Given the following binding energies:
1. 178O→131MeV
2. 5626Fe→493MeV
3. 23892U→1804MeV Compare the binding energies per nucleon (Δ=BA)of the three nuclei
A
ΔO>ΔFe>ΔU
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B
ΔFe>ΔO>ΔU
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C
ΔU>ΔO>ΔFe
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D
ΔU>ΔFe>ΔO
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Solution
The correct option is BΔFe>ΔO>ΔU The binding energy per nucleon is just the average binding energy shared by each nucleon, given as, NetbindingenergyNumberofnucleons=NetbindingenergyBMassnumberA The upper left number,i , in the representation ijXtells us the mass number, or total number of nucleons in the nucleus. Therefore, for 178O,A=17.For5626Fe,A=56.Andfor23892U,A=238. [BA]170=13117MeV=7.7MeV [BA]56Fe=49356MeV=8.8MeV [BA]238U=1804238MeV=7.6MeV, Therefore, representing the binding energy by Delta ΔFe>Δ0>ΔU Interesting, don't you think?
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P.S ∴To save a step, we could have calculated just Δ0 and ΔU , since ΔFeis bound to be higher than both, being close to the top of the Deltaversus Agraph