## E=mc2 and nuclear fission

1. The problem statement, all variables and given/known data[/b]

The overall question is:
For this assignment explain how the E=mc2 equation applies to nuclear fission. In your answer illustrate your explanation with an example, being sure to distinguish between mass and mass number, and explain how a nuclear equation differs from a chemical equation. In addition compare the energy released during fission with energy produced from a typical chemical reaction (such as fossil fuel oxidation). It may be useful for you to consider that the combustion of methane releases 50.1 kJ/g - how much mass is lost to produce 50.1 kJ?

I put it all here so hopefully someone can make sense of my anwer below. I welcome any suggestions!!

3. The attempt at a solution[/b]

Started by inventing a reaction: 92U235 + 0n1 -> 37Rb90 + 55Cs143 + 3 0n1. (typical of a nuclear fission reaction.) The actual atomic masses of these are:
Rb90 = 89.91481
Cs143 = 142.92732.
2 0n1 = 2.01732
Sum = 234.85945. Now subtract:
U235 = 235.04392 to get
loss = 0.18447. One AMU = 931 Mev, so 171.74 Mev liberated by the fission. Chemical reactions produce energies on the order of a few electron volts.

This is where I stop...do I need a table of energy conversion values to convert AMU per mole to joules per mole, and then do a bit of arithmetic? Thanks for helping me out!

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Homework Help Science Advisor You already have the energy released (MeV) per U235 atom. The energy release per mole is that times Avogadro's number. Once you have the result in MeV, you don't need a table. 1 electron volt = 1.60217646 × 10^(-19) joules. I just pasted that out of Google.