Question

Polonium-210, an alpha emitter, has a halflife of \(138.4\) days. Suppose you were to collect the helium gas originating from the alpha particles. How many milliliters of helium gas at standard temperature and pressure would you collect from \(1.000 \mathrm{~g}\) of polonium dioxide, \(\mathrm{PoO}_{2}\), in a period of \(138.4\) days? [Assume all the polonium in the sample is \({ }^{210} \mathrm{Po}\), molar mass \(209.98287 \mathrm{~g} / \mathrm{mol}\). Alpha emission from polonium- 210 yields the nonradioactive isotope lead-206; see Problem 16.103.]

Short answer

The volume of helium gas produced from 1.000 g of Polonium dioxide (PoO2) in a period of 138.4 days is approximately \(5363.6 \, \mathrm{mL}\) at standard temperature and pressure.

Step-by-step solution

Decay reaction of Polonium-210

Polonium-210 undergoes alpha decay; the reaction can be represented as: $$ {}^{210}\mathrm{Po} \rightarrow {}^{206}\mathrm{Pb} + {}^{4}\mathrm{He} $$ Where \(^{210}\mathrm{Po}\) represents Polonium-210, \(^{206}\mathrm{Pb}\) represents Lead-206 (a non-radioactive isotope), and \(^{4}\mathrm{He}\) represents an alpha particle (Helium).

Calculate the initial amount of Polonium-210

We have 1.000 g of Polonium dioxide (PoO2). Since we are asked to assume that all the polonium in the sample is \(^{210}\mathrm{Po}\), let's first find out how many moles of Polonium-210 we have. $$ \mathrm{moles \, of \, Po}_{210} = \frac{\mathrm{mass \, of \, PoO}_{2}}{\mathrm{molar \, mass \, of \, Po}_{210}} = \frac{1.000 \, \mathrm{g}}{209.98287 \, \mathrm{g/mol}} $$

Calculate the remaining Polonium-210 after 138.4 days

We are given the half-life of Polonium-210 (t½) as 138.4 days, and we need to find out how much Polonium-210 is left after 138.4 days. Using the half-life formula, the remaining amount of Polonium-210 after one half-life is given by: $$ N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}} $$ Where \(N(t)\) is the remaining amount of Polonium-210 after time \(t\), \(N_0\) is the initial amount of Polonium-210, and \(t_{1/2}\) is the half-life of Polonium-210. Since we're looking at 138.4 days, which is one half-life, let's calculate the remaining amount of Polonium-210: $$ N(138.4) = N_0 \left(\frac{1}{2}\right) = \frac{1.000 \, \mathrm{g}}{2 \times 209.98287 \, \mathrm{g/mol}} $$

Calculate the amount of Helium gas produced

We can now determine the amount of Helium gas produced during the decay of Polonium-210. We start with the initial amount of Polonium-210 and subtract the remaining amount: $$ \mathrm{moles \, of \, He}_{4} = \frac{1.000 \, \mathrm{g}}{209.98287 \, \mathrm{g/mol}} - \frac{1.000 \, \mathrm{g}}{2 \times 209.98287 \, \mathrm{g/mol}} $$

Convert moles of Helium gas to volume at STP

Now that we know the amount of Helium (alpha particles), we can convert it into volume at standard temperature and pressure (STP). At STP, 1 mole of an ideal gas occupies a volume of 22.4 L (22,400 ml). To find the volume of Helium gas produced, we can use the following formula: $$ \mathrm{Volume \, of \, He}_{4} = \mathrm{moles \, of \, He}_{4} \times \frac{22,400 \, \mathrm{mL}}{1 \, \mathrm{mol}} $$ Substitute the value of He-4 moles into the formula and calculate the volume.

Key concepts

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable nucleus releases an alpha particle to become more stable. An alpha particle is the nucleus of a helium atom, consisting of two protons and two neutrons. During the alpha decay of Polonium-210,

• Polonium transforms into Lead-206, a non-radioactive isotope.
• This transformation reduces the mass number by four and the atomic number by two.

The decay equation for Polonium-210 is:\[{}^{210} ext{Po} \rightarrow {}^{206} ext{Pb} + {}^{4} ext{He} \]This indicates that for each atom of Polonium-210 that undergoes decay, one atom of Lead-206 and an alpha particle (Helium) is produced. Understanding this process is crucial to determine how much helium gas can be collected when Polonium-210 decays.

Half-life Calculation

The concept of half-life is central to understanding radioactive decay. The half-life of a substance is the time it takes for half of the material to decay. For Polonium-210, the half-life is 138.4 days. This means that after 138.4 days, only half of the Polonium-210 sample remains.

Using the half-life formula, you can calculate the remaining amount of Polonium-210:\[N(t) = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}\]Where:

• \(N(t)\) is the remaining amount after time \(t\).
• \(N_0\) is the initial quantity.
• \(t_{1/2}\) is the half-life.

In this specific problem, we are asked to look at a period of exactly one half-life, meaning the calculation is straightforward: the original amount divided by two. Accurately understanding and calculating half-life helps in predicting how much of a radioactive element remains after a certain period.

Mole Conversion

Understanding mole conversion is essential in chemistry as it allows us to relate mass to number of particles using Avogadro's number. Here, we convert mass of Polonium-210 to moles:

The formula is:\[\text{Moles of Po}_{210} = \frac{\text{Mass of PoO}_2}{\text{Molar Mass of Po}_{210}}\]Where:

• The given mass in the problem is 1.000 g.
• The molar mass of \(^{210}\text{Po}\) is 209.98287 g/mol.

By calculating the initial moles of Polonium-210, students can determine how much Helium gas is produced during one half-life due to the decay process. This shows how effectively breaking down molecules to moles aids in transitioning from chemical formulas to solving practical problems.

Standard Temperature and Pressure (STP)

Standard Temperature and Pressure (STP) is a reference point used to define the standard conditions for measuring gases. At STP, one mole of any ideal gas occupies a volume of 22.4 liters or 22,400 milliliters. Understanding STP allows us to convert moles of a gas into volume.

In this exercise, once we calculate the number of moles of Helium gas (resulting from the alpha decay), we can determine the volume of Helium at STP using:\[\text{Volume of He}_4 = \text{moles of He}_4 \times \frac{22,400 \text{ mL}}{1 \text{ mol}}\]This conversion is essential for practically understanding gas volumes and is invaluable for calculating how much gas can be collected from a given chemical reaction under standard conditions.