Question

If the activity of a radioactive mineral falls from 800 counts/s to 500 counts/s in 80 minutes, the half-life of the mineral is ________ minutes. (Round off to two decimal places.)

Hint:

Use: \(t_{1/2} = 0.693 / \lambda\), and \(\lambda = \frac{\ln(A_0/A)}{t}\).

Answer 1

Correct Answer: 117.4

Step 1: Radioactive decay relation.
\[ A = A_{0} e^{-\lambda t} \] where \(A\) = final activity, \(A_0\) = initial activity, \(t\) = time, and \(\lambda\) = decay constant.
Step 2: Substitute known values.
\[ \frac{A}{A_0} = \frac{500}{800} = 0.625 \] \[ \ln(0.625) = -\lambda \times 80 \] \[ \lambda = -\frac{\ln(0.625)}{80} = 0.00586 \, \text{min}^{-1} \] Step 3: Calculate half-life.
\[ t_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.00586} = 118.3 \, \text{minutes.} \] Step 4: Conclusion.
The half-life of the mineral ≈ 118.3 minutes.