Despite some fundamental differences between **chemical **and **nuclear reactions**, they both share a **common feature** which you heard so many times – **the equation must be balanced**!

For chemical reactions, the equation is balanced by equalizing the number of given atoms on both sides of the equation. For example:

Ca(AlO_{2})_{2 } + 8HCl → 2AlCl_{3} + CaCl_{2} + 4H_{2}O

So, how do we balance a nuclear equation?

You need to keep in mind that the **sum** **of the mass numbers and atomic numbers are equal** on both sides of the equation.

In other words, the sum of the superscripts and subscripts on both sides of the equation must be equal. For example:

This is the equation for the alpha decay of the uranium-238 isotope to the thorium-234 isotope. The sum of the protons on the right side (90 + 2) is equal to the atomic number of uranium, and the sum of the mass numbers of Th and α (234 + 4) is equal to the atomic mass of the ^{238}_{92}U isotope.

This **principle applies to all the radioactive decay and transformations** such beta decay, positron emission, and electron capture.

You check this article for the details of these processes, but here is a summary for these radioactive processes which we are going to use when balancing different nuclear reactions:

You need to know what each of these terms mean because balancing a nuclear equation often means identifying the daughter nuclide (product) of the reaction. So, remember the following definitions from the atomic structure.

** **

**Proton (**or*p*– a subatomic particle with a positive electric charge of 1+.*p*^{+})**Neutron (**or*n*a neutral subatomic particle slightly heavier than a proton.*n*^{o}) –**Electron (**a subatomic particle with a negative electric charge of 1-.*e*^{–}) –**Positron**– the antiparticle of the electron as it has the same mass as an electron, but the opposite charge

**Atomic number**(**Z**)**–**the number of protons in an atom. Always the same for the given element**Mass number**(**A**) – the sum of the number of neutrons and protons*Every atom has an***equal number of electrons and protons**, so atoms have no net*electrical charge.***Isotopes**– atoms with the same number of protons (atomic number) but different numbers of- Because of a different number of neutrons, isotopes have different mass numbers.
**Nucleons**– a general term for both protons (p) and neutrons (n)**Nuclide**– the nucleus of a given isotope

Before addressing the specific types of nuclear reactions and strategies to balance them, let me mention right from the beginning that **balancing nuclear equations means identifying the unknown particle(s) **in the reaction. This is because most of the time, we do not have coefficients in front of the nuclides since we do not have the same element appearing on both sides of the equation. Instead, when balancing nuclear equations, we need to **identify the correct particle(s) to make sure the sum of the atomic numbers and mass numbers are balanced**.

**Balancing Equations for Alpha ****(***α*) **Decay**

*α*)

**For example**, let’s say the question asks you to identify the daughter nuclide for the alpha decay of ^{224}Ra.

Now, if you don’t know what alpha decay means then you are stuck in this problem. In that case, feel free to go over the main radioactive pathways in this article before proceeding forward.

Remember, alpha (α) decay occurs when an unstable nucleus **loses a particle** composed of **two protons and two neutrons**. These are called alpha particles symbolized as α, ^{2}_{4}α, or ^{2}_{4}He^{2+} which are identical to the helium-4 nuclei.

Therefore, we know that the **mass number **and** the atomic number **of ^{224}Ra are **going to decrease**. We can start by writing the symbol of ^{224}Ra on the left side of the nuclear equation and adding an alpha particle on the right side of the equation. Meanwhile, look up the atomic number of Ra in the periodic table, and write it as well since it will help identify the daughter nuclide:

^{224}_{88}Ra → X + ^{4}_{2}He

To determine the number of protons of the daughter nuclide, we subtract 2 (the two protons of the alpha particle) from 88 (the atomic number of Ra):

X (atomic number) = 88 – 2 = 86

Therefore, the atomic number of the daughter nuclide is 86, and using the periodic table, we determine that it is radon (Rn):

^{224}_{88}Ra → _{86}Rn + ^{4}_{2}He

At this point, we only **need to determine the mass number** of the specific Rn isotope that is formed in this reaction.

^{224}_{88}Ra → ^{?}_{86}Rn + ^{4}_{2}He

This, again, is determined based on the fact that the sum of the mass numbers must be balanced on both sides of the nuclear equation. Therefore, the mass number of Rn is the difference between those of ^{224}_{88}Ra and ^{4}_{2}He:

M (_{86}Rn) = M (^{224}_{88}Ra) – M (^{4}_{2}He)

M (_{86}Rn) = 224 – 4 = 220

Therefore, the balanced equation for the alpha decay of ^{224}_{88}Ra can be written as:

^{224}_{88}Ra → ^{220}_{86}Rn + ^{4}_{2}He

**Balancing Equations for Beta ****(***β*) **Decay**

*β*)

**Beta (**B**) decay **is the emission of an electron from an unstable nucleus. For example, carbon transforms into nitrogen when **one of the neutrons** in the ^{14}C nucleus is **converted into a proton** because losing a negatively charged particle with negligible charge converts a neutral particle into a positively charged particle. Therefore, there is a new atom formed with 7 protons which corresponds to the ^{14}N isotope.

