Question

The activity of a radioactive material is 2.56 × 10^–3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 × 10^–5 Ci?

• A. 30 days
• B. 35 days
• C. 40 days
• D. 25 days

Answer 1

The Correct Option is B

To solve the given problem, we need to calculate the time it will take for the activity of a radioactive material to decrease from an initial value of \(2.56 \times 10^{-3}\) Ci to \(2 \times 10^{-5}\) Ci, given that its half-life is 5 days.

1. First, let's use the formula for radioactive decay, which relates the remaining activity \(A\) at time \(t\) to the initial activity \(A_0\) and the half-life \(T_{1/2}\):
   A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}
2. Given:
   • \(A_0 = 2.56 \times 10^{-3}\) Ci
   • \(A = 2 \times 10^{-5}\) Ci
   • \(T_{1/2} = 5\) days
3. Substitute the given values into the decay formula:
   2 \times 10^{-5} = 2.56 \times 10^{-3} \left(\frac{1}{2}\right)^{\frac{t}{5}}
4. To solve for \(t\), divide both sides by the initial activity:
   \frac{2 \times 10^{-5}}{2.56 \times 10^{-3}} = \left(\frac{1}{2}\right)^{\frac{t}{5}}
   \Rightarrow 0.0078125 = \left(\frac{1}{2}\right)^{\frac{t}{5}}
5. Take the logarithm of both sides to solve for \(t\):
   \log_{10}(0.0078125) = \frac{t}{5} \log_{10}\left(\frac{1}{2}\right)
6. Calculate the logarithms:
   \log_{10}(0.0078125) \approx -2.1072
   \log_{10}\left(\frac{1}{2}\right) \approx -0.3010
7. Substitute these values back to find \(t\):
   -2.1072 = \frac{t}{5} \times -0.3010
   t = 5 \times \frac{-2.1072}{-0.3010} \approx 35

Therefore, the activity of the radioactive material will become \(2 \times 10^{-5}\) Ci after approximately 35 days. The correct answer is 35 days.