Respuesta :
Answer: The energy released in the given nuclear reaction is 1.3106 MeV.
Explanation:
For the given nuclear reaction:
[tex]_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}[/tex]
We are given:
Mass of [tex]_{19}^{40}\textrm{K}[/tex] = 39.963998 u
Mass of [tex]_{20}^{40}\textrm{Ca}[/tex] = 39.962591 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=(39.963998-39.962591)=0.001407u[/tex]
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(0.001407u)\times c^2[/tex]
[tex]E=(0.001407u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=1.3106MeV[/tex]
Hence, the energy released in the given nuclear reaction is 1.3106 MeV.
