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Q.
A radioactive element has half-life period $1600 \,yr$. After $6400 \,yr$, what part of element will remain?
Jharkhand CECEJharkhand CECE 2004
Solution:
From Rutherford Soddy law, the number of atoms left after $n$ half-lives is
$N=N_{0}\left(\frac{1}{2}\right)^{n}$
where, $n=\frac{\text { time }(t)}{\text { half-life(T }_{1/ 2} ) }$
Given, $t=6400\, y r, \,\,T=1600 \,y r $
$\therefore n=\frac{6400}{1600}=4$
So, $\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$
Hence, $\frac{1}{16}$ part of element will remain.