Question

Technetium-99 has been used as a radiographic agent in bone scans $({}_{43}{}^{99}\mathrm{Tc}$is absorbed by bones). If ${}_{\mathbf{43}}{}^{\mathbf{99}}\mathbf{Tc}$has a half-life of hours, what fraction of an administered dose of $\mathbf{100}\mathbf{}\mathit{\mu }\mathit{g}\mathbf{}{}_{\mathbf{43}}{}^{\mathbf{99}}\mathbf{Tc}$remains in a patient's body after 2.0 days?

Short answer

The remaining of${}_{43}{}^{99}\mathrm{Tc}$ is $0.39\mu g$.

Step-by-step solution

Decay constant.

The decay constant of a radioactive element is defined as the “inverse of the time” it takes for the number of atoms remaining in the sample to fall to $\frac{\mathbf{1}}{\mathbf{e}}$ times the original number.

Calculating the decay constant of Tc4399.

The half-life of ${}_{43}{}^{99}\mathrm{Tc}$is 6.0 hours .

The initial amount of technetium-99 is $N_0=100.0\mu \mathrm{g}$.

The time elapsed is:

$2.0\left(days\right)\times \frac{24\mathrm{hours}}{1\mathrm{day}}=48hours$

Decay constant can be evaluated as:

$\begin{array}{c}\mathrm{k}=\frac{\ln \left(2\right)}{{\mathrm{t}}_{1/2}}\\ =\frac{0.693}{6.0\mathrm{h}}\\ =0.1155{\mathrm{h}}^{-1}\end{array}$

 Step 3: Fraction of administered dose.

Now, using the integrated first order equation, the remaining ${}_{43}{}^{99}\mathrm{Tc}$is calculated.

$\begin{array}{c}\ln \left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)=-\mathrm{kt}\\ \ln \left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)=-0.1155{\mathrm{h}}^{-1}\times 48\mathrm{h}\\ \ln \left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)=-5.544\\ \left(\frac{\mathrm{N}}{{\mathrm{N}}_0}\right)={\mathrm{e}}^{-5.544}\\ \mathrm{N}=0.0039\times 100\mu \mathrm{g}\\ =0.39\mu \mathrm{g}\end{array}$

Thus, the remaining of ${}_{43}{}^{99}\mathrm{Tc}$is $0.39\mu \mathrm{g}$.