ChemStd 11d:

Students know the three most common forms of radioactive decay (alpha, beta, and gamma) and know how the nucleus changes in each type of decay


Alpha: ( α) is a type of radioactive decay in which an atomic nucleus emits an alpha particle, and thereby transforms (or 'decays') into an atom with a mass number 4 less and atomic number 2 less. An alpha particle is the same as a helium-4 nucleus, and both mass number and atomic number are the same.
Alpha 's mass is 4.0 units.
Their charge is +2.
Basically, alpha decay affects the nucleus by a loss of 4 in the mass (reducing by 4) and a loss of 2 in the atomic number (reduces by 2)

For example:

23892U	→	23490Th	+	42He2+|[1]

Beta:a type of radioactive decay in which a beta particle (an electron or a positron) is emitted. In the case of electron emission, it is referred to as beta minus (β^⁻), while in the case of a positron emission as beta plus (β⁺)

Beta's mass is .0005 units.
Their charge is -1.
Basically, this type of decay affects the nucleus by having an increase in the atomic number, of one(+1). Their mass does not change.

Gamma: Gamma decay is one of the three radioactive decay modes available to atomic nuclei, along with alpha and beta decay. In alpha and beta decay, the atomic number of the nucleusenergy state decays to a lower-energy state by emitting a high-energy photon. Following an alpha or beta decay, the number of protons and neutrons in the nucleus has changed, and the resulting nucleus may not be in its lowest energy state (called the ground state). As a result, the nucleus will decay to the ground state by emitting one or more gamma-ray photons.
changes, but in gamma decay the atomic number does not change. In gamma decay, a nucleus in an excited Gamma has no mass.
Their charge is neutral.
Basically, in gamma decay, the atomic number does not change.



RADIOACTIVE DECAY DEMONSTRATION

Steps to solve this nuclear equation:
²⁷
₁₃Al + ⁴₂He ---> ³⁰₁₅P + ?
1)The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products .

Using this rule, you can figure out any missing particles in the equation.

2)

To find the answer, you must add the mass and atomic numbers of the reactants. So, 27 + 4 = 31 , and 13 + 2 = 15 . This means that the products must have a total mass number of 31, and a total atomic number of 15. Subtract the known products from the sums of the reactants. So, 31 - 30 = 1 , and 15 - 15 = 0 . So, the missing particle would be ¹₀N .

Still having trouble? Here are two useful sites!

Here's a link that taught me how to balance nuclear equations!

Print this out, and practice practice practice! It's a great strategy to build up those balancing nuclear equation skills :]