Question

The activity of a radioactive material is 6.4 × 10^-4 curie. Its half life is 5 days. The activity will become 5 × 10^-6 curie after

• A. 7 days
• B. 15 days
• C. 25 days
• D. 35 days

Answer 1

The Correct Option is D

To determine how long it will take for a radioactive material's activity to decrease from \(6.4 \times 10^{-4}\) curie to \(5 \times 10^{-6}\) curie, we need to use the concept of radioactive decay and the formula for exponential decay of activity.

The formula for the activity \(A\) at any time \(t\) is given by:

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

where:

• \(A_0\) is the initial activity: \(6.4 \times 10^{-4}\) curie
• \(A\) is the final activity: \(5 \times 10^{-6}\) curie
• \(T_{1/2}\) is the half-life: 5 days
• \(t\) is the time in days that we need to find

Rearranging the formula to find \(t\), we get:

\(t = T_{1/2} \times \frac{\log\left(\frac{A_0}{A}\right)}{\log(2)}\)

Substitute the values:

\(t = 5 \times \frac{\log\left(\frac{6.4 \times 10^{-4}}{5 \times 10^{-6}}\right)}{\log(2)}\)

Simplify the fraction inside the logarithm:

\(\frac{6.4 \times 10^{-4}}{5 \times 10^{-6}} = \frac{6.4}{5} \times 10^{2} = 1.28 \times 10^{2}\)

Now substitute back into the equation:

\(t = 5 \times \frac{\log(128)}{\log(2)}\)

\(\log(128) \approx 2.1072\) and \(\log(2) \approx 0.3010\)

Substitute these into the equation:

\(t = 5 \times \frac{2.1072}{0.3010} \approx 5 \times 7 = 35\) days

Therefore, the activity will become \(5 \times 10^{-6}\) curie after 35 days.