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Cesium 137 (Cs137)is a Short-Lived Radioactive Isotope e(ln230)t e^{-\left(\frac{\ln 2}{30}\right) t}

Question 68

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Cesium 137 (Cs137) is a short-lived radioactive isotope.It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs137 remaining t years after A0 mg of the radioactive isotope is released is given by  Cesium 137 (Cs<sup>137</sup>) is a short-lived radioactive isotope.It decays at a rate proportional to the amount of itself present and has a half-life of 30 years (i.e., the amount of Cs<sup>137</sup> remaining t years after A<sub>0</sub> mg of the radioactive isotope is released is given by     e^{-\left(\frac{\ln 2}{30}\right)  t}  ) .As a result of its operations, a nuclear power plant releases Cs<sup>137</sup> at a rate of 0.12 mg per year.The plant began its operations in 1990, which we will designate as t = 0.Assume there is no other source of this particular isotope.Which of the following differential equations have a solution R(t) , the amount (in mg) of Cs<sup>137</sup> in t years? (We are assuming R(0) = 0.)  A)   \frac{d R}{d t}=0.12+e^{-\left(\frac{\ln 2}{30}\right)  R}  B)   \frac{d R}{d t}=0.12-e^{-\left(\frac{\ln 2}{30}\right) }+R  C)   \frac{d R}{d t}=0.12-\frac{\ln 2}{30} R  D)   \frac{d R}{d t}=0.12 R-\frac{\ln 2}{30} e(ln230) t e^{-\left(\frac{\ln 2}{30}\right) t} ) .As a result of its operations, a nuclear power plant releases Cs137 at a rate of 0.12 mg per year.The plant began its operations in 1990, which we will designate as t = 0.Assume there is no other source of this particular isotope.Which of the following differential equations have a solution R(t) , the amount (in mg) of Cs137 in t years? (We are assuming R(0) = 0.)


A) dRdt=0.12+e(ln230) R\frac{d R}{d t}=0.12+e^{-\left(\frac{\ln 2}{30}\right) R}
B) dRdt=0.12e(ln230) +R\frac{d R}{d t}=0.12-e^{-\left(\frac{\ln 2}{30}\right) }+R
C) dRdt=0.12ln230R\frac{d R}{d t}=0.12-\frac{\ln 2}{30} R
D) dRdt=0.12Rln230\frac{d R}{d t}=0.12 R-\frac{\ln 2}{30}

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