Question

The half life of a radioactive substance is 20 minutes. In how much time the activity of substance drops to \((\frac{1}{16})^{th}\) of its initial value?

• A. 80 minutes
• B. 20 minutes
• C. 40 minutes
• D. 60 minutes

Answer 1

The Correct Option is A

The half-life of a radioactive substance is the time it takes for half of the substance to decay. Given that the half-life is 20 minutes, we need to determine the time it takes for the activity to reduce to \((\frac{1}{16})^{th}\) of its initial value.

To find this, we use the property that activity reduces exponentially, meaning after one half-life, the activity is halved. After \( n \) half-lives, the activity is reduced to \( \left(\frac{1}{2}\right)^n \) times its initial value.

We need the activity to be \(\frac{1}{16}\), which can be expressed as:

\(\left(\frac{1}{2}\right)^n=\frac{1}{16}\)

Simplifying this gives:

\(n=\log_{1/2}(\frac{1}{16})\)

Recognizing that:

\(\frac{1}{16}=(\frac{1}{2})^4\)

So we have:

\(n=4\)

This means it takes 4 half-lives for the activity to become \((\frac{1}{16})^{th}\) of the original amount. Given one half-life is 20 minutes, the total time \( T \) needed is:

\(T=4 \times 20 = 80\)

Thus, the correct time for the activity of the substance to drop to \((\frac{1}{16})^{th}\) is 80 minutes.