# Energy released in nuclear fission

1. May 11, 2007

### AznBoi

1. The problem statement, all variables and given/known data
Find the energy released of the follow fission reaction:

$$\displaystyle{^{1}_0}n\; +\; ^{235}_{92}U\; \rightarrow \; ^{88}_{38}Sr\; +\; ^{136}_{54}Xe \;+\; 12 ^{1}_{0}n}$$

2. Relevant equations
E=mc^2

3. The attempt at a solution

How come the masses between the leftside and the right side are different if they all contain the same number of nucleons? Does one usually need to calculate the masses independent of the periodic table (by adding the individual nucleon masses up) or do these problems generally require a periodic table for mass substitution?

Thanks.

Last edited: May 11, 2007
2. May 11, 2007

### Curious3141

Yes, you will need a periodic table or other authoritative source for the precise masses. The "atomic mass" that adds up is just a nucleon number, not precise enough for a fission calculation; moreover, nuclear binding energy needs to be taken into account (this is already factored into the "true" atomic mass).

Last edited: May 11, 2007
3. May 11, 2007

### AznBoi

Okay, but I still don't get why the masses on both sides aren't equal. Also, how come the mass on the left hand side is greater than the mass on the right side? I thought that the individual nucleons weighed much more than the bound nucleons (which gives way to "mass defect"). How come this principle isn't applied here? Is binding energy different from the energy released by nuclear reactions (fission/fusion)??

4. May 11, 2007

### Curious3141

Because there is net energy released in the reaction. The combined products on the right hand side will have a mass deficit when compared to those on the left hand side. This mass deficit is related to the energy produced from fission by $$\Delta E = \Delta m c^2$$. This energy is manifested as the combined kinetic energy of the fission products.

At the level of stable individual nuclei, you will observe the mass-defect. If you take the precise mass of a stable nucleus comprising p protons and n neutrons, it will be less than the sum of the masses of p protons and n neutrons individually. This is "mass-defect" and it's a consequence of the fact that energy (biding energy) is released (often as gamma rays) in the putative fusion reaction of the nucleons in forming the nuclide. The product (stable nuclide) has less mass/energy than the constituent nucleons, so it is more stable. You need to add energy to the nuclide somehow to get it to become unstable enough to break apart into nucleons. This is the intuitive, albeit unrealistic, way to think of mass-defect. The mass-defect of a single nucleon is zero, of course.

The binding energy is the very source of the energy released in fission. Some of the input energy (kinetic energy of the incident neutron) has to go into overcoming the strong nuclear force in order to make the fission process happen. But when the process gets going, it ends up with a net release of energy because the masses of the stable products are less than those of the reactants. The mass deficit I spoke of earlier is, in fact, the net change in the summed mass-defects between the products and the reactants. That mass deficit is released as energy.

Total energy is, of course, conserved : when you sum up the kinetic energy of the neutron on the left hand side with the "mass-energies" of the reactants, you will get the same as the kinetic energy of the products plus the "mass-energies" of the products.