A) \[3\text{ }months\]
B) \[4\text{ }months\]
C) \[~8\text{ }months\]
D) \[\text{12 }months\]
Correct Answer: C
Solution :
From Rutherford-Soddy's law \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] Given, \[N=1-\frac{3}{4}=\frac{1}{4}{{N}_{0}},n=\frac{t}{T}=\frac{t}{4}\] \[\therefore \] \[\frac{1}{4}={{\left( \frac{1}{2} \right)}^{t/4}}\] \[\Rightarrow \] \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{t/4}}\] \[\Rightarrow \] \[2=\frac{t}{4}\] \[\Rightarrow \] \[~t=8\text{ }months\]
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