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B Binding Energy in a nucleus

  1. Apr 14, 2017 #1
    The mass of the nucleus is calculated as follows:

    mass (nucleus) = [Number of protons * Mass of proton] + [Number of Neutrons * mass of neutron] - [Binding energy/c^2]

    Why is the contribution to mass from the binding energy subtracted from the mass of the nucleus rather than being added to the nuclear mass ?
  2. jcsd
  3. Apr 14, 2017 #2


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    Because that is the definition of binding energy. If you had to spend energy in order to make a system bound, that system would be unstable and decay to the original constituents.
  4. Jun 11, 2017 #3
    Two hydrogen atoms come together to form a stable helium atom. Surely the mass of the helium atom's nucleus should be the sum of the masses of the two protons and neutrons i.e.

    mnucleus = (mproton * 2) + (mneutron * 2)

    But, this does not add up to the actual mass of a helium atom. There is a loss of ≈0.0303 amu which is the mass defect. Using Einstein's equation relating mass and energy, E = mc2, we find that the nuclear binding energy is 4.53 x 10-12 J.

    Thus in the equation which you stated above, the mass defect is calculated by dividing by c2, thus getting mass back. And to obtain the actual mass of the nucleus it has to be subtracted.

    NOTE: By switching the positions of 'binding energy' and 'mass(nucleus)' we get the mass defect equation.
    Δm = Zmproton + (A - Z)mneutron - M
    Where, Δm = mass defect
    Z = number of protons
    A = number of nucleons
    M = mass of the nucleus.
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