A) \[~\text{26}.\text{7 MeV}\]
B) \[\text{25}.\text{3 MeV}\]
C) \[~\text{22 MeV}\]
D) \[\text{24}.\text{5 MeV}\]
Correct Answer: A
Solution :
The proton-proton cycle by which this occurs is represented by the following sets of reactions \[_{\text{1}}^{\text{1}}\text{H + }_{\text{1}}^{\text{1}}\text{H}\to _{\text{1}}^{\text{2}}\text{H + }{{\text{e}}^{\text{+}}}\text{+ v + 0}\text{.42 Mev}\] ? (i) \[{{\text{e}}^{\text{+}}}\text{+ }{{\text{e}}^{\text{-}}}\to \text{ }\!\!\gamma\!\!\text{ + }\!\!\gamma\!\!\text{ + 1}\text{.02 MeV}\] ? (ii) \[_{\text{1}}^{\text{2}}\text{H + }_{\text{1}}^{\text{1}}\text{H}\to _{\text{2}}^{\text{3}}\text{He + }\!\!\gamma\!\!\text{ + 5}\text{.49 Mev}\] ? (iii) \[_{\text{2}}^{\text{3}}\text{He + }_{\text{2}}^{\text{3}}\text{He}\to _{\text{2}}^{\text{4}}\text{He + }_{\text{1}}^{\text{1}}\text{H +}_{\text{1}}^{\text{1}}\text{H +12}\text{.86 MeV}\] ?(iv) For the fourth reaction to occur, the first three reactions must occur twice, if we consider the combination 2(i) + 2 (ii) + 2 (iii) +2 (iv), the net effect is \[\text{4}_{\text{1}}^{\text{1}}\text{H + 2}{{\text{e}}^{\text{-1}}}\to \frac{\text{4}}{\text{2}}\text{He + 2v + 6v + 26}\text{.7 MeV}\] Thus, four hydrogen atom combines to form an \[_{2}^{4}\]He atom with a release of 26.7 MeV of energy.
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