What volume of radon will be produced per hour at STP from 1.000 g of mass of ${}^{226}Ra(t_{1/2}=1599yr;1yr=8766h;$one ²²⁶Ra atom =226.025402 amu)

The process by which an unstable atomic nucleus loses energy through radiation is known as radioactive decay (also known as nuclear decay, radioactivity, radioactive disintegration, or nuclear disintegration). A radioactive substance is one that contains unstable nuclei.

The half-life of a radioactive substance is defined as the amount of time required for the decay of half the initial amount of a substance. Half-life is denoted by t₁₂ .

Radioactive decay is a first-order reaction. The expression for half-life is given below.

${\text{t}}_{\text{1/2}}\text{=}\frac{\text{0.693}}{\text{k}}$

Where, k is the decay constant.

Using the above equation we will find the decay constant for the reaction.

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{1599yr}\times \left(\frac{1yr}{8766hr}\right)=4.945\times {10}^{-8} h{r}^{-}$

Using stoichiometry and Avogadro's number, we will determine the amount Ra atoms.

$$\begin{gathered} N=1gRa\left(\frac{1molRa}{226.025402gRa}\right)\left(\frac{6.022\times {10}^{23}atoms}{1molRa}\right) \\ N=2.6643\times {10}^{21}Ra-atoms \end{gathered}$$

Then using the formula for activity –

$$\begin{gathered} A=kN=4.945\times {10}^{-8} h{r}^{-}\left(2.6643\times {10}^{21}Ra-atoms\right) \\ A=1.3175\times {10}^{14} Ra-atoms/hr \end{gathered}$$

Using the value of for activity of Ra atoms, we will apply stoichiometry again to determine the amount of Rn atoms.

$$\begin{gathered} Rn=1.3175\times {10}^{14}\frac{Ra-atoms}{hr}\times \frac{1 mol Rn }{1 mol Ra}\left(\frac{1mol}{6.022\times {10}^{23}atoms}\right) \\ Rn=2.1878\times {10}^{-10}\frac{molRn}{hr} \end{gathered}$$

Using the ideal gas equation, we will compute for the rate of production of radon gas.

$$\begin{gathered} V=\frac{nRT}{P}=\frac{\left(2.1878\times {10}^{-10}\frac{molRn}{hr}\right)}{\left(0.08206\frac{L·atm}{mol·K}\right)(273.15K)1atm} \\ V=4.9039·{10}^{-9}\frac{L}{hr} \end{gathered}$$

Therefore, the value for volume is obtained as $V=4.9039·{10}^{-9}\frac{L}{hr}$.