Question

A sample of \({ }^{28} \mathrm{Mg}\) decays initially at a rate of 53,500 disintegrations \(/ \mathrm{min}\), but the decay rate falls to 10,980 disintegrations/min after \(48.0\) hours. What is the half-life of \({ }^{28} \mathrm{Mg}\) in hours?

Short answer

The half-life of \({ }^{28} \mathrm{Mg}\) is approximately 21.9 hours.

Step-by-step solution

Identify Initial and Remaining Decay Rate

The initial decay rate is 53,500 disintegrations/min and the remaining decay rate after 48.0 hours is 10,980 disintegrations/min. We use these values to determine the decay constant.

Use Exponential Decay Formula

The exponential decay formula is given by \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the number of disintegrations at time \( t \), \( N_0 \) is the initial number of disintegrations, and \( k \) is the decay constant. First, we set up the equation using the initial and final rates: \( 10,980 = 53,500 e^{-48k} \).

Solve for Decay Constant \( k \)

Rearrange the equation to solve for \( k \): \[ e^{-48k} = \frac{10,980}{53,500} \]. Calculate \( \frac{10,980}{53,500} \) to get the result 0.2051. Now, take the natural logarithm of both sides: \( -48k = \ln(0.2051) \). Solve for \( k \): \( k = -\frac{\ln(0.2051)}{48} \). Compute \( k \); \( k \approx 0.0317 \) per hour.

Use Half-Life Formula

The half-life \( T_{1/2} \) is defined as \( T_{1/2} = \frac{\ln(2)}{k} \). Using the value of \( k \) from Step 3, substitute \( k = 0.0317 \) into the formula: \[ T_{1/2} = \frac{\ln(2)}{0.0317} \]. Compute \( T_{1/2} \); \( T_{1/2} \approx 21.9 \) hours.

Key concepts

Half-Life Calculation

In nuclear physics, the concept of half-life is crucial. It represents the time required for a quantity to reduce to half its initial amount due to radioactive decay. Calculating the half-life helps determine how long a substance remains active.
To find the half-life, we use the formula: \( T_{1/2} = \frac{\ln(2)}{k} \). Here, \( \ln(2) \) is the natural logarithm of 2, which approximates to 0.693, and \( k \) is the decay constant, representing the rate of decay.
In the given exercise, after determining the value of \( k \) as 0.0317 per hour, we substitute it into the half-life formula to find that \( T_{1/2} \approx 21.9 \) hours. This means that every 21.9 hours, the radioactive sample of \(^{28} \text{Mg}\) is halved.

Exponential Decay Formula

Exponential decay describes how a quantity diminishes over time at a rate proportional to its current value. The exponential decay formula captures this pattern: \( N(t) = N_0 e^{-kt} \).
Here, \( N(t) \) is the amount remaining at time \( t \), \( N_0 \) is the initial quantity, and \( k \) is the decay constant. Exponential decay is characteristic of radioactive processes, where the rate of change is not constant but decreases over time as the sample size decreases.
In our exercise, we use this formula to relate the initial and final decay rates. By setting up \( 10,980 = 53,500 e^{-48k} \), we solve for \( k \) by manipulating the equation to isolate the decay constant, illustrating how exponential decay is used to model and predict the behavior of decaying substances.

Radioactive Decay Constant

The decay constant, symbolized as \( k \), is a fundamental component in understanding radioactive decay. It indicates the probability per unit time that a nucleus will decay.
A larger decay constant implies a faster decay rate. In other words, the substance will reach its half-life more quickly. The decay constant is determined using the formula \( k = -\frac{\ln\left(\frac{N(t)}{N_0}\right)}{t} \).
In the exercise, the decay constant \( k \) was calculated by plugging in the known values from the initial and remaining decay rates over a period of 48 hours. Upon solving, \( k \approx 0.0317 \) per hour. Thus, the decay constant provides a direct measure of the decay speed, essential for calculating half-lives and predicting future activity.