Problem 93 - Chemistry The Molecular Science

Question

A sample of the alpha emitter \({ }_{86}^{222} \mathrm{Rn}\) had an initial activity, \(A_{0},\) of \(7.00 \times 10^{4}\) Bq. After 10.0 days its activity, \(A\) had fallen to \(1.15 \times 10^{4} \mathrm{~Bq} .\) Calculate the decay constant and half-life of radon-222.

Step-by-Step Solution

Answer

The decay constant is approximately \(2.1 \times 10^{-6} \) s\(^{-1}\) and the half-life is about 3.82 days.

Step 1: Understand the Relationship

The radioactive decay equation is given by \( A = A_0 e^{- \lambda t} \), where \( A \) is the current activity, \( A_0 \) is the initial activity, \( \lambda \) is the decay constant, and \( t \) is the time elapsed. To find the decay constant, we'll rearrange this equation.

Step 2: Rearrange for Decay Constant

Rearranging the decay equation for \( \lambda \) gives: \( \lambda = \frac{1}{t} \ln \left(\frac{A_0}{A}\right) \). Now, we can use this formula to find the decay constant.

Step 3: Substitute Values

Substitute the given values into the equation: \( A_0 = 7.00 \times 10^4 \) Bq, \( A = 1.15 \times 10^4 \) Bq, and \( t = 10.0 \) days. This needs to be converted to seconds, so \( t = 10 \times 24 \times 3600 = 864000 \) seconds. The equation becomes \( \lambda = \frac{1}{864000} \ln \left( \frac{7.00 \times 10^4}{1.15 \times 10^4} \right) \).

Step 4: Compute Decay Constant

Calculate \( \lambda \): \( \lambda = \frac{1}{864000} \ln \left( \frac{7.00}{1.15} \right) \). This gives approximately \( \lambda \approx 2.1 \times 10^{-6} \) s\(^{-1}\).

Step 5: Calculate the Half-Life

The half-life \( t_{1/2} \) is related to the decay constant by \( t_{1/2} = \frac{\ln 2}{\lambda} \). Substitute the calculated \( \lambda \): \( t_{1/2} = \frac{0.693}{2.1 \times 10^{-6}} \approx 3.3 \times 10^5 \) seconds.

Step 6: Convert Half-Life to Days

Convert the half-life from seconds to days: \( t_{1/2} = \frac{3.3 \times 10^5}{86400} \approx 3.82 \) days.

Key Concepts

Decay Constant

The decay constant, often denoted by the Greek letter lambda (\( \lambda \)), is a critical value in the study of radioactive decay. This constant represents the probability per unit time that a nucleus will decay. In essence, it provides an understanding of how quickly a radioactive substance is disintegrating. To find the decay constant, we use the formula derived from the basic radioactive decay equation:

• \( A = A_0 e^{-\lambda t} \)

In this equation:

• \( A \) is the observed activity after time \( t \)
• \( A_0 \) is the initial activity
• \( \lambda \) is the decay constant
• \( t \) is the time elapsed

By rearranging the decay equation, we arrive at the formula for the decay constant: 

• \( \lambda = \frac{1}{t} \ln \left(\frac{A_0}{A}\right) \)

This equation shows that the decay constant depends on the time it takes for a certain activity level to be achieved, and the ratio of initial to current activity. Knowing \( \lambda \) can help predict the future behavior of the radioactive material, making it essential for both theoretical and practical applications.

Half-Life Calculation

The half-life of a radioactive isotope is the time required for half of the substance to decay. During this period, the activity or amount of the radioactive isotope decreases to half its initial value. The half-life is often symbolized as \( t_{1/2} \).Half-life is directly related to the decay constant through the convenient relationship:

• \( t_{1/2} = \frac{\ln 2}{\lambda} \)

Here, \( \ln 2 \) is a natural logarithm of 2, approximately 0.693. This relationship implies that the greater the decay constant, the shorter the half-life as the substance decays more quickly.For practical calculations, if the decay constant \( \lambda \) is known, substituting it into the formula will give the half-life. For radon-222 in our exercise:

• With \( \lambda \approx 2.1 \times 10^{-6} \) \( \text{s}^{-1} \), the half-life \( t_{1/2} = \frac{0.693}{2.1 \times 10^{-6}} \approx 3.82 \text{ days} \)

This calculation shows how the half-life is derived from the decay constant, offering an intuitive grasp of how long the decay process might last.

Alpha Emission

Alpha emission is a type of radioactive decay where an unstable nucleus emits an alpha particle. Alpha particles are essentially helium nuclei, consisting of two protons and two neutrons, and they are symbolized as \( \prescript{4}{2}{\alpha} \) or \( \prescript{4}{2}{\text{He}} \).In the context of our exercise involving radon-222, alpha emission is the process by which this element reduces its instability:

• When radon-222 undergoes alpha decay, it produces radium-218, a different element, and in the process loses two protons and two neutrons.

This process of emission occurs because the alpha particle clusters together within the nucleus and breaks away. Due to their relatively large mass, alpha particles do not travel far and can be stopped by barriers as thin as paper or human skin. However, they deposit energy in a small volume, making them highly damaging to living tissues if the source is ingested or inhaled.Alpha emission is crucial in explaining changes in the identity of a radioactive isotope and plays a significant role in nuclear reactions and applications such as smoke detectors and cancer treatments.