Question

When radioactive tracers are used for medical scans, the tracers both decay in the body via radiation, as well as being excreted from the body. The effective half-life of the radioactive tracer is due to the combination of both effects. Medical experiments show that a stable (nonradioactive) isotope of a particular element have an excretion half-life of 6 days. A radioactive isotope of the same element has a half-life of 9 days. What is the effective half-life of the radioactive tracer due to both effects?

Answer 1

Answer:

The effective half-life is  $(t_{\frac{1}{2} })_T = 8.6 \ days$

Explanation:

From the question we are told that

   The excretion half-life is  $t__{\frac{1}{2} } excretion} = 6 \ days$

    The  radiation half-life is  $t__{\frac{1}{2} } radiation} = 9 \ days$

The decay due to excretion is mathematically represented as

          $\frac{dN_a}{dt} = \lambda_e N_i$

Where $Ni$ is the original number of tracers

    The decay due to excretion is mathematically represented as

          $\frac{dN_b}{dt} = \lambda_r N_i$

Now from the question we are total decay is as result of the combined decay of both processes

  We have that

            $\frac{dN_T}{dt} =\lambda_T N_i= \frac{dN_a}{dt} + \frac{dN_b}{dt}$

Substituting for the formula above

           $\lambda_T = \lambda _e + \lambda_r$

Generally the formula for half-life is

          $t__{\frac{1}{2} }}= \frac{0.693}{\lambda }$

So     $\lambda = \frac{0.693}{t_{\frac{1}{2} }}$

Substituting this into the above equation

         $[\frac{0.693}{t_\frac{1}{2} } ]_T =[\frac{0.693}{t_\frac{1}{2} } ]_e + [\frac{0.693}{t_\frac{1}{2} } ]_r$

         $[\frac{1}{t_\frac{1}{2} } ]_T =\frac{[\frac{1}{t_\frac{1}{2} } ]_e + [\frac{1}{t_\frac{1}{2} } ]_r}{[\frac{1}{t_\frac{1}{2} } ]_e * [\frac{1}{t_\frac{1}{2} } ]_r}$

Substituting values

       $[\frac{1}{t_\frac{1}{2} } ]_T =\frac{ 9+6}{9 * 6}$

     $[\frac{1}{t_\frac{1}{2} } ]_T =\frac{ 15}{54}$

     $(t_{\frac{1}{2} })_T =\frac{ 54 }{15}$

     $(t_{\frac{1}{2} })_T = 8.6 \ days$

       

Answer 2

Answer:

Effective half-time of the tracer is 3.6 days

Explanation:

The formula for calculating the decay due to excretion for the first process is ;

$\frac{dN_1}{dt } = - \lambda _e N_o$

here ;

$N_o$ = initial number of tracers

Then to the second process ; we have :

$\frac{dN_2}{dt } = - \lambda _e N_o$

The total decay is as a result of the overall process occurring ; we have :

$\frac{dN_{total}}{dt } = \frac{dN_1}{dt} + \frac{dN_2}{dt}$   ------ (1)

here ;

$\frac{dN_{total}}{dt } = \lambda _{total} N_o$

Putting the values in (1);we have :

$- \lambda _{Total} N_o = - \lambda_e N_o + ( -\lambda r N_o})$

$\lambda _{Total} = \lambda_e + \lambda r$

As we also know that:

$\frac{1}{t_{1/2}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{[t_{1/2}]_{radiation}+[t_{1/2}]_{excretion}}{[t_{1/2}]_{radiation}*[t_{1/2}]_{excretion}}$

$\frac{1}{t_{1/2}}_{effective}} = \frac{9+6}{9*6}$

$\frac{1}{t_{1/2}_{effective}}}=\frac{15}{54}$

$t_{1/2}_{effective}} = \frac{54}{15}$

= 3.6 days