A radium source contains 1.00mg of ${}^{\mathbf{226}}\mathit{R}\mathit{a}$, which decays with a half-life of 1600y to produce ${}^{\mathbf{222}}\mathit{R}\mathit{n}$, a noble gas. This radon isotope in turn decays by alpha emission with a half-life of 3.82d . If this process continues for a time much longer than the half-life of ${}^{\mathbf{222}}\mathit{R}\mathit{n}$, the${}^{\mathbf{222}}\mathit{R}\mathit{n}$ decay rate reaches a limiting value that matches the rate at which${}^{\mathbf{222}}\mathit{R}\mathit{n}$ is being produced, which is approximately constant because of the relatively long half-life of$\mathbf{}{}^{\mathbf{226}}\mathit{R}\mathit{a}$ . For the source under this limiting condition, what are (a) the activity of$\mathbf{}{}^{\mathbf{226}}\mathit{R}\mathit{a}$ (b) the activity of${}^{\mathbf{222}}\mathit{R}\mathit{n}$ , and (c) the total mass of${}^{\mathbf{222}}\mathit{R}\mathit{n}$ ?

a) Mass of radium source${}^{226}Ra$ ,$m_{Ra}=1.00mg$

b) Half-life of radium source${}^{226}Ra$ ,$T_{1/2_{Ra}}=1600y$

c) Radium undergoes alpha decay to give radon.

d) Half-life of radon${}^{222}Rn$ ,$T_{1/2_{Rn}}=38.2d$

e) Decay rate matches the limiting value of production rate at a time and is constant because of long half-life of radium

In a secular equilibrium condition, the activity or the production rate of the sample is equal to that of the decay rate of the sample. The rate gives the exponential function of decay constant and time for the given decay process.

The disintegration constant is as follows:

$\lambda =\frac{\ln 2}{T_{1/2}}$ …… (i)

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

The rate of decay of a radioactivity is as follows:

$R=\lambda N$ …… (ii)

The number of atoms in a given mass of a substance,

$N=\frac{M}{m}{N}_A$ ……(iii)

Here, M is the given mass in the sample and m is the molar mass of the substance and$N_A=6.022\times {10}^{23}\frac{atoms}{mol}$ .

Substituting equations (i) and (ii) with the given data in equation (ii), we can get the activity of the radium source ${}^{226}Ra$as follows:

$\begin{array}{rcl}R& =& \left(\frac{\ln 2}{T_{1/2}}\right)\left(\frac{M}{m}{N}_A\right)\\ & =& \left(\frac{\ln 2}{\left(1600y\right)\left(3.15\times {10}^7{\displaystyle \frac{s}{y}}\right)}\right)\left(\frac{1.0}{226{\displaystyle \frac{g}{mol}}}\right)\left(6.022\times {10}^{23}mo{l}^{-1}\right)\left(\because m=226\frac{g}{mol}forradium\right)\\ & =& 3.66\times {10}^7{s}^{-1}\end{array}$

Hence, the activity is$\begin{array}{rcl}& & 3.66\times {10}^7{s}^{-1}\end{array}$ .

The decay rate is approximately constant as it gets equal to the value of production rate.

Hence, the activity of radon source is also$\begin{array}{rcl}& & 3.66\times {10}^7{s}^{-1}\end{array}$ .

As,$R_{Ra}=R_{Rn}$

Thus, using equation (ii), we can get that

${\lambda }_{Ra}{N}_{Ra}={\lambda }_{Rn}{N}_{Rn}$

Now, using equations (i) and (ii) in the above equation, we can get the total mass of radon in the sample as follows:

$$\begin{gathered} \left(\frac{\ln 2}{T_{1/2_{Ra}}}\right)\left(\frac{M_{Ra}}{m_{Ra}}{N}_A\right)=\left(\frac{\ln 2}{T_{1/2_{Rn}}}\right)\left(\frac{M_{Rn}}{m_{Rn}}{N}_A\right) \\ {M}_{Rn}=\left(\frac{T_{1/2_{Rn}}}{T_{1/2_{Ra}}}\right)\left(\frac{M_{Rn}}{m_{Ra}}\right)M_{Ra} \\ {M}_{Rn}=\left(\frac{3.82d}{\left(1600y\right)\left(365{\displaystyle \frac{d}{y}}\right)}\right)\left(\frac{222u}{226u}\right)\left(1.00\times {10}^{-3}g\right) \\ {M}_{Rn}=6.42\times {10}^{-9}g \end{gathered}$$

Hence, the value of the mass is$6.42\times {10}^{-9}g$ .