To find how long it takes for 0.150 mol of silver-107 to form from 1.25 mol of palladium-107, we need to understand the radioactive decay process and apply the appropriate half-life equations.
Step-by-Step Solution:
1. Initial Conditions:
- Initial amount of palladium-107 ([tex]\(N_0\)[/tex]) = 1.25 mol
- Amount of silver-107 formed = 0.150 mol
- This means the amount of palladium-107 after the formation of silver-107 is:
[tex]\[
N(t) = N_0 - \text{amount of silver-107 formed} = 1.25 \text{ mol} - 0.150 \text{ mol} = 1.1 \text{ mol}
\][/tex]
2. Fraction of Palladium Remaining:
- The fraction of palladium-107 remaining after the time [tex]\(t\)[/tex] is:
[tex]\[
\text{fraction remaining} = \frac{N(t)}{N_0} = \frac{1.1}{1.25} = 0.88
\][/tex]
3. Exponential Decay Formula:
- Palladium-107 decays following the exponential decay law:
[tex]\[
N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\][/tex]
- Where [tex]\(t_{1/2}\)[/tex] is the half-life of palladium-107.
- We can write the fraction remaining in terms of the half-life equation:
[tex]\[
\left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} = 0.88
\][/tex]
4. Solving for Time [tex]\(t\)[/tex]:
- Taking the natural logarithm on both sides gives:
[tex]\[
\ln{\left( \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \right)} = \ln{(0.88)}
\][/tex]
- Simplifying the left-hand side:
[tex]\[
\frac{t}{t_{1/2}} \ln{\left( \frac{1}{2} \right)} = \ln{(0.88)}
\][/tex]
- Since [tex]\(\ln{\left( \frac{1}{2} \right)} = -\ln{2}\)[/tex], we get:
[tex]\[
\frac{t}{t_{1/2}} (-\ln{2}) = \ln{(0.88)}
\][/tex]
- Solving for [tex]\(t\)[/tex]:
[tex]\[
t = \frac{\ln{(0.88)}}{-\ln{2}} \times t_{1/2}
\][/tex]
5. Plugging in the Values:
- Given the half-life [tex]\(t_{1/2}\)[/tex] of palladium-107 is [tex]\(6.5 \times 10^5\)[/tex] years:
[tex]\[
t = \frac{\ln{(0.88)}}{-\ln{2}} \times 6.5 \times 10^5 \text{ years}
\][/tex]
6. Numerical Calculation:
- Computing the natural logarithms, we get:
[tex]\[
\ln{(0.88)} \approx -0.1278 \quad \text{and} \quad \ln{2} \approx 0.6931
\][/tex]
[tex]\[
t = \frac{-0.1278}{-0.6931} \times 6.5 \times 10^5 \text{ years}
\][/tex]
[tex]\[
t \approx 0.1844 \times 6.5 \times 10^5 \text{ years}
\][/tex]
[tex]\[
t \approx 119875.97 \text{ years}
\][/tex]
Hence, the time required for 0.150 mol of silver-107 to form from 1.25 mol of palladium-107 is approximately 119875.97 years.
Therefore, the correct answer is:
[tex]\[
\boxed{1.2 \times 10^5 \text{ years}}
\][/tex]
