Question

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?

Short answer

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

Step-by-step solution

The given data

1. Half-life of ${}^{99}Tc$,$T_{1/2_{\mathrm{\pi }}}=67h$
2. De-excitation half-life,$R=8.2x{10}^7\mathrm{Bq}$
3. Activity of the sample, $T_{1/2}=6hor6\times 3600s$
4. Emission rate of the gamma-ray photons, R'=38 /s

Understanding the concept of decay  

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:

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

Here, $T_{1/2}$is the half-life of the substance, Nis the number of undecayed nuclei.

a) Calculate the decay process

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.

b) Calculate the production of gamma-ray photons each second

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 $R=8.2x{10}^7\mathrm{Bq}$.

c) Calculate the number of excited states of technetium

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:

$$\begin{gathered} N=\frac{R{T}_{{\displaystyle \frac{1}{2}}}}{\ln 2} \\ =\frac{\left(38s^{-1}\right)\left(3600s\right)}{\ln 2} \\ =1.2\times {10}^{6}states \end{gathered}$$

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