A nucleus emits a $ \alpha $ particle : $ _2^4$ He The nucleus loses 2 protons $p^+$ and two neutrons $n^0$: $_Z^A$X $\rightarrow $ $_2^4$He + $_{Z-2}^{A-4}$Y Ex: Alpha decay of uranium 238 into thorium: $_{92}^{238}$U $\rightarrow $ $_2^4$He + $_{90}^{234}$Th
In the nucleus a neutron decays into a proton and an electron $ _{-1}^0$e ($ \beta^- $): $_0^1$n $\rightarrow$ $_1^1p^+$ + $_{-1}^0$e So: $_Z^A$X $\rightarrow$ $_{Z+1}^{A}$Y + $_{-1}^0$e Ex: Carbon 14 decay : $_6^{14}$C $\rightarrow$ $_7^{14}N$ + $_{-1}^0$e
In the nucleus a proton decays into a neutron and a positron $_1^0$e ($\beta^+$): $_1^0$p $\rightarrow$ $_0^1$n + $_1^0$e So: $_Z^A$X $\rightarrow$ $_{Z-1}^{A}$Y + $_1^0$e Ex : Phosphorus decay: $_{15}^{30}P$ $\rightarrow$ $_{14}^{30}Si$ + $_1^0$e
This is the emission of extremely penetrating very high frequency electromagnetic radiation ($ \nu $> 1018 Hz). Indeed, during the $ \alpha $, $ \beta^+ $ and $ \beta^- $ decays, the child nuclei formed are very often in an excited state of energy. Let us denote these nuclei $ Y^* $. They regain their stable ground state by emitting energy in the form of $ \gamma $ radiation. $_Z^AY^*$ $\rightarrow$ $_Z^AY$ + $\gamma$
Give the formulas of the following radioactive decays: $_{88}^{226}Ra$ ($\alpha$)→ ??? $_{88}^{226}Ra$ $\rightarrow $ $_2^4$He + $_{86}^{222}Rn$
$_{90}^{234}Th$ ($\beta^-$)→ ??? $_{90}^{234}Th$ $\rightarrow $ $_{-1}^0$e + $_{91}^{234}Pa$
$_{9}^{18}F$ ($\beta^+$)→ ??? $_{9}^{18}F$ $\rightarrow $ $_1^0$e + $_{8}^{18}O$