Nuclei within make up virtually all the mass of matter in the universe today, so it's lucky for us that almost all nuclei are incredibly stable, or else everything would fall apart. However, some nuclei are too large for their own good and can spontaneously disintegrate. After all, in nature, nothing can be completely stable! When these nuclei disintegrate, they go off with a bang, releasing magnitudes of energy larger than anything else on Earth. It's this disintegration energy that can be used to power whole cities in nuclear power stations, but also to produce some of the greatest atrocities ever seen, such as the dropping of the atomic bombs on Hiroshima and Nagasaki in 1945 by the US Air Force. In this article, we'll get to grips with the physics behind the fearsome power of nuclear energy.
Radioactive decay results in the release of enormous amounts of energy, an example of this is atomic bombs, Wikimedia commons Public Domain
Disintegration Energy Definition
To understand what exactly disintegration energy is, we first need to re-cap what radioactive decay is.
Radioactive decay is a natural process by which large unstable nuclei reduce their internal energy by splitting and emitting radiation in the form of either an alpha (\(\alpha\)), beta (\(\beta\)), or gamma (\(\gamma\)) particle.
Not all nuclei are radioactive, most commonly radioactive decay occurs in the heaviest elements in the periodic table, such as uranium and plutonium.
Some elements are radioactive due to their nucleus being unstable, these are highlighted in yellow in the table, Wikimedia commons Public Domain
Radioactive decay occurs in large nuclei because the attractive force holding the protons and neutrons together, known as the strong nuclear force, can no longer outweigh the repulsive electrostatic force due to so many positively charged protons being packed together. In order to become stable, the nuclei must reduce their internal energy either by:
- releasing two neutrons and two protons (\(\alpha\) decay), transforming a proton into a neutron,
- or vice versa (\(\beta\) decay), which emits an energetic electron,
- or by emitting a high-energy photon (\(\gamma\) decay).
During this process, the excess internal energy of the unstable nucleus is released as disintegration energy.
Disintegration energy is the energy change of a nucleus when it undergoes radioactive decay.
Binding Energy
To understand why nuclei become more stable, we need to understand the binding energy of a nucleus. Nuclei are formed of nucleons (protons and neutrons) bonded together by the strong nuclear force, this means that they are more stable, with lower internal energy, than they would be as a collection of separate particles. This energy difference is due to the equivalence between mass and energy, given by Einstein's famous equation \(E=mc^2\), this lower internal energy means that the mass of a nucleus is lower than the combined mass of the individual nucleons. It's this difference in mass, or mass defect, that determines the binding energy of a nucleus.
The Binding Energy, \(\Delta E\), of a nucleus is the energy associated with the difference in mass between a nucleus and its constituent nucleons, \(\Delta m\).
\[\Delta E=\Delta mc^2\]
where \(\Delta E\) is the binding energy measured in electron-volts \(\mathrm{eV}\), \(\Delta m\) is the change in the mass of the nucleus before and after decay measured in mass units \(\mathrm{u}\), and \(c\) is the speed of light, measured in \(\mathrm{m/s}\). Binding energy is usually defined to be negative.
In nuclear physics, a special set of units are often used. Mass is usually measured in the uniform mass unit \(\mathrm{u}\) where \(1\,\mathrm{u}=1.66\times10^{-27}\,\mathrm{kg}\) and energy is measured in mega electron-volts where \(1\,\mathrm{MeV}=1.6\times 10^{-13}\,\mathrm{J}\). Let's try to make sense of all of these equations by looking at an example.
The mass of one proton is \(m_p=1.00728\,\mathrm{u}\) and the mass of one neutron is \(m_n=1.00866\,\mathrm{u}\). What is the mass defect and binding energy of an \(\alpha\) particle \(\ce{^{4}_{2}\alpha}\), formed from two protons and two neutrons, if its mass is \(m_{\alpha}=4.00153\,\mathrm{u}\)?
A mass defect is a common way of describing the difference predicted mass of a nucleus, and the actual mass of a nucleus, due to the presence of binding energy.
\[\Delta m=m_{\alpha}-m_{\text{nucleons}}=m_{\alpha}-(2m_p+2m_n)\]
\[\Delta m=4.00153\;\mathrm{u}-(2\cdot1.00728\;\mathrm{u}-2\cdot1.00866\;\mathrm{u})=-0.03035\,\mathrm{u}\]
To calculate the binding energy associated with the mass defect, convert the mass defect into kilograms and then use Einstein's equation.
\[\Delta m_{\mathrm{kg}}=-0.03035\,\mathrm{u}\cdot1.66\times10^{-27}\,\frac{\mathrm{kg}}{\mathrm{u}}=-5.04\times10^{-29}\,\mathrm{kg}\]
Plugging this into Einstein's equation gives:
\[\begin{align}\Delta E&=\Delta m_{\mathrm{kg}}c^{2} \\ &=-5.04\times10^{-29}\,\mathrm{kg}\cdot(3.0\times10^{8}\,\mathrm{m/s})^{2} \\&=-4.5\times10^{-12}\,\mathrm{J} \\ &=-28\,\mathrm{MeV} \end{align}\]
Recall that before we emphasized that our binding energy is usually negative. This is because energy must be put into the nucleus to break it up into its constituents.
