Question 28.QE.4: Fusion energy in the Sun The energy released by the Sun come......
Fusion energy in the Sun
The energy released by the Sun comes from several sources, including the so-called proton-proton chain of fusion reactions. This chain of reactions can be summarized as follows:
2{ }_1^1 \mathrm{H}+2{ }_0^1 \mathrm{n} \rightarrow{ }_2^4 \mathrm{He}+\text { energy }
Determine the energy released in this chain of reactions in MeV. Use the masses { }_1^{\mathrm{C}} \mathrm{H}(1.007825 \mathrm{u}),{ }_0^1 \mathrm{n}(1.008665 \mathrm{u}) \text { and }{ }_2^4 \mathrm{He}(4.002602 \mathrm{u}) to determine the rest energy converted to other forms.
Represent mathematically The energy released, the reaction energy Q, is determined using Eq. (28.5):
Q=\left(\sum_{\text {reactants }} m-\sum_{\text {products }} m\right) c^2
Solve and evaluate Inserting the appropriate values and converting to MeV gives
Q=[2(1.007825 \mathrm{u})+2(1.008665 \mathrm{u})
-4.002602 \mathrm{u}] c^2\left(\frac{931.5 \mathrm{MeV}}{\mathrm{u} \cdot c^2}\right)
= 28 MeV
This value is approximately a million times more energy than is released in a typical chemical reaction.
Try it yourself: What fraction of the total reactant rest energy in the proton-proton chain reactions is converted to other forms of energy during the above fusion reaction?
Answer: About (0.03 \mathrm{u}) /(4.0 \mathrm{u})=0.008, \text { or } 0.8 \%