Find BE per nucleon of ^56Fe where m(^56Fe) = 55.936u m_n = 1.00727u, m_p = 1.007274 u.

Correct option is B. 8.52 MeV
We know that ,Binding Energy ,BE=Δmc²
where Δm is the mass defect
Here ,Δm=Mass of nucleon- Mass of nucleus=26 mp+30 mn−m⁵⁶Fe
Hence BE=[26 mp+30 mn−m⁵⁶Fe]c
Where ,mn=1.00866 ump=1.00727 u
BE=[26×1.00727+30×1.00866−55.936]×931=0.51282×931=477.435 MeV
Binding energy per nucleon=477.435/56 MeV=8.52 MeV.