We can write a balanced nuclear equation for this reaction as:

Let’s do a **practice example**: Identify the product X by writing a balanced nuclear equation:

^{137}_{55}Cs → ^{137}_{56}Ba + X

Because the mass number is the same on both sides of the equation, **X must have a mass number of 0**. One the other hand, the **atomic number of the product is 1 more** than that of the reactant which there is an **additional proton formed** during this process. Remember, a proton is formed when a neutron loses an electron, and therefore, **X is a beta particle**. Thus, the balanced equation would be:

^{137}_{55}Cs → ^{137}_{56}Ba + ^{0}_{-1}*β*

**Balancing Equations for Positron Emission**

**Positron Emission **is the ejection of a **positron ( ^{0}_{+1}β) **from a neutron-poor nucleus. Positron is a high-velocity particle emitted from the nucleus and it has the

**same mass as an electron**, but the

**opposite charge**, it is the

**of the electron.**

*antiparticle*We can visualize positron emission as shown below:

**For example**, write the nuclear equation for the position emission of the ^{30}_{15}P isotope:

^{30}_{15}P → X + ^{0}_{+1}β

Positron is a positively charged particle emitted from a proton, and therefore, it **decreases the atomic number by 1** as one proton is converted into a neutron. Therefore, the product is going be an isotope of the element with the atomic number 14 which is Si.

^{30}_{15}P → ^{30}_{14}Si + ^{0}_{+1}β (or ^{0}_{+1}e)

Notice that the mass number does not change as protons and neutrons have almost identical masses.

**Balancing Equations for Electron Capture**

Electron capture occurs when a neutron-poor **nucleus absorbs an own electron from an inner shell atomic orbital**. Like positron emission, the electron capture **decreases the atomic number by 1** as one proton is converted into a neutron.

For example, Ru-92 undergoes electron capture, converting into Tc-92:

Let’s a **practice example** of electron capture as well. Identify the product X in the following equation for the electron capture of the Au-195 isotope:

^{195}_{79}Au + ^{0}_{-1}e→ X

Capturing an electron converts one of the protons to a neutron, and therefore, the atomic number of the daughter nucleus is going to be 78. From the periodic table, we determine that it is platinum:

^{195}_{79}Au + ^{0}_{-1}e→ ^{?}_{78}Pt

At this point, we are only missing the mass number of Pt, and since electron capture does not change it because the proton and neutron have nearly identical masses, we can write the balanced nuclear equation as:

^{195}_{79}Au + ^{0}_{-1}e→ ^{195}_{78}Pt

**Other Nuclear Reactions**

Some nuclear reactions do not occur by themselves, external triggers such as bombarding their nuclei with appropriate particles can facilitate processes like nuclear fission when the parent nuclide is split into other elements.

**For example**, the U-235 nuclide can be split into isotopes of Ba and Kr when bombarded with neutrons:

^{235}_{92}U + ^{1}_{0}n→ ^{140}_{56}Ba + ^{93}_{36}Kr + 3 ^{1}_{0}n + energy

Notice that despite looking more complex than the reactions we discussed earlier, the nuclear equation is balanced as the sum of the atomic numbers and the mass numbers are equal on both sides of the equation. For the sum of the atomic numbers on the left, we have 92 + 0 = 92, and on the left side is 56 +36 + 0 = 92. The sum of the mass numbers on the left is: 235 +1 = 236, and on the right side, it is 140 + 93 + 3 x 1 = 236.

**Practice Problems On Radioactive Decay and Nuclear Reactions**

Use this summary of radioactive processes and how they affect the number of protons and neutrons in the nucleus to solve the **practice problems** on **balancing nuclear reactions** such as alpha decay, beta decay, positron emission, electron capture, nuclear fission, and nuclear fusion.

Remember, out of all the radioactive processes discussed today, it is** only the alpha decay that decreases the number of protons and neutrons** and, as a result, the **atomic mass** of the daughter nuclide also **decreases**.

In **position emission** and** electron capture**, the **number of protons decreases by one** and that of **neutrons increases by one** because one proton is converted into a neutron. Therefore, an isotope of the **element preceding the mother nuclide in the periodic table is formed** which has the same atomic mass.

In **beta decay,** a negatively charged particle (electron) is emitted from a **neutron** which **converts into a proton**, and therefore, the **daughter nuclide** is an isotope of the element that **follows the mother nuclide **in the periodic table.

**Gamma radiation by itself does not change the number of protons or neutrons** unless accompanied by another type of radioactive decay. So, unless there is another type of radioactive decay, gamma radiation does not alter the nuclide, and therefore, the **mass number stays the same** as well.

#### Practice

Complete the following nuclear equations and identify X in each reaction:

The atomic number of X = 41 + (-1) = 40, therefore, the element is Zr. The mass number is 97 + 0 = 97.

When the **atomic numbers are not given**, the first step is to** look them up** in the periodic table. The atomic number of Bi is 83, and that of Pb is 82. So, the atomic number decreased by 1 which indicates either electron capture or positron emission, and because X is on the right side, it means it is emitted from Bi rather than absorbed by it. Therefore, the correct answer is that X is a positron, and we have a **positron emission**.

