Question

An atom of 206Pb has a mass of 205.974440 amu. Calculate its binding energy in MeV per nucleon.
Use the masses:mass of 1H atom = 1.007825 amu mass of a neutron = 1.008665 amu 1 amu = 931.5 MeV

Answer 1

Answer: The binding energy of the given nucleus is 7.88 MeV/nucleon

Explanation:

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

We are given a nucleus having representation:  $_{82}^{206}\textrm{Pb}$  

Number of protons = 82

Number of neutrons = 206 - 82 = 124  

To calculate the mass defect of the nucleus, we use the equation:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)-M$

where,

$n_p$ = number of protons = 82  

$m_p$ = mass of one proton  = 1.007825 amu

$n_n$ = number of neutrons  = 124  

$m_n$ = mass of one neutron = 1.008665 amu

M = nuclear mass = 205.974440 amu  

Putting values in above equation, we get:  

$\Delta m=[(82\times 1.007825)+(124\times 1.008665)]-205.974440\\\\\Delta m=1.74167amu$

To calculate the binding energy of the nucleus, we use the equation:

$E=\Delta mc^2\\E=(1.74167u)\times c^2$

$E=(1.74167u)\times (931.5MeV)=1622.36MeV$    (Conversion factor:  $1u=931.5MeV/c^2$  )

Number of nucleons in $_{26}^{56}\textrm{Fe}$ atom = 206

To calculate the binding energy per nucleon, we divide the binding energy by the number of nucleons, we get:  

$\text{Binding energy per nucleon}=\frac{\text{Binding energy}}{\text{Nucleons}}$

$\text{Binding energy per nucleon}=\frac{1622.36MeV}{206}=7.88MeV/nucleon$

Hence, the binding energy of the given nucleus is 7.88 MeV/nucleon