Binding energy per nucleon
As a nucleus is made up of neutrons and protons, there are forces of repulsion between the positive charges of the protons. It takes energy, called binding energy, to hold General term for components of a nucleus. Neutrons and protons are the two types of nucleons. together as a nucleus.
Iron has a mass number of 56 and is one of the most stable of all the elements. We say that iron has a high binding energy per nucleon.
Elements with lower and higher mass numbers per nucleon are less stable.
The total mass of a nucleus is less than the total mass of the nucleons that make up the nucleus. This difference is known as the mass defect. It is equivalent to the binding energy of the nucleus, using \(E = mc^{2}\).
In fission, an unstable nucleus is converted into more stable nuclei with a smaller total mass. This difference in mass, the mass defect, is the binding energy that is released.
In fusion, the mass of the nucleus that is created is slightly less than the total mass of the original nuclei. Again the mass defect is the binding energy that is released, since the nucleus that is formed is more stable.
Question
Looking at the graph of binding energy per nucleon above which process, fission or fusion, is likely to release most energy per nucleon?
Fusion as the difference in energy per nucleon for the lighter elements is far greater than the heavy elements. However the technical difficulties in releasing this energy mean successful energy production using fusion is still some way off.
Use the following steps in all calculations using \(E = mc^{2}\)
Step one: Check that the mass numbers and atomic numbers are balanced for the reaction
Step two: Calculate the total mass before the reaction
Step three: Calculate the total mass after the reaction
Step four: Calculate the mass defect
Step five: Calculate the energy released using \(E = mc^{2}\)
For steps two and three you will be given the masses of the reactants and the products as needed. Some masses are given in the table.
| Particle | \(_{0}^{1}\textrm{n}\) | \(_{92}^{235}\textrm{U}\) | \(_{37}^{93}\textrm{Rb}\) | \(_{55}^{141}\textrm{Cs}\) | \(_{1}^{3}\textrm{H}\) | \(_{1}^{2}\textrm{H}\) | \(_{2}^{4}\textrm{He}\) |
| Mass \(( \times 10^{-27}kg)\) | 1.675 | 390.173 | 154.248 | 233.927 | 5.007 | 3.343 | 6.645 |
| Particle |
|---|
| \(_{0}^{1}\textrm{n}\) |
| \(_{92}^{235}\textrm{U}\) |
| \(_{37}^{93}\textrm{Rb}\) |
| \(_{55}^{141}\textrm{Cs}\) |
| \(_{1}^{3}\textrm{H}\) |
| \(_{1}^{2}\textrm{H}\) |
| \(_{2}^{4}\textrm{He}\) |
| Mass \(( \times 10^{-27}kg)\) |
|---|
| 1.675 |
| 390.173 |
| 154.248 |
| 233.927 |
| 5.007 |
| 3.343 |
| 6.645 |
Question
Calculate the energy released in the following fission reaction.
\(_{92}^{235}\textrm{U}+_{0}^{1}\textrm{n} \rightarrow _{37}^{93}\textrm{Rb}+_{55}^{141}\textrm{Cs}+2_{0}^{1}\textrm{n}\)
Step one:
Atomic number before reaction \(=92\)
Atomic numbers after reaction \(=37+55=92\)
Mass numbers before reaction \(=235-1=236\)
Mass numbers after reaction \(=93+141+(2\times1)=236\)
Atomic numbers and mass numbers are balanced.
Step two:
Total mass before the reaction
\(= (390.173 + 1.675) \times 10^{-27}\)
\(=391.848\times10^{-27}kg\)
Step three:
Total mass after the reaction \(=(154.248+233.927+(2\times 1.675))\times10^{-27}\)
\(=391.525 \times 10^{-27}kg\)
Step four:
Mass defect \(=(391.848-391.525)\times10^{-27}kg\)
\(=3.23\times10^{-28}kg\)
Step five:
Energy released \(E=mc^{2}\)
\(=3.23\times10^{-28}\times(3\times 10^{8})^{2}\)
\(=2.9\times10^{-11}J\)
Question
Calculate the energy released in the following fusion reaction.
\(_{1}^{3}\textrm{H}+\,_{1}^{2}\textrm{H}\rightarrow\,_{2}^{4}\textrm{He}+\,_{0}^{1}\textrm{n}\)
Step one:
Atomic numbers before reaction \(=1+1=2\)
Atomic numbers after reaction \(=2\)
Mass numbers before reaction \(=3+2=5\)
Mass numbers after reaction \(=4+1=5\)
Atomic numbers and mass numbers are balanced.
Step two:
Total mass before the reaction \(=(5.007+3.343)\times10^{-27}\)
\(=8.350\times10^{-27}kg\)
Step three:
Total mass after the reaction \(=(6.645+1.675)\times10^{-27}\)
\(=8.320\times10^{-27}kg\)
Step four:
Mass defect \(=(8.350-8.320)\times10^{-27}\)
\(=3.0\times10^{-29}kg\)
Step five:
Energy released \(E=mc^{2}\)
\(=3.0\times10^{-29}\times(3\times10^{8})^{2}\)
\(=2.7\times10^{-12}J\)