カリウム-アルゴン法
K 40 + e − ⟶ Ar 40 + γ {\displaystyle {\ce {{}^{40}K{}+{\mathit {e}}^{-}->{}^{40}Ar\ +\gamma }}} ( 1 , 505 M e V ) {\displaystyle {\rm {(1,505MeV)}}} , λ ε = 0 , 581 ⋅ 10 − 10 a n − 1 {\displaystyle {\rm {\lambda \varepsilon =0,581\cdot 10_{-10}an_{-1}}}}
K 40 ⟶ Ca 40 + β − {\displaystyle {\ce {{}^{40}{K}-> {}^{40}Ca\ + \beta^-}}} (1,311 MeV) {\displaystyle {\rm {\mbox{(1,311 MeV)}}}} , λ ε = 4 , 962.10 − 10 a n − 1 {\displaystyle {\rm {\lambda \varepsilon =4,962.