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I was doing a multiple choice problem that asked about when the following nuclear decay sequence reaches steady state (When does the activity of $\ce{^99Mo}$ roughly equal the activity of $\ce{^{99m}Tc}$ ?)

decay sequence

The correct answer choice said after approximately 20 hours.

Let's denote $\ce{^99Mo}$ by $\ce{R}$ and $\ce{^{99m}Tc}$ by $\ce{I}$.

By definition, the time we're looking for is the time when $\frac{\ln 2}{65}N_R=\frac{\ln 2}{6}N_I$, which is when $\frac{N_R}{N_I}=10.83$ I don't have the formula for the exact amount of the intermediate vs. time memorized, but I don't think the problem required us to know the formula. I tried estimating the amounts of the various species present after 20 hours to see if steady state is actually reached then:

Let the initial ${N_R}=1$. I know that at $t=20$, $N_R=\frac{1}{2}^{20/65}=0.8079$. To estimate $N_I$, I did $N_I\approx(1-0.8079)*\frac{1}{2}^{20/6}=0.0191$. But using this approximation, it seems like the steady state condition is not met. Thus, I was wondering, what is a better way to estimate $N_I$ to get that the decay sequence is in steady state after 20 hours?

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This type of consecutive reaction does not really come to steady state for very long, as seen in the figure, i.e there is only a small time over which $d[mTC]/dt \approx 0$. The decay rate of mTC is faster than that of Mo and so mTc appears quickly and decays slowly. (The amount of mTC is multiplied by 10 in the figure is the amount formed is small].

The figure is drawn assuming $k_1 = \ln(2)/65, \, k_2 =\ln(2)/6 $ and $\ce{[Mo]_0} = 1$ and that the other species are zero at $t=0$ The equations are $\displaystyle \ce{[Mo]=[Mo]}_0e^{-k_1t}$ and $$\displaystyle \ce{[mTc] = [Mo]}_0\frac{k_1}{k_1-k_2}\left( e^{-k_1t}-e^{-k_2t} \right)$$ The amount of Tc is found as $\ce {[Tc] = [Mo]_0-[Mo]-[mTc]}$

The maximum amount of mTC, which is assumed to be 'steady state' is found by differentiating.

The time at maximum is $\displaystyle t_m= \ln(k_2/k_1)/(k_2-k_1)$ which is 22.7 hours so the 'steady state' is established just before this and lasts a little past this time. How much in either way is somewhat subjective unless you use a definition of say $\pm 5$ % of the maximum etc.
ABC kinetics

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  • $\begingroup$ Thanks for the answer. I was asking more about how to estimate [m Tc] without using the actual formula, because I don't think I'm expected to have the formula memorized for this test. $\endgroup$ – carbenoid Jun 4 '17 at 16:19
  • $\begingroup$ As the mTc decays rapidly compared to Mo then at the smallest time t when $\exp(-k_2t) \lt 1$ the ratio $[Mo]/[mTC] = k_2/k_1 -1$ but I can't see how this time gives anything more than a very approximate estimate at which the transient steady state is present. $\endgroup$ – porphyrin Jun 4 '17 at 21:16

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