Question

A sample of argon-39 initially weighed 1578 grams. After 538 years, the sample's mass reduced to 394.5 grams. What is the half-life of argon-39?(1 point)
Responses

135 years
135 years

538 years
538 years

269 years
269 years

180 years
180 years

Answer 1

To determine the half-life of argon-39, we can use the formula for exponential decay based on its mass reduction over time:

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

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

Given:
- \(N_0 = 1578\) grams,
- \(N_t = 394.5\) grams,
- \(t = 538\) years.

We rearrange the formula to solve for the half-life \(t_{1/2}\):

\[
394.5 = 1578 \left( \frac{1}{2} \right)^{\frac{538}{t_{1/2}}}
\]

First, divide both sides by 1578:

\[
\frac{394.5}{1578} = \left( \frac{1}{2} \right)^{\frac{538}{t_{1/2}}}
\]

Calculating the left side:

\[
\frac{394.5}{1578} \approx 0.25
\]

This can be rewritten as:

\[
0.25 = \left( \frac{1}{2} \right)^{\frac{538}{t_{1/2}}}
\]

Recognizing that \(0.25\) can be expressed as \(\left(\frac{1}{2}\right)^2\), we set the exponent equal to 2:

\[
\frac{538}{t_{1/2}} = 2
\]

Now, solve for \(t_{1/2}\):

\[
t_{1/2} = \frac{538}{2} = 269 \text{ years}
\]

Thus, the half-life of argon-39 is **269 years**.

The correct answer is: **269 years**.