How Do You Calculate the Energy Released in a Fission Reaction?

For IB Physics SL and HL, calculate the mass defect between all reactants and products, then apply mass and energy equivalence, $E=\Delta mc^2$. If masses are given in unified atomic mass units, use $1\text{ u}=931.5\text{ MeV}/c^2$.

The Calculation Method

In nuclear fission, the total mass of the products is slightly less than the total mass of the reactants. This mass difference is converted into energy because the fission products have a greater average binding energy per nucleon than the original heavy nucleus.

Follow these steps:

1. Write a balanced nuclear equation, conserving nucleon number and proton number.
2. Add the masses of all reactants, including any incident neutron.
3. Add the masses of all products, including emitted neutrons.
4. Calculate $\Delta m=m_{\text{reactants}}-m_{\text{products}}$.
5. Use $$, or multiply a mass defect in by .

Consider this fission channel:

$${}^{235}_{92}\text{U}+{}^{1}_{0}\text{n}\rightarrow{}^{141}_{56}\text{Ba}+{}^{92}_{36}\text{Kr}+3{}^{1}_{0}\text{n}$$

Using atomic masses:

$$m_{\text{reactants}}=235.04393+1.008665=236.052595\text{ u}$$ $$m_{\text{products}}=140.91441+91.92616+3(1.008665)=235.866565\text{ u}$$

Therefore,

$$\Delta m=236.052595-235.866565=0.186030\text{ u}$$

The energy released is

$$E=(0.186030)(931.5)=173\text{ MeV}$$

Using $1\text{ MeV}=1.602\times10^{-13}\text{ J}$:

$$E=173(1.602\times10^{-13})=2.77\times10^{-11}\text{ J}$$

The released energy appears mainly as kinetic energy of the fission fragments, with some carried by emitted neutrons and gamma radiation. In a reactor, repeated fission events transfer this energy to thermal energy in the fuel and coolant.

As a useful scale, one mole of identical fissions would release $2.77\times10^{-11}\times6.02\times10^{23}\approx1.67\times10^{13}\text{ J}$, showing why nuclear fuels have very high energy density.

This result applies to this particular fission channel; other product combinations release slightly different energies. Atomic masses are valid because the total number of electrons is 92 on each side, so their masses cancel.

A common misconception is to compare only the uranium nucleus with the products. The incoming neutron is also a reactant and must be included. Another error is reversing the subtraction, which produces a negative value even though energy is released.

Exam Technique

For an IB Calculate question, show the balanced equation, both mass totals, the positive mass defect, and the energy conversion. Include units throughout and avoid rounding intermediate values too early.