# Writing and Balancing Nuclear Equations Nuclides change to form one or more new nuclides in a nuclear reaction. In a balanced nuclear equation, proton and neutron numbers are balanced.
In a nuclear reaction, one or more nuclides change to form one or more new nuclides. Radioactive decay is an example of a nuclear reaction. Balancing the equation for a nuclear reaction is different from balancing a chemical equation. In balancing chemical equations, the number of atoms on each side of the equation and the electric charge must be balanced. Balancing nuclear equations involves balancing proton and neutron numbers.
Step-By-Step Example
Writing the Nuclear Equation for a Positron Emission Reaction
Oxygen-15 (${}_{\;\;8}^{15}\rm {O}$) undergoes positron emission. Write the nuclear equation for this reaction.
Step 1
Oxygen has an atomic number of 8. Oxygen-15 indicates the mass number of the oxygen in this example is 15. Therefore, oxygen can be written as ${}_{\;\;8}^{15}\rm {O}$. Since the oxygen emits a particle, it will be alone on the left side of the equation:
${}_{\;\,8}^{15}\rm {O}\rightarrow$
Step 2
A positron is an antiparticle, shown as ${}_{+1}^{\;\,\,0}\rm{e}$. The nucleus changes after positron emission. Add these to the equation:
${}_{\;\,8}^{15}{\rm O\rightarrow{}}_{+1}^{\;\,\,0}\rm{e}+{}{}_Z^A\rm {X}$
Step 3
Positron emission changes a proton to the neutron. In effect, it reduces the proton number by one but does not affect the mass number. Note the 0 and the +1 on the symbol for the positron. The 0 on the top indicates that the mass number does not change. The +1 on the bottom indicates that the atomic number changes. The positron's symbol is written as ${}_{+1}^{\;\,\,0}\rm{e}$ because the numbers 0 and +1 can be used to balance the nuclear reaction.
Step 4
To determine A in ${}_Z^A\rm {X}$, the following equation can be written:
$15=0+A$
So $A=15$.
Step 5
To determine Z in ${}_Z^A\rm {X}$, the following equation can be written:
$8=1+Z$
So $Z=7$.
Solution
The X in ${}_Z^A\rm {X}$ is the symbol for the element with atomic number 7. From the periodic table, the element with atomic number 7 is nitrogen (N). Therefore the final reaction is:
${}_{\,\,8}^{15}\rm{O}\rightarrow{}_{+1}^{\;\,\,0}\rm{e}+{}_{\,\,7}^{15}\rm{N}$
Note that both proton and neutron numbers balance on both sides of the equation.
Step-By-Step Example
Balancing a Nuclear Equation for an Alpha Decay Reaction
Balance the nuclear equation for the alpha decay of radon-222 (${}_{\;\,86}^{222}\rm{Rn}$).
${}_{\;\;86}^{222}{\rm{Rn}}\rightarrow{}_Z^A\rm {X}+{}_2^4{\rm{He}}$
Step 1
A can be found using an equation for the mass number.
\begin{aligned}222&=A+4\\A&=218\end{aligned}
Step 2
Z can be found using an equation for the atomic number.
\begin{aligned}86&=Z+2\\Z&=84\end{aligned}
Step 3
X is the symbol for the element with the atomic number, or proton number, of 84. According to the periodic table, the element with atomic number 84 is polonium (Po).
Solution
Therefore, the balanced reaction is
${}_{\;\;86}^{222}{\rm{Rn}}\rightarrow{}_{\;\;84}^{218}{\rm{Po}}+{}_2^4{\rm{He}}$