Disintegration Energy Formula
Nuclei with higher binding energies are more stable, this means that when an unstable nucleus decays into one or more stable nuclei the new nuclei have a lower mass overall than the unstable nucleus. It's this difference in mass between the 'parent' and 'daughter' nuclei that determines the disintegration energy, as the excess mass-energyto is released in the process. From this, we can establish the formula for disintegration energy.
\[E_{\text{disintegration}}=(m_{\text{parent}}-m_{\text{daughter}})\times c^2\]
where \(E_{\text{disintegration}}\) is the disintegration energy measured in electron-volts \(\mathrm{eV}\), \(m_{\text{daughter}}\) is the mass of the resultant daughter nucleus after decay measured in mass units \(\mathrm{u}\), \(m_{\text{parent}}\) is the mass of the initial parent nucleus before decay measured in mass units \(\mathrm{u}\), and \(c\) is the speed of light measured in \(\mathrm{m/s}\). That is, the disintegration energy is the difference in mass between the nuclei produced by the decay and the decaying nucleus multiplied by the speed of light squared. As we can see the more stable the products, the lower their mass will be and the larger the disintegration energy will be. It is important to note that when we refer to the daughter nucleus, we refer to all the resultant particles produced by the radioactive decay.
Disintegration Energy of an Alpha Particle
When we began our discussion on disintegration energy, we established that one of the forms of radioactive decay is the release of alpha particles from the parent nucleus.
An alpha particle is a radioactive decay particle consisting of two protons and two neutrons.
Let us consider the radioactive decay equation involving an alpha particle
\[\ce{^232_92X} = \ce{^228_90Y} + \ce{^4_2\alpha}\]
where the parent nucleus \(\mathrm{X}\) on the left-hand side has a mass number of 232 and an atomic number of 92. On the other right-hand side, the daughter nucleus \(\mathrm{Y}\) has a mass number of 228 and an atomic number of 90, whilst the alpha \(\alpha\) particle has a mass number of 4 and an atomic number of 2. This equation tells us that the radioactive decay of nucleus \(\mathrm{X}\) results in nucleus \(\mathrm{Y}\) being formed and an extra alpha \(\alpha\) particle. It is important to note that the total atomic and mass number is always conserved throughout the equation. This is because the total number of nucleons (protons and neutrons) and electrons must be conserved.
The process of alpha decay from parent nucleus Americium-241 to Neptunium-237, Wikimedia Commons CC BY-SA 4.0
Now in order to calculate the disintegration energy as the result of the alpha \((\alpha\)) decay, we must work out the difference in mass between parent and daughter nuclei due to the decay.
Then to work out the disintegration energy of this nuclear decay, we first need to know the masses of all the nuclei involved. As the disintegration energy of these particles is very small numbers, we need to be very specific about the weights of the particles we are considering. The exact mass of particle \(\mathrm{X}\) is given by \(232.037\;\mathrm{u}\), particle \(\mathrm{Y}\) is \(228.028\;\mathrm{u}\), and the alpha \(\alpha\) particle is \(4.001\;\mathrm{u}\).
We now plug these mass values into our equation for disintegration energy,
\[\begin{align} \Delta E_{\text{disintegration}} &= (m_{\mathrm{X}} - m_{\mathrm{Y}} - m_{\alpha})\cdot c^2 \\ &= (232.0371 - 228.028 - 4.001)\cdot(1.66\times10^{-27})\cdot(3\times10^8)^2 \\ &= 1.195\times10^{-12}\;\mathrm{J}.\end{align}\]
To make this slightly easier to read, we can convert this to electron-volts \(\mathrm{eV}\),
\[ 1.195\times10^{-12}\;\mathrm{J} \cdot 6.242\times10^{18}\;\mathrm{\frac{eV}{J}} = 7.459\times10^6\;\mathrm{eV} = 7.459\;\mathrm{MeV}.\]
So the disintegration energy of an alpha particle is \(7.459\;\mathrm{MeV}\).
Disintegration Energy of a Beta Particle
Now that we have covered the disintegration energy of an alpha \(\alpha\) particle, we turn our attention to another form of radioactive decay, beta \(\beta\) decay. First, let us define a formal definition.
A beta particle is a radioactive decay particle comprising a singular electron.
The process of beta \(\beta\) decay Caesium-137 to Barium-137, adapted from Wikimedia Commons CC BY-SA 4.0
Again, let us look at a nuclear decay equation that involves beta \(\beta\) decay.
\[ \ce{^137_55X} = \ce{^137_56Y} + \ce{^0_-1\beta}\]
where the parent nucleus \(\mathrm{X}\) on the left-hand side has a mass number of 137 and an atomic number of 55. On the right-hand side, the daughter nucleus \(\mathrm{Y}\) has a mass number of 137 and an atomic number of 56, while the beta \(\beta\) particle has a mass number of 0 and an atomic number of -1.
