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

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

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Answer 1 · 2014-06-28T18:20:26

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

$α$ $alpha$

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

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

$P u - 238 \to U - 234 +$ $Pu-238 -> U-234 +$ $α$ $alpha$

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

$β$ $beta$

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

$I - 131 \to X e - 131 +$ $I-131 -> Xe-131 +$ $β$ $beta$

Nothing is being conserved in this equation.

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

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