To find the constant of proportionality \( k \) for the decay of calcium-41, we can use the half-life formula. The relationship between the half-life and \( k \) is given by:
\[ k = \frac{\ln(0.5)}{t_{1/2}} \]
where \( t_{1/2} \) is the half-life. Given that the half-life of calcium-41 is 106 years, we can substitute to find \( k \):
\[ k = \frac{\ln(0.5)}{106} \]
Calculating \( \ln(0.5) \):
\[ \ln(0.5) \approx -0.69315 \]
Now substituting this value into the equation for \( k \):
\[ k = \frac{-0.69315}{106} \approx -0.00654 \]
Rounding this to five decimal places:
\[ k \approx -0.00654 \]
Next, we can write the exponential decay equation for the amount \( P(t) \) of calcium-41 as follows:
\[ P(t) = P_0 e^{kt} \]
where \( P_0 \) is the initial amount (which is 100 grams), and \( k \) is the proportionality constant we just calculated.
Substituting the values, we have:
\[ P(t) = 100 e^{-0.00654t} \]
Summary
- The constant of proportionality \( k \) is \( -0.00654 \).
- The exponential decay equation is:
\[ P(t) = 100 e^{-0.00654t} \]
Question 2
To determine how much calcium-41 will be present after 40 years, we substitute \( t = 40 \) into the equation:
\[ P(40) = 100 e^{-0.00654 \times 40} \]
Calculating the exponent:
\[ -0.00654 \times 40 \approx -0.2616 \]
Now, using this result in the equation:
\[ P(40) = 100 e^{-0.2616} \]
Calculating \( e^{-0.2616} \):
\[ e^{-0.2616} \approx 0.77060 \quad \text{(using a calculator)} \]
Now plug this value back into the equation:
\[ P(40) = 100 \times 0.77060 \approx 77.06 \]
Final Answer
- After 40 years, the amount of calcium-41 present is:
\[ P(40) \approx 77.06 \text{ grams} \]