Question

The rate at which a radioactive tracer is lost from a patient’s body is the rate at which the isotope decays plus the rate at which the element is excreted from the body. Medical experiments have shown that stable isotopes of a particular element are excreted with a 6.0 day half-life. A radioactive isotope of the same element has a half-life of 9.0 days. What is the effective half-life of the isotope, in days, in a patient’s body?

Short answer

There fore the effective half life of the isotope is 3.6days

Step-by-step solution

Step 1. Introduction

In radioactive decay the unstable nuclie decay first.. if ${\lambda }_1\mathrm{and}{\mathrm{\lambda }}_2$are the decay constant representing each decay processes, then

The total decay rate is

$\frac{-dN}{dt}={\lambda }_{1}N+{\lambda }_{2}N$

The solution is,

$$\begin{gathered} N=N_0{e}^{-({\lambda }_1+{\lambda }_2)t} \\ \mathrm{Here},\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{atoms}\mathrm{at}\mathrm{the}\mathrm{time}t=0is{N}_0 \end{gathered}$$

Step 2. Explanation

Half lifetime(t_1/2) is the time within which the nuclei decay to half or it original amount

That is

$$\begin{gathered} N=\frac{No}{2}ast\to t\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{The}\mathrm{equation}\left(1\right)\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{written} \\ N=No{\left(\frac{1}{2}\right)}^{-\left(\frac{1}{{\left(t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}^{}\right)}_1}+\frac{1}{{\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{}\right)}_2}\right)} \end{gathered}$$

Step 3

Here the half lives of the two process are

The half lives of the process can then be combined in the following fashion to obtain the effective half life

$\frac{1}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\left(\frac{1}{{\left(t_{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}^{}\right)}_1}+\frac{1}{{\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{}\right)}_2}\right)$

Step 4

The medical experiments

$$\begin{gathered} \frac{1}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\left(\frac{1}{6days}+\frac{1}{9days}\right) \\ \frac{1}{t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{\left(6\right)\left(9\right)}{6+9}days \\ =3.6days \end{gathered}$$

From the medical experiments it has been revealed that the iscope of certain element

There fore the effective half life of the isotope is 3.6days