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(Nuclear) Binding Energy

  1. Oct 20, 2013 #1
    I understand what nuclear binding energy is and its importance in nuclear weaponry an the fueling of stars, but why, in other types of bonds, such as chemical bonds, Einstein's equation e=mc2 is not applied and the bonding of two substances is not assumed to have energy as mass? But when an atomic nuclei forms, its energy is assumed to have mass and the energy released can be calculated with the equation?
  2. jcsd
  3. Oct 20, 2013 #2
    The mass of an electron bound in an atom is less than that of a free electron. However, the energy equivalence of this mass difference is negligible compared to Coulomb binding.
  4. Oct 20, 2013 #3


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    1. Look up the energy released in some simple chemical reaction. It will probably be given on a per-mole or per-gram basis, or something like that.
    2. Calculate the mass equivalent of that energy using E = mc2.
    3. Compare that mass equivalent with the mass of a mole of reactant, or with one gram, or whatever basis was used for specifying the amount of energy.
    Last edited: Oct 21, 2013
  5. Oct 21, 2013 #4

    When in a chemical reaction is mass lost? If two elements form a molecule in a ionic bond, one electron being transferred will result in less mass then a free electron, but if that electron becomes free of that bond, will it gain mass and therefore energy? When you say "reactant," what if both elements move their electrons like in a covalent bond--what happens to those electrons?
    And what about kinetic energy? This is what my initial confusion was based on. Since kinetic energy is still energy, why can't kinetic energy be released, instead of energy in the form of mass?
  6. Oct 21, 2013 #5


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    Yes. When the two atoms bond, energy is released, which means that mass has been removed from the molecules. (Since energy has mass) To overcome this bond energy must be supplied, which means that breaking the molecule into its component atoms requires energy (and thus mass) be added to the particles. Simply put, the sum of the masses of the atoms before the bond is greater than the mass of the molecule after the bond.

    When looking at a system and not just a lone object, kinetic energy has mass that adds to the mass of the system as a whole. Consider two atoms coming together and bonding. Lets say 100% of the energy is converted to kinetic energy. The system (the molecule and whatever environment it is in) still has the same mass as before the bond, as the mass lost from the particles is equal to the mass gained in the form of kinetic energy. If that molecule now collides with another molecule and radiates 50% of that kinetic energy out of the system as EM radiation, then the mass of the system is now smaller.

    Does that make sense?
  7. Oct 21, 2013 #6
    Ahh okay.
    So energy released, like kinetic energy, can compensate for the loss of mass during the bond.
    But, since the formula for kinetic energy is Ke = mv^2, how do you know mass is lost/gained, changing the value of Ke, but not the velocity? Just by looking at the equation, it seems to me as if energy can be lost and mass can stay the same, while v simply lowers n value.
  8. Oct 21, 2013 #7


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    That's not the right equation to use in this case. If you want to account for both mass and velocity changes, then you need to use E2=(mc2)2+(pc)2.
    The variable for momentum, p, is where the the velocity factors in.

    Plugging in the momentum and mass of the particles before and after bonding would show that the energy is exactly the same in our hypothetical situation above.

    Let's say you know the mass and momentum of the particles before bonding. Plugging them into the equation gives you the total energy. Then, knowing the total energy and that it must remain the same, you could measure either the mass or the momentum after bonding and solve for the missing variable.
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