There are three types of radiation:

 Type of Radiation What is it? What can stop it? Alpha (α) a Helium nucleus (2 neutrons, 2 protons) a thin piece of paper Beta (ß) a fast moving electron (₋₁e) a few millimetres of aluminium Gamma (γ) electromagnetic radiation few centimetres of lead

 Type of Radiation Atomic number Mass number Alpha (α) 2 4 Beta (ß) -1 0 Gamma (γ) 0 0

In the exam, you may be asked to identify the correct properties of each type of radiation.

Question can also be based on certain nuclear reactions.

Example: Alpha decay
Consider the following reaction:

²⁴⁰Pu → ˣU + ⁴He
(atomic number of Pu = 94, U = y, He = 2)
Find the value of ‘x’ and ‘y’.

Solution:
We know ‘Pu’ decays by alpha decay since a He nucleus is emitted.

The mass numbers must be equal on the LHS and RHS.
Mass number on LHS = 240
Mass number on RHS = x + 4
Mass number on LHS = Mass numbers on RHS
240 = x + 4
x = 240 – 4
x = 236

The atomic numbers must be equal on the LHS and RHS.
Atomic number on LHS = 94
Atomic numbers on RHS = y + 2
Atomic number on LHS = Atomic numbers on RHS
94 = y + 2
y = 94 – 2
y = 92

Example: Beta decay
Consider the following reaction:

²³⁴Th → ˣPa + ₋₁ß
[atomic number of (Th = 90), (Pa = y), (ß = 0)]
Find the value of ‘x’ and ‘y’.

Solution:
We know ‘Th’ decays by beta decay since a beta particle (electron) is emitted.

The mass numbers must be equal on the LHS and RHS.
Mass number on LHS = 234
Mass number on RHS = x + 0
Mass number on LHS = Mass numbers on RHS
234 = x + 0
x = 234 – 0
x = 234

The atomic numbers must be equal on the LHS and RHS.
Atomic number on LHS = 90
Atomic numbers on RHS = y + (-1) = (y – 1)
Atomic number on LHS = Atomic numbers on RHS
90 = y – 1
y = 90 + 1
y = 91

It is very unlikely that you will be given a reaction with gamma decay as a gamma particle has 0 atomic number and 0 mass number.

You may also be asked to identify the type of decay that occurs in a reaction.

If the mass number decreases by 4 and the atomic number decreases by 2, the decay is Alpha (α)

If the mass number remains the same on the LHS and RHS but the atomic number increases by 1, the decay is Beta (ß)

If there is no change in the mass number and the atomic number, the decay is likely to be Gamma (γ)