\(\lambda=\frac{\ell \mathrm{n} 2}{20}\)
\(\frac{0.8 \mathrm{N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t_{1}}}{0.2 \mathrm{N}_{0}=\mathrm{N}_{0} \mathrm{e}^{-\lambda\left(t_{2}\right)}}\)
\(4=e^{+\lambda\left(t_{2}-t_{1}\right)}\)
\(\ln 4=\lambda\left(t_{2}-t_{1}\right)\)
\(\left(t_{2}-t_{1}\right)=40 \mathrm{\,min}\)
(${T_{1/2}}=$ અર્ધઆયુ સમય $\lambda =$ ક્ષય નિયતાંક)
$X \stackrel{a}{\longrightarrow} Y$
$Y \underset{2 \beta}{\longrightarrow} Z$
, ત્યારે