Respuesta :
Answer: The correct answer is 2.24 MeV.
Explanation:
The chemical reaction for the formation of deuterium from proton and neutron follows:
[tex]_{1}^1\textrm{H}+_0^1\textrm{n}\rightarrow _1^2\textrm{H}[/tex]
We are given:
Mass of [tex]_{1}^{1}\textrm{H}[/tex] = 1.00784 u
Mass of [tex]_{0}^{1}\textrm{n}[/tex] = 1.008665 u
Mass of [tex]_{1}^{2}\textrm{H}[/tex] = 2.014102 u
To calculate the mass defect, we use the equation:
[tex]\Delta m=\text{Mass of reactants}-\text{Mass of products}[/tex]
[tex]\Delta m=(m_{_1^2H}+m_{n})-(m_{_1^2H})\\\\\Delta m=(1.00784+1.008665)-(2.014102)=0.002403u[/tex]
To calculate the energy released, we use the equation:
[tex]E=\Delta mc^2\\E=(0.002403u)\times c^2[/tex]
[tex]E=(0.002403u)\times (931.5MeV)[/tex] (Conversion factor: [tex]1u=931.5MeV/c^2[/tex] )
[tex]E=2.24MeV[/tex]
Hence, the energy released in the given nuclear reaction is 2.24 MeV.
