A radioactive isotope has a half life of 100 years. How long will it take the activity to reduce to $$1\% $$  of its original valve?

Solution :
$$\eqalign{ & N = {N_0}{\left( {\frac{1}{2}} \right)^n} \cr & {\text{so,}}\,{\text{for}}\,1\% \,{\text{decay}} \cr & \frac{1}{{100}}{N_0} = {N_0}{\left( {\frac{1}{2}} \right)^n} \cr & \Rightarrow \frac{1}{{100}} = {\left( {\frac{1}{2}} \right)^n} \cr & \Rightarrow n = \frac{2}{{\log 2}} \cr & \Rightarrow \frac{t}{T} = \frac{2}{{\log 2}} \cr & \Rightarrow t = 6.6T \cr & = 6.6 \times 100 = 660\,{\text{years}} \cr} $$
If $$\alpha $$ and $$\beta $$ are emitted simultaneously.