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Q.
The half-life of a radioactive isotope is $3\, h$. If the initial mass of the isotope was $300\, g$, the mass which remains undecayed in $18\, h$ would be
AMUAMU 2001
Solution:
(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, $t_{1 / 2}=3\, h,\,\, T=18\, h$
$\therefore n =\frac{18}{3}=6$
Now using, $\frac{N}{N_{0}} =\left(\frac{1}{2}\right)^{n}$
$N_{0} =300\, g$
$\frac{N}{300} =\left(\frac{1}{2}\right)^{6}$
$\frac{N}{300}=\frac{1}{64}$
$N =\frac{300}{64}$
$N =4.68\, g$