It is important to note the difference between the nuclear equation for alpha \(\alpha\) decay and beta \(\beta\) decay; while the mass number and atomic number are still conserved in both equations, the actual numbers vary. As a beta \(\beta\) particle is only made up of a single electron, we say it has a mass number of 0 because electrons are extremely light in comparison to nucleons. They weigh in at about \(9.05\times10^{-28}\;\mathrm{g}\), for comparison, a pin weighs \(10\;\mathrm{g}\)!
Furthermore, the beta \(\beta\) decay particle has an atomic number of -1. That seems odd, why is it negative? This is because the atomic number is also sometimes referred to as the charge number. Electrons have a negative charge, thus the charge number of a beta \(\beta\) particle is -1.
Now to work out the disintegration energy of beta \(\beta\) decay. From the above radioactive decay equation, the exact mass of particle \(\mathrm{X}\) is \(136.907\;\mathrm{u}\), the mass of particle \(\mathrm{Y}\) is \(136.905\;\mathrm{u}\), and for the beta \(\beta\) particle it is \(0.0005\;\mathrm{u}\). Note that the mass of the beta \(\beta\) particle is extremely small! Plugging these numbers into the disintegration energy formula gives,
\[\begin{align} \Delta E_{\text{disintegration}} \\&= (m_{\mathrm{X}} - m_{\mathrm{Y}} - m_{\beta})\cdot c^{2} \\&= (136.907-136.905- 0.0005)\cdot (1.66\times 10^{-27})\cdot (3\times10^8)^2 \\&= 2.241\times10^{-13}\;\mathrm{J}.\end{align}\]
Again we can convert this to electron-volts \(\mathrm{eV}\),
\[ 2.241\times10^{-13}\;\mathrm{J} \cdot 6.242\times10^{18}\;\mathrm{\frac{eV}{J}} = 1.399 \times 10^6\;\mathrm{eV} = 1.399\;\mathrm{MeV}.\]
So the disintegration energy of one beta \(\beta\) particle is \(1.399\;\mathrm{MeV}\).
Disintegration Energy Examples
We have now seen how to calculate the disintegration energy of an alpha \(\alpha\) and a \(\beta\) particle. Let's look at a more complicated example question.
Consider the decay equation
\[ \ce{^210_84Po} = \ce{^?_?X} + \ce{^4_2\alpha}\]
which describes the decay of Polonium into a mystery element \(\mathrm{X}\) through the emission of an alpha \(\alpha\) particle.
Firstly, let's work out the mystery element so we can work out the disintegration energy for the decay. Recall that the atomic and mass number in decay equations must always be conserved so we can use the information about Polonium and the alpha \(\alpha\) particle to work it out.
The mass number \(mn_X\) is given by
\[ 210 = mn_X + 4\]
\[mn_X = 210 - 4 = 206\]
and the atomic number \(an_X\) is given by
\[ 84 = an_X + 2\]
\[ an_X = 84 - 2 = 82.\]
So our mystery element \(X\) has a mass number of 206 and an atomic number of 82, meaning that it is Lead (or denoted as Pb).
Using the fact that the precise mass of a Polonium particle is \(209.983\;\mathrm{u}\), a Lead particle is \(205.974\;\mathrm{u}\), and an alpha \(\alpha\) particle is \(4.001\;\mathrm{u}\), we can now work out the disintegration energy of this decay.
Once again we plug this into our equation
\[\begin{align}\Delta E_{\text{disintegration}} &= (m_{\mathrm{Po}} - m_{\mathrm{Pb}} - m_{\alpha})\cdot c^2 \\&= ( 209.983-205.974- 4.001)\cdot (1.66\times10^{-27})\cdot(3\times10^8)^2 \\&= 1.195\times10^{-12}\;\mathrm{J}.\end{align}\]
Which we can then convert to electron-volts \(\mathrm{eV}\) as
\[ 1.195\times10^{-12}\;\mathrm{J}\cdot 6.242\times10^{18}\;\mathrm{\frac{eV}{J}} = 7.459\times10^6\;\mathrm{eV} = 7.459\;\mathrm{MeV}.\]
So the disintegration energy for the alpha decay of Polonium-210 is \(7.459\;\mathrm{MeV}\).
Disintegration Energy - Key takeaways
- Unstable particles go through a process of radioactive decay where a new element is formed and radioactive particles are emitted.
- Mass and energy are interlinked through Einstein's equation.
- Sometimes mass is not conserved in radioactive dec and is emitted as disintegration energy.
- The exact masses of the particles must be known in order to calculate disintegration energy.
- Mass number and atomic number is always conserved in decay equations.
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