3) Time constant and radioactive decay (SPECT scans) Single Photon Emission Comp

Question

3) Time constant and radioactive decay (SPECT scans) Single Photon Emission Computed Tomogra Small quantities of a radioactive isotope are injected into a patient's arteries. The decay of the unstable nucleus emits gamma rays which are detected by SPECT imaging instruments to provide three-dimensional (tomographic) images of the distribution of these radioactive tracer molecules in the patient's body. The 3D images are computer generated from alarge number.of projection images of the body recorded at different angles. The cameras are mounted on a rotating gantry that allows the detectors to be moved in a tight circle aroumd a patient who is lying motionless on a pallet. The metastable state of Technetium-99 (99mTc, mass-98.9AMU, t12-360-35 min) is the most commonly used medical radioisotope. What is the radioactive decay time constant (t) ofTc? 4) Time constant and radioactive decay (PET scans) Positron emission tomography (PET): Small quantities of a radioactive isotope such as Fluorine-18 are njected nto a patient's arteries. The nuclear decay of the unstable isotope emits positrons, the antimatter counterparts of electrons. The e &e; meet inside the patient's body, annihilate and produce a pair of gamma rays directions. PEI produces an image with high spatial resolution by mapping gamma rays that arive at the same time, this extra information relative to SPECT yields enhanced spatial resolution (better images). An advantage of PET over procedures that employ gamma emitting tracers is the greater availability of (biologically) suitable isotopes. Positron emitting isotopes of travelling in opposite biologically active elements such as fluorine, carbon and oxvgen are all available. Fluorine-18 in. paticular, can be used to make a radioactive analogue of glucose which is preferentially taken up by brain and cancer cells making an ideal tool for detecting tumours, PET can also be used to map brain function and the diagnosis of conditions such as Alzheimer's disease. (18F, m= 18.000938 AMU, ti 2= 109.771 min). What fraction of the original radioactive lsF will be left in the patient's body 24 hours after they were injected?

Explanation / Answer

mass of Fluorine -18 m= 18.000938 AMU
Half-life t1/2 = 109.771 min
Decay-rate is
? = - ln(2) / t1/2....(1),

ln(2) = 0.693 (approx)

The decay-rate is always negative as the element mass gets reduced

Now, substituting the value of "t1/2" in equation 1,

? = - (0.693) / (109.771 * 60) s= - 1.0522 * 10-4 s-124
24 hours =(24 * 3600) = 86400 seconds = t (say)
thus, the decay in mass after 24 hours = ? * t = -1.0522 * 10-4 * 86400 = - 9.09 (approx)

Therefore, Mass of Fluorine-18 left after 24 hours = Original mass of Fluorine -18 + Decay in mass

= 18.000938 + (? * t) = 18.000938 - 9.09 = 8.910938 AMU

Thus, fraction of 18F (Fluorine -18) left after 24 hour = Mass of Fluorine-18 left after 24 hours / Original mass of Fluorine -18

Or, fraction of 18F (Fluorine -18) left after 24 hour = 8.910938 / 18.000938 = 0.495 (approx)

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