Question

A human body excretes (removes by waste discharge, sweating etc) certain materials by a law similar to radioactivity. If technetium is injected in some form in a human body, the body excretes half the amount in $24$ hours. A patient is given an injection containing ${}^{99}\text{Tc}$. This isotope is radioactive with a half-life of $6$ hours. The activity from the body just after the injection is $6\mu \text{Ci}$. How much time will elapse before the activity falls to $3\mu \text{Ci}$ ?

• A. $24\text{hours}$
• B. $6.4\text{hours}$
• C. $4.8\text{hours}$
• D. $2\text{hours}$

Answer 1

The correct option is C $4.8\text{hours}$

Here, the technetium injected disintegrates by two different processes, one by human body with half-life $t_1=24\text{hours}$ and the other by natural radioactivity with half-life $t_2=6\text{hours}$.

So, the effective half-life is given by,

$t_{\frac{1}{2}}=\frac{t_1{t}_2}{t_1+t_2}=\frac{24\times 6}{24+6}=4.8\text{h}.$

Given that,

$A_0=6\mu \text{Ci};A=3\mu \text{Ci}$

$\therefore A=\frac{A_0}{2^{t/t_{1/2}}}$

$\Rightarrow 3\mu \text{Ci}=\frac{6\mu \text{Ci}}{2^{t/4.8\text{h}}}$

$\Rightarrow \frac{t}{4.8h}=1\Rightarrow t=4.8\text{h}$

Hence, option $\left(C\right)$ is correct.

Why this question? Key concept: If a given sample disintegrates simultaneously by different processes, the effective half-life is given by,