Problem 20 - Algebra and Trigonometry

Question

After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a) What is the half-life of radon-222? (b) How long will it take the sample to decay to 20% of its original amount?

Step-by-Step Solution

Answer

(a) Half-life: 3.8 days (b) Time for 20%: 7.4 days

Step 1: Determine the Decay Constant

The decay of a substance can be described by the formula \( N(t) = N_0 e^{-kt} \), where \( N_0 \) is the initial amount, \( N(t) \) is the amount after time \( t \), and \( k \) is the decay constant. We know that after 3 days, the sample is 58% of \( N_0 \), i.e., \( N(3) = 0.58N_0 \). Therefore, \( 0.58N_0 = N_0 e^{-3k} \). Divide both sides by \( N_0 \) to get \( 0.58 = e^{-3k} \). Taking the natural logarithm of both sides gives \( \ln(0.58) = -3k \), so \( k = -\frac{\ln(0.58)}{3} \).

Step 2: Calculate the Half-Life

The half-life \( T_{1/2} \) is the time it takes for the sample to decay to 50% of its original amount. This can be found using the formula \( T_{1/2} = \frac{\ln(2)}{k} \). Substitute \( k = -\frac{\ln(0.58)}{3} \) into the equation to get \[ T_{1/2} = \frac{\ln(2)}{-\frac{\ln(0.58)}{3}} \]. Calculate this value to find the half-life.

Step 3: Solve for Time to Reach 20%

We use the same decay formula \( N(t) = N_0 e^{-kt} \) for part (b). Now, \( N(t) = 0.20N_0 \). Thus, \( 0.20N_0 = N_0 e^{-kt} \), which simplifies to \( 0.20 = e^{-kt} \). Taking the natural logarithm gives \( \ln(0.20) = -kt \), which rearranges to \[ t = -\frac{\ln(0.20)}{k} \]. Substitute the value of \( k \) from Step 1 to compute \( t \).

Key Concepts

Half-life Calculation

In radioactive decay, the half-life of a substance is a crucial concept. It represents the time it takes for half of the radioactive atoms in a sample to decay. Understanding half-life helps us predict how long it will take for a particular amount of substance to reduce to half its initial quantity.

Here's how to calculate it:

• The formula for half-life is given by: \[ T_{1/2} = \frac{\ln(2)}{k} \]where \( k \) is the decay constant.
• In the original problem, we had used the decay constant derived from the relationship \( N(t) = N_0 e^{-kt} \). When the sample decays to 58% after 3 days, the decay constant \( k \) was determined by solving the equation \( 0.58 = e^{-3k} \).
• Using a calculated decay constant, you can plug this value into the half-life formula to find out how long it takes for the sample to decay to 50% of its original quantity.

Calculating the half-life provides a clearer picture of the rate at which a radioactive isotope decays and assists in predicting future decay periods.

Exponential Decay

Exponential decay describes a process wherein the quantity of a substance reduces at a rate proportional to its current value. This is a common pattern found in radioactive decay, where the rate of decay slows down over time as there is less material to decay.

The formula capturing exponential decay is:

• \( N(t) = N_0 e^{-kt} \)
• Here, \( N_0 \) is the initial quantity of the substance, \( N(t) \) is the remaining quantity at time \( t \), \( k \) is the decay constant, and \( e \) is the base of the natural logarithm.

In our example, exponential decay is demonstrated through the process of a radon-222 sample decaying to 58% of its original amount after 3 days. By using exponential decay, you can determine how much of the substance will remain after a given period. This behavior of substances is predictable, allowing us to understand the decay pattern over time. It also complements the concept of half-life, providing a mathematical framework to analyze decay rates.

Decay Constant

The decay constant, denoted by \( k \), is a fundamental part of understanding radioactive decay. It represents the probability of decay per unit time and provides insight into how quickly a substance is decaying.

Here's how it's applied:

• The decay constant is derived from the equation \( N(t) = N_0 e^{-kt} \), where it defines the rate at which the decay process occurs.
• To find \( k \), we can rearrange the formula to \( k = -\frac{\ln(N(t)/N_0)}{t} \).
• In our situation with radon-222, knowing that 58% of the original amount remains after 3 days allowed us to solve for \( k \) by using the equation \( 0.58 = e^{-3k} \).

Understanding the decay constant enables you to quantitate adjustments in the decay rate. It serves as a bridge between the initialization of a sample's decay and the calculation of its changes over time. By calculating \( k \), predictions on how quickly a sample will decay can be accurately made.