In Nuclear Imaging, radioactive isotopes, called radiotracers, are injected into

Question

In Nuclear Imaging, radioactive isotopes, called radiotracers, are injected into the body, where they emit radiation that is detected by an external camera system. These radiotracers are designed to bind to specific tissues (e.g. tumors) so that the cameras can be used to locate regions where the target tissue is prevalent. Radiotracers are often generated through a three-stage radioactive decay process. An unstable "parent isotope undergoes radioactive decay and produces the radiotracer isotope. The radiotracer isotope undergoes its own simultaneous decay process that produces an unusable by-product. This decay process is illustrated by the expression below: Parent (P) Radiotrace (R)Product (B) Where the decay rates between P to R and R to B are proportional to the amount of radioactive material in P and R, respectively, with proportionality constants 1 and 22 (6 pts) Define a system of ODEs that describes this three-stage decay process. Write your system in the standard matrix form. a. b. (12 pts) Using the system defined in 'a', use the eigenvalue/vector approach to develop expressions for the amount of Parent (P) and Radiotracer (R) available over time. Assume that initially only Parent material is available PCO) Po C. (12 pts) Suppose a new system was developed to generate the Parent isotope. The amount of Parent isotope generated follows the function f(t) 20(-e 005t). If all else remains the same, then the new system becomes Isotope Generation-Parent (P) T Radiotracer (R)By-Product (B) Let ,-0.0 105, 2-0.1 155, and Po-0. Use the eigenvalue/vector approach to develop expressions for the amount of radioactive material available in the P and R stages.

Explanation / Answer

P(0)=0,so Nr/Np=e^-(lemda) 1t

NbNp=e^-(lemda) 2t

Finally we find relation between Nr and Nb

Nr =e^(lemda 2-lemda1)tNb Put lemda 1&lemda2 value

Nr^9*Nb=Np^10 ,N is indicates amount

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