# When writing a balanced nuclear equation, what must be conserved?

Jun 28, 2014

Nothing needs to really be conserved in a nuclear equation: let me just illustrate one $\alpha$ and one $\beta$ equation to emphasise this.

$\alpha$

$P u$-238 (Plutonium, 238) decays by $\alpha$ emission to form an atom, which atom is this?

In an $\alpha$ decay equation, we lose an atomic number of $2$ an a mass number of $4$ - this is the equivalent of a Helium ($H e$) atom. So,

$P u - 238 \to U - 234 +$$\alpha$

Uranium is formed because it is element number $92$ - Plutonium is element number $94$, so if we take two away from $94$ we get $92$ which is the atomic number of $U$. There is nothing conserved in this reaction.

$\beta$

When writing a $\beta$ equation, remember that in the nucleus, a neutron ($n$) decays into a proton (${p}^{+}$) and a high energy electron which is known as the beta ($\beta$) particle. Because a new proton has formed, the atomic number of the original atom will increase by $1$.

$I - 131 \to X e - 131 +$$\beta$

Nothing is being conserved in this equation.