Nuclear Chemistry - Result Question 10
####13. Half-life for radioactive ${ }^{14} C$ is 5760 years. In how many years, $200 mg$ of ${ }^{14} C$ will be reduced to $25 mg$ ?
[1995]
(a) 5760 years
(b) 11520 years
(c) 17280 years
(d) 23040 years
Show Answer
Solution:
- (c) As we know that
$\frac{N}{N_0}=(\frac{1}{2})^{n}$
where $N_0$ original amount of radioactive sustacnce
$N=$ Amount of substance remain after $n$ half lives
$\therefore \quad \frac{25}{200}=(\frac{1}{2})^{n}$ or $\frac{1}{8}=(\frac{1}{2})^{n}$
or $(\frac{1}{2})^{3}=(\frac{1}{2})^{n} ; \therefore n=3$
now $T=n \times t _{1 / 2}$
where $T=$ total time
$T=3 \times 5760$ years $=17280$ years
