A radioactive isotope having a half-life period of 3 days was received after 12 days. If 3 g of the isotope is left in the container, what would be the initial mass of the isotope?

A radioactive isotope decays over time, and we are given its half-life is 3 days. This means that every 3 days, the amount of the isotope reduces to half of what it was at the start of that period. The isotope was received after 12 days, and at that time, 3 grams are left. We need to find the initial mass.

The formula for the remaining amount of a radioactive substance after time $$ t $$ is:

$$ N = N_0 \left( \frac{1}{2} \right)^{t / T_{1/2}} $$

where $$ N $$ is the amount left, $$ N_0 $$ is the initial amount, $$ t $$ is the time elapsed, and $$ T_{1/2} $$ is the half-life period.

Here, $$ N = 3 $$ grams, $$ t = 12 $$ days, and $$ T_{1/2} = 3 $$ days. We substitute these values into the formula:

$$ 3 = N_0 \left( \frac{1}{2} \right)^{12 / 3} $$

First, simplify the exponent: $$ 12 / 3 = 4 $$, so the equation becomes:

$$ 3 = N_0 \left( \frac{1}{2} \right)^4 $$

Now, calculate $$ \left( \frac{1}{2} \right)^4 $$:

$$ \left( \frac{1}{2} \right)^4 = \frac{1^4}{2^4} = \frac{1}{16} $$

So the equation is:

$$ 3 = N_0 \times \frac{1}{16} $$

To solve for $$ N_0 $$, multiply both sides by 16:

$$ N_0 = 3 \times 16 $$

$$ N_0 = 48 $$

Therefore, the initial mass of the isotope was 48 grams.

Looking at the options:

A. 12 g

B. 36 g

C. 48 g

D. 24 g

Hence, the correct answer is Option C.