Respuesta :
Answer: The energy released in the given nuclear reaction is 3.526 MeV.
Explanation:
For the given nuclear reaction:
[tex]_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}[/tex]
We are given:
Mass of [tex]_{19}^{42}\textrm{K}[/tex] = 41.962403 u
Mass of [tex]_{20}^{42}\textrm{Ca}[/tex] = 41.958618 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=(41.962403-41.958618)=0.003785u[/tex]
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(0.003785u)\times c^2[/tex]
[tex]E=(0.003785u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=3.526MeV[/tex]
Hence, the energy released in the given nuclear reaction is 3.526 MeV.
