Question

question 9
the half-life of radium-226 is 1590 years. use the the formula $a(t) = a_0e^{kt}$ to model the situation.
(a) find the exact value of $k$, the rate of decay per year. round to 6 decimal places or enter an exact answer.
$k = \square$
(b) if a sample contains 400 mg, how many mg will remain after 1000 years? round to 2 decimal places.
amount: $\square$ mg
(c) if a sample contains 400 mg, how long will it take to decay to 325 mg? round to the nearest year.
time: $\square$ years

Explanation:

Step1: Find k (Part a)

Half - life means when \(t = 1590\), \(A(t)=\frac{A_0}{2}\). Substitute into \(A(t)=A_0e^{kt}\):
\(\frac{A_0}{2}=A_0e^{k\times1590}\)
Divide both sides by \(A_0\) (assuming \(A_0
eq0\)): \(\frac{1}{2}=e^{1590k}\)
Take natural logarithm of both sides: \(\ln(\frac{1}{2}) = 1590k\)
Since \(\ln(\frac{1}{2})=-\ln(2)\), then \(k=\frac{-\ln(2)}{1590}\approx\frac{- 0.693147}{1590}\approx - 0.000436\)

Step2: Find amount after 1000 years (Part b)

We know \(A_0 = 400\), \(t = 1000\), \(k\approx - 0.000436\). Use \(A(t)=A_0e^{kt}\):
\(A(1000)=400\times e^{-0.000436\times1000}\)
First calculate the exponent: \(-0.000436\times1000=-0.436\)
Then \(e^{-0.436}\approx0.65\) (more accurately, \(e^{-0.436}\approx e^{-0.436}\approx0.65\) (using calculator: \(e^{-0.436}\approx0.65\))
\(A(1000)=400\times e^{-0.436}\approx400\times0.6507\approx260.28\) (using more precise calculation of \(e^{-0.436}\))

Step3: Find time to decay to 325 mg (Part c)

We have \(A_0 = 400\), \(A(t)=325\), \(k = \frac{-\ln(2)}{1590}\). Substitute into \(A(t)=A_0e^{kt}\):
\(325 = 400e^{kt}\)
Divide both sides by 400: \(\frac{325}{400}=e^{kt}\)
Simplify: \(\frac{13}{16}=e^{kt}\)
Take natural logarithm: \(\ln(\frac{13}{16})=kt\)
We know \(k=\frac{-\ln(2)}{1590}\), so \(t=\frac{\ln(\frac{13}{16})}{k}=\frac{\ln(\frac{13}{16})}{\frac{-\ln(2)}{1590}}=\frac{1590\times\ln(\frac{13}{16})}{-\ln(2)}\)
\(\ln(\frac{13}{16})=\ln(13)-\ln(16)\approx2.5649 - 2.7726=-0.2077\)
\(t=\frac{1590\times(- 0.2077)}{-0.6931}\approx\frac{1590\times0.2077}{0.6931}\approx\frac{330.243}{0.6931}\approx476\)

Answer:

s:
(a) \(k\approx\boldsymbol{-0.000436}\)
(b) amount \(\approx\boldsymbol{260.28}\) mg
(c) time \(\approx\boldsymbol{476}\) years