(a) Charge : Charge of the nucleus = +Ze
Where e is the charge on the electron = $1.6 \times 10^{-19}$ coulomb
(b) Mass : Approximately equal to

$Z = (A- Z)m_n + Zm_p$

Where $m_n$ & $m_p$ are mass of neutron & proton $m_n \approx 1840 m_e, \ m_p \approx 1836 m_e$

(c) Size : The volume is proportional to total number of nucleons in it

$\dfrac{4}{3} \pi R^3 \propto A$

$\Rightarrow \ \ \ \ R \propto A^{1/3}$

So, radius of nuclear $R = R_\circ A^{1/3}$ where $R_\circ = 1.1 - 1.2 \mathrm{fm} (1 \mathrm{fm} = 10^{-15}m)$

Nuclear surface is defined as the surface outside which there is negligible probability of finding any of the nuclear constituents. The electric charge distribution within the nucleus is approximately same as nuclear mass distribution. The electrical radius of the nucleus and nuclear matter radius are approximately the same.

Distriabution of charge = $\dfrac{\rho_{ch} (r)}{\rho_\circ}$

$\rho_{ch}(r) = \dfrac{\rho_\circ}{1 + \exp\left [ \dfrac{(r - R)}{a} \right ]}$ where, a = 0.5 fm

(d) Density

Nuclear density, $\rho_N = \dfrac{\mathrm{Nuclear \ mass}}{\mathrm{Nuclear \ volume}}$

Nuclear mass = $Am_n$

Where A = mass number

$m_n$ = mass of nucleon

$= 1.67 \times 10^{-27} kg$

Nuclear volume = $= \dfrac{4}{3} \pi R^3$

$= \dfrac{4}{3} \pi (R_\circ A^{1/3})^3$

$= \dfrac{4}{3} \pi R_{\circ}^{3} A$

$\rho_N = \dfrac{Am_n}{\dfrac{4}{3} \pi R_{\circ}^{3} A} = \dfrac{m_n}{\dfrac{4}{3} \pi R_{\circ}^{3}} = 2 \times 10^{17} kg/m^3$

As the density of the nucleus is independent of A, its value is almost same for all nuclei

(e) Angular momentum

(i) Spin angular momentum : It is due to particles spinning motion about its own axis through centre of mass magnitude of spin angular momentum $\left| {\vec S} \right| = \sqrt{{\not S(\not S + 1)}}\hbar$ where $\not S = \dfrac{1}{2}$ is the spin angular momentum quantum number.
Z-component of angular momentum is $S_Z = m_s \hbar$ , where $m_s$ is magnetic spin quantum number $m_s = \pm \dfrac{1}{2}$
+ ve for spin axis parallel to Z-axis
– ve for anti parallel spin

$S_z = \pm \dfrac{1}{2} \hbar$

(ii) Orbital Angular Momentum : Each individual nucleon may be pictured as having as angular momentum associated with orbital motion. This is called orbital angular momentum. Its magnitude is $|\vec l| = \sqrt{l(l + 1)\hbar},$ where l is orbital angular momentum quantum number l = 0, 1, 2, ….

Z-component of orbital angular momentum

$L_z = m_l \hbar$ where $m_l$ is orbital magnetic quantum number $m_l = -l, -(l - 1),\cdots 1, -1, 0, 1, \cdots (l - 1), l$

(iii) Total angular momentum : Total angular momentum of a nucleus J is vector sum of its orbital & spin angular momenta

$\vec j = \vec l + \vec s$

$\left | \vec J \right | = \sqrt{j(j + 1) \hbar}$

where j is total angular momentum quantum number and for both proton & neutron, $S = \dfrac{1}{2}$

So, J is always half integral

Z component of total angular momentum $J_z = m_j \hbar$ where $m_j$ is total magnetic angular momentum quantum number $m_j = j,(j - 1), \cdots -(j - 1), -j$

Total angular momentum of nucleus

Total angular momentum of the nucleus $\vec I$ is the resultant of the individual total angular momentum of all the constituent nucleons in the nucleus.

