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Q.
Half-life of radium is $1600$ years. If the initial mass is $1\, kg$, what is the amount of radium left after $4800$ years ?
BHUBHU 2002
Solution:
From Rutherford and Soddy law for radioactive decay, if $N$ be the number of atonıs of radioactive substance lift at some instant of time, then
$N=N_{0}\left(\frac{1}{2}\right)^{n}$
where $N_{0}$ is original number of atoms and $n$ is number of half-lives.
Given, $T_{1 / 2}=1600$ years,
$N_{0}=1 \,kg$
$n =\frac{4800}{1600}=3 $
$\therefore N =\left(\frac{1}{2}\right)^{3} \times 1 $
$=\frac{1}{8} \times 1=0.125 \,kg$
Note : The number of atoms of a radioactive substance that is, the radioactive activity of the substance continuously decreases with time.