
     
  2. jcsd
  3. May 30, 2009 #2

    Hao

    User Avatar

    This is a tricky one. The answer is in the definitions.

    Atomic mass includes the mass of Protons, Neutrons, and Electrons!

    When isotopic mass is quoted, it includes the mass of the electrons. The mass of the nuclei needs to be corrected for the mass of the electrons.

    Using the values you quoted here, the reaction that you are calculating for is

    [tex]p + ^{3}_{1}H + e \rightarrow ^{3}_{2}He + n + 2e[/tex].

    The first thing that should occur to you is the violation of conservation of charge.
     
    Last edited: May 30, 2009
  4. May 30, 2009 #3
    No, it's nuclear reaction, and nuclear does not include electron. U don't need to count the mass of the electron. In fact, it's too small comparing to the mass of proton or neutron, so u usually just count it out.
     
  5. May 30, 2009 #4

    Hao

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    Usually, the mass of the electron may be approximated to be negligible, but in this particular scenario, the rest energy of an electron = 0.511 MeV, which is pretty much the difference between -1.272MeV and -0.763MeV, except for rounding errors.
     
    Last edited: May 30, 2009
  6. May 30, 2009 #5
    It's true that if there's Beta radiation, u must count the mass of electron. But there's no Beta rad here, just neutron.

    (1,1)p + (3,1)H > (3,2)He + (1,0)n

    U just can't put e in.
     
  7. May 30, 2009 #6

    Hao

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    Ah, I think I was not clear about what I meant by the reaction
    [tex]
    p +^{3}_{1}H + e \rightarrow^{3}_{2}He + n + 2e
    [/tex].

    I meant it as what Astrofield was calculating - I'm not proposing it as the actual reaction that occurs in nature.

    I agree with what you say - what I'm pointing out is that Q-value calculations are based on nuclear masses - not atomic masses. As atomic masses are the masses that are typically quoted, electron mass needs to be subtracted to obtain the nuclear mass.

    In some cases, this correction is not needed. However, in this case, p refers to a bare proton, and not a [tex]^{1}_{1}H[/tex] atom, and atomic values quoted for a [tex]^{1}_{1}H[/tex] atom differ from the mass of a bare proton due to one additional electron.

    Here is a reference:
    "ctaps.yu.edu.jo/physics/Courses/Phys441/PDF_Files/Phys441_Chapter9.pdf"[/URL]
     
    Last edited by a moderator: Apr 24, 2017
  8. May 30, 2009 #7
    Thanks a lot Hao - forgot about those little suckers! That reference you provided even has the exact calculation as an example!

    Cheers!
     
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