Question

The noble gas radon has been the focus of much attention because it may be 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) Calculate the time required for a sample of radon to decrease to \(10.0 \%\) of its original activity.

Short answer

(a) \( ^{222}_{86}\text{Rn} \rightarrow \, ^4_2\text{He} \, + \, ^{218}_{84}\text{Po} \); (b) 12.7 days.

Step-by-step solution

Identify the Nuclear Subprocess

Identify the type of radioactive decay involved which is the alpha decay of Radon-222. An alpha particle consists of 2 protons and 2 neutrons, so it can be represented as \( ^4_2\text{He} \) or simply \( \alpha \).

Write the Balanced Nuclear Equation

Radon-222 undergoes alpha decay by losing an alpha particle. This means that Radon's atomic number decreases by 2 and its mass number decreases by 4. Therefore, the balanced nuclear equation is \( ^{222}_{86}\text{Rn} \rightarrow \, ^4_2\text{He} \, + \, ^{218}_{84}\text{Po} \) where Polonium-218 is the resulting daughter nucleus.

Understand Half-life Concept

The half-life of Radon-222 is 3.82 days, which represents the time it takes for half of the substance to decay. This is fundamental in calculating the remaining amount after a certain period.

Determine Remaining Quantity

We are tasked with finding the time it takes for a sample to decrease to 10% of its original activity. Using the formula \( N = N_0 \left(\frac{1}{2}\right)^{t/T} \), we set \( N = 0.1N_0 \), where \( t \) is the time period and \( T = 3.82 \) days.

Solve for Time

Rearranging the formula gives \( 0.1 = \left(\frac{1}{2}\right)^{t/3.82} \). Taking the natural logarithm of both sides gives \( \ln(0.1) = \frac{t}{3.82} \ln(0.5) \). Solving for \( t \) yields \( t = 3.82 \times \frac{\ln(0.1)}{\ln(0.5)} \approx 12.7 \) days.

Key concepts

Alpha Decay

In the world of radioactive decay, alpha decay is a crucial process worth understanding. It is characterized by the emission of an alpha particle, which consists of two protons and two neutrons. This makes it similar to a helium nucleus, often represented as \( ^4_2\text{He} \) or simply denoted by the Greek letter \( \alpha \). When a radioactive atom undergoes alpha decay, it loses an alpha particle, resulting in a reduction of its atomic number by 2 and its mass number by 4. This process transforms the original parent nucleus into a different daughter nucleus. For example, when radon-222 decays, it loses an alpha particle, resulting in the formation of polonium-218. Alpha decay is commonly observed in heavy elements like uranium, thorium, and their decay products, such as radon and polonium. It plays a significant role in the chain reactions found in radioactive series.

Half-life Calculation

Half-life is a key concept in understanding radioactive decay, referring to the time it takes for half of a radioactive substance to decay. Different isotopes have different half-lives, depending on their inherent stability. Radon-222, for instance, has a half-life of 3.82 days. This means that if you start with a given amount of radon-222, only half of it will remain after 3.82 days.When calculating how much of a substance remains after a given period, the half-life formula can be very useful. The formula is:

• \( N = N_0 \left(\frac{1}{2}\right)^{t/T} \)
• Where \( N \) is the remaining quantity, \( N_0 \) is the initial quantity, \( t \) is the elapsed time, and \( T \) is the half-life.

For example, if you want to know how long it takes for a sample of radon-222 to decay to 10% of its original activity, you rearange the formula to solve for \( t \).

Nuclear Equation Balancing

Writing balanced nuclear equations is an essential skill when dealing with radioactive decay. Much like chemical equations, nuclear equations must be balanced in terms of both mass number and atomic number. This means ensuring the total mass number and atomic number on the reactant side equals those on the product side.For radon-222 undergoing alpha decay, the balanced nuclear equation is: \( ^{222}_{86}\text{Rn} \rightarrow \, ^4_2\text{He} \, + \, ^{218}_{84}\text{Po} \).Here's a breakdown:

• The mass number decreases from 222 to 218 after losing a mass of 4 (the alpha particle).
• The atomic number decreases from 86 to 84 after expelling 2 protons along with the alpha particle.
• This results in the formation of polonium-218 as the daughter nucleus.

Balancing nuclear equations helps in understanding the outcome of the decay process and the new element that forms.

Radon-222 Decay

Radon-222 is a significant isotope to study due to its environmental and health implications. As a noble gas, it is colorless, odorless, and radioactive. Radon-222 is part of the uranium-238 decay series and is of concern because it can accumulate in homes, particularly in basements. The decay process of Radon-222 involves alpha decay, producing polonium-218. Here’s why it matters:

• Radon gas can seep through cracks in floors and walls, potentially leading to lung cancer with prolonged exposure.
• Understanding radon-222 decay can help in designing better ventilation systems to minimize risk.
• Its relatively short half-life of 3.82 days makes it a useful subject in studies related to radioactive decay dynamics and safety assessments.

Knowledge of Radon-222's properties and decay behavior is crucial for health physics and environmental protection.