Question

A radioactive isotope having $t_{1/2}$ = 3 days was read after 12 days. If 3g of isotope is contained in container, the initial weight of isotope was

Answer 1

48 g

Answer 2

Core Solution:

The half-life ($t_{1/2}$) of the radioactive isotope is 3 days.
The time elapsed ($t$) is 12 days.
The remaining weight ($N$) is 3 g.

First, calculate the number of half-lives ($n$) that have occurred: $n = \frac{\text{Total time elapsed}}{\text{Half-life}} = \frac{t}{t_{1/2}}$ $n = \frac{12 \text{ days}}{3 \text{ days}} = 4$

Next, use the radioactive decay formula to find the initial weight ($N_0$): $N = N_0 \left(\frac{1}{2}\right)^n$ Substitute the known values: $3 \text{ g} = N_0 \left(\frac{1}{2}\right)^4$ $3 = N_0 \times \frac{1}{16}$

Solve for $N_0$: $N_0 = 3 \times 16$ $N_0 = 48 \text{ g}$

Answer:

The initial weight of the isotope was 48 g.