# Nuclear Chemistry

Important Relationships

$1~\mathrm{amu} = 1.66\times 10^{-27}~\mathrm{kg}$ $1~\mathrm{eV} = 1.602\times 10^{-22}~\mathrm{kJ}$ $1~\mathrm{MeV} = 1\times 10^{6}~\mathrm{eV}$

$1~\mathrm{J} \equiv 1~\mathrm{kg~m^2~s^{-2}}$

## Atomic structure notation

Much of chemistry focuses on chemical reactions which is governed by the behavior of electrons. Nuclear chemistry focuses on nuclear reactions which involve a change in an atom’s nucleus (referred to as a nuclide). Recall the notation used to describe isotopes which gives the

• atomic number (number of protons)
• mass number (number of protons + neutrons)

The same notation is used to describe a nuclide. For example, the notation for the most common isotope of helium is

$^{4}_{2}\mathrm{He}$

where “2” is the number of protons (directly from the periodic table) and the mass number is “4” indicating this isotope of helium has 2 neutrons. The notation for carbon-14 would be

$^{14}_{\phantom{1}6}\mathrm{C}$ indicating that carbon-14 has 8 neutrons.

Protons and neutrons are collectively called nucleons. These subatomic particles are held together by a the very powerful strong nuclear force that is strong over very short distances (about 1 fm) and rapidly decays in strength beyond around 2.5 fm. There is an incredible amount of energy here that holds the nucleus together and is called the nuclear binding energy.

## Mass Defect and Energy

Consider the masses of subatomic particles:

• proton = 1.00727 amu
• neutron = 1.00866 amu
• electron = 0.00055

If we wanted to calculate the mass of a helium-4 atom (2 protons, 2 neutrons, and 2 electrons), one might perform the following calculation:

$m_{\mathrm{He~atom}} = (2 \times 1.00727~\mathrm{amu}) + \left ( 2\times 1.00866~\mathrm{amu} \right ) + \left ( 2 \times 0.00055~\mathrm{amu} \right ) = 4.03296~\mathrm{amu}$

However, careful mass spectrometry measurements indicate that the mass of a helium-4 atom is 4.002602 amu (see NIST), one of the most precisely known masses (see Table 3 in 2003 IUPAC, Pure and Applied Chemistry 75, 683–800). We quickly conclude that the measured mass of a helium-4 atom is lighter than the sum of its parts! This difference between these two masses is called the mass defectm). If we think about this in terms of a typical reaction, we have

$2~^1_1\mathrm{p} + 2~^1_0\mathrm{n} + 2~^{\phantom{-}0}_{-1}e ~\longrightarrow ~^{4}_{2}\mathrm{He}$

to give the following mass defect

\begin{align*} \Delta m &= \mathrm{mass~of~products} - \mathrm{mass~of~reactants}\\ &= 4.002602~\mathrm{amu} - 4.03296~\mathrm{amu}\\ &= -0.030358~\mathrm{amu} \end{align*}

There is nothing incorrect with the measured values of the subatomic particles or the mass of the atom. So what is going on? Albert Einstein demonstrated that the mass of an object is directly proportional to its energy and is described using the mass-energy equivalence equation.

$E = mc^2 \quad or \quad \Delta E=\Delta mc^2$ where

• m is the mass defect
• c is the (two-way) speed of light in a vacuum (approximately 3.00×108 m s–1)
• E is the energy equivalent to the mass defect

In short, any energy change is accompanied by a change in mass and vice-versa. This applies to everything from nuclear reactions to celestial bodies. Mass that is lost is converted to energy and energy that is lost is converted to mass.

We can imagine the formation of the He-4 nucleus to be analogous to that of forming a bond between two atoms. As the subatomic particles come together, the energy of the nucleus drops as it becomes more stable (ΔE is negative). This energy change is associated with the mass defect (the mass that was lost; Δm < 0). The energy given off is equivalent to the mass lost.

### Mass defect in chemical reactions

Mass changes can be applied to chemical reactions, though the lost mass is too small to have any non-trivial effect on the properties of the reaction. Consider a chemical reaction such as the combustion of graphite to produce carbon dioxide (Exercise adopted from LibreTexts).

