Question

which equation demonstrates that nuclear fusion forms elements that are heavier than helium?

Explanation:

Nuclear fusion is a process where two or more light - nuclei combine to form a heavier nucleus. Let's analyze each equation:

• The first equation \(\ce{^{235}_{92}U + ^{1}_{0}n

ightarrow^{95}_{42}Mo + ^{139}_{57}La + 2^{1}_{0}n + 7^{0}_{- 1}e}\) is a nuclear fission reaction (a heavy nucleus splits into lighter nuclei).

• The second equation \(\ce{^{16}_{8}O+^{4}_{2}He

ightarrow^{20}_{10}Ne}\) shows two nuclei (\(\ce{O}\) and \(\ce{He}\)) combining to form a heavier nucleus (\(\ce{Ne}\)).

• The third equation \(\ce{^{235}_{92}U + ^{1}_{0}n

ightarrow^{131}_{53}I+^{89}_{39}Y + 16^{1}_{0}n}\) is a nuclear fission reaction.

• The fourth equation \(\ce{^{2}_{1}H+^{3}_{1}H

ightarrow^{4}_{2}He + ^{1}_{0}n}\) forms \(\ce{He}\) (not heavier than \(\ce{He}\) as the question asks for elements heavier than \(\ce{He}\)).

Answer:

\(\ce{^{16}_{8}O+^{4}_{2}He
ightarrow^{20}_{10}Ne}\)