Question

A nuclear technician was accidentally exposed to potassium- 42 while doing some brain scans for possible tumors. The error was not discovered until \(36 \mathrm{~h}\) later when the activity of the potassium- 42 sample was \(2.0 \mu \mathrm{Ci}\). If potassium- 42 has a halflife of \(12 \mathrm{~h}\), what was the activity of the sample at the time the technician was exposed?

Short answer

The initial activity was 16.0 μCi.

Step-by-step solution

Understanding the Problem

The problem provides the half-life of potassium-42 and the activity of a sample after a certain time. The goal is to find the initial activity of the sample.

Apply the Half-Life Formula

Use the formula for radioactive decay: \[ A = A_0 \times \frac{1}{2}^{\frac{t}{t_{1/2}}} \]Where \(A\) is the final activity, \(A_0\) is the initial activity, \(t\) is the time that has elapsed, and \(t_{1/2}\) is the half-life.

Substitute the Given Values

Substitute \(A = 2.0 \, \text{μCi}\), \(t = 36 \, \text{h}\), and \(t_{1/2} = 12 \, \text{h}\) into the formula: \[ 2.0 = A_0 \times \frac{1}{2}^{\frac{36}{12}} \]

Simplify the Exponent

Calculate the exponent: \[ \frac{36}{12} = 3 \]Hence, the equation becomes: \[ 2.0 = A_0 \times \frac{1}{2}^3 \]

Solve for the Initial Activity

Simplify further: \[ 2.0 = A_0 \times \frac{1}{8} \]To find \(A_0\), multiply both sides by 8: \[ A_0 = 2.0 \times 8 = 16.0 \, \text{μCi} \]

Key concepts

half-life

Half-life is a key concept in nuclear chemistry. It represents the time it takes for half of a radioactive substance to decay. This means that over each half-life period, the amount of the substance will be reduced by 50%. The concept of half-life helps predict how long a radioactive material will remain active.
For example, if a substance has a half-life of 12 hours, in 12 hours, only half of the substance will remain. After another 12 hours, half of that remaining half will decay, leaving only a quarter of the original substance. This continual halving process is what makes the concept of half-life critical in understanding radioactive decay.

potassium-42

Potassium-42 is a radioactive isotope of potassium. It decays through beta emission and has a half-life of approximately 12 hours.
In the field of nuclear medicine, potassium-42 is used for its radioactive properties. Specifically, it's involved in brain scans and other diagnostics. If you understand the half-life of potassium-42, you can calculate how its radioactivity changes over time.
In our exercise, the technician was accidentally exposed to potassium-42. Its half-life is crucial in determining how much of the material had decayed and what the initial activity was when the exposure occurred.

nuclear chemistry

Nuclear chemistry is a branch of chemistry that deals with the study of radioactive elements and their reactions. It involves understanding the changes in the nucleus of atoms, unlike regular chemistry that involves electron interactions.
Key topics in nuclear chemistry include radioactivity, nuclear reactions, and the properties of radioisotopes. Understanding these concepts helps in various applications like medical imaging, cancer treatment, and energy production.
The concept of half-life and the behavior of isotopes like potassium-42 fall under the umbrella of nuclear chemistry. It allows us to predict how radioactive materials behave over time, which is essential for both safety and practical applications.

radioisotopes

Radioisotopes, or radioactive isotopes, are atoms that have an unstable nucleus. They emit radiation as they decay to become more stable. This process releases energy in the form of alpha, beta, or gamma rays.
Different radioisotopes have different half-lives and decay modes. Some are naturally occurring, while others are man-made. Radioisotopes have numerous applications, from medical imaging to power generation.
In our example, potassium-42 is a radioisotope. Its decay and half-life are key to solving the problem. By understanding how radioisotopes behave, we can tackle questions like calculating the activity of a sample over time.