Respuesta :
Answer: The energy released for the decay of 3 grams of 230-Thorium is [tex]2.728\times 10^{-15}J[/tex]
Explanation:
First we have to calculate the mass defect [tex](\Delta m)[/tex].
The equation for the alpha decay of thorium nucleus follows:
[tex]_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}[/tex]
To calculate the mass defect, we use the equation:
Mass defect = Sum of mass of product - Sum of mass of reactant
[tex]\Delta m=(m_{Ra}+m_{He})-(m_{Th})[/tex]
[tex]\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg[/tex]
(Conversion factor: [tex]1amu=1.66\times 10^{-27}kg[/tex] )
To calculate the energy released, we use Einstein equation, which is:
[tex]E=\Delta mc^2[/tex]
[tex]E=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2[/tex]
[tex]E=2.0916\times 10^{-13}J[/tex]
The energy released for 230 grams of decay of thorium is [tex]2.0916\times 10^{-13}J[/tex]
We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:
As, 230 grams of Th release energy of = [tex]2.0916\times 10^{-13}J[/tex]
Then, 3 grams of Th will release energy of = [tex]\frac{2.0916\times 10^{-13}}{230}\times 3=2.728\times 10^{-15}J[/tex]
Hence, the energy released for the decay of 3 grams of 230-Thorium is [tex]2.728\times 10^{-15}J[/tex]
