Question

If the binding energy of ${}^{50}Cr$ (in V/nucleon) for the following nuclides is approximately $x$ Mev. Find the value of $\frac{x}{112}$.

Given : Atomic mass $=49.94605$ amu

Answer 1

${}_{24}^{50}Cr$ contains 24 protons, 24 electrons and 26 neutrons.
The mass defect is $(24\times 1.00758+24\times 0.000543+26\times 1.00893)-49.94605amu=0.481082amu$
The mass defect in g is $0.481082amu\times 1.66\times {10}^{-24}=7.98\times {10}^{-25}g$
This is the mass defect for 1 nucleus. The mass defect for 1 mole is
$7.98\times {10}^{-25}g\times 6.023\times {10}^{23}=0.48g/mol$.
The binding energy is $0.481082amu\times 931.5=448MeV$

$\therefore$ value of $\frac{448}{112}=4$