Answer :
Let’s identify and understand the third step involved in the hydrogen fusion process by looking at each reaction given:
1. [tex]\( \ce{^1_1H + ^1_1H -> ^2_1H + e^{+} + v + \text{energy}} \)[/tex]
- Here two protons ([tex]\(^1_1H\)[/tex]) fuse to form deuterium ([tex]\(^2_1H\)[/tex]), a positron ([tex]\(e^+\)[/tex]), a neutrino ([tex]\(v\)[/tex]), and release energy.
2. [tex]\( \ce{^3_2He + ^3_2He -> ^4_2He + 2(^1_1H) + \text{energy}} \)[/tex]
- In this reaction, two helium-3 nuclei ([tex]\(^3_2He\)[/tex]) fuse to form a helium-4 nucleus ([tex]\(^4_2He\)[/tex]), two protons ([tex]\(^1_1H\)[/tex]), and release energy.
3. [tex]\( \ce{^2_1H + ^1_1H -> ^3_2He + \text{energy}} \)[/tex]
- This reaction involves a deuterium nucleus ([tex]\(^2_1H\)[/tex]) fusing with a proton ([tex]\(^1_1H\)[/tex]) to form a helium-3 nucleus ([tex]\(^3_2He\)[/tex]) and release energy.
4. [tex]\( \ce{6(^1_1H) + 2(^-1_0e) -> ^4_2He + 2(^1_1H) + \text{energy} + 2v} \)[/tex]
- In this step, six protons ([tex]\(^1_1H\)[/tex]) and two electrons ([tex]\(e^-\)[/tex]) combine to form helium-4 ([tex]\(^4_2He\)[/tex]), two protons ([tex]\(^1_1H\)[/tex]), releasing energy and two neutrinos ([tex]\(v\)[/tex]).
Focusing on the third reaction, we notice that the process described involves combining a deuterium nucleus ([tex]\(^2_1H\)[/tex]) with a proton ([tex]\(^1_1H\)[/tex]) to form a helium-3 nucleus ([tex]\(^3_2He\)[/tex]) and release energy. This matches the given description of the third step.
Thus, the third step of the hydrogen fusion process is:
[tex]\[ \ce{^2_1H + ^1_1H -> ^3_2He + \text{energy}} \][/tex]
1. [tex]\( \ce{^1_1H + ^1_1H -> ^2_1H + e^{+} + v + \text{energy}} \)[/tex]
- Here two protons ([tex]\(^1_1H\)[/tex]) fuse to form deuterium ([tex]\(^2_1H\)[/tex]), a positron ([tex]\(e^+\)[/tex]), a neutrino ([tex]\(v\)[/tex]), and release energy.
2. [tex]\( \ce{^3_2He + ^3_2He -> ^4_2He + 2(^1_1H) + \text{energy}} \)[/tex]
- In this reaction, two helium-3 nuclei ([tex]\(^3_2He\)[/tex]) fuse to form a helium-4 nucleus ([tex]\(^4_2He\)[/tex]), two protons ([tex]\(^1_1H\)[/tex]), and release energy.
3. [tex]\( \ce{^2_1H + ^1_1H -> ^3_2He + \text{energy}} \)[/tex]
- This reaction involves a deuterium nucleus ([tex]\(^2_1H\)[/tex]) fusing with a proton ([tex]\(^1_1H\)[/tex]) to form a helium-3 nucleus ([tex]\(^3_2He\)[/tex]) and release energy.
4. [tex]\( \ce{6(^1_1H) + 2(^-1_0e) -> ^4_2He + 2(^1_1H) + \text{energy} + 2v} \)[/tex]
- In this step, six protons ([tex]\(^1_1H\)[/tex]) and two electrons ([tex]\(e^-\)[/tex]) combine to form helium-4 ([tex]\(^4_2He\)[/tex]), two protons ([tex]\(^1_1H\)[/tex]), releasing energy and two neutrinos ([tex]\(v\)[/tex]).
Focusing on the third reaction, we notice that the process described involves combining a deuterium nucleus ([tex]\(^2_1H\)[/tex]) with a proton ([tex]\(^1_1H\)[/tex]) to form a helium-3 nucleus ([tex]\(^3_2He\)[/tex]) and release energy. This matches the given description of the third step.
Thus, the third step of the hydrogen fusion process is:
[tex]\[ \ce{^2_1H + ^1_1H -> ^3_2He + \text{energy}} \][/tex]
