a
\({\text{N}} = {{\text{N}}_{\text{0}}}{\left( {\frac{1}{2}} \right)^{t/{\tau _{1/2}}}}\, \Rightarrow \,{N_A} = 10{\left( {\frac{1}{2}} \right)^{t/1}}\)  અને

\({{\text{N}}_{\text{B}}} = 1{\left( {\frac{1}{2}} \right)^{t/2}}\)

\({N_A} = {N_B} \Rightarrow \,10{\left( {\frac{1}{2}} \right)^t} = {\left( {\frac{1}{2}} \right)^{t/2}}\,\,\)   આપેલ છે 

\( \Rightarrow \,\,{\text{10}} = {\left( {\frac{{\text{1}}}{{\text{2}}}} \right)^{ - t/2}}\, \Rightarrow \,\,10 = {2^{t/2}}\)

બંને બાજુ \(log\) લેતાં,

\({\log _{10}}10 = \frac{t}{2}{\log _{10}}2\,\)

\(\Rightarrow \,1 = \frac{t}{2} \times 0.3010\,\)

\(\Rightarrow \,t = 6.62\,year\)