$\mathrm{C(graphite, s)} + \frac{1}{2}\mathrm{O_2}(g) \longrightarrow \mathrm{CO_2}(g) \qquad \Delta H^{\circ} = -393.5~\mathrm{kJ~mol^{-1}}$ We can determine the mass change in this reaction by employing our mass-energy equivalence equation.

\begin{align*} \Delta E &= \Delta mc^2 \\[1.25ex] \Delta m &= \dfrac{\Delta E}{c^2} \\[1.25ex] &= \dfrac{-393.5~\mathrm{kJ~mol^{-1}}}{\left ( 3.00\times 10^{8}~\mathrm{m~s^{-1}} \right ) ^2}\\[2ex] &= \dfrac{-3.935\times 10^5~\mathrm{J~mol^{-1}}}{\left ( 3.00\times 10^{8}~\mathrm{m~s^{-1}} \right ) ^2}\\[2ex] &= \dfrac{-3.935\times 10^5~\mathrm{kg~m^2~s^{-2}~mol^{-1}}}{\left ( 3.00\times 10^{8}~\mathrm{m~s^{-1}} \right ) ^2}\\[2ex] &= -4.37\times 10^{-12}~\mathrm{kg~mol^{-1}}\\ &= -4.37\times 10^{-9}~\mathrm{g~mol^{-1}} \end{align*}

We see that for every mole (12.011 g) of graphite consumed, about a billionth of a gram of mass is lost. This loss of mass for chemical reactions is too small to measure.

### Mass defect in nuclear reactions

Let’s look at our He-4 forming nuclear reaction again. Fusing the subatomic particles together causes a loss in mass. This mass is converted into energy.

$2~^1_1\mathrm{p} + 2~^1_0\mathrm{n} + 2~^{\phantom{-}0}_{-1}e ~\longrightarrow ~^{4}_{2}\mathrm{He} \qquad \Delta E < 0 \propto \Delta m < 0$

We can determine the energy of this nuclear reaction per mole of He created using the known mass defect.

First, convert the mass defect of one helium-4 nucleus into the mass defect of one mol of helium-4 nuclei.

$\left( \dfrac{-0.030358~\mathrm{amu}}{\mathrm{atom}} \right ) \left ( \dfrac{1.66\times 10^{-27}~\mathrm{kg}}{\mathrm{amu}} \right) \left ( \dfrac{6.022\times 10^{23}~\mathrm{atom}}{\mathrm{mol}} \right ) = -3.035\times 10^{-5}~\mathrm{kg~\mathrm{mol^{-1}}}$

\begin{align*} \Delta E &= \Delta mc^2 \\[1.25ex] &= (-3.035\times 10^{-5}~\mathrm{kg~mol^{-1}})( 3.00\times 10^{8}~\mathrm{m~s^{-1}} )^2\\ &= -2.73\times 10^{12}~\mathrm{kg~m^2~s^{-2}~mol^{-1}}\\ &= -2.73\times 10^{12}~\mathrm{J~mol^{-1}}\\ &= -2.73\times 10^{9}~\mathrm{kJ~mol^{-1}} \end{align*}

When one mole (or approximately 4 grams) of helium-4 is created from its constituent subatomic particles, a staggering 2.73×109 kJ of energy is released. Compare this to Castle Bravo, the largest thermonuclear weapon ever detonated by the US, which had an energy yield of about 15 Megatons of TNT (or about 6×1013 kJ).

Castle Bravo Blast

Comparison of some nuclear bomb yields

### Nuclear Stability

Due to the very small energies associated with nuclear reactions on a per atom basis, energy is usually expressed in units of megaelectronvolts (MeV).

For our reaction above, the mass defect for one helium-4 nucleus is equivalent to –4.535×10–15 kJ. Convert this to MeV.

$\Delta E = -4.535\times 10^{-15}~\mathrm{kJ} \left ( \dfrac{1~\mathrm{eV}}{1.602\times 10^{-22}~\mathrm{kJ}} \right ) \left ( \dfrac{1~\mathrm{MeV}}{10^6~\mathrm{eV}} \right ) = -28.3~\mathrm{MeV}$

This is equivalent to the nuclear binding energy of the nucleus of a helium-4 atom (i.e. the energy required to split the He-4 nucleus into its constituent subatomic particles is 28.3 MeV).

