Disintegration Energy

Disintegration Energy Definition

To understand what exactly disintegration energy is, we first need to re-cap what radioactive decay is.

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.

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 $E=m{c}^2$, 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.

In nuclear physics, a special set of units are often used. Mass is usually measured in the uniform mass unit $\mathrm{u}$ where $1\mathrm{u}=1.66\times {10}^{-27}\mathrm{kg}$ 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.

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 $\mathrm{m}/\mathrm{s}$. 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.

Let us consider the radioactive decay equation involving an alpha particle

$${\phantom{A}}_{\phantom{92}}^{\phantom{232}}{\phantom{A}}_{\phantom{2}92}^{\phantom{2}232}\mathrm{X}={\phantom{A}}_{\phantom{90}}^{\phantom{228}}{\phantom{A}}_{\phantom{2}90}^{\phantom{2}228}\mathrm{Y}+{\phantom{A}}_{\phantom{2}}^{\phantom{4}}{\phantom{A}}_{\phantom{2}2}^{\phantom{2}4}\alpha$$

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,

$$\begin{array}{rl}\mathrm{\Delta }{E}_{\text{disintegration}}& =(m_{\mathrm{X}}-m_{\mathrm{Y}}-m_{\alpha })\cdot c^2\\ & =(232.0371-228.028-4.001)\cdot (1.66\times {10}^{-27})\cdot (3\times {10}^8{)}^2\\ & =1.195\times {10}^{-12}\mathrm{J}.\end{array}$$

To make this slightly easier to read, we can convert this to electron-volts $\mathrm{eV}$,

$$1.195\times {10}^{-12}\mathrm{J}\cdot 6.242\times {10}^{18}\frac{\mathrm{eV}}{\mathrm{J}}=7.459\times {10}^6\mathrm{eV}=7.459\mathrm{MeV}.$$

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.

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.

$${\phantom{A}}_{\phantom{55}}^{\phantom{137}}{\phantom{A}}_{\phantom{2}55}^{\phantom{2}137}\mathrm{X}={\phantom{A}}_{\phantom{56}}^{\phantom{137}}{\phantom{A}}_{\phantom{2}56}^{\phantom{2}137}\mathrm{Y}+{\phantom{A}}_{\phantom{-1}}^{\phantom{0}}{\phantom{A}}_{\phantom{2}-1}^{\phantom{2}0}\beta$$

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 $9.05\times {10}^{-28}\mathrm{g}$, 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,

$$\begin{array}{r}\mathrm{\Delta }{E}_{\text{disintegration}}\\ & =(m_{\mathrm{X}}-m_{\mathrm{Y}}-m_{\beta })\cdot c^2\\ & =(136.907-136.905-0.0005)\cdot (1.66\times {10}^{-27})\cdot (3\times {10}^8{)}^2\\ & =2.241\times {10}^{-13}\mathrm{J}.\end{array}$$

Again we can convert this to electron-volts $\mathrm{eV}$,

$$2.241\times {10}^{-13}\mathrm{J}\cdot 6.242\times {10}^{18}\frac{\mathrm{eV}}{\mathrm{J}}=1.399\times {10}^6\mathrm{eV}=1.399\mathrm{MeV}.$$

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.

References

1. Fig.3 Alpha decay of Americium (https://commons.wikimedia.org/wiki/File:Alpha-decay-example.svg) Licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/)
2. Fig.4 Beta decay of Caesium (https://commons.wikimedia.org/wiki/File:Beta-decay-example.svg) Licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/)