Sure, let's solve this step by step.
First, summarize what's given in the problem:
- Initial mass ([tex]\(N_0\)[/tex]) = 16.0 grams
- Final mass ([tex]\(N(t)\)[/tex]) = 0.250 grams
- Half-life ([tex]\(t_{\frac{1}{2}}\)[/tex]) = 33.0 hours
We need to find the time [tex]\(t\)[/tex] it takes for the sample to decay from 16.0 grams to 0.250 grams.
To solve this, we use the decay formula:
[tex]\[ N(t) = N_0 \cdot \left( \frac{1}{2} \right)^{\frac{t}{t_{\frac{1}{2}}}} \][/tex]
Here's a step-by-step method to find [tex]\(t\)[/tex]:
### Step 1: Set up the equation with the known values
[tex]\[ 0.250 = 16.0 \cdot \left( \frac{1}{2} \right)^{\frac{t}{33.0}} \][/tex]
### Step 2: Isolate the decay term
[tex]\[ \frac{0.250}{16.0} = \left( \frac{1}{2} \right)^{\frac{t}{33.0}} \][/tex]
Calculate the left-hand side:
[tex]\[ \frac{0.250}{16.0} = 0.015625 \][/tex]
So, the equation becomes:
[tex]\[ 0.015625 = \left( \frac{1}{2} \right)^{\frac{t}{33.0}} \][/tex]
### Step 3: Take the natural logarithm of both sides to solve for [tex]\(t\)[/tex]
[tex]\[ \ln(0.015625) = \ln \left( \left( \frac{1}{2} \right)^{\frac{t}{33.0}} \right) \][/tex]
Using the property of logarithms [tex]\(\ln(a^b) = b \cdot \ln(a)\)[/tex]:
[tex]\[ \ln(0.015625) = \frac{t}{33.0} \cdot \ln\left( \frac{1}{2} \right) \][/tex]
### Step 4: Calculate the natural logarithms
[tex]\[ \ln(0.015625) \approx -4.158883 \][/tex]
[tex]\[ \ln\left( \frac{1}{2} \right) \approx -0.693147 \][/tex]
So the equation becomes:
[tex]\[ -4.158883 = \frac{t}{33.0} \cdot -0.693147 \][/tex]
### Step 5: Solve for [tex]\(t\)[/tex]
[tex]\[ t = \frac{-4.158883}{-0.693147} \cdot 33.0 \][/tex]
[tex]\[ t \approx 6 \cdot 33.0 \][/tex]
[tex]\[ t \approx 198.0 \][/tex]
### Conclusion
It takes approximately 198 hours for a 16.0 g sample of Sr to decay to 0.250 g when the half-life is 33.0 hours.
