If \({\rm{1}}{\rm{.000 g}}\) of \({}_{{\rm{88}}}^{{\rm{226}}}{\rm{Ra}}\) produces \({\rm{0}}{\rm{.0001 mL}}\) of the gas \({}_{{\rm{88}}}^{{\rm{222}}}{\rm{Rn}}\) at \({\rm{STP}}\) (standard temperature and pressure) in \({\rm{24 h}}\), what is the half-life of \({}^{{\rm{226}}}{\rm{Ra}}\) in years?

There is a simple formula to calculate the ratio of the amount of the substance –

\(\left( {\frac{{\rm{n}}}{{{{\rm{n}}_{\rm{1}}}}}} \right){\rm{ = }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\frac{{\rm{t}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}}}\)

Where\(\frac{{\rm{n}}}{{{{\rm{n}}_{\rm{1}}}}}\)is the ratio of the amount of the substance left to the amount of the substance initially present,\({{\rm{t}}_{{\rm{1/2}}}}\)is the half-life of the substance, and\({\rm{t}}\)is the time after which the ratio is determined.

But first, get the number of moles:

\({{\rm{n}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{volume}}}}{{{\rm{molar volume}}}}\)

The volume of\({\rm{1}}{\rm{.0\;mole}}\)of the gas is\({\rm{22400\;mL}}\).

\(\begin{array}{l}{{\rm{n}}_{\rm{2}}}{\rm{ = 0}}{\rm{.00001\;mL}} \cdot \frac{{{\rm{1\;mol}}}}{{{\rm{22400\;mL}}}}\\{{\rm{n}}_{\rm{2}}}{\rm{ = 4}}{\rm{.46 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;mol}}\end{array}\)

The molar mass of\(Ra\)is\({\rm{226 g mo}}{{\rm{l}}^{{\rm{ - 1}}}}\). From that, we can get\({{\rm{n}}_{\rm{1}}}\).

\(\begin{array}{l}{{\rm{n}}_{\rm{1}}}{\rm{ = }}\frac{{\rm{m}}}{{\rm{M}}}\\{{\rm{n}}_{\rm{1}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.000\;g}}}}{{{\rm{226gmo}}{{\rm{l}}^{{\rm{ - 1}}}}}}\\{{\rm{n}}_{\rm{1}}}{\rm{ = 4}}{\rm{.42}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;mol}}\end{array}\)

\(\begin{array}{l}{\rm{n = }}{{\rm{n}}_{\rm{1}}}{\rm{ - }}{{\rm{n}}_{\rm{2}}}\\{\rm{n = 4}}{\rm{.42}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;mol - 4}}{\rm{.46}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;mol}}\\{\rm{n = 4}}{\rm{.42}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;mol}}\end{array}\)

From the first formula, the ratio is:

\(\frac{{\rm{n}}}{{{{\rm{n}}_{\rm{1}}}}}{\rm{ = }}\frac{{{\rm{4}}{\rm{.42}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;mol}}}}{{{\rm{4}}{\rm{.42}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{\;mol}}}}{\rm{ = 1}}\)

Substituting the values into the formula :

\(\begin{array}{c}\left( {\frac{{\rm{n}}}{{{{\rm{n}}_{\rm{1}}}}}} \right){\rm{ = }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\frac{{\rm{t}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}}}\\{\rm{1 = }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\frac{{\rm{t}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}}}\end{array}\)

The value for time is\({\rm{24 h}}\):

\(\begin{array}{c}{\rm{1 = }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\frac{{{\rm{24h}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}}}\\{\rm{log1 = log}}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\frac{{{\rm{24h}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}}}\\{\rm{log1 = }}\frac{{{\rm{24\;h}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}{\rm{log}}\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)\\\frac{{{\rm{log1}}}}{{{\rm{log}}\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)}}{\rm{ = }}\frac{{{\rm{24\;h}}}}{{{{\rm{t}}_{{\rm{1/2}}}}}}\end{array}\)

And if\(\log 1\)is divided by\({\rm{log(1/2)}}\), it gives\({\rm{0}}\).

Therefore, the value of the half-life is infinite.