Question

Radiogenic atoms of \(^{40}K\) have a half-life of \(1.25 \times 10^9\) years. The percentage of \(^{40}K\) atoms left after six half-lives will be .............. (Round off to two decimal places)

Hint:

To calculate the percentage remaining after \(n\) half-lives, use the formula \(\left( \frac{1}{2} \right)^n \times 100\), where \(n\) is the number of half-lives.

Answer 1

Step 1: Understand the half-life process.
After each half-life, half of the remaining atoms are decayed. After \(n\) half-lives, the percentage of remaining atoms is given by: \[ \text{Remaining percentage} = \left( \frac{1}{2} \right)^n \times 100 \] Step 2: Calculate the percentage after six half-lives.
For \(n = 6\), we have: \[ \text{Remaining percentage} = \left( \frac{1}{2} \right)^6 \times 100 = \frac{1}{64} \times 100 = 1.5625% \] Final Answer: \[ \boxed{1.56} \]