Question

A human body required \(0.01 \mu\) activity of radioactive substance after 24 hours. Half-life of radioactive substane is 6 hours. Then injection of maximum activity of a radioactive substance that can be injected is (a) \(0.08\) (b) \(0.04\) (c) \(0.16\) (d) \(0.32\)

Short answer

The correct answer is (c) \( 0.16 \).

Step-by-step solution

Understanding the Concept

First, we need to understand that the activity of a radioactive substance decreases over time according to its half-life. The half-life of a substance is the time required for its activity to reduce to half of its initial value. In this problem, the half-life of the substance is 6 hours.

Using the Half-life Formula

We use the formula for radioactive decay: \( A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \), where \( A \) is the final activity, \( A_0 \) is the initial activity, \( t \) is the time elapsed, and \( T \) is the half-life of the substance.

Setting Up the Equation

Given that after 24 hours the activity needs to be \(0.01\ \,\mu\) and the half-life is 6 hours, we set up the equation: \( 0.01 = A_0 \times \left( \frac{1}{2} \right)^{\frac{24}{6}} \). Simplifying \( \frac{24}{6} \) gives us 4 half-lives.

Solving for Initial Activity \( A_0 \)

Substitute the values into the equation to solve for \( A_0 \): \[ 0.01 = A_0 \times \left( \frac{1}{2} \right)^4 \]. This simplifies to \[ 0.01 = A_0 \times \frac{1}{16} \]. Multiply both sides by 16 to solve for \( A_0 \): \[ A_0 = 0.01 \times 16 = 0.16 \].

Conclusion

The initial maximum activity of the substance that can be injected is \( 0.16\, \mu \). Therefore, the correct answer to the problem is option (c) \( 0.16 \).

Key concepts

Half-life

The term 'half-life' refers to the time it takes for half of a radioactive substance to decay. When a substance undergoes radioactive decay, it loses its radioactivity over time. The concept of half-life aids us in determining how quickly this process occurs.

In our original exercise, the half-life of the given substance is 6 hours. This means that every six hours, the amount of the substance that is radioactive is reduced by half. Understanding this concept is crucial because it helps us predict how much of the radioactive substance will be left after a given time period.

For example, if we start with 100 grams of a substance, after one half-life of 6 hours, 50 grams will remain radioactive. After another 6 hours, only 25 grams will be left as radioactive. The formula to calculate the remaining activity after a certain time is needed here, and is typically given by:

• \( A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \)

where \( A \) is the final activity, \( A_0 \) is the initial activity, \( t \) is the elapsed time, and \( T \) is the half-life.

Radioactive Activity

Radioactive activity refers to the rate at which a radioactive source decays. It's dependent on the half-life and the amount of the substance initially present. In simple terms, activity measures how many radioactive decays occur over a period of time.

In the exercise, we aim to keep the radioactive activity at a specific level after 24 hours. To calculate this, we consider how the activity diminishes over time due to the half-life of the substance. This understanding can be applied using the radioactive decay formula:

• \( A = A_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \)

To achieve a desired final activity, we need to adjust the initial activity to compensate for this decay. This process involves using the half-life information to ensure that the radioactive substance will still fulfill its required role after a given period.

Initial and Final Activity

The initial and final activities of a radioactive substance are important factors when determining safety and effectiveness. Initial activity, denoted as \( A_0 \), is the activity of the substance at the beginning, while final activity, denoted as \( A \), is the remaining activity after a specified time.

In calculations, one starts with the desired final activity (in this case \(0.01 \mu\) after 24 hours) and uses the known half-life to determine what the initial activity \(A_0\) must have been. Our exercise shows:

• \( 0.01 = A_0 \times \left( \frac{1}{2} \right)^4 \)
• Solving, yields \(A_0 = 0.16 \mu\)

Knowing both the initial and final activity enables one to predict how long a radioactive substance will remain effective. This relationship is vital in fields like medicine where radioactivity is used for treatments and diagnostics.