Solved: PROBLEM 15RPPotassium-40 Decay One of the most

Chapter , Problem 15RP

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Potassium-40 Decay One of the most abundant metals found throughout the Earth’s crust and oceans is potassium. Although potassium occurs naturally in the form of three isotopes, only the isotope potassium-40 (K-40) is radioactive. This isotope is a bit unusual in that it decays by two different nuclear reactions. Over time, by emitting a beta particle, a great percentage of an initial amount of K-40 decays into the stable isotope calcium-40 (Ca-40), whereas by electron capture a smaller percentage of K-40 decays into the stable isotope argon-40 (Ar-40).* Because the rates at which the amounts C(t) of Ca-40 and A(t) of Ar-40 increase are proportional to the amount K(t) of potassium present, and the rate at which potassium decreases is also proportional to K(t) we obtain the system of linear first-order equation

\(\frac{d C}{d t}=\lambda_{1} K\)

\(\frac{d A}{d t}=\lambda_{2} K\)

\(\frac{d K}{d t}=-\left(\lambda_{1}+\lambda_{2}\right) K\)

Where \(\lambda_{1} \text { and } \lambda_{2}\) are positive constants of proportionality.

(a) From the foregoing system of differential equations find \(K(t) \text { if } K(0)=K_{0}\) if Then find and if C(0) = 0 and A(0) = 0

(b) It is known that \(\lambda_{1}=4.7526 \times 10^{-10}\) and \(\lambda_{2}=0.5874 \times 10^{-10}\). Find the half-life of K-40.

(c) Use your solutions for C(t) and to determine the percentage of an initial amount of \(K_{0}\) of K-40 that decays into Ca-40 and the percentage that decays into Ar-40 over a very long period of time.

Text Transcription:

frac{d C}{d t}=\lambda_{1} K

frac{d A}{d t}=\lambda_{2} K

frac{d K}{d t}=-\left(\lambda_{1}+\lambda_{2}\right) K

lambda_{1} \text { and } \lambda_{2}

K(0)=K_{0}

lambda_{1}=4.7526 \times 10^{-10}

lambda_{2}=0.5874 \times 10^{-10}

K_{0}