To find the constant of proportionality \( k \) for the decay of calcium-41, we use the half-life formula:
\[ k = \frac{\ln(2)}{t_{1/2}} \]
where \( t_{1/2} \) is the half-life. For calcium-41, the half-life is 106 years.
First, we calculate \( k \):
\[ k = \frac{\ln(2)}{106} \]
Calculating \( \ln(2) \):
\[ \ln(2) \approx 0.693147 \]
Now substituting in the value:
\[ k = \frac{0.693147}{106} \approx 0.00654 \]
Rounding \( k \) to five decimal places:
\[ k \approx 0.00654 \]
Next, we can set up the exponential decay equation given the initial amount \( P_0 = 100 \) grams. The general formula for exponential decay is:
\[ P(t) = P_0 e^{-kt} \]
Substituting the values we have:
\[ P(t) = 100 e^{-0.00654 t} \]
Now, to determine how much calcium-41 will be present after 40 years, we calculate \( P(40) \):
\[ P(40) = 100 e^{-0.00654 \cdot 40} \]
Calculating the exponent:
\[ -0.00654 \cdot 40 = -0.2616 \]
Now calculate \( e^{-0.2616} \):
\[ e^{-0.2616} \approx 0.77019 \]
Now multiplying by 100:
\[ P(40) = 100 \cdot 0.77019 \approx 77.019 \]
Rounding to the hundredths place:
\[ P(40) \approx 77.02 \]
Final Answers
- \( k \) = 0.00654
- \( P(t) \) = \( 100 e^{-0.00654 t} \)
- \( P(40) \approx 77.02 \) grams.