Problem 35 - Chemistry and Chemical Reactivity

Question

Radon has been the focus of much attention recently because it is often found in homes. Radon-222 emits \(\alpha\) particles and has a half-life of 3.82 days. (a) Write a balanced equation to show this process. (b) How long does it take for a sample of \(^{22} \mathrm{Rn}\) to \(\mathrm{de}\) crease to \(20.0 \%\) of its original activity?

Step-by-Step Solution

Answer

(a) \( ^{222}_{86}\text{Rn} \rightarrow ^{218}_{84}\text{Po} + ^{4}_{2}\text{He} \). (b) 8.87 days.

Step 1: Understanding Alpha Decay

Alpha decay involves the emission of an alpha particle, which is equivalent to a helium nucleus with 2 protons and 2 neutrons. Its notation is \(^{4}_{2}\text{He}\). For Radon-222, this means that the atomic number will decrease by 2 and the mass number will decrease by 4.

Step 2: Balancing the Equation

Radon-222 \((^{222}_{86}\text{Rn})\) emits an alpha particle \((^{4}_{2}\text{He})\). The resulting nucleus must have an atomic number of 84 and mass number of 218. The element with atomic number 84 is Polonium \((^{218}_{84}\text{Po})\). The balanced equation is: \[ ^{222}_{86}\text{Rn} \rightarrow ^{218}_{84}\text{Po} + ^{4}_{2}\text{He} \]

Step 3: Using the Half-Life Formula

The half-life formula is \( N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \), where \( N \) is the final concentration, \( N_0 \) is the initial concentration, \( t \) is time, and \( t_{1/2} \) is the half-life. Given the final activity is 20% of the original, \( \frac{N}{N_0} = 0.20 \). Set \( t_{1/2} = 3.82 \) days.

Step 4: Solving for Time

Substitute the known values into the half-life formula to solve for \( t \): \[ 0.20 = \left( \frac{1}{2} \right)^{\frac{t}{3.82}} \] Take the logarithm of both sides:\[ \log(0.20) = \frac{t}{3.82} \cdot \log(0.5) \]Rearrange to solve for \( t \):\[ t = \frac{\log(0.20)}{\log(0.5)} \cdot 3.82 \] Calculate \( t \) to find the number of days.

Key Concepts

Half-Life Calculation

Understanding the concept of half-life is essential when dealing with radioactive substances. The half-life is the time it takes for half of the radioactive atoms in a sample to decay. It's a crucial factor in determining how long a radioactive sample remains active. In the case of Radon-222, its half-life is 3.82 days, meaning that every 3.82 days, half of the Radon-222 in a sample will decay into another element.

This concept allows us to calculate how long it takes for a substance to reduce to a given level of radioactivity. For example, if we need to know how long it takes for Radon-222 to decay to 20% of its original activity, we use the half-life formula:

• \( N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \)

Here, \( N \) is the amount remaining, \( N_0 \) is the initial amount, \( t \) is the time elapsed, and \( t_{1/2} \) is the half-life.

Substituting the known values helps solve these types of problems effectively.

Balancing Nuclear Equations

Nuclear equations represent radioactive decay processes. They require careful balancing of both the atomic and mass numbers. This ensures that both the type and number of particles are conserved.

When Radon-222 undergoes alpha decay, an alpha particle, which is a helium nucleus \( ^{4}_{2}\text{He} \), is emitted. This emission results in the transformation of the parent element, Radon-222 \(^{222}_{86}\text{Rn} \), into a new element. The atomic number decreases by two and the mass number by four, leading to Polonium-218 \(^{218}_{84}\text{Po} \). The balanced equation for this reaction is:

• \[ ^{222}_{86}\text{Rn} \rightarrow ^{218}_{84}\text{Po} + ^{4}_{2}\text{He} \]

This demonstrates how an initially non-stable element can morph into a different element entirely.

Radioactive Decay

Radioactive decay is the natural process through which unstable atomic nuclei lose energy. This process results in the emission of particles or electromagnetic waves. There are several types of decay, with alpha decay being one of the most common.

In alpha decay, an alpha particle is released from the nucleus. This particle consists of two protons and two neutrons, resembling a helium nucleus. Over time, this emission alters the original unstable nucleus, transforming it into a different element.

This transformation changes the atomic and mass numbers of the original element. For instance, Radon-222 turns into Polonium-218 after undergoing alpha decay. Radioactive decay is critical in various fields, including environmental science, medicine, and archaeology. Understanding these processes is key, as they have significant applications and implications in real-world scenarios.