# Calculating energy released in nuclear fission

Consider the neutron induced fission $\text{U-235} + n \to \dots \to \text{La-139} + \text{Mo-95} + 2n$, where $\dots$ denotes intermediate decay steps.

I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ($B$):

$$\Delta E = B(139,57) + B(95,42) - B(235,92) \approx 202,3 \, \mathrm{MeV}$$

(Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).

Another way is:

$$\Delta E = (m(\text{U-235}) + m_{\text{Neutron}} - m(\text{Mo-95}) - m(\text{La-139}) - 2m_{\text{Neutron}} )c^2 \approx 211,3 \, \mathrm{MeV}$$

Which one gives the correct result? Why?

You notice that $57+42 \neq 92$, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?

A slightly other point of view: What different questions do both calculations answer?

Below I show how the discrepancy $(202.3 \, \mathrm{MeV})$ and $(211.3 \, \mathrm{MeV})$ between your two methods has arisen.

$B(139,57) = \rm {57p +82n+57e} -m(139,57)$
$B(95,42) = \rm {42p +53n+42e} -m(95,42)$
$B(235,92) = \rm {92p +143n+92e} -m(235,92)$

$B(139,57) + B(95,42) - B(235,92) = m(235,92) - m(139,57)\rm -n +7e +[7p-7n]$

$\rm 7p-7n = 7(938.272-939.565)= - 9.051$

Note that your original equation was unbalanced if you used the masses of the atoms because there is a difference of seven electrons between the left hand side $(\rm U235 + n)$ and the right hand side $(\rm La139 + Mo95+2n)$ of your equation.
• Thanks, is $202.2 \mathrm{MeV}$ or $211.3\mathrm{MeV}$ the correct answer? – Julia May 19 '17 at 7:26