Its magnitude $\vec I = \sqrt{l(l + 1)\hbar}$ where I is total angular momentum quantum number for the nucleus. The value of I depends on the type of interaction or coupling in the nucleus.

(f) Nuclear Spin

Nuclear magnetic moment is proportional to angular momentum

$\vec \mu_I = \dfrac{g_I \mu_N}{\hbar} \vec I$

Where g is called nuclear g factor

$\mu_I = g_I \mu_N \sqrt{I(I + 1)}$

where $\mu_N$ is called the nuclear magneton

$\mu_B = 5.79 \times 10^{-5} eV/T$

$= 9.27 \times 10^{-24} J/T$

$\dfrac{\mu_n}{\mu_B} = \dfrac{m_e}{m_p} = \dfrac{1}{1836}$

$\mu = 3.15 \times 10^{-8} ev/\tau$

$= 5.05 \times 10^{-27} J/T$

Nuclear g-factor varies from nucleus to nucleus

$\mu_{\mathrm{proton}} = \pm 2.792 \mu_N$ in Z-direction

$\mu_{\mathrm{neutron}} = \mp 1.913 \mu_N$

$(\pm)$ sign used for proton because $\mu_{pz} || \vec S$

$(\mp)$ sign used for neutron because $\mu_{nz}$ is anti-parallel to $\vec S$

Negative nuclear magnetic moment of the neutron indicates that negative charge on the average is father from the axis.

The total angular momentum of the whole atom

$\vec F = \vec L + \vec S + \vec I$

$= \vec J + \vec I$

Vector $\vec I$ & $\vec J$ process around their resultant $\vec F$

(g) Quadrupole moment :

Quadrupole moment q is a measure of the departure from spherical symmetry of the nuclear charge distribution.
The electric quadrupole moment of a nuclear charge distribution which is symmetric about axis is given by

$q = \int_{v} (3z^2 - r^2) \rho(x, y, z)d \tau$ where $\rho$ is average nuclear charge density in terms of terms of proton charges & $r^2 = x^2 + y^2 + z^2$ for uniformly charged ellipsoid of revolution defined by the equation

$\dfrac{x^2 + y^2}{a^2} + \dfrac{z^2}{b^2} = 1$

The electric quadrupole moment reduces to $q = \dfrac{2Ze}{5} (b^2 - a^2)$ where Ze is total nuclear charge

q = 0 for spherically symmetric charge distribution i.e. a = b

q > 0 for charged distribution stretched in z-direction (b > a) i.e. Oblate spheroid.

q < 0 for charged distribution stretched in perpendicular to z-direction (b < a) i.e. Oblate spheroid.

(h) Wave Mechanical Properties

Nucleus has two have wave mechanical properties

(i) Statistics (ii) parity

(i) Statistics : Quantum mechanical description of system with number of particles like nucleus is given by either Bose – Einstein or Fermi – Dirac Statistics.

Bose Einstein Statistics : All the particles with integral spin (in unit of $\hbar$ ) or zero obey B.E. statistics & are called bosons e.g. Photon $\pi$ -meson, deuteron

All nuclei with even mass number A obey B.E. statistics

The wave function of system obeying B.E. statistics is symmetric.

Fermi-Dirac Statistics : The particle with half integral spin $(\hbar/2)$ obey Fermi – dirac (F.D.) statistics, are called fermions eg. electrons, proton & neutrons.

All the nuclei with odd mass number A obey F.D. statistics.

The wave functions of system obeying F.D. statistics are anti-symmetric. This means that if all the co-ordinates of any pair of identical particles are intercharged, the new system will be identical with the original except for the charge of sign in wave function $\psi(x_1, x_2,\cdots x_i, x_j, \cdots, x_n) = -(x_1, x_2 \cdots x_j, x_i, \cdots x_n)$