Chemistry - Radioactivity Question with Solution | TestHub

$${}_{29}^{64}Cu\stackrel{{\lambda }_1}{\to }{}_{-1}^{0}e+{}_{30}^{64}Zn$$

$${}_{29}^{64}Cu{\stackrel{{\lambda }_2}{\to }}_{+1}^{0}e{+}_{28}^{64}Ni$$

$${}_{29}^{64}Cu{+}_{-1}^{0}e{\stackrel{{\lambda }_3}{\to }}_{}^{}{+}_{28}^{64}Ni$$

$\begin{array}{rl}& \\ ({\lambda }_1+{\lambda }_2+{\lambda }_3)=\frac{\ln 2}{(t_{1/2})}=\frac{\ln 2}{12.8}& \\ \frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=\left(\frac{38}{100}\right)& \\ {\lambda }_1=\left(\frac{38}{100}\right)\frac{\ln 2}{12.8}& \\ t_{1/2}=\frac{\ln 2}{{\lambda }_1}& \\ t_{1/2}=\frac{\ln 2\times 100\times 12.8}{38\times \ln 2}& \\ t_{1/2}=\frac{1280}{38}=33.68hrs& \end{array}$