Answer: The energy released in the given nuclear reaction is 94.99 MeV.
Explanation:
For the given nuclear reaction:
[tex]_{92}^{235}\textrm{U}+_0^1\textrm{n}\rightarrow _{53}^{131}\textrm{I}+_{39}^{89}\textrm{Y}+16_{0}^1\textrm{n}[/tex]
We are given:
Mass of [tex]_{92}^{235}\textrm{U}[/tex] = 235.043924 u
Mass of [tex]_{0}^{1}\textrm{n}[/tex] = 1.008665 u
Mass of [tex]_{53}^{131}\textrm{I}[/tex] = 130.9061246 u
Mass of [tex]_{39}^{89}\textrm{Y}[/tex] = 88.9058483 u 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_U+m_{n})-(m_{I}+m_{Y}+16m_{n})\\\\\Delta m=(235.043924+1.008665)-(130.9061246+88.9058483+16(1.008665))=0.1019761u[/tex]
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(0.1019761u)\times c^2[/tex]
[tex]E=(0.1019761u)\times (931.5MeV)[/tex] Â (Conversion factor: Â [tex]1u=931.5MeV/c^2[/tex] Â )
[tex]E=94.99MeV[/tex]
Hence, the energy released in the given nuclear reaction is 94.99 MeV.
