When a given material is radioactive, the tendency is for it to eliminate alpha, beta and gamma radiation . These radiations are eliminated from the nucleus of the atom due to the nuclear instability of the material’s atoms.

Knowing the radioactive materials a little, we can calculate, for example, the number of alpha and beta particles that will be eliminated from the nucleus of an atom. For this, it is important to know the compositions of each type of radiation:

- Alpha radiation:composed of two protons (atomic number 2) and two neutrons, resulting in mass number 4, thus:
_{2}α^{4} - Beta radiation:composed of an electron, resulting in atomic number -1 and mass number 0, thus:
_{-1}β^{0}

Knowing the particles, we realize that: when an atom eliminates alpha radiation (Soddy’s first law), it forms a new element whose atomic number will be two units smaller and the mass number will be four units smaller. By eliminating beta radiation (Soddy’s second law), the atom will form a new element whose atomic number will have one more unit and its mass will remain the same.

♦ First law : _{Z} X ^{A} → _{2} α ^{4} + _{Z-2} Y ^{A-4}

♦ Second law : _{Z} X ^{A} → _{-1} β ^{0} + _{Z + 1} Y ^{A}

It is worth remembering that the elimination of alpha and beta particles is simultaneous and always a new element will be originated. If this originating element is radioactive, radiation elimination will continue until a stable atom is formed.

With all this information given, we can now calculate the number of alpha and beta particles that have been eliminated by radioactive material until a stable atom has been formed.

For this, we use the following equation:

_{Z} X ^{A} → c _{2} α ^{4} + d _{-1} β ^{0} + _{b} Y ^{a}

Z = Atomic number of the initial radioactive material;

A = Initial mass number of the initial radioactive material;

c = Number of alpha particles eliminated;

d = Number of beta particles eliminated;

a = Mass number of the formed stable element;

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b = Atomic number of the formed stable element.

Since the **sum of the mass numbers before and after the arrow are equal **, we have to:

A = c.4 + d.0 + a

**A = 4c + a**

(knowing A and a, we can determine the number of alpha particles eliminated)

Since the **sum of the atomic numbers before and after the arrow are equal **, we have to:

Z = c.2 + d. (- 1) + b

**Z = 2c – d + b**

(knowing Z, c and b, we can determine the number of beta particles eliminated)

Here’s an **example **:

*Determine the number of alpha and beta particles that were eliminated by a radio atom ( *_{86}* Rn *^{226}* ) so that it was transformed into an *_{84}* X *^{210}* atom .*

Exercise data: the initial radioactive atom is Rn and the formed one is X, like this:

Z = 86

A = 226

c =?

d =?

a = 210

b = 84

**Initially we determine the number of alpha particles **:

A = 4c + a

226 = 4c + 210

4c = 226 -210

4c = 16

c = __16__

4

c = 4 (alpha particles)

**Next, we calculate the number of beta particles **:

Z = 2c – d + b

86 = 2.4 – d + 84

86 – 84 – 8 = – d. (- 1 to eliminate the negative of d)

d = 6 (beta particles)