Question

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)

Short answer

The volume of radon that will be produced per hour is$V=4.9039·{10}^{-9}\frac{L}{hr}$ .

Step-by-step solution

Concept Introduction

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.

Half-life

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.

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

Number of  Atoms

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

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

Calculation for Volume

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 $V=4.9039·{10}^{-9}\frac{L}{hr}$.