Question

The binding energy of an imaginary iron ${}_{36}^{56}Fe$ is $\underset{–}{}$
(Given atomic mass of Fe is 55.9349 amu and that of hydrogen is 1.00783 amu. Mass of neutron is 1.00876 amu)

• A. $3.49MeV$
• B. $4.31MeV$
• C. $6.49MeV$
• D. $931.49MeV$

Answer 1

The correct option is D $931.49MeV$

The number of proton of Fe nucleus $p=36$ and number of neutron, $n=56-36=20$

The combined mass of nucleus $=$ total mass of proton $+$ total mass of neutron

$=(36\times 1.00783)+(20\times 1.00876)=56.4571$ amu

Mass defect $\Delta m=56.457-55.9349=0.522$ amu

Binding energy, $BE=\Delta m{c}^2$

So, $BE=0.522c^2$ amu $=0.522\times 931.49MeV=486.24$ MeV where, $1amu=931.49MeV/c^2$