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?


2 Answers 2


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

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

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

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

There is your difference.

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.

  • $\begingroup$ @annav Thank you. I have added an opening sentence to my answer. Also the "missing" 7 electrons represent an energy of approximately 3.5 MeV. $\endgroup$
    – Farcher
    Commented May 19, 2017 at 6:37
  • $\begingroup$ Thanks, is $202.2 \mathrm{MeV}$ or $211.3\mathrm{MeV}$ the correct answer? $\endgroup$
    – Julia
    Commented May 19, 2017 at 7:26
  • $\begingroup$ You find the mass defect ie the difference in mass between the parent nucleus + mass of neutron and the total mass of the products of the fission and then convert that mass into an energy. In this case it is your 211.3 but note that if you are using atomic masses rather than nuclear masses you have 7 electrons unaccounted for. These 7 electrons represent an energy of about 3.5 MeV. $\endgroup$
    – Farcher
    Commented May 19, 2017 at 7:35
  • $\begingroup$ Is it possible to get the correct answer relying only on the binding energy concept? Since there are several beta decays in the intermediate steps, you get some transformed to a proton. Since the binding energy of a proton is higher some additional energy (in my case in total 9 MeV) is released. $\endgroup$
    – Julia
    Commented May 19, 2017 at 11:58
  • $\begingroup$ Your example shows that it with care it can be done but why would you bother? $\endgroup$
    – Farcher
    Commented May 19, 2017 at 12:53

The released energy is the difference in energy of intitial/end products. For a nucleus, the energy is given by its mass, which in turn can be calculated as the difference of "naive mass", i.e. all single constituents' masses summed up, and the binding energy. This is the calculation you'll have to do. Obviously, all these calculations neglect kinetic energy terms. Edit: this link might clear things up.

  • $\begingroup$ Please could you make your answer a bit more specific to my question? $\endgroup$
    – Julia
    Commented May 19, 2017 at 5:33

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