##### Example \(\PageIndex{1}\)

Write a balanced nuclear equation to describe each reaction.

- the beta decay of \(^{35}_{16}\textrm{S}\)
- the decay of \(^{201}_{80}\textrm{Hg}\) by electron capture
- the decay of \(^{30}_{15}\textrm{P}\) by positron emission

**Given:** radioactive nuclide and mode of decay

**Asked for:** balanced nuclear equation

**Strategy:**

**A** Identify the reactants and the products from the information given.

**B** Use the values of *A* and *Z* to identify any missing components needed to balance the equation.

**Solution**

a.

**A** We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\):

\[^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta\]

**B** Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of *A* = 35 − 0 = 35 and an atomic number of *Z* = 16 − (−1) = 17. The element with *Z* = 17 is chlorine, so the balanced nuclear equation is as follows:

\[^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta\]

b.

**A** We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{e}\). The reaction is as follows:

\(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X}\)

**B** Both protons and neutrons are conserved, so the mass number of the product must be *A* = 201 + 0 = 201, and the atomic number of the product must be *Z* = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus

\(^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au}\)

c.

**A** As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore

\(^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta\)

**B** The mass number of the second product is *A* = 30 − 0 = 30, and its atomic number is *Z *= 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows:

\(^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta\)