# Calculate Energy during Nuclear Fission

1. Oct 13, 2009

### wilson_chem90

1. The problem statement, all variables and given/known data
Another possible form of the fission of U-235 is:
$$\begin{array}{cc}235&92\[\tex] U + [tex]\begin{array}{cc}1&0\[\tex] n [tex]/rightarrow$$ $$\begin{array}{cc}141&56\[\tex] Ba + [tex]\begin{array}{cc}92&36\[\tex] Kr + 3 [tex]\begin{array}{cc}1&0\$$ n

a) Given the masses of the particles in the table below, calculate the amount of energy released in the fission of a U-235 nucleus.

U = 234. 993 u
1 n = 1.008 u
Ba = 140.883 u
Kr = 91.905 u

2. Relevant equations

3. The attempt at a solution
Before i start, im not sure what to do with the last product formed from the U-235 reaction. I'mm not sure if its 3 protons or 2 protons and 1 neutron. anyways i just assumed it was 3 neutrons though.

Mass or reactants :
234.993 u + 1.008 u = 236.001 u
Mass of products:
140.883 u + 91.905 u + 1.008 u(3) = 235.812 u

Mass difference:
236.001 u
- 235.812 u
= 0.189 u
Energy released:

m = (0.189 u)(1.6605 x 10^-27 kg/u)
=3.138345 x 10^-28 kg

E = mc^2
= (3.138345 x 10^-28 kg)(2.998 x 10^8 m/s)^2
= 2.82 x 10^-11 J

Im honestly not sure that this is correct.

2. Oct 13, 2009

### wilson_chem90

i couldn't do the laTeX image, but its 235/92 U + 1/0 n = 141/56 Ba + 92/63 Kr + 3 1/0 n