Find the missing value for the radioactive isotope. (Round your answer to two decimal places.)

\begin{tabular}{|c|c|c|c|}
\hline Isotope & \begin{tabular}{c}
Half-life \\
(years)
\end{tabular} & \begin{tabular}{c}
Initial \\
Quantity
\end{tabular} & \begin{tabular}{c}
Amount After \\
1000 Years
\end{tabular} \\
\hline [tex]${}^{14} C$[/tex] & 5715 & [tex]$\times$[/tex] g & 3 g \\
\hline
\end{tabular}

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Answer :

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.