AP EAMCET · PHYSICS · Nuclear Physics

If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleon and that of $N^{15}$ is $7.7 MeV$ per nucleon. then the energy required to remove a neutron from $N^{15}$ is

A $5.25 MeV$
B $0.2 MeV$
C $10.5 MeV$
D $0.4 MeV$

Verified Solution

Answer & Solution

Correct Answer

(C) $10.5 MeV$

Step-by-step Solution

Detailed explanation

The total binding energy, N14=7.5×14 MeV N15=7.7×15 MeV The energy required to remove a neutron from N15, E=BE N15-BE N14 E=7.7×15-7.5×14 MeV ⇒ E=10.5 MeV

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If the binding energy of N14 is 7.5 MeV per nucleon and that of N15 is7.7 MeV per nucleon. then the energy required to remove a neutron from N15 is

Step-by-step Solution

If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleon and that of $N^{15}$ is $7.7 MeV$ per nucleon. then the energy required to remove a neutron from $N^{15}$ is