Question

Nuclear power comes from the conversion of mass to energy by the equation E=mc². In nuclear reactions the energy released is found by taking the total of the mass of the reactants minus the total mass of the products and mulitplyig by c².

We are going to look at the enrgy produced by a typical fusion reaction and a typical fission reaction and compare that to the energy content of chemical reactions Fission is splitting an atom into two smaller atoms and fusion is comibing two lighter atoms to make a heavier atom. Lighter elements (lighter than iron) tend to produce energy when conbined, fusion. Heavier elements make energy when split, fission.

• The fusion reaction ²H+³H→ n+⁴He is a typical reaction taking place in stars, H-Bombs and is being researched as a source of energy to replace fossil fuels
• The fission reaction n+²³⁵U→ +3n+¹⁴⁴Ba+⁹²Kr a typical reaction taking place in current nuclear power plants and a-Bombs.

Using the table below calculate the energy released in the fusion reaction as:

• [(mass of (²H+³H) - mass of (n+⁴He)]×c2  where c=3×10⁸m. J?
• Now lets express this in joules per kilogram =(joules per reaction from above)/mass of (²H+³H)  J?

Using the table below calculate the energy released in the fission reaction as:

• [(mass of (n+²³⁵U) - mass of (3 n's+⁹²Kr+¹⁴⁴Ba)]×c2  where c=3×10⁸ m/s²  
  J?
• Now lets express this in joules per kilogram =  (joules per reaction from above)/mass of (n+²³⁵U)
  • J?

Compare this to the energy released in burning gasoline 45×107Jkg45×107Jkg

Nucleus	Mass m in kg
n a neutron	1.6749×10−271.6749×10-27
2H Deuterium	4.8390×10−274.8390×10-27
3H Tritium	5.0081×10−275.0081×10-27
4He Helium	6.6463×10−276.6463×10-27
89Kr	1.4615×10−251.4615×10-25
144Ba	2.3898×10−252.3898×10-25
235U	3.9029×10−253.9029×10-25

Answer 1