In 1992, Swiss police arrested two men who were attempting to smuggle osmium out of Eastern Europe for a clandestine sale. However, by error, the smugglers had picked up ${}^{\mathbf{137}}\mathit{C}\mathit{s}$. Reportedly, each smuggler was carrying a1.0gsample of${}^{\mathbf{137}}\mathit{C}\mathit{s}$in a pocket! In (a) Becquerel and (b) curies, what was the activity of each sample? The isotope${}^{\mathbf{137}}\mathit{C}\mathit{s}$has a half-life of$\mathbf{30}\mathbf{.}\mathbf{2}\mathbf{}\mathit{y}$. (The activities of radioisotopes commonly used in hospitals range up to a few millicuries.)

The mass of the sample ${}^{137}\mathrm{Cs}$, m = 1.0 g

The molar mass of the sample ${}^{137}\mathrm{Cs}$, A = 137 g/mol

The half-life of the isotope ${}^{137}\mathrm{Cs}$,$T_{1/2}=30.2y=9.53\times {10}^{3}s$

The radioactive decay is due to the loss of the elementary particles from an unstable nucleus to convert them into a more stable one. Here, the activity of the sample is the initial disintegration rate of the given isotope after the steal.

Formulae:

The rate of decay,

$R=\left(\frac{\ln 2}{T_{1/2}}\right)N$ ….. (i)

Where, $T_{1/2}$is the half-life of the substance, N is the number of undecayed nuclei.

The conversion unit from Curie to Becquerel,

$1\mathrm{Ci}=3.7x{10}^{10}\mathrm{Bq}$ ….. (ii)

The number of undecayed nuclei in a given mass of a substance,

$N=\frac{m}{A}{N}_A$ ….. (iii)

Here, the Avogadro number is,$N_A=6.022\times {10}^{23}mo{l}^{-1}$

Using the given data and equation (iii) for number of undecayed nuclei in equation (i), we can get the activity of the sample in Becquerel as follows:

$$\begin{gathered} R=\left(\frac{\ln 2}{T_{1/2}}\right)\frac{m}{A}{N}_A \\ =\frac{\left(\ln 2\right)\left(1g\right)}{\left(9.53\times {10}^{8}s\right)\left(137g/mol\right)}\left(6.022\times {10}^{23}mo{l}^{-1}\right) \\ =3.2\times {10}^{12}{s}^{-1} \\ =3.2\times {10}^{12}Bq \end{gathered}$$

Hence, the value of the activity is $3.2\times {10}^{12}Bq$.

Using the conversion relation from equation (ii), the above activity of the sample can be given in Curie as follows:

$\begin{array}{ll}R=\frac{3.2\times {10}^{12}}{3.7\times {10}^{10}}& C_1\\ =86Ci& \end{array}$

Hence, the value of the activity is 86 Ci .