Physics > Advanced Modern Physics > 3.0 Nuclear Structure
Advanced Modern Physics
1.0 X-Rays
2.0 Moseley's Law.
3.0 Nuclear Structure
4.0 Nuclear binding energy
4.1 Binding energy per nucleon
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability
5.0 Radioactivity
6.0 Radioactive decay law
3.1 Atomic Mass Number:
4.2 Variation of Binding energy per nucleon with mass number
4.3 Nuclear stability
It is the nearest integer value of mass represented in $amu$ (atomic mass unit). $$1\;amu = 1.656 \times {10^{ - 27}}Kg$$
Example: Atomic mass number of aluminum $(Al)$ is 13, chlorine $(Cl)$ is 17 etc.
Each nuclear species with a given $Z$ and $A$ is called a nuclide.
The number of protons in the nucleus characterizes the species of the atom, means the element to which it belongs.
The electron can be added or removed from an atom, but it doesn't change its species.
Here are some notations that will be used in further discussions-
$$\begin{equation} \begin{aligned} Z \to {\text{ Atomic number}} \\ N \to {\text{ Number of neutrons}} \\ A\left( { = Z + N} \right) \to {\text{ Atomic mass number}} \\ {}_Z{X^A} \to {\text{ Atom}} \\ {}_Z^AX \to {\text{ Nucleus}} \\ {}_1^1H \to {\text{ Proton}} \\ {}_{ - 1}^0e \to {\text{ electron}} \\ {}_0^1n \to {\text{ neutron}} \\\end{aligned} \end{equation} $$
| Particle | Electric charge(C) | Mass | |
| In kilograms(Kg) | In atomic mass unit(u) | ||
| Electron | $ - 16.0 \times {10^{ - 19}}$ | $9.109390 \times {10^{ - 31}}$ | $5.485799 \times {10^{ - 4}}$ |
| Proton | $ - 16.0 \times {10^{ - 19}}$ | $1.672623 \times {10^{ - 27}}$ | $1.007276$ |
| Neutron | 0 | $1.674929 \times {10^{ - 27}}$ | $1.008665$ |
Isotopes
The nuclei having the same number of protons but different number of neutrons are called isotopes.
Example. Isotopes of carbon ${}_6^{11}C,{\text{ }}{}_6^{12}C,{\text{ }}{}_6^{13}C,{\text{ }}{}_6^{14}C$
Isotones
Nuclei having the same number of neutrons but different number of protons is called isotones.
Example: ${}_6^{14}C,{\text{ }}{}_7^{15}N,{\text{ }}{}_8^{16}C,{\text{ }}{}_9^{17}F$
Isobars
Nuclei having same mass number but different atomic number are called isobars.
e.g. ${}_6^{16}C,{\text{ }}{}_7^{16}N,{\text{ }}{}_8^{16}O,{\text{ }}{}_9^{16}F$
Many experiments have shown that nuclei are approximately spherical and average radius is given by
$$r = {r_0}{A^{\frac{1}{3}}}$$
where $r_0$ is a constant equal to $1.2 \times {10^{ - 15}}m$.
Question 7. Find the numerical value of density of a nucleus.
Solution:
Let the mass of proton be $M$. So, mass of the nucleus=$AM$
Assuming the nucleus is spherical and using Eq. (i), we find the volume is
$$V = \frac{4}{3}\pi {r^3}$$ So, $$V = \frac{4}{3}\pi r_0^3A\quad \left( {As,r = {r_0}{A^{\frac{1}{3}}}} \right)$$
The nuclear density can be calculated as,
$${\rho _n} = \frac{{{\text{mass}}}}{{{\text{volume}}}} = \frac{{AM}}{{\frac{4}{3}\pi r_0^3A}} = \frac{{3M}}{{4\pi r_0^3}}{\text{ }}$$
The fact that $A$ cancels out of our expression for nuclear density proves our statement above, that all nuclei have roughly the same density.
As we know, ${r_0} = 1.2 \times {10^{ - 15}}m$ and $M = 1.67 \times {10^{ - 27}}{\text{ }}Kg$.
So, $$\begin{equation} \begin{aligned} {\rho _n} = \frac{{3\left( {1.67 \times {{10}^{ - 27}}{\text{ }}Kg} \right)}}{{4\pi {{\left( {1.2 \times {{10}^{ - 15}}{\text{ }}m} \right)}^3}}} \\ {\rho _n} = 2.3 \times {10^{17}}{\text{ }}Kg\;{m^{ - 3}} \\\end{aligned} \end{equation} $$
Question 8. Calculate the electric potential energy of interaction due to the electric repulsion between two nuclei of ${}^{12}C$ when they "touch" each other at the surface.
Solution:
The radius of a ${}^{12}C$ nucleus is,
$$r = {r_0}{A^{\frac{1}{3}}}$$
The separation between the centers of the nuclei is, $$2R = 5.50fm$$
The potential energy of the pair is given by,
$$\begin{equation} \begin{aligned} U = \frac{{{q_1}{q_2}}}{{4\pi {e_0}r}} \\ U = \left( {9 \times {{10}^9}N{m^2}{C^{ - 2}}} \right)\frac{{{{\left( {6 \times 1.6 \times {{10}^{ - 19}}C} \right)}^2}}}{{5.50 \times {{10}^{ - 15}}m}} \\ U = 1.50 \times {10^{ - 12}}J \\\end{aligned} \end{equation} $$
As, $1\;MeV = 1.60 \times {10^{ - 13}}J$ So,
$$U = 9.39MeV$$
