Question

The half life period of ${}_{84}^{210}Po$ is $140$ days. In how many days $1g$ of this isotope is reduced to $0.25g$?

Answer 1

Original quantity of the isotope $\left(N_0\right)=1g$
Final quantity of the isotope $N=0.25g$
We know that, $N={\left(\frac{1}{2}\right)}^n{N}_0$
So, $\frac{1}{4}={\left(\frac{1}{2}\right)}^n\times 1$
or ${\left(\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^n$
or $n=2$
Time taken $T=n\times t_{1/2}=2\times 140=280$ days