Question

In the total binding energies of ${}_1{H}^{2}an{d}_{2}H{e}^4{,}_{26}F{e}^{56}$ and ${}_{92}{U}^{235}$ nuclei are 2.22, 28.3, 492 and 1786 MeV respectively. Identify the most stable nucleus out of the following

• A. ${}_1{H}^2$
  
• B. ${}_{2}H{e}^4$
  
• C. ${}_{26}F{e}^{56}$
  
• D. ${}_{92}{U}^{235}$

Answer 1

The correct option is C ${}_{26}F{e}^{56}$


BE per nucleon for ${}_1{H}^2{,}_{2}H{e}^4{,}_{26}F{e}^{56}$ and ${}_{92}{U}^{235}$ are respectively,
$\frac{2.22}{2},\frac{28.3}{4},\frac{492}{56},\frac{1786}{235}$
Out of them all $\frac{492}{56}$ is the greatest. Therefore, ${}_{26}F{e}^{56}$ is the most bound or most stable nucleus.