The last stage in the fusion of hydrogen to form helium in the Sun involves the formation of an ƒ¿-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
To calculate the amount of energy released in this reaction, you can use the concept of mass-energy equivalence. According to Einstein's famous equation E=mc^2, energy (E) is equal to mass (m) multiplied by the speed of light in a vacuum (c) squared.
First, let's find the mass difference between the reactants (two 3He nuclei) and the product (4He nucleus).
The mass of two 3He nuclei is:
(2 x mass of 3He) = (2 x 3 nucleons x 2.57 MeV/nucleon)
= 15.42 MeV
The mass of 4He nucleus is:
(1 x mass of 4He) = (1 x 4 nucleons x 7.07 MeV/nucleon)
= 28.28 MeV
Now, let's calculate the mass difference:
Mass difference = mass of reactants - mass of product
Mass difference = 15.42 MeV - 28.28 MeV
= -12.86 MeV (negative sign indicates the mass difference is released as energy)
Therefore, the amount of energy released in this reaction is approximately 12.86 MeV. So, 12.9 MeV is indeed close, but you should consider the negative sign and round it correctly to 12.9 MeV (since it is an absolute value).
Please note that this calculation assumes that there are no other energy contributions such as kinetic energy or potential energy.