Answer:
[tex]4.3\cdot 10^{-12} J[/tex]
Explanation:
The fusion reaction in this problem is
[tex]4^1_1H \rightarrow ^4_2He +2e^+[/tex]
The total energy released in the fusion reaction is given by
[tex]\Delta E = c^2 \Delta m[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\Delta m[/tex] is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products
For this fusion reaction we have:
[tex]m(^1_1H)=1.007825u[/tex] is the mass of one nucleus of hydrogen
[tex]m(^4_2 He)=4.002603u[/tex] is the mass of one nucleus of helium
So the mass defect is:
[tex]\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u[/tex]
The conversion factor between atomic mass units and kilograms is
[tex]1u=1.66054\cdot 10^{-27}kg[/tex]
So the mass defect is
[tex]\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg[/tex]
And so, the energy released is:
[tex]\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J[/tex]
