E=mc2 and nuclear fission

  • Thread starter bjoyful
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1. Homework Statement [/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!
 

Answers and Replies

  • #2
Dick
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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.
 

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