which equation demonstrates that nuclear fusion forms elements that are heavier than helium?\n$_{1}^{2}h +…

which equation demonstrates that nuclear fusion forms elements that are heavier than helium?\n$_{1}^{2}h + _{1}^{3}h\\longrightarrow_{2}^{4}he + _{0}^{1}n$\n$_{8}^{16}o + _{2}^{4}he\\longrightarrow_{10}^{20}ne$\n$_{92}^{235}u + _{0}^{1}n\\longrightarrow_{53}^{131}i + _{39}^{89}y + 16_{0}^{1}n$\n$_{92}^{235}u + _{0}^{1}n\\longrightarrow_{42}^{95}mo + _{57}^{139}la + 2_{0}^{1}n+7_{-1}^{0}e$
Answer
Explanation:
Step1: Recall nuclear fusion concept
Nuclear fusion is the combination of two or more light - nuclei to form a heavier nucleus.
Step2: Analyze each option
- Option 1: ${1}^{2}H+{1}^{3}H\rightarrow_{2}^{4}He + _{0}^{1}n$ is a hydrogen - hydrogen fusion reaction resulting in helium, not an element heavier than helium.
- Option 2: ${8}^{16}O+{2}^{4}He\rightarrow_{10}^{20}Ne$ is a nuclear fusion reaction where oxygen and helium combine to form neon (a heavier element than helium).
- Option 3: ${92}^{235}U+{0}^{1}n\rightarrow_{53}^{131}I + {39}^{89}Y+16{0}^{1}n$ is a nuclear fission reaction of uranium - 235, not fusion.
- Option 4: ${92}^{235}U+{0}^{1}n\rightarrow_{42}^{95}Mo+{57}^{139}La + 2{0}^{1}n+7_{ - 1}^{0}e$ is also a nuclear fission reaction of uranium - 235.
Answer:
${8}^{16}O+{2}^{4}He\rightarrow_{10}^{20}Ne$