A) 87.5%
B) 52.5%
C) 25.5%
D) 8.5%
Correct Answer: A
Solution :
From Rutherford and soddy law, the rate of decay of a radioactive substance is proportional to number of atoms left at that instant, using this we can arrive at \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] where \[{{N}_{0}}\] is original number of atoms, n is number of half-lives \[n=\frac{t}{{{T}_{1/2}}}=\frac{180}{60}=3\] \[\therefore \] \[\frac{N}{{{N}_{0}}}={{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{3}}=\frac{1}{8}\] \[\therefore \] \[N=\frac{{{N}_{0}}}{8}=0.125\,{{N}_{0}}=12.5%N\] Amount decayed \[=100-12.5=87.5%N\]
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