To solve this problem, we need to determine the initial quantity of the isotope [tex]\(^{14}C\)[/tex] given its half-life, the time elapsed, and the amount remaining after that time.
Here’s a step-by-step guide to approach the solution:
1. Understand the Exponential Decay Formula:
The quantity of a radioactive substance can be modeled by the exponential decay formula:
[tex]\[
A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T}}
\][/tex]
Where:
- [tex]\( A \)[/tex] is the remaining quantity after time [tex]\( t \)[/tex].
- [tex]\( A_0 \)[/tex] is the initial quantity.
- [tex]\( t \)[/tex] is the elapsed time.
- [tex]\( T \)[/tex] is the half-life of the substance.
2. Identify the Given Values:
- Half-life ([tex]\( T \)[/tex]) = 5715 years
- Time elapsed ([tex]\( t \)[/tex]) = 1000 years
- Remaining quantity after [tex]\( t \)[/tex] years ([tex]\( A \)[/tex]) = 3 grams
3. Rearrange the Formula to Solve for Initial Quantity [tex]\( A_0 \)[/tex]:
We need to find [tex]\( A_0 \)[/tex]. Rearranging the formula for [tex]\( A_0 \)[/tex]:
[tex]\[
A_0 = \frac{A}{\left( \frac{1}{2} \right)^{\frac{t}{T}}}
\][/tex]
4. Plug in the Known Values:
Substituting the values into the rearranged formula:
[tex]\[
A_0 = \frac{3}{\left( \frac{1}{2} \right)^{\frac{1000}{5715}}}
\][/tex]
5. Compute the Exponential Term:
First, calculate the exponent:
[tex]\[
\frac{1000}{5715} \approx 0.175
\][/tex]
Then, compute the decay factor:
[tex]\[
\left( \frac{1}{2} \right)^{0.175} \approx 0.8825
\][/tex]
6. Calculate the Initial Quantity:
Plugging this back into the equation for [tex]\( A_0 \)[/tex]:
[tex]\[
A_0 = \frac{3}{0.8825} \approx 3.399
\][/tex]
7. Round to Two Decimal Places:
Finally, rounding the result to two decimal places:
[tex]\[
A_0 \approx 3.39 \text{ grams}
\][/tex]
So, the initial quantity of the [tex]\(^{14}C\)[/tex] isotope is approximately [tex]\( 3.39 \)[/tex] grams.
