Question

The mass of the atom\(_{11}^{23}Na\)is\(22.9898\)amu.

(a) Calculate its binding energy per atom in millions of electron volts.

(b) Calculate its binding energy per nucleon.

Short answer

a)\(E = 178.105\,{\rm{MeV}}\)

b) \(E = 7.74\,{\rm{MeV}}\)

Step-by-step solution

a) The binding energy per atom in millions of electron volts.

Nuclear binding energy is the energy which is required to break the nucleus into protons and neutrons.

The atomic number of sodium is 11, which means number of protons and electrons is \(11.\)

We can get the number of neutrons by subtracting atomic mass with atomic number: \(23 - 11 = 12\).

We can get the mass of the atom when we multiply the number of protons, electrons and neutrons with their masses and then we sum them up.

Mass of proton is\(1.00728{\rm{amu}}\), mass of neutron is \(1.008{\rm{amu}}\) and mass of electron is\(0.00055{\rm{amu}}\).

From that we get:

\(\begin{aligned}{}Mass{\rm{ }} &= {\rm{ }}\left( {mass{\rm{ }}of{\rm{ }}proton \times 11} \right) + \left( {mass{\rm{ }}of{\rm{ }}neutron \times 12} \right) + \left( {mass{\rm{ }}of{\rm{ }}electron \times 11} \right)\\{\mathop{\rm mass}\nolimits} &= (1.00728\,{\rm{amu}} \times 11) + (1.008\,{\rm{amu}} \times 12) + (0.00055\,{\rm{amu}} \times 11)\\ &= 23.182\,amu\end{aligned}\)

But, mass of sodium is \(22.9898{\rm{amu}}\)

From that we can calculate the mass defect:

\(\begin{aligned}{}Mass\,defect &= 23.182{\rm{amu}} - 22.9898{\rm{amu}}\\{\rm{ = 0}}{\rm{.193}}\,{\rm{amu}}\end{aligned}\)

The energy required is:

\(E = m{c^2}\)

But for that we need to convert amu into\({\rm{kg}}\).

\(1\,{\rm{amu}} = 1.66 \times {10^{ - 27}}\;{\rm{kg}}\),

from which we get:

\(\begin{aligned}{}Mass\,defect &= 0.193\,{\rm{amu}} \times 1.66 \times {10^{ - 27}}\;{\rm{kg}}\\ &= 3.19 \cdot {10^{ - 27}}\;{\rm{kg}}\end{aligned}\)

\(\begin{aligned}{l}E = m{c^2}\\E &= 3.19 \times {10^{ - 27}}\;{\rm{kg}} \cdot {\left( {2.99 \times {{10}^8}\;{\rm{m}}{{\rm{s}}^ - }1} \right)^2}\\E &= 2.852 \times {10^ - }^{11\;}{\rm{J}}\end{aligned}\)

Now we need to convert the energy from J to\({\rm{MeV}}\):

\(\begin{aligned}{}1\,\,{\rm{MeV}} &= {1.60210^{ - 13}}\;{\rm{J}}\\E &= 2.852 \times {10^ - }^{11}\;{\rm{J}} \times \frac{{1\,{\rm{MeV}}}}{{1.602 \cdot {{10}^{ - 13}}\;{\rm{J}}}}\\ &= 178.105\,MeV\end{aligned}\)

Hence,\(E = 178.105{\rm{MeV}}\)

b) The binding energy per nucleon

The binding energy per nucleon is obtained by dividing energy with mass number (nucleon number).

\(\begin{aligned}{}{E_{nucleon}} &= \frac{E}{A}\\{E_{nucleon}} &= \frac{{178.105{\rm{MeV}}}}{{23}}\\{E_{nucleon}} &= 7.74{\rm{MeV}}\end{aligned}\)