Question

the half - life of radium - 226 is 1590 years. use the the formula ( a(t)=a_{0}e^{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=)
(b) if a sample contains 100 mg, how many mg will remain after 1000 years? round to 2 decimal places.
amount: ( ) mg
(c) if a sample contains 100 mg, how long will it take to decay to 18.75 mg? round to the nearest year.
time: ( ) years

Explanation:

Step1: Find the value of \( k \)

Given the formula \( A(t)=A_{0}e^{kt} \), for half - life \( t = 1590 \) years and \( A(1590)=\frac{A_{0}}{2} \).
Substitute into the formula: \( \frac{A_{0}}{2}=A_{0}e^{k\times1590} \).
Divide both sides by \( A_{0} \) (since \( A_{0}
eq0 \)): \( \frac{1}{2}=e^{1590k} \).
Take the natural logarithm of both sides: \( \ln(\frac{1}{2})=\ln(e^{1590k}) \).
Using the property \( \ln(e^{x}) = x \), we get \( -\ln(2)=1590k \).
Solve for \( k \): \( k =-\frac{\ln(2)}{1590}\approx - 0.000436 \).

Step2: Find the amount after \( t = 1000 \) years

We know \( A_{0}=100 \), \( k\approx - 0.000436 \), and \( t = 1000 \).
Substitute into \( A(t)=A_{0}e^{kt} \): \( A(1000)=100e^{-0.000436\times1000} \).
First calculate the exponent: \( - 0.000436\times1000=-0.436 \).
Then \( A(1000)=100e^{- 0.436}\approx100\times0.646=64.60 \).

Step3: Find the time \( t \) when \( A(t) = 18.75 \)

We have \( A_{0}=100 \), \( A(t)=18.75 \), and \( k\approx - 0.000436 \).
Substitute into \( A(t)=A_{0}e^{kt} \): \( 18.75 = 100e^{-0.000436t} \).
Divide both sides by \( 100 \): \( 0.1875=e^{-0.000436t} \).
Take the natural logarithm of both sides: \( \ln(0.1875)=\ln(e^{-0.000436t}) \).
Using \( \ln(e^{x}) = x \), we get \( \ln(0.1875)=-0.000436t \).
Solve for \( t \): \( t=-\frac{\ln(0.1875)}{0.000436}\).
Since \( \ln(0.1875)\approx - 1.673 \), then \( t=\frac{1.673}{0.000436}\approx3837 \).

Answer:

a. \( k\approx - 0.000436 \)
b. \( 64.60 \) mg
c. \( 3837 \) years