The sum of atomic numbers on the left is 12 + 1 = 13, and therefore, the atomic number of X = 13 – 2 = 11 which means X is an isotope of Na. The mass number of Na would be (26 + 1) – 4 = 23.

The sum of atomic numbers on the left is 92 + 6 = 98, and therefore, the atomic number of X = 98 – 6 x 0 = 98 which means X is an isotope of Cf. The mass number of Cf would be: (238 + 12) – 6 x 1 = 244.

The sum of atomic numbers on the left is 27 + 1 = 28, and therefore, the atomic number of X = 28 – 27 = 1 which means X is an isotope of hydrogen. The mass number of H would be: (59 + 2) – 60 = 1, and therefore, X is a proton.

When the **atomic numbers are not given**, the first step is to **look them up** in the periodic table. The atomic number of Tc is 43, and that of Mo is 42. So, the **atomic number decreased by 1** which indicates either electron capture or positron emission, and because the X is on the right side, it means it is emitted from Tc rather than absorbed by it. Therefore, the correct answer is that X is a **positron**.

The sum of atomic numbers on the left is 24 + 2 = 26, and therefore, the atomic number of X = 26 – 0 = 26 which means X is an isotope of Fe. The mass number of Fe would be: (53 + 4) – 1 = 56.

When the atomic numbers are not given, the first step is to look them up in the periodic table. The atomic number of Pu is 94, and that of U is 92. So, the **atomic number decreased by 2** which indicates **alpha decay**. Notice that the mass number sum decreased by 4 which confirms that X is ^{4}_{2}He.

The sum of atomic numbers on the left is 8, and because F has an atomic number of 9, there must have been **beta decay** which converts a neutron into a proton.

The sum of atomic numbers on the right is 8 + 1 = 9, and therefore, the atomic number of X = 9 – 2 = 7 which means X is an isotope of N. The mass number of N would be: (17 + 1) – 4 = 14.

The sum of atomic numbers on the right is 11 + 2 = 13, and therefore, the atomic number of X = 13 – 12 = 1 which means X is an isotope of H. The mass number of H would be: (24 + 4) – 26 = 2.

The sum of atomic numbers on the left is 98, and therefore, the atomic number of X = 98 – (56 + 4 x 0) = 42 which means X is an isotope of Mo. The mass number of Mo would be: 252 – (142 + 4 x 1) = 106.

The sum of atomic numbers on the left is 92, and therefore, the atomic number of X = 92 – [7 x 2 + 4 x (-1)] = 82 which means X is an isotope of Pb. The mass number of Pb would be: 235 – (7 x 4 + 4 x 0) = 207.

Identify the missing species in the following nuclear reaction:

Alpha decay decreases the atomic number by 2, and therefore, the daughter nuclide is Ac. The mass number of Ac is 231 – 4 = 227. Beta decay increases the atomic number by one, so Ac converts into _{90}Th which then undergoes another beta decay forming _{91}Pa. Notice that the mass number does not change during beta decay.

Alpha decay decreases the atomic number by 2, and therefore, the daughter nuclide is Pu. The mass number of Pu is 247 – 4 = 243. Beta decay increases the atomic number by one, so Pu converts into _{95}Am which then undergoes alpha decay forming ^{239}_{93}Np as the mass number decreases by 4, and the atomic number decreases by 2.

Going to the right, the first step is beta decay which increases the atomic number of ^{232}_{90}Th by 1 forming ^{232}_{91}Pa. Notice again, that the mass number is not changing during beta decay. Next, ^{232}_{91}Pa undergoes alpha decay decreasing the atomic number by 2, and the mass number by 4, therefore, the daughter nuclide is ^{228}_{89}Ac. To identify the first nuclide, we need to increase the atomic number of ^{232}_{90}Th by 2, and therefore, it must have been _{92}U. The mass number must be 4 more than that of ^{232}_{90}Th, so it is the ^{236}_{92}U isotope.

Going to the right, the first step is beta decay which increases the atomic number of ^{209}_{83}Bi by 1 forming ^{209}_{84}Po. Notice again, that the mass number is not changing during beta decay. Going to the left, ^{209}_{83}Bi is formed by alpha decay, and therefore, we need to increase the atomic number by 2, and therefore, the preceding nuclide must have been _{85}At. The mass number must be 4 more than that of ^{209}_{83}Bi, so it is the ^{213}_{85}At isotope. This nuclide is formed when the first one did an electron capture. Now, electron capture decreases the atomic number by 1 because one proton is converted into a neutron. Therefore, the first nuclide must have an atomic number that is greater by one compared to ^{213}_{85}At, therefore, it is ^{213}_{86}Rn_{.}

The first step is a positron emission which means one proton is converted into a neutron, and therefore, the second element is ^{238}_{91}Pa. Again, the mass number does not change since protons and neutrons have almost identical masses. The next step is alpha decay which increases the atomic number and mass number by 2 and 4 respectively. Therefore, the resulting nuclide is the ^{234}_{89}Ac isotope. The last step is beta decay which leaves he mass number intact while increasing the atomic number by 1. Therefore, the final product is the ^{234}_{90}Th nuclide.