Question

The half-life of oxygen-15 is \(124 \mathrm{~s}\). If a sample of oxygen-15 has an activity of \(4000 \mathrm{~Bq}\), how many minutes will elapse before it reaches an activity of \(500 \mathrm{~Bq}\) ?

Short answer

6.2 minutes.

Step-by-step solution

- Understanding the half-life

The half-life of a radioactive substance is the time required for half of the radioactive atoms to decay. For oxygen-15, the half-life is given as 124 seconds.

- Activity Decay Formula

The decay of a radioactive substance can be described using the formula: \( A = A_0 \times \frac{1}{2}^{\frac{t}{T}} \), where \(A\) is the final activity, \(A_0\) is the initial activity, \(t\) is the elapsed time, and \(T\) is the half-life of the substance. In this case, \(A = 500 \mathrm{~Bq}\), \(A_0 = 4000 \mathrm{~Bq}\), and \(T = 124 \text{ seconds} \).

- Solve for elapsed time \(t\)

First, rearrange the equation to solve for \(t\): \[ 500 = 4000 \times \frac{1}{2}^{\frac{t}{124}} \] Divide both sides by 4000: \[ \frac{500}{4000} = \frac{1}{2}^{\frac{t}{124}} \] Simplify the fraction: \[ \frac{1}{8} = \frac{1}{2}^{\frac{t}{124}} \]

- Using logarithms

To solve for \(t\), take the logarithm of both sides of the equation. For simplicity, use the natural logarithm (ln): \[ \text{ln}\bigg(\frac{1}{8}\bigg) = \text{ln}\bigg(\frac{1}{2}^{\frac{t}{124}}\bigg) \] Use the property of logarithms \( \text{ln}(a^b) = b \text{ln}(a) \): \[ \text{ln}\bigg(\frac{1}{8}\bigg) = \frac{t}{124} \times \text{ln}\bigg(\frac{1}{2}\bigg) \]

- Solving the logarithmic equation

Calculate the natural logarithms: \[ \text{ln}\bigg(\frac{1}{8}\bigg) = \text{ln}(8^{-1}) = -\text{ln}(8) = -\text{ln}(2^3) = -3 \text{ln}(2) \] and \[ \text{ln}\bigg(\frac{1}{2}\bigg) = -\text{ln}(2) \] So the equation becomes: \[ -3 \text{ln}(2) = \frac{t}{124} \times -\text{ln}(2) \]

- Isolate \(t\)

Divide both sides by \(-\text{ln}(2)\): \[ t = 3 \times 124 \] Thus, \[ t = 372 \text{ seconds} \]

- Convert seconds to minutes

Since 1 minute is equal to 60 seconds, convert the time from seconds to minutes: \[ \frac{372}{60} = 6.2 \text{ minutes} \]

Key concepts

half-life

The half-life of a radioactive substance is the time it takes for half of the radioactive atoms in a sample to decay. In simpler terms, it's the time needed for the activity of the substance to reduce to half its initial value. For oxygen-15, the half-life is given as 124 seconds. This means that every 124 seconds, the activity of oxygen-15 will be halved.

decay constant

The decay constant (\text{λ}) is a measure of how quickly a substance undergoes radioactive decay. It is defined as the probability per unit time that a given atom will decay. The relationship between the half-life (\text{T}_{1/2}) and the decay constant is given by the formula: \( \text{λ} = \frac{\ln(2)}{\text{T}_{1/2}} \). For oxygen-15, where the half-life is 124 seconds, the decay constant can be calculated as: \( \text{λ} = \frac{\ln(2)}{124 \text{ seconds}} \).

logarithms

Logarithms are the opposite of exponentiation. When dealing with equations involving exponential decay, logarithms help to solve for the unknown variable. For a given equation in the form \( \text{A} = \text{A}_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \), taking the natural logarithm (ln) of both sides simplifies solving for the time \(t\). For instance, using the properties of logarithms: \( \ln\bigg(\frac{1}{8}\bigg) = \ln\bigg(\left(\frac{1}{2}\right)^{\frac{t}{124}}\bigg) \), which further expands using the power rule: \( \ln\bigg(\frac{1}{8}\bigg) = \frac{t}{124} \times \ln\bigg(\frac{1}{2}\bigg) \).

activity decay formula

The activity decay formula describes how the activity of a radioactive substance decreases over time. The formula is: \( \text{A} = \text{A}_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \), where \( \text{A} \) is the final activity, \( \text{A}_0 \) is the initial activity, \( t \) is the elapsed time, and \( T \) is the half-life. This formula allows us to calculate the activity remaining after a certain period. For example, if you start with 4000 Bq of oxygen-15 and want to know how long it will take to reach 500 Bq, you rearrange and solve the formula for \( t \). Through logarithmic manipulation and solving, the time \(t\) calculated would be 372 seconds, which converts to 6.2 minutes.