The last stage in the fusion of hydrogen to form helium in the Sun involves the formation of an alpha particle (4He) from two 3He nuclei:
3He + 3He 4He + p + p
The binding energy for 3He is 2.57 MeV per nucleon and that for 4He is
7.07 MeV per nucleon. Calculate the amount of energy released in this reaction.
Is 12.9 MeV close? Thanks
3He + 3He>> 4He + p + p
energy released= 7.07*4-2*2.57*3
approx 28-15= 13MEV which is close to your answer.
check my thinking.
Thanks- it makes sense, since i had to choose the closest to my answer.
To calculate the amount of energy released in the fusion reaction, you need to find the difference in binding energy between the reactants (two 3He nuclei) and the product (4He nucleus). This energy difference is the energy released during the reaction.
Step 1: Determine the total binding energy of the reactants (two 3He nuclei):
Total binding energy of 3He = 2.57 MeV/nucleon * 3 nucleons = 7.71 MeV
Step 2: Determine the binding energy of the product (4He nucleus):
Binding energy of 4He = 7.07 MeV/nucleon * 4 nucleons = 28.28 MeV
Step 3: Calculate the energy released:
Energy released = Binding energy of the reactants - Binding energy of the product
= 7.71 MeV - 28.28 MeV
= -20.57 MeV (negative because energy is released)
So the correct answer is -20.57 MeV, not 12.9 MeV. This means that 20.57 MeV of energy is released during the fusion reaction.