Respuesta :
Answer: The amount of energy released in the process is 4.042 MeV.
Explanation:
The chemical reaction for the fusion of deuterium nucleus follows the equation:
[tex]_1^2\textrm{H}+_1^2\textrm{H}\rightarrow _1^3\textrm{H}+_1^1\textrm{H}[/tex]
Atomic mass of the nucleus also contains some mass of the electrons.
Mass of electron in [tex]MeV/c^2[/tex] is [tex]0.511MeV/c^2[/tex]
- Calculating the mass of deuterium nucleus:
[tex]m(_1^2\textrm{H})=(2.014102u)\times (931.494MeV/c^2.u)-\text{Mass of electron}[/tex]
So, initial mass of the reaction = [tex]2[(1876.124MeV/c^2)-(0.511MeV/c^2)]=3751.226MeV/c^2[/tex]
- Calculating the mass of tritium nucleus:
[tex]m(_1^3\textrm{H})=(3.106049u)\times (931.494MeV/c^2.u)-\text{Mass of electron}[/tex]
[tex]m(_1^3\textrm{H})=(3.106049u)\times (931.494MeV/c^2.u)-(0.511MeV/c^2)=2803.921MeV/c^2[/tex]
So, final mass of the reaction = [tex]2803.921-[(1.007267u\times 931.494MeV/c^2.u)]=3747.184[/tex]
Difference between the masses of the nucleus:
[tex]\Delta m=m_{initial}-m_{final}\\\\\Delta m=(3751.226-3747.184)MeV/c^2=4.042MeV/c^2[/tex]
Energy released in the process is calculated by using Einstein's equation, which is:
[tex]E=\Delta mc^2[/tex]
Putting value of [tex]\Delta m[/tex] in above equation, we get:
[tex]E=(4.042MeV/c^2)\times c^2\\\\E=4.042MeV[/tex]
Hence, the amount of energy released in the process is 4.042 MeV.
