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(Nuclear) Binding Energy 
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#1
Oct2013, 09:48 PM

P: 16

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
Oct2013, 11:00 PM

P: 252

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.



#3
Oct2013, 11:03 PM

Mentor
P: 11,632

1. Look up the energy released in some simple chemical reaction. It will probably be given on a permole or pergram basis, or something like that.
2. Calculate the mass equivalent of that energy using E = mc^{2}. 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. 


#4
Oct2113, 03:10 PM

P: 16

(Nuclear) Binding Energy
jtbell:
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 bondwhat 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? 


#5
Oct2113, 06:34 PM

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P: 11,629

Does that make sense? 


#6
Oct2113, 09:47 PM

P: 16

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. 


#7
Oct2113, 10:02 PM

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P: 11,629

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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