ક્ષયનો દર એટલે કે \(R\) એ \(R \propto N\) અથવા \(R ={ \lambda }N\)
\(N = N_0e^{-\lambda t} \, ,R = {\lambda} N 0e^{-\lambda t}= R_0e^{-\lambda t}\) જ્યાં \(R_0 = {\lambda} N_0\) પ્રારંભિક ક્ષય દર
આથી \(\,\frac{{{R_0}}}{R}\,\, = \,\,{e^{ - \lambda t}}\,\,.....\,\,(i)\,\,\)
તેથી \(\frac{R}{{{R_0}}}\, = \,\,\frac{9}{{15}}\, = \,\,\frac{3}{5}\,\,\,......\,\,(ii)\)
સમી \((I)\) અને \((II)\) પરથી, \(e^{-\lambda t} = 3/5 ⇒ e^{\lambda t} = 5/3\)
\( \Rightarrow \,\,\,\,\lambda t\,\, = \,\,\ln \,\,(5/3)\,\,\,\,\, \Rightarrow \,\,\,\,t\,\, = \,\,\frac{{\ln \,(5/3)}}{\lambda }\,\,\,\,\,\)
\( \Rightarrow \,\,\,\,t\,\, = \,\,\frac{{\ln \,\,(5/3)}}{{\ln 2}}\,\, \times \,\,{T_{1/2}}\)
\(=\,\,\frac{{\log \,\,(5/3)}}{{\log 2}}\,\, \times \,\,5730\,\,y\,\,\,\,\left( {as\,\,{T_{1/2}} = \,\,\frac{{\ln \,2}}{\lambda }} \right)\)
\(\therefore\) \(t\,\, = \,\,\frac{{0.2219}}{{0.3010}}\,\, \times \,\,5730\,\,y\,\, = \,\,4224\,\,year\)