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Question

Given the following binding energies:
1. 178O 131 MeV
2. 5626Fe 493MeV
3. 23892U 1804 MeV
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 B ΔFe>ΔO>ΔU
The binding energy per nucleon is just the average binding energy shared by each nucleon, given as,
Net binding energyNumber of nucleons=Net binding energy BMass number A
The upper left number,i , in the representation ijX tells us the mass number, or total number of nucleons in the nucleus. Therefore, for 178O,A=17.For 5626Fe,A=56 .And for 23892U,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 ΔFe is bound to be higher than both, being close to the top of the Delta versus Agraph


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