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 ${ }_{26}^{56} X = 55.934939 u$)

8.8 MeV

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 = $\frac{total\,binding\,energy}{Number \,of\,Nucleon}$

$=\frac{492.3}{56}≃8.79MeV$