Question 14.2: We have seen the nuclear equation for the beta decay involve......
We have seen the nuclear equation for the beta decay involved in radiocarbon dating. Now complete the equations for each of the following \beta^{-} decay reactions, using _{-1}^0\beta to represent the beta particle.
\begin{aligned}& { }_{90}^{234} Th \rightarrow{ }_{91}^{234} Pa + ? \\& { }_{91}^{234} Pa \rightarrow ?+{ }_{-1}^0 \beta+\bar{\nu}\end{aligned}
Strategy We can use both the mass numbers and the charges given to determine the missing particles in the equations, much as in the previous example for alpha decay.
Consider each equation separately, starting with the thorium-234 decay:
{ }_{90}^{234} Th \rightarrow{ }_{91}^{234} Pa +\text { ? }
The fact that both Th and Pa have mass numbers of 234 indicates that the other particle in the equation must have a mass number of 0 . This is consistent with beta decay. Balancing the charges on the reactant and product sides requires that the unknown has a charge of 1 – (so that 91-1=90). Because this is beta decay, an antineutrino is also emitted. The equation is
{ }_{90}^{234} Th \rightarrow{ }_{91}^{234} Pa +{ }_{-1}^0 \beta+\bar{\nu}
Next turn to the decay of protactinium-234:
{ }_{91}^{234} Pa \rightarrow ?+{ }_{-1}^0 \beta+\bar{\nu}
Because the beta particle given as a product has no mass number, we know that the unknown particle has a mass of 234 . The beta particle has a charge of 1-, so the unknown particle must have an atomic number of 92 (so 92-1=91). The missing particle is { }^{234} U, so the equation is
{ }_{91}^{234} Pa \rightarrow{ }_{92}^{234} U +{ }_{-1}^0 \beta+\bar{\nu}
Check Your Understanding The only remaining way to look at this type of question results when the unknown is the reactant. Suppose you observe beta decay and then determine that the product nuclide is { }^{218} At. What was the reactant nuclide?