Question

Find the binding energy of ${}_{26}^{56}Fe$. Atomic mass of ${}^{56}Fe$ is 55.9349 u and that of ${}^{1}H$ is 1.00783 u. Mass of neutron = 1.00867 u.

• A. 292 MeV
• B. 392 MeV
• C. 492 MeV
• D. 592 MeV

Answer 1

The correct option is C 492 MeV

We know that when a nucleus is formed, the energy is released.( binding energy ). Here for ${}_{26}^{56}Fe$ Nuclei
Binding energy will be = [30 (Mass of neutron) + 26(mass of proton) - mass of Fe nucleus] $c^2$
Fairly simple right. Find mass defect between nucleons and the nuclei and they multiply with $c^2$ to find the energy equivalence of it.
If only life had been so simple :(
We have been given the mass of Iron atom. Which also contains the mass of its electrons.
But a good thing is that we have been also given the mass of Hydrogen atom, which also contains the mass of electrons.
Let's try using this fact.
So we have been given the mass of neutron = 1.00867 u ------ (i)
Hydrogen has 1 proton and an electron.
So mass of a proton = (mass of hydrogen atom - mass of electron) = $(1.00783-m_e)u$ -------(ii)
Now we need to find the mass of Fe nuclei. It has 26 electrons in it.
So it’s nuclear mass = (mass of Iron atom - 26x mass of an electron) = $(55.9349–26m_e)u$ -------- (iii)
So now we know all the variables. Let's put it all in and see.
[30 (Mass of neutron) + 26(mass of proton) - mass of Fe nucleus] $c_2$
$=\left[30\right(1.00867)+26(1.00783-m_e)-(55.9349-26m_e\left)\right]c^2$
$=\left[\right(26\times 1.00783u)+(30\times 1.00867u)-55.9349u]c^2$
$=\left(0.52878u\right)c^2.$
Now,
$\left(1u\right)c^2=931MeV$
Hence,
$\left(0.52878u\right)c^2=0.52878\times 931MeV=492MeV.$