Chemistry Labs

Problem 1

Molecular imaging is a powerful tool in medical diagnostics. The nuclear isomer 99mTc^{99m}\ce{Tc} (m = metastable) is an excellent γ-emitter (t1/2=6.015t_{1/2} = 6.015 h) obtained by β− decay of a mother nuclide in a technetium generator as [99mTcOX4]−[^{99m}\ce{TcO4}]^-. (a) Identify the mother nuclide A and the emitted particle B in A→X99mX2299mTc+B\ce{A -> ^{99m}Tc + B}. (b) An eluate from a 99mTc^{99m}\ce{Tc} generator has an activity of 12.5 GBq (1 GBq = 10910^{9} decays per second). Calculate how many moles of 99mTc^{99m}\ce{Tc} are present. (c) For standard imaging about 200 MBq are administered to a patient. Assuming no activity is lost through excretion, calculate how many hours the patient has to wait until the injected activity decreases to under 1% of the starting activity.
Step 3 of 3: Decay to 1 %
t=ln⁡(A0/A)λ=ln⁡1003.20×10−5=1.44×105 s=39.9 ht = \dfrac{\ln(A_0/A)}{\lambda} = \dfrac{\ln 100}{3.20\times 10^{-5}} = 1.44\times 10^{5}\ \text{s} = 39.9\ \text{h}
Analysis

A=A0e−λtA = A_0 e^{-\lambda t} gives t=ln⁡(200/2)/λ=ln⁡100/λt = \ln(200/2)/\lambda = \ln 100/\lambda. With λ=3.209×10−5\lambda = 3.209\times 10^{-5} s−1^{-1}, t=1.435×105t = 1.435\times 10^{5} s ≈39.9\approx 39.9 h, i.e. about ln⁡100/ln⁡2=6.64\ln 100/\ln 2 = 6.64 half-lives.