After long effort, in 1902 Marie and Pierre Curie succeeded in separating from uranium ore the first substantial quantity of radium, one decigram of pure $\mathit{R}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}$. The radium was the radioactive isotope${}^{\mathbf{226}}\mathit{R}\mathit{a}$, which has a half-life of 1600 y. (a) How many radium nuclei had the Curies isolated? (b) What was the decay rate of their sample, in disintegrations per second?

Mass of the pure${\mathrm{RaCl}}_2$,role="math" localid="1661925792054" ${\mathrm{m}}_{\mathrm{Ra}{\mathrm{Cl}}_2}=0.1\mathrm{g}$

Half-life of radioactive isotope${}^{226}R$,$T_{1/2}=1600y$

Molar mass of the isotope ${}^{226}R$, A = 226 g/mol

In a mixture, a radioactive isotope of radium is found. Being radioactive, it has an unstable nature and thus undergoes disintegrations per second which is its decay rate. Now, using the given mass of the substantial product, we can get the mass of the isotope present in that mass value using the average concept. Now, using this mass we can obtain the number of atoms of the isotope in the amount. Thus, this gives the rate of decay using appropriate values in the basic formula. Again, the decay rate in disintegrations per second gives the activity of the sample.

Formulae:

The mass fraction of Ra in$RaC{l}_2$ is given by,

$Massfractionofradium=\frac{M_{Ra}}{M_{Ra}+2M_{Cl}}........\left(1\right)$

where,$M_{Ra}$is the molar mass of Radium.

$M_{Cl}$is the molar mass of chlorine.

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

$N=\frac{m}{A}{N}_A.........\left(2\right)where{N}_A=6.022\times {10}^{23}nuclei/mol$

The rate of decay, $R=\left(\frac{\ln 2}{T_{1/2}}\right)N.............\left(3\right)$

where, $T_{1/2}$is the half-life of the substance.

Nis the number of undecayed nuclei.

We assume that the chlorine in the sample had the naturally occurring isotopic mixture, so the average molar mass is $M_{Cl}=35.453g/mol$, as given in Appendix F.

Then, the mass ofrole="math" localid="1661926269542" ${}^{226}\mathrm{Ra}$can be given using equation (1) as follows:

$$\begin{gathered} m=MassFraction\times MassofRaC{l}_2 \\ =\frac{226g/mol}{226g/mol+2\left(35.453g/mol\right)}\left(0.1g\right) \\ =76.1\times {10}^{-3}g \end{gathered}$$

Now, using the mass of the isotope${}^{226}\mathrm{Ra}$in equation (2), we can get the number of radium nuclei in the isotope as follows:

$$\begin{gathered} N=\frac{76.1\times {10}^{-3}g}{226g/mol}\left(6.022\times {10}^{23}nuclei/mol\right) \\ =2.0\times {10}^{20}nuclei \end{gathered}$$

Hence, the number of nuclei is $2.0\times {10}^{20}$.

Now, using the value of radium nuclei in equation (3), we can get the decay rate of the radium isotope${}^{226}\mathrm{Ra}$as follows:

$$\begin{gathered} R=\left(\frac{1n2}{1600y\times 3.156\times {10}^{7}s/y}\right)\left(2.03\times {10}^{20}\right) \\ =2.8\times {10}^9{s}^{-1} \end{gathered}$$

Hence, its decay rate is $2.8\times {10}^9{s}^{-1}$.