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Q46P

Expert-verifiedFound in: Page 1304

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The radioactive nuclide**** ${}^{{\mathbf{99}}}\mathbf{T}\mathbf{c}$ 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 ${}^{{\mathbf{99}}}{\mathit{M}}{\mathit{o}}$****, a radioactive nuclide that decays to ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$**** with a half-life of 67h****. Once a day, the cow is “milked” for its ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$****, which is produced in an excited state by the ${}^{{\mathbf{99}}}{\mathit{M}}{\mathit{o}}$****; the ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$**** 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 6.0h****. (a) By what process does ${}^{{\mathbf{99}}}{\mathit{M}}{\mathit{o}}$**** decay to ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$****? (b) If a patient is injected with a ${\mathbf{8}}{\mathbf{.}}{\mathbf{2}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{7}}}{\mathbf{}}{\mathit{B}}{\mathit{q}}$**** sample of ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$****, 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**** ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$ is 38 per second at a certain time, how many excited states ${}^{{\mathbf{99}}}{\mathit{T}}{\mathit{c}}$**** are located in the tumor at that time?**

- The nuclide${}^{99}\mathrm{Mo}$ decays into nuclide ${}^{99}Tc$ through beta decay.
- The gamma-ray photons produced within the patient each second is $8.2\times {10}^{7}Bq$.
- The number of excited states ${}^{99}Tc$ located in the tumor at that time is $1.2\times {10}^{6}$.

- Half-life of ${}^{99}Tc$, ${T}_{1/{2}_{\mathrm{\pi}}}=67h$
- De-excitation half-life, $R=8.2x{10}^{7}\mathrm{Bq}$
- Activity of the sample, ${T}_{1/2}=6hor6\times 3600s$
- Emission rate of the gamma-ray photons, R'=38 /s

**For decay from Molybdenum element to technetium, the process must be beta decay as the reaction indicates the increase of a proton number that is an atomic number. Now, as the half-life of the de-excitation process is lower than that of the isotope technetium, we get that the activity rate is the rate of emission of gamma-ray photons. Technically, for a given emission rate and de-excitation half-life, one can determine the number of photons emitted.**

The rate of decay is as follows:

$R=\left(\frac{\mathrm{ln}2}{{T}_{1/2}}\right)N$ (i)

Here, ${T}_{1/2}$ is the half-life of the substance, *N *is the number of undecayed nuclei.

Molybdenum beta decays into technetium:

${}_{42}{}^{99}Mo\to {}_{42}{}^{99}Tc+{e}^{-}+\nu $

For the given problem, here the decay results in the addition of a proton that $\beta +$ is decay with production of electron and neutrino.

Each decay of the sample 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 that is given as: $R=8.2x{10}^{7}\mathrm{Bq}$

Hence, the value of the production of gamma-ray photons per second is $R=8.2x{10}^{7}\mathrm{Bq}$.

The number of remaining nuclei of the nuclide gives us the excited states. Thus, using the given data in equation (i), determine the number of excited states ${}^{99}T\mathrm{c}$ located in the tumor at that time as follows:

$N=\frac{R{T}_{\frac{1}{2}}}{\mathrm{ln}2}\phantom{\rule{0ex}{0ex}}=\frac{\left(38{s}^{-1}\right)\left(3600s\right)}{\mathrm{ln}2}\phantom{\rule{0ex}{0ex}}=1.2\times {10}^{6}states$

Hence, the number of excited states is $1.2\times {10}^{6}$.

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