We can break this down even more into MeV per nucleon. The helium-4 atom has 4 nucleons.

$28.3~\mathrm{MeV}/\mathrm{4~nucleons} = 7.1~\mathrm{MeV~nucleon}$

This MeV/nucleon notation is a measure of the stability of an atomic nucleus. The higher the binding energy (per nucleon), the more stable the nucleus.

### Fusion

We have seen with our helium-4 example that combining subatomic particles to create an atomic nucleus creates (releases) energy due to a loss of mass. Fusion is the process by which nuclei of atoms are fused together to create a heavier nucleus. This can occur when two very fast moving nuclei slam into each other with sufficient energy. These nuclear processes generate a tremendous amount of energy on the macroscale (see Hydrogen Bomb). Light atoms tend to undergo fusion to increase the stability of its nucleus (and release energy in the process).

Fusion of deuterium and tritium to create helium-4 and a neutron (Image from Wikipedia).

### Fission

Fission is the opposite of fusion. It is the nuclear process by which an atomic nucleus is split into smaller, lighter nuclei. Heavy atoms readily undergo fission as their nuclei are rather unstable (higher in energy than they want to be). Energy is released when this process occurs (see Atomic Bomb)

(Image from openStax)

### Example: Fission

Uranium-238 is unstable and radioactive. It undergoes the following decay process (called an α decay).

$^{238}_{\phantom{0}92}\mathrm{U} ~\longrightarrow ~^{234}_{\phantom{0}92}\mathrm{Th} +^{4}_{2}\mathrm{He}$

The mass defect (Δm) is -0.004584 amu which means that mass is lost and converted to energy. Determine the energy (in kJ and eV) per atom and the energy (in kJ) per mole of uranium-238 decay.

Energy released per uranium-238 atom

Convert the mass defect from amu to kg.

$-0.004584~\mathrm{amu} \left ( \dfrac{1.66\times 10^{-27}~\mathrm{kg}}{\mathrm{amu}} \right ) = -7.6094\times 10^{-30}~\mathrm{kg}$ Convert the mass (in kg) into kJ and MeV.

\begin{align*} \Delta E &= \Delta m c^2\\ \Delta E &= (-7.6094\times 10^{-30}~\mathrm{kg})(3.00\times 10^{8}~\mathrm{m~s^{-1}})^2\\ &= -6.850\times 10^{-13}~\mathrm{kg~m^2~s^{-2}} \\ &= -6.850\times 10^{-13}~\mathrm{J}\\ &= \textcolor{red}{-6.850\times 10^{-16}~\mathrm{kJ}} \left ( \dfrac{\mathrm{eV}}{1.602\times 10^{-22}~\mathrm{kJ}} \right ) \left ( \dfrac{\mathrm{MeV}}{10^6~\mathrm{eV}} \right ) \\ &= \textcolor{red}{-4.276~\mathrm{MeV}} \end{align*}

Energy released per mole of uranium-238

Now we know how much energy is released per atom. Use Avogadro’s number to determine the energy released (in kJ) per mole of uranium-238.

$\dfrac{-6.850\times 10^{-16}~\mathrm{kJ}}{\mathrm{atom}} \left ( \dfrac{6.022\times 10^{23}~\mathrm{atoms}}{\mathrm{mol}} \right ) = -4.125\times 10^{8}~\mathrm{kJ~mol^{-1}}$

We see that when one U-238 atom undergoes an alpha decay, a very small amount of energy (6.850×10–13 kJ) is released. When one mole of U-238 atoms undergoes an alpha decay (approximately 238 g), a staggering amount of energy (4.125×108 kJ) is released.

## Stability of Nuclei

We can see from the image below (adapted from LibreTexts.org) that light elements undergo fission to produce energy (you see their binding energies per nucleon increase) and heavy elements undergo fusion to produce energy (their binding energies per nucleon increase).

Iron-56 is one of the most stable nuclei as it has one of the largest nuclear binding energies (its nucleus is very tightly bound). Its binding energy is about 8.8 MeV/nucleon.

## Practice

See Examples 21.2 and 21.3 in the textbook.

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