The binding energy per nucleon of the nucleus ${ }_{26}^{56} X$ in units of MeV is:
(Given that $m_{H}=1.007825 u, m_{n}=1.008665 u$ and mass of ${ }_{26}^{56} X = 55.934939 u$)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8.8 MeV
The mass deficit (Δm) is,
$Δm=Zm_H+(A-Z)m_n-m_{nucleons}$
$=26(1.007)+30×1.008-55.934$
$=0.528u$
The energy equivalent of mass deficit is,
$E=Δmc^2$
$=0.528×931.V$
$≃492.3MeV$
Binding energy per nucleon = $\frac{total\,binding\,energy}{Number \,of\,Nucleon}$
$=\frac{492.3}{56}≃8.79MeV$