Question 42.46: The radioactive nuclide ^99Tc can be injected into a patient......
The radioactive nuclide ^{99}Tc can be injected into a patient’s bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a “cow” containing ^{99}Mo, a radioactive nuclide that decays to ^{99}Tc with a half-life of 67 h. Once a day, the cow is “milked” for its ^{99}Tc, which is produced in an excited state by the ^{99}Mo; the ^{99}Tc de-excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half-life of 6.0 h. (a) By what process does ^{99}Mo decay to 99Tc? (b) If a patient is injected with an 8.2 × 10^7 Bq sample of ^{99}Tc, how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected ^{99}Tc is 38 per second at a certain time, how many excited-state ^{99}Tc are located in the tumor at that time?
(a) Molybdenum beta decays into technetium:
{ }_{42}^{99} Mo \rightarrow{ }_{43}^{99} Tc +e^{-}+v
(b) Each decay corresponds to a photon produced when the technetium nucleus de-excites (note that the de-excitation half-life is much less than the beta decay half-life). Thus, the gamma rate is the same as the decay rate: 8.2 × 10^7/s.
(c) Equation 42-20 leads to
R=\frac{N_{40} \ln 2}{T_{1 / 2}} (42-20)
N=\frac{R T_{1 / 2}}{\ln 2}=\frac{(38 / \mathrm{s})(6.0 \mathrm{~h})(3600 \mathrm{~s} / \mathrm{h})}{\ln 2}=1.2 \times 10^6 .