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Calculating the weight of 1 atomic mass unit

  1. Feb 28, 2017 #1
    I'm trying to deduce the weight of 1 atomic mass unit (##1u##) in ##kilograms## from the following scenario:

    One atomic mass unit is 1/12 of the weight of a ##^{12}C## atom in its ground state. A ##^{12}C## atom consists of 6 protons, 6 neutrons and 6 electrons. This means that
    $$1u = \frac{1}{12} \cdot mass(6 protons + 6 neutrons + 6 electrons) = \frac{1}{2} \cdot (m_p + m_n + m_e)$$
    The masses of the proton, neutron and electron in kg are:

    Proton: ##1.67262189821 \cdot 10^{-27} kg##
    Neutron: ##1.67492747121 \cdot 10^{-27} kg##
    Electron: ##9.1093835611 \cdot 10^{-31} kg##

    Filling these weights in the above formula shows that ##1u = 1.67423015389 \cdot 10^{-27} kg##. However, Wiki shows that ##1u = 1.66053904020 \cdot 10^{−27} kg## which already differs from the second decimal. What is the cause of this difference?
     
  2. jcsd
  3. Feb 28, 2017 #2

    Drakkith

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    The sum of the masses of 6 free particles of each type is more than the mass of the Carbon-12 atom. This is because the binding together of these particles releases energy from the system (atom), requiring that it have less mass than the total mass of the free particles. This is known as binding energy. I haven't crunched the numbers, but I'd bet that's the reason for the discrepancy.
     
  4. Feb 28, 2017 #3
    So this would be an example of mass->energy conversion through
    E2=p2c2 + m2c4
    ?
     
  5. Mar 1, 2017 #4

    Borek

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    [itex]E=mc^2[/itex] is perfectly enough.
     
  6. Mar 1, 2017 #5
    Thanks for the clarification. Is there a way to prove that my concluded difference in mass is exactly the binding energy release? For example, by looking at how much energy is needed per binding?
     
  7. Mar 1, 2017 #6

    Borek

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    Doable, but not straightforward, as it would require finding energies released in several reactions yielding the C12 nucleus (it is not a single step 6p + 6n → C12, google for nuclear synthesis of C12). Typically we do it the other way around - we calculate the binding energy from the mass deficit, as this is much easier.
     
  8. Mar 2, 2017 #7
    Thanks, I'll try and see what I can get. One other question, does the weight of a proton, neutron or electron differ even further depending on how many bindings there are in an atom? For example, a proton has a lower mass than its rest mass when it's bounded in an atom with 6 protons, 6 neutrons and 6 electrons. Does a proton have an even lower mass when it's bounded in an atom with 12 protons, 12 neutrons and 12 electrons?
     
  9. Mar 2, 2017 #8

    Borek

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    Number of electrons doesn't matter for the nucleus mass.

    Yes, the mass deficit is a function of the nucleus size, it has maximum value around nuclei containing 56 nucleons. For larger nuclei the binding energy per nucleon becomes lower again (which is why large nuclei can undergo fission).

    500px-Binding_energy_curve_-_common_isotopes.svg.png
    (image from wikipedia)
     
  10. Mar 2, 2017 #9
    Very interesting to read its Wiki page! So I've found the binding energy of a Carbon-12 atom on Wiki which is 92161.753 ± 0.014 keV. I'd assume that this is the energy-mass equivalent to my calculated mass difference.

    My calculated mass difference in my OP is:

    $$1.67423015389 \cdot 10^{−27} kg - 1.66053904020 \cdot 10^{-27} kg = 1.369111369 \cdot 10^{-29} kg$$

    The released bindingenergy has a mass-equivalent of:

    $$\frac{92161753 eV \cdot (1.60217656535 \cdot 10^{-19} J/eV)}{c^2} = 1.640660010 \cdot 10^{-28} kg$$

    This is already a factor of 10 difference. What am I missing? Should I instead use the net energy release during the triple alpha process?

    EDIT: When using the net energy release during the triple alpha process instead, which is 7.273 MeV, I'd get a mass difference of ##1.296529960 \cdot 10^{-27} kg##, which is even a factor of 100 difference. Am I doing something wrong?
     
    Last edited: Mar 2, 2017
  11. Mar 2, 2017 #10
    Your calculated difference is what you worked out for 1u. 12C is 12u. So multiply your value by 12...
     
  12. Mar 2, 2017 #11
    Of course! Didn't pay attention to that. Dividing the mass-equivalent of the binding energy 92161753 eV by 12 would give ##1.367216675 \cdot 10^{-29} kg## while my calculated difference is ##1.369111369 \cdot 10^{-29} kg##.
    I might be very nitpicky here but is there an explanation other than inacurracy why there's a difference of ##0.001894694 \cdot 10^{-29}##? I can see that the measured binding energy has an inacurracy range of +/- 0.014 keV. Perhaps this is the cause..
     
  13. Mar 2, 2017 #12
    I have one question regarding if we're talking about the total mass of an atom (not just the nucleus);

    Doesn't the mass of an electron change depending on how many electrons there are in an atom and how massive the nucleus is? The cause being that these 2 factors would result in different velocities for an electron and thus a different (relativistic) mass?
     
  14. Mar 2, 2017 #13

    DrClaude

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    The mass of an electron doesn't change ("relativistic mass" is an outdated concept). But yes, different electrons in the atom have different momenta, so they contribute unequally to the mass of the atom.
     
  15. Mar 2, 2017 #14

    DrClaude

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    Try and calculate that. It's a good exercise.
     
  16. Mar 2, 2017 #15
    Aren't these different momenta caused by the individual velocities, though?
     
  17. Mar 2, 2017 #16

    Borek

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    Velocity is a classic concept, which doesn't apply here easily. It made sense in a planetary model of an atom, but it doesn't work for an orbital (being better described as a standing wave).
     
  18. Mar 2, 2017 #17
    Was about to ask the same question as Comeback City. So does a planetary model of an atom, wherein the mass of an electron is calculated relativistically, not give an accurate total mass of the atom?

    Another question that I'm curious about: Does the released binding energy quantity include the compensation for the increase of total atom mass because of the orbiting electrons having momenta?
     
  19. Mar 2, 2017 #18

    Borek

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    Try to estimate what fraction of the atom mass is given by its electrons, and try to estimate by how much would the mass of an atom change when the electron gets excited (you can use ionization energies for that - first ionization is in a way equivalent to the maximum amount of energy that the atom can absorb while getting excited). Compare these numbers with the atom mass.
     
  20. Mar 2, 2017 #19
    Before I attempt this, do different isotopes of elements have different ionization energies?
     
  21. Mar 2, 2017 #20
    I take it you mean that the increase in atom mass because of the electrons developing momenta is negligible compared to the mass loss from the protons and neutrons when they bind?
     
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