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A specific radioactive isotope has a half-life of 12,000 years.

\(\textbf{1)}\) What is the decay constant k?

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The answer is k\(\approx-.0000577623\)

Show Work

\(\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\,\,\,\,\,\,.5=1e^{k12{,}000}\)

\(\,\,\,\,\,\,\ln{.5}=1\ln{e^{k12{,}000}}\)

\(\,\,\,\,\,\,\ln{.5}=k12{,}000\)

\(\,\,\,\,\,\,\displaystyle\frac{\ln{.5}}{12{,}000}=k\)

\(\,\,\,\,\,\,\)The answer is k\(\approx-.0000577623\)

\(\textbf{2)}\) How long will it take 5 grams to decay to 1 gram?

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The answer is \(27{,}863.14\) years

Show Work

\(\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\,\,\,\,\,\,1=5e^{-.0000577623t}\)

\(\,\,\,\,\,\,\frac{1}{5}=e^{-.0000577623t}\)

\(\,\,\,\,\,\,\ln{\frac{1}{5}}=\ln{e^{-.0000577623t}}\)

\(\,\,\,\,\,\,\ln{.2}=-.0000577623t\)

\(\,\,\,\,\,\,\displaystyle\frac{\ln{.2}}{-.0000577623}=t\)

\(\,\,\,\,\,\,27{,}863.14\approx t\)

\(\,\,\,\,\,\,\)The answer is \(27{,}863.14\) years

\(\textbf{3)}\) How much of a 30 gram sample will remain after 20,000 years?

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The answer is \(9.449\) grams

Show Work

\(\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\,\,\,\,\,\,A_{20{,}000}=30e^{-.0000577623 \cdot (20{,}000)}\)

\(\,\,\,\,\,\,A_{20{,}000}=30e^{-1.155246}\)

\(\,\,\,\,\,\,A_{20{,}000}=30 \cdot (0.31498)\)

\(\,\,\,\,\,\,A_{20{,}000}=9.4494\)

\(\,\,\,\,\,\,\)The answer is \(9.449\) grams

See Related Pages\(\)

\(\bullet\text{ Exponential Functions}\)
\(\,\,\,\,\,\,\,\,y=a(b)^x\)

\(\bullet\text{ Compound Interest}\)
\(\,\,\,\,\,\,\,\,A=P\left(1+\frac{r}{n} \right)^{nt}\,\,\,\,\,\,\,\,A=P e^{rt}\)

\(\bullet\text{ Half Life Questions}\)
\(\,\,\,\,\,\,\,\,A_t=A_0e^{kt}\)

\(\bullet\text{ Graphing Exponential Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=2^{x-1}-2\,\, \)

\(\bullet\text{ Inverse of Exponential Functions}\)
\(\,\,\,\,\,\,\,\,f^{-1}(x)=\log_5(x-1)\)

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