Let's solve the problem step-by-step.
#### Problem:
Selenium-83 decays exponentially and has a half-life of 25 minutes. How long would it take for a 10 mg sample to decay and have only 1 mg remaining?
#### Given:
- Initial amount [tex]\(A_0 = 10\)[/tex] mg
- Final amount [tex]\(A = 1\)[/tex] mg
- Half-life of Selenium-83 [tex]\( t_{\frac{1}{2}} = 25\)[/tex] minutes
#### Step 1: Find the decay constant [tex]\(k\)[/tex]
The decay constant [tex]\(k\)[/tex] can be found using the formula:
[tex]\[ k = \frac{\ln (\frac{A}{A_0})}{t_{\frac{1}{2}}} \][/tex]
Plugging in the values:
[tex]\[ A_0 = 10 \text{ mg} \][/tex]
[tex]\[ A = 1 \text{ mg} \][/tex]
[tex]\[ t_{\frac{1}{2}} = 25 \text{ minutes} \][/tex]
[tex]\[ k = \frac{\ln (\frac{1}{10})}{25} \][/tex]
[tex]\[ k = \frac{\ln (0.1)}{25} \][/tex]
Using properties of logarithms, we have:
[tex]\[ \ln (0.1) \approx -2.302585 \][/tex]
So,
[tex]\[ k = \frac{-2.302585}{25} \][/tex]
[tex]\[ k \approx -0.0921 - 0.0223 \][/tex]
Thus,
[tex]\[ k = -0.0223 \][/tex]
#### Step 2: Find the time [tex]\( t \)[/tex]
The time [tex]\( t \)[/tex] can be found using the formula:
[tex]\[ t = \frac{\ln (\frac{A}{A_0})}{k} \][/tex]
Plugging in the values again:
[tex]\[ A_0 = 10 \text{ mg} \][/tex]
[tex]\[ A = 1 \text{ mg} \][/tex]
[tex]\[ k = -0.0223 \][/tex]
[tex]\[ t = \frac{\ln (\frac{1}{10})}{-0.0223} \][/tex]
[tex]\[ t = \frac{\ln (0.1)}{-0.0223} \][/tex]
Again, using the value of [tex]\(\ln (0.1)\)[/tex]:
[tex]\[ \ln (0.1) \approx -2.302585 \][/tex]
So,
[tex]\[ t = \frac{-2.302585}{-0.0223} \][/tex]
[tex]\[ t \approx 103.25493690556257 \][/tex]
Thus, the time [tex]\( t \)[/tex] it takes for the 10 mg sample of Selenium-83 to decay to 1 mg is approximately 103.25 minutes.
