PX262 - I3 - liquid drop model of nucleus

based on observation that all nuclei have the same density
individual nucleons appear to be analogous to molecules in a liquid
it treats the nucleus as a droplet of incompressible nuclear material
eg: a fission process

_{92}^{235} U + slow neutrons \to +_{40}^{97} Zr +_{52}^{137} Te

a semi-empirical formula for the binding energy, $E_{B}$ , has five terms in terms of $Z$ , $N$ (and $A = Z + N$ )
- since the nuclear interactions show saturation, there is a term $\propto A$ , $C_{1} A$ , where $C_{1}$ is extracted from experimental data
- nucleons on the surface of the nucleus are less tightly bound than those in the interior, negative terms $\propto 4 π R^{2}$ , $- C_{2} A^{2 / 3}$ , ( $R \propto A^{1 / 3}$ )
- each proton repels the other protons, electrostatic interaction $\propto 1 / R$ term, $- C_{3} Z (Z - 1) / A^{1 / 3}$
- from experiments, nuclei appear to need a balance between energies associated with neutrons and protons so that $N ≃ Z$ from small nuclei, and $N$ is slightly larger than $Z$ for larger nuclear term: $- C_{4} (N - Z)^{2} / A = - C_{4} (A - 2 Z)^{2} / A$
- nuclear forces favour pairing for both protons and neutrons. positive if both $Z$ and $N$ are even, $+ C_{5} / A^{4 / 3}$ , $- C_{5} / A^{4 / 3}$ if both are odd, or zero otherwise

E_{B} = C_{1} A - C_{2} A^{2 / 3} - C_{3} \frac{Z (Z - 1)}{A^{1 / 3}} - C_{4} \frac{(A - 2 Z)^{2}}{A} + (-) C_{5} A^{- 4 / 3}

all atomic masses are less than the sum of their parts owing to the forces keeping the nucleons together in terms of energy:

M_{atomic nucleus} c^{2} < Z m_{p} c^{2} + N m_{n} c^{2}

the binding energy:

E_{B} = (Z m_{p} + N m_{n}) c^{2} -_{Z}^{A} M c^{2} > 0

is there is a theoretical limit to the number of protons in a nucleus of an atom (and periodic table extent)?

estimating from the bohr model
electron orbiting a nucleus:

\begin{aligned} \frac{m_{e} v^{2}}{r} & = \frac{Z e^{2}}{4 π ε_{0} r^{2}} \\ (m_{e} v r) v & = \frac{Z e^{2}}{4 π ε_{0}} \end{aligned}

in the above equation, $m_{e} v r$ is the angular momentum, $l = n ℏ$

v = \frac{Z e^{2}}{4 π ε_{0} ℏ} < c

the speed of the electron has to be less than the speed of light

Z_{m a x} = {(\frac{e^{2}}{4 π ε_{0} ℏ c})}^{- 1} = \frac{1}{α}

where, $α$ is the fine structure constant, $α = 1 / 137$

so, $Z < 137$
the heaviest element identified so far in experiment has $Z = 118$ , oganesson, $_{188}^{294}$ Og