Question

If a sample of Bismuth-212 weighs 4.65 grams (g) after 121 seconds (s), what was the original mass of the sample given that Bismuth-212 has a half-life of 60.5 seconds (s)?

Answer 1

To find the original mass of the sample of Bismuth-212, we can use the formula that relates the remaining mass after a period of time to the original mass and the half-life of the substance:

\[
N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]

Where:
- \( N \) is the remaining mass after time \( t \).
- \( N_0 \) is the original mass.
- \( t \) is the elapsed time.
- \( t_{1/2} \) is the half-life.

In this case:
- \( N = 4.65 \, \text{g} \)
- \( t = 121 \, \text{s} \)
- \( t_{1/2} = 60.5 \, \text{s} \)

First, we need to determine how many half-lives have passed in the 121 seconds:

\[
\text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{121 \, \text{s}}{60.5 \, \text{s}} \approx 2.0
\]

This means that approximately 2 half-lives have passed.

Now we can relate the remaining mass to the original mass:

\[
4.65 \, \text{g} = N_0 \left( \frac{1}{2} \right)^{2}
\]

Calculating \( \left( \frac{1}{2} \right)^{2} \):

\[
\left( \frac{1}{2} \right)^{2} = \frac{1}{4}
\]

Thus, we have:

\[
4.65 \, \text{g} = N_0 \cdot \frac{1}{4}
\]

To find \( N_0 \), we can rearrange this equation:

\[
N_0 = 4.65 \, \text{g} \times 4 = 18.6 \, \text{g}
\]

Therefore, the original mass of the sample of Bismuth-212 was:

\[
\boxed{18.6 \, \text{g}}
\]