In beta decay, a neutron is converted into a proton and an electron.

$^1_0$n $rightarrow$ $^1_1$P $+$ e$^{-}$ $+$ $bar{gamma_e}$; $bar{gamma_e}$ is anti-neutrino.

So the atomic number ($Z$) of daughter nucleus increases by 1 and the mass number ($A$) remains same.

In $^{234}_{90}$Th, $Z = 90$. So when $^{234}_{90}$Th undergoes beta decay, the atomic number of daughter nucleus is $Z = 90 + 1 = 91$. It is protactinium (Pa) element.

So the daughter nucleus is $^{234}_{91}$Pa.

$^{234}_{90}$Th $rightarrow$ $^{234}_{91}$Pa $+$ e$^{-1}$ + $bar{gamma_e}$

(a) In beta decay of $^{234}_{90}$Th, the mass of the system decreases. The decreased mass is released as energy. The energy released is equal to the decrease in mass divided by the square of the speed of light ($dfrac{Delta m}{c^2}$).

(b) $1 :u = 931.5:MeV$

So the change in mass is given by

$0.274:cancel{MeV} times dfrac{1:u}{931.5:cancel{MeV}} = dfrac{0.274}{931.5} :u = 0.00029415:u$

In beta decay, a neutron is converted into a proton and an electron.

$^1_0$n $rightarrow$ $^1_1$P $+$ e$^{-}$ $+$ $bar{gamma_e}$; $bar{gamma_e}$ is anti-neutrino.

So the atomic number ($Z$) of daughter nucleus increases by 1 and the mass number ($A$) remains same.

In $^{238}_{94}$Pu, $Z = 94$. So when $^{238}_{94}$Pu undergoes beta decay, the atomic number of daughter nucleus is $Z = 94 + 1 = 95$. It is Americium (Am) element.

So the daughter nucleus is $^{238}_{95}$Am.

$^{238}_{94}$Pu $rightarrow$ $^{238}_{95}$Am $+$ e$^{-1}$ + $bar{gamma_e}$

$^{238}_{95}$Am