Answer :
To determine which equation demonstrates that nuclear fusion forms elements that are heavier than helium, let's carefully consider each reaction.
1. [tex]\(\ce{^2_1H + ^3_1H -> ^4_2He + ^1_0n}\)[/tex]:
- This reaction involves deuterium ([tex]\(\ce{^2_1H}\)[/tex]) and tritium ([tex]\(\ce{^3_1H}\)[/tex]), which are both isotopes of hydrogen.
- These two hydrogen nuclei combine (fusion) to form a helium nucleus ([tex]\(\ce{^4_2He}\)[/tex]) and a neutron ([tex]\(\ce{^1_0n}\)[/tex]).
While this is a fusion reaction, helium is not heavier than helium since the output is just helium and a neutron.
2. [tex]\(\ce{^16_8O + ^4_2He -> ^20_{10}Ne}\)[/tex]:
- This reaction involves an oxygen nucleus ([tex]\(\ce{^16_8O}\)[/tex]) and a helium nucleus ([tex]\(\ce{^4_2He}\)[/tex]).
- These nuclei combine (fusion) to form a neon nucleus ([tex]\(\ce{^20_{10}Ne}\)[/tex]).
This is a fusion reaction that results in the formation of neon, an element that is heavier than helium.
3. [tex]\(\ce{^235_{92}U + ^1_0n -> ^131_{53}I + ^89_{39}Y + 16 ^1_0n}\)[/tex]:
- This reaction involves uranium-235 ([tex]\(\ce{^235_{92}U}\)[/tex]) and a neutron ([tex]\(\ce{^1_0n}\)[/tex]).
- The uranium nucleus undergoes fission, splitting into iodine ([tex]\(\ce{^131_{53}I}\)[/tex]) and yttrium ([tex]\(\ce{^89_{39}Y}\)[/tex]), along with multiple neutrons.
This is a fission reaction, not a fusion reaction.
4. [tex]\(\ce{^235_{92}U + ^1_0n -> ^95_{42}Mo + ^139_{57}La + 2 ^1_0n + 7 ^0_{-1}e}\)[/tex]:
- This reaction also involves uranium-235 ([tex]\(\ce{^235_{92}U}\)[/tex]) and a neutron ([tex]\(\ce{^1_0n}\)[/tex]).
- The uranium nucleus again undergoes fission, resulting in the formation of molybdenum ([tex]\(\ce{^95_{42}Mo}\)[/tex]) and lanthanum ([tex]\(\ce{^139_{57}La}\)[/tex]), along with several neutrons and electrons.
This is another example of a fission reaction, not a fusion reaction.
Based on the analysis, the equation that demonstrates nuclear fusion forming an element that is heavier than helium is:
[tex]\[ \ce{^16_8O + ^4_2He -> ^20_{10}Ne} \][/tex]
Therefore, the correct answer is the second equation.
1. [tex]\(\ce{^2_1H + ^3_1H -> ^4_2He + ^1_0n}\)[/tex]:
- This reaction involves deuterium ([tex]\(\ce{^2_1H}\)[/tex]) and tritium ([tex]\(\ce{^3_1H}\)[/tex]), which are both isotopes of hydrogen.
- These two hydrogen nuclei combine (fusion) to form a helium nucleus ([tex]\(\ce{^4_2He}\)[/tex]) and a neutron ([tex]\(\ce{^1_0n}\)[/tex]).
While this is a fusion reaction, helium is not heavier than helium since the output is just helium and a neutron.
2. [tex]\(\ce{^16_8O + ^4_2He -> ^20_{10}Ne}\)[/tex]:
- This reaction involves an oxygen nucleus ([tex]\(\ce{^16_8O}\)[/tex]) and a helium nucleus ([tex]\(\ce{^4_2He}\)[/tex]).
- These nuclei combine (fusion) to form a neon nucleus ([tex]\(\ce{^20_{10}Ne}\)[/tex]).
This is a fusion reaction that results in the formation of neon, an element that is heavier than helium.
3. [tex]\(\ce{^235_{92}U + ^1_0n -> ^131_{53}I + ^89_{39}Y + 16 ^1_0n}\)[/tex]:
- This reaction involves uranium-235 ([tex]\(\ce{^235_{92}U}\)[/tex]) and a neutron ([tex]\(\ce{^1_0n}\)[/tex]).
- The uranium nucleus undergoes fission, splitting into iodine ([tex]\(\ce{^131_{53}I}\)[/tex]) and yttrium ([tex]\(\ce{^89_{39}Y}\)[/tex]), along with multiple neutrons.
This is a fission reaction, not a fusion reaction.
4. [tex]\(\ce{^235_{92}U + ^1_0n -> ^95_{42}Mo + ^139_{57}La + 2 ^1_0n + 7 ^0_{-1}e}\)[/tex]:
- This reaction also involves uranium-235 ([tex]\(\ce{^235_{92}U}\)[/tex]) and a neutron ([tex]\(\ce{^1_0n}\)[/tex]).
- The uranium nucleus again undergoes fission, resulting in the formation of molybdenum ([tex]\(\ce{^95_{42}Mo}\)[/tex]) and lanthanum ([tex]\(\ce{^139_{57}La}\)[/tex]), along with several neutrons and electrons.
This is another example of a fission reaction, not a fusion reaction.
Based on the analysis, the equation that demonstrates nuclear fusion forming an element that is heavier than helium is:
[tex]\[ \ce{^16_8O + ^4_2He -> ^20_{10}Ne} \][/tex]
Therefore, the correct answer is the second equation.
