Re: Why is energy released during a fission reaction?

Joseph,

The problems you are experiencing arise from applying the concepts you 
learned for chemical reactions to nuclear reactions. The change in mass 
between products and reactants in a chemical reaction is so small that 
using the rule of “Conservation of Mass” produces negligible error. 

This is not true for nuclear reactions; you have to use the rule 
of “Conservation of Mass-Energy”. Mass and energy are related by the most 
famous equation in Physics: Albert Einstein’s 

	E = mc2  where Energy is equal to mass times (the speed of light) 
squared.

We can use the definition:  1 Joule (J) = 1 kg x (2.99792 x 108 
meters/second)2

While Joules are the SI unit for energy, nuclear physicists and chemists 
use the units of Million Electron Volts (MeV) for nuclear reactions. One 
electron volt is the energy an electron has after being accelerated 
through a potential (voltage) difference of one volt. 

This is because much data is from the measurement of radioactive decay, 
and MeV is a very common unit for the energy of alpha and beta particles 
and gamma rays and X-rays. Accurate masses have been measured (isotope 
ratio mass spectrometry can give more than 7 significant figures with 
great precision). Accuracy is dependent on the measurement of Carbon-12 
as the standard. On the Scale of Atomic Weights, Carbon-12 (6 protons and 
6 neutrons) has the atomic weight of 12 Daltons or atomic mass units.

	1 eV = 1.602 x 10-19 J         or        1 MeV = 1.602 x 10-13 J

The most useful conversion for nuclear energy calculations is

	1 Dalton (amu)  = 931.5 MeV  

These accurate weights are available on the Chart of the Nuclides 15th 
Edition Revised 1996 by Parrington et al. and published by GE Nuclear 
Energy, San Jose, CA. There are also standard nuclear data available on 
the internet—tables of Atomic Masses (nuclide masses), Decay Energies, 
and Binding Energies. Note that we need nuclide masses, not the atomic 
masses given on the Periodic Table. Those are weighted averages of all 
the stable isotopes of an element. Here we are interested in the weight 
of each of those isotopes.
 http://ie.lbl.gov/toi2003/
MassSearch.asp 

This site is the Lawrence Berkeley Laboratory, and they have been 
publishing nuclide mass and decay data for over 50 years.

Now we are ready to tackle your dilemma: Nuclei are held together by the 
strong nuclear force. This force plus the dilution of the positive 
charges of the protons by the neutrons overcomes the repulsion of the 
plus charges of the protons. All stable and radioactive nuclei must have 
this strong force or they would fly apart. This binding energy is 
calculated by taking the sum of the number of protons times the mass of a 
proton (1.0072765 amu) plus the number of neutrons times the mass of a 
neutron (1.0086649 amu) and subtracting the measured mass of the nuclide 
in amu. 

For example, the decay of the neutron to a proton:

	1n  1p + β- (plus a negligible neutrino)

Using the data above, E (transition) = (1.0086641 – 1.0078250) 931.5 MeV 
= 0.7816 MeV
The atomic mass of a proton includes the mass of the electron (the beta 
particle is an electron emitted from the nucleus; its energy equivalent 
is 0.511 MeV). 

Knowing that for all nuclides, the mass is less than the mass of the 
protons and neutrons, we now look at how binding energy behaves as a 
function of Z, the atomic number, or equivalently, the number of protons 
in a nucleus. For easy comparison, the Binding Energy Per Nucleon is 
calculated for each nuclide:

atomic Binding Energy Graph.© NMM London
 

 
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 http://www.nmm.ac.uk/server.php?
show=conMediaFile.5919

Now we can see the answer to our quest. The largest binding energy per 
nucleon is at Fe-56 (Iron with 26 protons and 30 neutrons). Note that the 
graph is monotonically decreasing after iron. All heavier nuclei are 
energetically favored to decay towards iron. This is the driving force 
for nuclear fission of the heavy elements (as well as alpha decay for the 
heaver nuclides and beta decay for almost all others—note that beta decay 
includes beta minus (an electron), beta plus (a positron), and electron 
capture (the nucleus captures one of the inner shell electrons and 
changes a proton into a neutron)

The energy release in fission comes from the increased binding energy of 
the fission products, so the total mass will be less than the starting 
nucleus that is fissioned by absorbing a neutron. 

One common fission reaction for U-235 (more than 200 different nuclides 
of 35 different elements have been found as fission products for U-235 
plus a neutron) is

	1n + 235U = 91Kr + 142Ba + 3 1n     

Looking up the numbers on the web table given above, and doing the math 
shows that the REACTANTS weighed 0.1866971 amu more than the products. 
Multiplying by 

	1 amu = 931.5 MeV gives 174 MeV energy released per fission and 
products that weigh less than the reactants. 

To get an idea of the magnitude of this amount of energy, 

	174 MeV x 1.602 x 10-13 J/MeV = 2.79 x 10-11 J per fission. 
Assuming we fission one mole of 235U, we would get 

	2.79 x 10-11 J per fission x 6.02 x 1023 fissions per mole = 1.68 
x 1013 J/mole

converting to calories we get

	1.68 x 1013 J/mole x 1 cal/4.184 J = 4.01 x 1012 cal/mole = 4.01 
x 109 kcal/mole

compared to 20 or so kcal/mole for chemical reactions, you can see the 
tremendous energy released in fission (roughly 4 Tcal per mole).

I hope this tour de force has helped to resolve your dilemma.