Question

The half-life of a radioactive isotope is $3h$. If the initial mass of the isotope was $300g$, the mass which remains undecayed in $18h$ would be

• A. 2.34 g
• B. 1.17 g
• C. 9.36 g
• D. 4.68 g

Answer 1

Answer: (D)

(i) $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$
where $n=$ number of half-life period
$N_0=$ initial concentration
$N=$ concentration left after $n$ half-life periods.
(ii) $T=n\times t_{1/2}$
where, $T=$ total time
$t_{1/2}=$ half-life period
Here, $t_{1/2}=3h,T=18h$
$\therefore n=\frac{18}{3}=6$
Now using, $\frac{N}{N_0}={\left(\frac{1}{2}\right)}^n$
$N_0=300g$
$\frac{N}{300}={\left(\frac{1}{2}\right)}^6$
$\frac{N}{300}=\frac{1}{64}$
$N=\frac{300}{64}$
$N=4.68g$