Question

The activity of a sample is 64 × 10^–5 Ci. Its half-life is 3 days. The activity will become 5 × 10^–6 Ci after.

• A. 12 days
• B. 7 days
• C. 18 days
• D. 21 days

Answer 1

Answer: (D)

21 days

Answer 2

$A = A_{0}\left( \frac{1}{2} \right)^{t/T_{1/2}} \Rightarrow 5 \times 10^{- 6} = 64 \times 10^{- 5}\left( \frac{1}{2} \right)^{t/3}$

$\Rightarrow \frac{1}{128} = \left( \frac{1}{2} \right)^{t/3} \Rightarrow t = 21$ days