Question

26Fe56 , 28Ni58 are example for _____​

Answer 1

Answer:

In .26Fe56, no. of protons 26,

no. of neutrons =56−26=30

∴ Mass defect =26mp+30mn−MFe

=26×1.007825+30×1.008665−55.934939

=26.20345+30.25995−55.934939

=0.528461u.

B.E/(nucleon)=0.528461×93156=8.79MeV/N

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