Answer : The energy released in first step of thorium-232 decay chain is [tex]7.974\times 10^{-13}J[/tex]
Explanation :
First we have to calculate the mass defect [tex](\Delta m)[/tex].
The balanced reaction is,
[tex]^{232}Th\rightarrow ^{228}Ra+^{4}He[/tex]
Mass defect = Sum of mass of product - sum of mass of reactants
[tex]\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})[/tex]
[tex]\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg[/tex]
conversion used : [tex](1amu=1.66\times 10^{-27}kg)[/tex]
Now we have to calculate the energy released.
[tex]Energy=\Delta m\times (c)^2[/tex]
[tex]Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2[/tex]
[tex]Energy=7.974\times 10^{-13}J[/tex]
The energy released is [tex]7.974\times 10^{-13}J[/tex]
Therefore, the energy released in first step of thorium-232 decay chain is [tex]7.974\times 10^{-13}J[/tex]