- #1
wilson_chem90
- 137
- 0
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 [tex]/rightarrow[/tex] [tex]\begin{array}{cc}141&56\[\tex] Ba + [tex]\begin{array}{cc}92&36\[\tex] Kr + 3 [tex]\begin{array}{cc}1&0\[/tex] n<br /> <br /> 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. <br /> <br /> U = 234. 993 u<br /> 1 n = 1.008 u<br /> Ba = 140.883 u<br /> Kr = 91.905 u <br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> 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. <br /> <br /> Mass or reactants :<br /> 234.993 u + 1.008 u = 236.001 u<br /> Mass of products:<br /> 140.883 u + 91.905 u + 1.008 u(3) = 235.812 u<br /> <br /> Mass difference:<br /> 236.001 u<br /> - 235.812 u<br /> = 0.189 u <br /> Energy released:<br /> <br /> m = (0.189 u)(1.6605 x 10^-27 kg/u)<br /> =3.138345 x 10^-28 kg<br /> <br /> E = mc^2<br /> = (3.138345 x 10^-28 kg)(2.998 x 10^8 m/s)^2<br /> = 2.82 x 10^-11 J<br /> <br /> Im honestly not sure that this is correct.[/tex][/tex][/tex][/tex]