Question

If $1.000\mathrm{g}$ of ${}_{88}^{226}\mathrm{R}\mathrm{a}$ produces $0.0001\mathrm{m}\mathrm{L}$ of the gas ${}_{88}^{222}\mathrm{R}\mathrm{n}$ at $\mathrm{S}\mathrm{T}\mathrm{P}$ (standard temperature and pressure) in $24\mathrm{h}$, what is the half-life of ${}^{226}\mathrm{R}\mathrm{a}$ in years?

Short answer

The half-life of ${}_{88}^{226}\mathrm{R}\mathrm{a}$ in years is infinite.

Step-by-step solution

Concept Introduction

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

$\left(\frac{\mathrm{n}}{{\mathrm{n}}_1}\right)={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{t}}_{1/2}}}$

Where$\frac{\mathrm{n}}{{\mathrm{n}}_1}$is the ratio of the amount of the substance left to the amount of the substance initially present,${\mathrm{t}}_{1/2}$is the half-life of the substance, and$\mathrm{t}$is the time after which the ratio is determined.

But first, get the number of moles:

${\mathrm{n}}_2=\frac{\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}$

Calculation for Ratio

The volume of$1.0\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}$of the gas is$22400\mathrm{m}\mathrm{L}$.

$\begin{array}{l}{\mathrm{n}}_2=0.00001\mathrm{m}\mathrm{L}\cdot \frac{1\mathrm{m}\mathrm{o}\mathrm{l}}{22400\mathrm{m}\mathrm{L}}\\ {\mathrm{n}}_2=4.46\times 10^{-9}\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

The molar mass of$Ra$is$226\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$. From that, we can get${\mathrm{n}}_1$.

$\begin{array}{l}{\mathrm{n}}_1=\frac{\mathrm{m}}{\mathrm{M}}\\ {\mathrm{n}}_1=\frac{1.000\mathrm{g}}{226\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}\\ {\mathrm{n}}_1=4.42\cdot 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

$\begin{array}{l}\mathrm{n}={\mathrm{n}}_1-{\mathrm{n}}_2\\ \mathrm{n}=4.42\cdot 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}-4.46\cdot 10^{-9}\mathrm{m}\mathrm{o}\mathrm{l}\\ \mathrm{n}=4.42\cdot 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}\end{array}$

From the first formula, the ratio is:

$\frac{\mathrm{n}}{{\mathrm{n}}_1}=\frac{4.42\cdot 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}}{4.42\cdot 10^{-3}\mathrm{m}\mathrm{o}\mathrm{l}}=1$

Calculating the Half-Life

Substituting the values into the formula :

$\begin{array}{c}\left(\frac{\mathrm{n}}{{\mathrm{n}}_1}\right)={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{t}}_{1/2}}}\\ 1={\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{{\mathrm{t}}_{1/2}}}\end{array}$

The value for time is$24\mathrm{h}$:

$\begin{array}{c}1={\left(\frac{1}{2}\right)}^{\frac{24\mathrm{h}}{{\mathrm{t}}_{1/2}}}\\ \mathrm{l}\mathrm{o}\mathrm{g}1=\mathrm{l}\mathrm{o}\mathrm{g}{\left(\frac{1}{2}\right)}^{\frac{24\mathrm{h}}{{\mathrm{t}}_{1/2}}}\\ \mathrm{l}\mathrm{o}\mathrm{g}1=\frac{24\mathrm{h}}{{\mathrm{t}}_{1/2}}\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{1}{2}\right)\\ \frac{\mathrm{l}\mathrm{o}\mathrm{g}1}{\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{1}{2}\right)}=\frac{24\mathrm{h}}{{\mathrm{t}}_{1/2}}\end{array}$

And if$\log 1$is divided by$\mathrm{l}\mathrm{o}\mathrm{g}(1/2)$, it gives$0$.

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