Half life of a sample is 20 minutes. In how much time will the acitivity of sample drop to `1//16` of its initial value?

Answer

To solve the problem, we need to determine how long it will take for the activity of a radioactive sample to drop to \( \frac{1}{16} \) of its initial value, given that its half-life is 20 minutes.

### Step-by-step Solution:

1. **Understand the Concept of Half-Life**:
The half-life of a radioactive substance is the time required for half of the sample to decay. In this case, the half-life \( T \) is given as 20 minutes.

2. **Determine the Fraction Remaining**:
We want to find the time when the activity drops to \( \frac{1}{16} \) of its initial value. This can be expressed in terms of half-lives.

3. **Relate the Remaining Activity to Half-Lives**:
The remaining fraction of the sample after \( n \) half-lives is given by:
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \]
where \( N \) is the remaining quantity, \( N_0 \) is the initial quantity, and \( n \) is the number of half-lives.

4. **Set Up the Equation**:
We need to find \( n \) such that:
\[ \left(\frac{1}{2}\right)^n = \frac{1}{16} \]
Since \( \frac{1}{16} = \left(\frac{1}{2}\right)^4 \), we can equate:
\[ n = 4 \]

5. **Calculate the Total Time**:
The total time \( t \) taken for the activity to drop to \( \frac{1}{16} \) of its initial value is:
\[ t = n \times T = 4 \times 20 \text{ minutes} = 80 \text{ minutes} \]

Thus, the time required for the activity of the sample to drop to \( \frac{1}{16} \) of its initial value is **80 minutes**.