Question

Find whether $\alpha -decay$ or any of the $\beta -decay$ are allowed for ${}_{89}^{226}Ac$

Answer 1

Our first step will be to write the reaction, then find the disintegration energy $Q$.

If $Q>0$, the decay is allowed.

$\alpha -decay{:}_{89}^{226}Ac{\to }_{87}^{222}Fr+\alpha$

$Q=\left[M{(}_{89}^{226}Ac\right)-M{(}_{87}^{222}Fr)-M{(}^{4}He)]c^2$
$=5.50MeV$ ($\alpha$ decay is allowed)

${\beta }^{-}decay{:}_{89}^{226}Ac{\to }_{90}^{226}Th+{\beta }^{-}+\overline{v}$

$Q=\left[M{(}_{89}^{226}Ac\right)-M{(}_{90}^{226}Th\left)\right]c^2$
$=1.12MeV$ (${\beta }^{-}$ decay is allowed)

${\beta }^{+}decay{:}_{89}^{226}Ac{\to }_{88}^{226}Ra+{\beta }^{+}+v$

$Q=\left[M{(}_{89}^{226}Ac\right)=M{(}_{88}^{226}Ra)-2m_e]c^2$
$=-0.38MeV$ (${\beta }^{+}$ decay is not allowed)

Electron capture: ${}_{89}^{226}Ac+e^{-}{\to }_{88}^{226}Ra+v$

$Q=[M{(}_{89}^{226}Ac=M{(}_{88}^{226}Ra)]c^2$
$=0.64MeV$ (Electron capture is allowed)