In nuclear chemistry and nuclear physics, a nuclear reaction is the process in which an atom and a subatomic particle or two nuclei like a high energy electron, neutron or proton form on the atom’s exterior and collide. This produces one nuclide or more different from the one that started the process. Thus, nuclear reactions need to cause a transformation of a minimum of one nuclide to another, which this Physics course explains even further.

**List of Nuclear Equations**

**List of Nuclear Equations**

Elements change from one to another when they go through radioactive decay. Nuclear decay occurs by positron emissions or when losing high energy beta or alpha particles. Transmutation is the name of this process. To track any change that occurs during transmutation, nuclear equations are written. When writing a nuclear equation, it is important to conserve mass and charge, which you might want to learn more about by reading this Physics article.

Here is a list of nuclear equations you might come across when studying the theory of particle physics and nuclear physics:

- Nuclear radius: r= r
_{0 }A^{1/3 }so that nuclear surface is ∝ A^{2/3}and nuclear volume is ∝A - Mass number: A= Z+ N
- Equivalent dose: H=DQ
- Radioisotope time constant, which is an atom’s mean lifetime before decay: r – 1/
*λ* - Number of half-lives: n = t/T
_{1/2} - Radioisotope half-life: t àt + T ½
- N à N/2

- Decay constant: λ= A/N
- Decay rate: A= dN/dt
- Number of atoms: N
_{0}= N + N_{D }where N is the number of atoms that remain at the time (t), N_{D }is the number of decayed atoms at the time (t) and N_{0 }is the initial atom number at the time (zero).

**Rules for Writing Nuclear Equations **

As mentioned earlier, on each side of the equation both the charges and the masses need to be equal. The atomic number is the nuclear charge and can be used for identifying new elements in this form. In general, the format for nuclear equations is:

Where:

- Original element = X
- New element = Y
- Radioactive emission = x
- Mass number = A or a
- Atomic number = Z or z

When writing nuclear reaction equations, a chemical symbol identifies every element involved. To this symbol, there are 2 numbers attached. The mass number is the upper right number, which is also known as number A. This describes and atom’s atomic weight and identifies how many protons and neutrons the nucleus has. The atomic number is found at the lower left side, and is also known as number Z. This number determines the type of atom and the number of protons contained in the nucleus.

For example, the Uranium 238 symbol is

238

U

92

The symbol for Uranium above tells you that Uranium has an atomic number of 92 and a mass number of 238.

**Nuclear Equation Notations**

Equations for nuclear reactions can be shown in the same way as chemical equations, for which on either side invariant mass needs to balance out, which you can learn more about in this Chemistry course. Also, there are a few conservation laws that particle transformations need to follow, such as the baryon number and the conservation of charge. The total atomic mass number is the baryon number. Here is an example of these notations:

^{6}_{3} Li + ^{2}_{1 }H à ^{4}_{2} He + ?

In order for the above equation for mass number, charge and mass to be balanced, the second nucleus on the right needs to have a mass number 4 and an atomic number 2. Thus, it is helium-4. The completed nuclear equation is then:

^{ 6}_{3} Li + ^{2}_{1 }H à ^{4}_{2} He + ^{4}_{2} He

Which can also be written like this:

^{6}_{3} Li + ^{2}_{1 }H à 2 ^{4}_{2} He

Rather than using full equations like this one here, may times a compact notation is used for describing nuclear reactions. In this shorthand, common light particles are frequently abbreviated, such as:

- Helium -4 or alpha particles represented by α
- Deuteron represented by d
- Neutron represented by n
- Proton represented by p

Knowing this, the above reaction would then be written like this:

Li-6 (d, α) α

**Baryons**

A baryon is a subatomic heavy particle that quarks being bound by gluons create. It is a hadron that contains 3 quarks. Baryons are fermions as they have half-odd integral spins. This category includes the atomic nucleus’ common neutron and proton.

**Nuclear Equation for Beta Decay**

Neutrons in beta decay change into a proton plus 1 electron. The proton remains in the nucleus and the electron leaves the atom with a beta particle high in energy. When it emits a beta particle, the nucleus has one less neutron and one more proton. This means that the atomic number is increased by one and the atomic mass number remains the same.

**Nuclear Equation for Alpha Decay**

A type of radioactive decay, alpha decay happens when an atomic nucleus emits alpha particles and thus ‘decays’ or transforms into atoms with an atomic number of two or less and a mass number of four or less.

**Nuclear Equation for Positron Emission**

A positron emission is a radioactive decay type where a proton within a radio-nuclide nucleus converts into neutrons as it releases electron neutrinos and a positrons.

### Nuclear Equation for Gamma Decay

Gamma decay occurs when there is a nuclei conservative rule violated by some form of the other modes of decay. There is an ‘odd’ nuclear parity if:

Y(x) = -y(-x), which conserves nuclear parity. When this happens, some of the nuclei is unable to go through the decaying process and instead, they go through what is referred to as ‘gamma decay.’ Aside from conservation of parity, nuclear reactions need to use lower nucleus energy and conserve angular momentum, charge and energy.

There is said to be an ‘even’ nuclear parity if the ‘y function’ has the property: y(x) = y(-x)

**Balancing Nuclear Equations**

In order for a nuclear equation to be balanced, there needs to be equal atomic numbers and mass numbers on either side of the arrow. Many times, nuclear reactions are described by compact notations, such as in the form A + b = c + D, which is equivalent to A (b,c) D.

Often, nuclear equation problems will have one missing particle. To find the particle missing, figure out what the atomic number and mass number is needed to balance the equation and provide this.

Here are examples of balanced nuclear equations:

^{237}_{93 }Np produces an alpha particle: ^{237}_{93 }Np à ^{4}_{2 }He + ^{233}_{91} Pa

^{214}_{83} Bi decays by beta particle emission: ^{214}_{83} Bi à ^{0}_{-1 } e + ^{11}_{5} B

^{11}_{6}C decays by positron emission: ^{11}_{6}C à ^{0}_{1 }e + ^{11}_{5 }B

**Nuclear Equations Example #1**

^{214}_{84}Po + 2 ^{4}_{2 }He + 2 ^{0}_{-1 }e à ?

Another way you can write this out is, ‘What is given by two electrons plus two alpha particles plus Polonium 214?’ For this nuclear equation’s solution, all you need to do is to add the mass numbers Polonium 214 plus helium (two alpha particles or 2 x4) plus the electrons’ zero resulting in the mass number of 222. For more practice problems such as these you might want to look into this Chemistry course.

To get the atomic number, take Polonium 84 and add four for helium (2 x 2) then minus 2 (2 x -1 for two beta emissions or electrons) which results in 86, Radon’s atomic number. Thus the nuclear equation should read:

^{214}_{84}Po + 2 ^{4}_{2 }He + 2 ^{0}_{-1 }e à ^{222}_{86} Rn

**Nuclear Equations Example#2**

^{235}_{92} U à ^{231 }_{90} Th + ?

Another way of writing this out would be, ‘What plus Thorium 231 is given by Uranium 235?’ For the solution, you will need to find the difference between the atomic numbers and the atomic masses in the product and the reactant.

^{235}_{92} U à ^{231 }_{90} Th + ^{4 }_{2} He

As you can see, this results in an atomic number difference of 2 and an atomic mass difference of 4, which fits an alpha particle’s description. Here is the answer, which you might want to study even further by taking this chemistry course: