Question

Technetium Dosage. Radioactive isotopes are often used for medical imaging. Technetium-99m (Tc-99m) is a particularly popular isotope since it has a short half-life and is easy to make. Also, Tc-99m can be chemically bonded to a number of different compounds and cell types for imaging different parts of the body. Tc-99m has a half-life of 6.02 hours. It decays into Tc-99 by emitting a gamma ray of energy 140 keV. For this problem, assume that Tc-99m is present in the body for 2.5 hours after being injected and then is immediately removed. This assumption is reasonable for certain techniques such as renal scans. (For others, a more complex model of technetium removal is needed.) a. A typical amount of Tc-99m injected for medical imaging is 15.9 mCi. Convert this to Bq. Initial activity = Bq b. What number No of Tc-99m atoms must be present to have an activity of 15.9 mCi? Initial amount = nuclei c. Using the value of No that you calculated in part b), find the number of decays AN that occur in 2.5 hours. Number of decays in 2.5 hours = d. The gamma ray released by each decay carries 140 keV of energy. Find the total energy E (in J) released by these decays in 2.5 hours. Total energy released = e. If the amount of Tc-99m that you calculated in part d) were administered to an 71 kg man, staying in his body for 2.5 hours as described, what would his absorbed dose be in rads? Absorbed dosage = rad f. Equivalent dose correlates strongly with the biological effects of radiation. Calculate the equivalent dose in rem, keeping in mind that Tc- 99m emits gamma rays. Equivalent dosage = rem g. How would the equivalent dose change if Tc-99m emitted a beta ray of the same energy? Greatly increase Slightly increase Greatly decrease Slightly decrease h. What would be the equivalent dosage if this quantity of Tc-99m had been administered to a small dog of mass 3 kg? Equivalent dosage = rem

Answer 1

o Initial activity = 15.9 mei = 15.9x103 ä .-15.9x10 X3.7 xlolo =588.3 x106. Bq. Bq activity Activity 0.693 112 3) 588.3 8106 0.693 6.02 x 3600 N 2 Nos 108441013 atoms C) The no. of atoms present after 2.5hr N= No 205/6.02 = 1.84x1013 1.33356 7.33356 - 1.38 x 1013 atoms. in No. of atoms decays LN = No-N = (1.84-138) x10 12 - 0.46X10 alons d) Energy in each decay = 140 ken = 140 x103 x 16 xol - 2.24x104 s. i. Total energy released - 2.24x1024 XAN = 2124 x 1024 x 0.46x1013 14 e) Absorbed dose = 1.0304x10 = 0.10304 75 = 1.3738 x103 Tlus. = 1.37 38 X10 3 Xloo = 0.13738 radi