Respuesta :
Using Einstein's equivalence, the energy released in the decay is equal to the variation of mass between the initial state and the final state:
[tex]\Delta E = \Delta m c^2[/tex] (1)
where c is the speed of light.
In the first step of the decay, an alpha particle (2 protons+2 neutrons) is emitted, with atomic mass 4.001506179 u. The variation of mass in the decay therefore is
[tex]\Delta m=232.038054 u -(228.0301069+4.001506179)u =0.006440921u[/tex]
and since 1 u is equal to the mass of the proton:
[tex]1 u = 1.66 \cdot 10^{-27} kg[/tex]
The mass difference of the decay in kg is
[tex]\Delta m=(0.006440921u)(1.66 \cdot 10^{-27}kg/u)=1.069 \cdot 10^{-29} kg[/tex]
And so by using (1) the energy released in the decay is
[tex]\Delta E = \Delta m c^2 = (6.65 \cdot 10^{-27} kg)(3 \cdot 10^8 m/s)^2=9.62 \cdot 10^{-13} J[/tex]
And since [tex]1 eV = 1.6 \cdot 10^{-19}J[/tex], we can convert this energy into eV:
[tex]\Delta E= \frac{9.62 \cdot 10^{-13}J }{1.6 \cdot 10^{-19} J/eV} =6.013 \cdot 10^6 eV = 6.013 MeV[/tex]
[tex]\Delta E = \Delta m c^2[/tex] (1)
where c is the speed of light.
In the first step of the decay, an alpha particle (2 protons+2 neutrons) is emitted, with atomic mass 4.001506179 u. The variation of mass in the decay therefore is
[tex]\Delta m=232.038054 u -(228.0301069+4.001506179)u =0.006440921u[/tex]
and since 1 u is equal to the mass of the proton:
[tex]1 u = 1.66 \cdot 10^{-27} kg[/tex]
The mass difference of the decay in kg is
[tex]\Delta m=(0.006440921u)(1.66 \cdot 10^{-27}kg/u)=1.069 \cdot 10^{-29} kg[/tex]
And so by using (1) the energy released in the decay is
[tex]\Delta E = \Delta m c^2 = (6.65 \cdot 10^{-27} kg)(3 \cdot 10^8 m/s)^2=9.62 \cdot 10^{-13} J[/tex]
And since [tex]1 eV = 1.6 \cdot 10^{-19}J[/tex], we can convert this energy into eV:
[tex]\Delta E= \frac{9.62 \cdot 10^{-13}J }{1.6 \cdot 10^{-19} J/eV} =6.013 \cdot 10^6 eV = 6.013 MeV[/tex]
Answer: The energy released in the given nuclear reaction is 4.98 MeV.
Explanation:
The equation for the alpha decay of Th-232 nucleus follows:
[tex]_{90}^{232}\textrm{Th}\rightarrow _{88}^{228}\textrm{Ra}+_{2}^{4}\textrm{He}[/tex]
We are given:
Mass of [tex]_{90}^{232}\textrm{Th}[/tex] = 232.038054 u
Mass of [tex]_{88}^{228}\textrm{Ra}[/tex] = 228.0301069 u
Mass of [tex]_{2}^{4}\textrm{He}[/tex] = 4.002602 u
To calculate the mass defect, we use the equation:
[tex]\Delta m=\text{Mass of reactants}-\text{Mass of products}[/tex]
Putting values in above equation, we get:
[tex]\Delta m=(m_{Th})-(m_{Ra}+m_{He})\\\\\Delta m=(232.038054)-(228.0301069+4.002602)=0.0053451u[/tex]
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(0.0053451u)\times c^2[/tex]
[tex]E=(0.0053451u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=4.98MeV[/tex]
Hence, the energy released in the given nuclear reaction is 4.98 MeV.
