Question
A sample of argon-39 had an original mass of 1578 grams. After 538 years, the sample is 394.5 grams. What is the half-life of argon-39?
Answers
GPT-4o mini
To determine the half-life of argon-39, we can use the information about the decay of the sample over time. The original mass of the argon-39 sample is 1578 grams, and after 538 years, only 394.5 grams remain. We can use the following formula related to half-life:
\[
N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]
where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life,
- \( t \) is the elapsed time.
First, we need to express the remaining quantity in terms of the original quantity:
\[
N(t) = 394.5 \text{ grams}
\]
\[
N_0 = 1578 \text{ grams}
\]
\[
t = 538 \text{ years}
\]
Substituting these values into the half-life formula, we get:
\[
394.5 = 1578 \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
Next, we will isolate the half-life term \( T_{1/2} \):
1. Divide both sides by 1578:
\[
\frac{394.5}{1578} = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
Calculating the left side:
\[
\frac{394.5}{1578} \approx 0.25
\]
So we have:
\[
0.25 = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
2. Recognize that \( 0.25 = \left( \frac{1}{2} \right)^2 \):
\[
\left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
3. Since the bases are the same, we can equate the exponents:
\[
2 = \frac{538}{T_{1/2}}
\]
4. 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**.
\[
N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]
where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( T_{1/2} \) is the half-life,
- \( t \) is the elapsed time.
First, we need to express the remaining quantity in terms of the original quantity:
\[
N(t) = 394.5 \text{ grams}
\]
\[
N_0 = 1578 \text{ grams}
\]
\[
t = 538 \text{ years}
\]
Substituting these values into the half-life formula, we get:
\[
394.5 = 1578 \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
Next, we will isolate the half-life term \( T_{1/2} \):
1. Divide both sides by 1578:
\[
\frac{394.5}{1578} = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
Calculating the left side:
\[
\frac{394.5}{1578} \approx 0.25
\]
So we have:
\[
0.25 = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
2. Recognize that \( 0.25 = \left( \frac{1}{2} \right)^2 \):
\[
\left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{\frac{538}{T_{1/2}}}
\]
3. Since the bases are the same, we can equate the exponents:
\[
2 = \frac{538}{T_{1/2}}
\]
4. 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**.