Problem 37 - Chemistry: The Central Science in SI Units, Global Edition

Question

Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt- 60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. If an original sample was purchased in June \(2016,\) when will it be necessary to replace the cobalt- \(60 ?\)

Step-by-Step Solution

Answer

The Cobalt-60 in the radiotherapy unit will need to be replaced in March 2018, as it takes approximately 1.84 years for its radioactivity to decrease to 75% of its original value.

Step 1: Identify the formula for radioactive decay.

The formula for radioactive decay is given by \[A_t = A_0 \cdot (1/2)^{\frac{t}{t_{1/2}}}\] Where, \(A_t\) = Activity at time t, \(A_0\) = Initial activity, \(t\) = Time (in years), \(t_{1/2}\) = Half-life of the substance (in years).

Step 2: Set up the equation using the given information.

We know that we have to replace Cobalt-60 when its radioactivity falls to \(75\%\) of the original sample. So, \(A_t = 0.75A_0\). The half-life of Cobalt-60 is given as 5.26 years. We need to find the time t when the activity reduces to 75%. So, the equation becomes: \[0.75A_0 = A_0 \cdot (1/2)^{\frac{t}{5.26}}\]

Step 3: Solve for the time t.

As we are interested in finding the time t, we can eliminate the term \(A_0\) from the equation: \[\frac{0.75A_0}{A_0} = (1/2)^{\frac{t}{5.26}}\] Now, take the natural logarithm of both sides of the equation: \[\ln{0.75} = \ln{(1/2)^{\frac{t}{5.26}}}\] Next, use the logarithm power rule to move the exponent to the front: \[\ln{0.75} = \frac{t}{5.26} \ln{\frac{1}{2}}\] Now, solve for t by isolating t on one side of the equation: \[t = 5.26 \cdot \frac{\ln{0.75}}{\ln{\frac{1}{2}}}\] Calculate t: \[t \approx 1.84\]

Step 4: Determine the replacement date of Cobalt-60.

The original sample was purchased in June 2016. It will take about 1.84 years for the radioactivity to decrease to 75% of its original value. Thus, we need to replace the Cobalt-60 after 1.84 years from June 2016. Let's convert 1.84 years to months: \[1.84 \mathrm{years} \times \frac{12 \mathrm{months}}{1 \mathrm{year}} \approx 22.08 \mathrm{months}\] Round this value to the nearest whole number, which is 22 months. Count 22 months from June 2016: June (1), July (2), August (3), ... , March (22) So, it will be necessary to replace the Cobalt-60 in March 2018.

Key Concepts

Cobalt-60

Cobalt-60 is a radioactive isotope of cobalt, commonly used in various scientific and medical applications. What makes Cobalt-60 particularly interesting is its ability to emit gamma rays. These gamma emissions occur because Cobalt-60 undergoes a type of change called radioactive decay.
When an unstable atomic nucleus, like that of Cobalt-60, releases energy by emitting radiation, it's known as radioactive decay. This process helps stabilize the atom. The emitted gamma rays have high energy and can penetrate materials, making them ideal for certain applications.
Cobalt-60 has a relatively long half-life of approximately 5.26 years. This makes it suitable for applications that require sustained gamma radiation over a period of time, such as radiotherapy. It's important to consider factors like decay rate when using Cobalt-60, as it must be replaced periodically to ensure effective functionality.

Half-life

Half-life is a fundamental concept in understanding radioactive decay. It's defined as the time required for a quantity of a radioactive substance to reduce to half its initial amount. Every radioactive element has its own unique half-life, reflecting how quickly or slowly it decays.
For Cobalt-60, the half-life is 5.26 years. This means that every 5.26 years, the activity or radioactivity of a Cobalt-60 sample will decrease by half. This predictable decrease helps gauge when the substance might become less effective and need replacement.
The mathematical representation of half-life in calculations is crucial for determining the period within which activities like replacement should occur. Using the decay formula, one can calculate how quickly a substance like Cobalt-60 will decay to a certain activity percentage, such as the problem scenario where the activity decreases to 75%.

Radiotherapy

Radiotherapy is a medical treatment that uses radiation to destroy cancer cells. It strategically damages the DNA inside cancer cells, inhibiting their ability to reproduce. Radiotherapy can be external, where the radiation source is outside the body, or internal, where radioactive materials are placed inside the body.
Cobalt-60 plays an instrumental role in external beam radiotherapy. Its gamma rays can penetrate deep into tissues, targeting tumors with precision. This makes it useful in treating various types of cancer safely and effectively. However, it is crucial that Cobalt-60's activity is closely monitored. If its radioactivity falls below the necessary level, it may not deliver a sufficient dose to effectively treat the cancer.
Replacing Cobalt-60 sources at the appropriate time, such as when they fall to 75% activity, ensures that treatments remain effective and safe for patients. This is an essential part of managing a successful radiotherapy treatment program.