Question

Question: Astatine is the rarest naturally occurring with ${}^{\mathbf{219}}\mathbf{At}$appearing as the product of a very minor side branch in the decay of ${}^{\mathbf{235}}\mathbf{U}$(itself not a very abundant isotope). It is estimated that the mass of all naturally occurring in${}^{\mathbf{219}}\mathbf{At}$ in the upper kilometer of the earth’s atmosphere has a steady state value of only 44 mg. Calculate the total activity (in disintegration per seconds) caused by all the naturally occurring astatine in this part of the earth. The half-life of ${}^{\mathbf{219}}\mathbf{At}$is 54 s and its atomic mass is 219.01 u.

Short answer

The total activity of${}^{219}\mathrm{At}$is$1.44\times {10}^{18}$disintegrations per second.

Step-by-step solution

Decay constant of At219

The half-life of a radioactive element is defined as the amount of time required by the element to decay to half of its initial amount. It is represented by ${\mathrm{t}}_{1/2}$. Since, radioactive decay is a first-order reaction, the half-life and decay constant are related by the following expression:

role="math" localid="1660795723407" ${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$

We will use the above expression to calculate the decay constant of ${}^{219}\mathrm{At}$.

$$\begin{gathered} k=\frac{0.693}{54}{s}^{-1} \\ =0.012s^{-1} \end{gathered}$$

Activity of At219

To calculate the activity of $44\mathrm{mg}$data-custom-editor="chemistry" ${}^{219}\mathrm{At}$, we have to find out the number of atoms present in the sample.

Number of atoms present in$44\mathrm{mg}{}^{219}\mathrm{At}$$=\frac{44\times {10}^{-3}\mathrm{g}}{219.01\mathrm{g}/\mathrm{mol}}\times \frac{6.023\times {10}^{23}\mathrm{atoms}}{1\mathrm{mol}}=1.2\times {10}^{20}\mathrm{atoms}$

The activity can be calculated from the following relation:

$$\begin{gathered} \mathrm{A}=\mathrm{kN} \\ =0.012{\mathrm{s}}^{-1}\times 1.2\times {10}^{20}\mathrm{atoms} \\ =1.44\times {10}^{18}\mathrm{disintegrations}\mathrm{per}\mathrm{second} \end{gathered}$$