Respuesta :
Answer: The energy absorbed in the given nuclear reaction is 1.2072 MeV.
Explanation:
For the given nuclear reaction:
[tex]_{7}^{14}\textrm{N}+_2^4\textrm{He}\rightarrow _{8}^{17}\textrm{O}+_{1}^{1}\textrm{H}[/tex]
We are given:
Mass of [tex]_{7}^{14}\textrm{N}[/tex] = 14.003074 u
Mass of [tex]_{8}^{17}\textrm{O}[/tex] = 16.999132 u
Mass of [tex]_{2}^{4}\textrm{He}[/tex] = 4.002602 u
Mass of [tex]_{1}^{1}\textrm{H}[/tex] = 1.00784 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_N+m_{He})-(m_{O}+m_{H})\\\\\Delta m=(14.003074+4.002602)-(16.999132+1.00784)=-0.001296u[/tex]
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(-0.001296u)\times c^2[/tex]
[tex]E=(-0.001296u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=-1.2072MeV[/tex] (negative sign indicates that energy is getting absorbed)
Hence, the energy absorbed in the given nuclear reaction is 1.2072 MeV.
