Calculate Energy during Nuclear Fission

[wilson_chem90]

Homework Statement

Another possible form of the fission of U-235 is:
[tex]\begin{array}{cc}235&92\[\tex] U + [tex]\begin{array}{cc}1&0\[\tex] n $$/rightarrow$$ [tex]\begin{array}{cc}141&56\[\tex] Ba + [tex]\begin{array}{cc}92&36\[\tex] Kr + 3 $$\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

Homework Equations

The Attempt at a Solution

Before i start, I am 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.

 

[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

 

[tomcue]

I am not familiar with the notation used in the problem, so I had to make some assumptions. Also, the last product formed from the reaction is most likely 3 neutrons, as stated in the problem. However, I am not sure how to calculate the energy released from the production of neutrons. It is possible that the mass difference calculated above already takes into account the mass of the 3 neutrons, but I am not certain. Further clarification on the problem and the notation used would be helpful in providing a more accurate response.