Problem 36 - 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 \(^{222} \mathrm{Rn}\) to decrease to \(20.0 \%\) of its original activity?

Step-by-Step Solution

Answer

(a) \( ^{222}_{86}\mathrm{Rn} \rightarrow ^{218}_{84}\mathrm{Po} + ^{4}_{2}\mathrm{He} \) (b) 8.45 days

Step 1: Understanding Alpha Decay

Alpha decay occurs when a radioactive nucleus emits an alpha particle, which is composed of 2 protons and 2 neutrons. This results in a decrease in the atomic number by 2 and a decrease in the mass number by 4.

Step 2: Writing the Balanced Equation

For Radon-222, which has an atomic number of 86, the balanced equation for alpha decay is:\[ ^{222}_{86}\mathrm{Rn} \rightarrow ^{218}_{84}\mathrm{Po} + ^{4}_{2}\mathrm{He} \]Here, Radon-222 decays into Polonium-218 while emitting an alpha particle \(^{4}_{2}\mathrm{He}\).

Step 3: Understanding Half-Life Decay Formula

The formula for decay based on half-life is:\[ N_t = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]where \(N_t\) is the remaining quantity, \(N_0\) is the initial quantity, \(t\) is the time, and \(T_{1/2}\) is the half-life.

Step 4: Setting Up the Equation for 20% Remaining Activity

We need to find the time \(t\) when the remaining activity is 20% of the initial activity, so \(N_t = 0.2 N_0\).\[ 0.2 N_0 = N_0 \left(\frac{1}{2}\right)^{\frac{t}{3.82}} \]Cancelling out \(N_0\), we get:\[ 0.2 = \left(\frac{1}{2}\right)^{\frac{t}{3.82}} \]

Step 5: Solving for Time Using Logarithms

To solve for time \(t\), we take the logarithm of both sides:\[ \log(0.2) = \frac{t}{3.82} \log\left(\frac{1}{2}\right) \]This leads to:\[ t = 3.82 \frac{\log(0.2)}{\log(0.5)} \]

Step 6: Calculating the Time

Using a calculator, compute:\[ \log(0.2) \approx -0.69897 \] and \[ \log(0.5) \approx -0.30103 \]Thus,\[ t = 3.82 \times \frac{-0.69897}{-0.30103} \approx 8.45 \text{ days} \]

Key Concepts

Alpha Decay

Alpha decay is a type of radioactive decay where an unstable atom increases its stability by releasing an alpha particle. An alpha particle consists of two protons and two neutrons. This process decreases the mass number of the original atom by 4 and its atomic number by 2. The loss of these particles changes the identity of the element, turning it into a different element with lower atomic mass. For example, when radon-222 undergoes alpha decay, it transforms into polonium-218.

• The emitted alpha particle is essentially a helium nucleus.
• Alpha decay reduces the element's atomic number, often producing a new element.

This form of decay is one of several ways unstable atoms can transform to achieve greater stability.

Half-life

Half-life is the time required for half of a sample of a radioactive substance to decay into another element or isotope. It is a measure of the rate at which a radioactive material disintegrates, unique to each isotope. For radon-222, the half-life is specifically 3.82 days.

• The concept of half-life helps predict how long a radioactive sample will remain active.
• It is essential for determining safe handling times for radioactive substances.

The exponential nature of decay means that after one half-life, 50% of the original radioactive nuclei have decayed. After two half-lives, only 25% of the nuclei remain, and so on.

Radon-222

Radon-222 is a radioactive isotope that is notable due to its prevalence and potential health risks in homes. As an alpha emitter, it can pose a significant health risk when it decays and is inhaled.

• It is produced naturally as a decay product of uranium-238.
• Radon is colorless, odorless, and tasteless, making it undetectable without specialized equipment.

Its decay initiates a chain reaction that can result in harmful radioactive substances in dwelling spaces. Mitigating radon levels in homes is crucial for safety.

Balanced Nuclear Equations

Balanced nuclear equations are used to represent nuclear reactions, such as alpha decay, and must account for changes in both mass and atomic numbers. The law of conservation of both mass and atomic number guides these equations.

• For radon-222 decaying into polonium-218, the balanced equation is:\[ ^{222}_{86}\mathrm{Rn} \rightarrow ^{218}_{84}\mathrm{Po} + ^{4}_{2}\mathrm{He} \]
• The summed atomic numbers and mass numbers on each side of the equation must match.

In each nuclear reaction, a balanced equation ensures accuracy in representing the identities and amounts of materials before and after the reaction.