11-16 July 2022
Europe/Moscow timezone
For authors of online talks: we will send videoconference links 1 day before the session to e-mail addresses specified in the Registration form

4G model of fitting RMS radius of proton

15 Jul 2022, 18:30
20m
Физический ф-т, 5-19

Физический ф-т, 5-19

Oral talk (15 min + 5 min questions) Nuclear structure: theory and experiment

Speaker

Prof. Laksminarayana S (Andhra University, Visakhapatnam-03,AP, INDIA)

Description

In our 4G model of final unification, along with three large atomic gravitational constants pertaining to electromagnetic, strong and weak interactions, we have proposed the existence of a strong nuclear charge of magnitude $e_s \cong 2.9463591e$ and existence of electroweak fermion of rest energy $M_wc^2\cong 584.725$ GeV. With reference to protons, pions and weak bosons, we noticed that, $\frac{m_p}{M_w}\cong \frac{\sqrt{\left(m_\pi\right)^{0} \left(m_\pi\right)^{ \pm }}}{\sqrt{ \left(m_z\right)^{0}\left(m_w\right)^{\pm}}}\cong 0.0016$ and this number seems to play a crucial role in understanding nuclear stability and binding energy. Increasing number of free nucleons, increasing nuclear radii and increasing asymmetry about stable mass number play important role in reducing the nuclear binding energy. Unified binding energy coefficient can be expressed with, $B_0 \cong \sqrt{\left(\frac{e_s^2}{8 \pi \epsilon_0 \left(\frac{\hbar}{m_pc}\right)}\right)\left(\frac{e^2}{8 \pi \epsilon_0 \left(\frac{\hbar}{m_pc}\right)}\right)}\cong 10.1$ MeV. A four term nuclear binding energy relation for $\left[\left(Z \geq 3\right) \textrm{and}\left(N\geq Z\right)\right]$ can be expressed as, $BE\cong \left[A-\left(1+0.0016\left(\frac{Z^2+A^2}{2}\right)\right)-A^{\frac{1}{3}}-\left(\frac{\left(A_s-A\right)^2}{A_s}\right)\right]\left(10.1\right)$ MeV where $A_s\cong \left(2Z\right)+0.0016\left(2Z\right)^2\cong \left(2Z\right)+0.0064Z^2$ = Estimated mass number close to stability zone. Based on these relations, we noticed a relation for fitting RMS radius of proton as, $R_p \cong \sqrt{\frac{4 \pi \epsilon_0 \hbar^3}{e_s^2 m_p^2 c}} \cong \sqrt{\frac{\alpha_s}{\alpha}}\left(\frac{\hbar}{m_p c}\right)\cong 0.835$ fm where $\alpha_s \cong 0.1152$ represents the strong coupling constant. It needs further study.

The speaker is a student or young scientist No
Section 1. Nuclear structure: theory and experiment

Primary authors

Seshavatharam UVS (I-SERVE, Survey no-42, Hitech city, Hyderabad-84,Telangana, India) Prof. Laksminarayana S (Andhra University, Visakhapatnam-03,AP, INDIA)

Presentation Materials