Radioactive decay follows first-order kinetics . The activity ($A$) of a sample after a period of time is related to its initial activity ($A_0$) by the number of half-lives ($n$) that have passed, using the formula $A = A_0 / 2^n$ .

Given the following data:

• Final activity required ($A$) = $0.01\text{ Curie}$
• Total time elapsed ($t$) = $24\text{ hours}$
• Half-life ($T_{1/2}$) = $6\text{ hours}$

1. Calculate the number of half-lives ($n$): $n = \frac{t}{T_{1/2}} = \frac{24}{6} = 4$ half-lives.

2. Calculate the initial activity ($A_0$): $0.01 = \frac{A_0}{2^4}$ $0.01 = \frac{A_0}{16}$ $A_0 = 0.01 \times 16 = 0.16\text{ Curie}$.

Therefore, the initial activity to be injected is $0.16\text{ Curie}$.