Question

complete the following decay equations you will need a periodic table
86 222 rn $xrightarrow{a}$ 2 4 he +84 218 po
43 99 tc $xrightarrow{\beta}$ -1 0 e +44 99 ru
92 238 u $xrightarrow{a}$ +
82 214 pb $xrightarrow{\beta}$ +

Explanation:

Step1: Recall alpha - decay rule

In alpha - decay, the mass number ($A$) of the parent nucleus decreases by 4 and the atomic number ($Z$) decreases by 2. For the alpha - decay of $^{238}_{92}U$, the mass number of the daughter nucleus is $238 - 4=234$ and the atomic number is $92 - 2 = 90$. From the periodic table, the element with $Z = 90$ is Thorium (Th). So the equation is $^{238}_{92}U
ightarrow^{4}_{2}He+^{234}_{90}Th$.

Step2: Recall beta - decay rule

In beta - decay ($\beta^-$), a neutron in the nucleus is converted into a proton, an electron, and an antineutrino. The mass number ($A$) remains the same and the atomic number ($Z$) increases by 1. For the beta - decay of $^{214}_{82}Pb$, the mass number of the daughter nucleus is 214 and the atomic number is $82+1 = 83$. From the periodic table, the element with $Z = 83$ is Bismuth (Bi). So the equation is $^{214}_{82}Pb
ightarrow^{0}_{ - 1}e+^{214}_{83}Bi$.

Answer:

$^{238}_{92}U
ightarrow^{4}_{2}He+^{234}_{90}Th$; $^{214}_{82}Pb
ightarrow^{0}_{ - 1}e+^{214}_{83}Bi$