Answer
${\bf 3.6}\;\rm day$
Work Step by Step
To determine the effective half-life of the isotope in a patient's body, we need to consider both the radioactive decay and the excretion processes, which work together to remove the isotope. This is similar to calculating the combined effect of two processes running in parallel.
The rate of removal from the body, $ \frac{dN}{dt} $, is the sum of the decay rate and the excretion rate:
$$
\frac{dN}{dt} = \left( \frac{dN}{dt} \right)_{\text{decay}} + \left( \frac{dN}{dt} \right)_{\text{excrete}} $$
So,
$$ \frac{dN}{dt} = -r_d N - r_e N = -(r_d + r_e) N
$$
- Here, $ r_d $ is the decay rate, and $ r_e $ is the excretion rate. The effective removal rate $ r_{\text{eff}} $ is the sum of $ r_d $ and $ r_e $, $$ r_{\text{eff}} = r_d + r_e \tag 1$$
$$ \frac{dN}{dt} = -r_{\text{eff}} N
$$
Recalling that
$$
r = \frac{\ln(2)}{t_{1/2}}
$$
Thus, the effective decay constant $ r_{\text{eff}} $ can be written as:
$$
r_{\text{eff}} = \frac{\ln(2)}{(t_{1/2})_\text{decay}} + \frac{\ln(2)}{(t_{1/2})_\text{excrete}}
$$
Recalling that $r=\ln(2)/t_{1/2}$, so from (1)
$$
\frac{\color{red}{\bf\not} \ln(2)}{(t_{1/2})_\text{eff}} = \frac{\color{red}{\bf\not} \ln(2)}{ (t_{1/2})_\text{decay}} + \frac{\color{red}{\bf\not} \ln(2)}{(t_{1/2})_\text{excrete}}
$$
$$
\frac{1}{(t_{1/2})_\text{eff}} = \frac{ 1}{ (t_{1/2})_\text{decay}} + \frac{1}{(t_{1/2})_\text{excrete}}
$$
Substitute the given values of $(t_{1/2})_\text{decay} = 9.0 $ days and $(t_{1/2})_\text{excrete}= 6.0 $ days)
$$
(t_{1/2})_\text{eff} = \left[\frac{1}{9.0 } + \frac{1}{6.0 }\right]^{-1}
$$
$$
(t_{1/2})_\text{eff} = \color{red}{\bf 3.6}\;\rm day
$$
