Binding Energy of the Deuteron

[joecoss]

It seems that for the case of the Electron in the Ground State of the Hydrogen Atom that when the value of the Binding Energy (13.6 eV), as calculated by Bohr, is divided by the separation distance (the Bohr Radius = 5.29 x 10^-11 m) we only get half of the Coulomb Force (4.11 x 10^-8 N) which exists between the Proton and the Electron (8.22 x 10^-8 N). Am I missing a factor of 2 when dividing the Binding Energy by the radius?
Is there a relationship between the value of the Binding Energy of the Deuteron (2.22 MeV) and the actual value of the Force that exists between the Proton and the Neutron at a distance of approximately 2.13 x 10^-15 m ? What is an approximate value of this Force between the Nucleons(in Newtons)? Thanks.

 

[Orion1]

half-integrals


Am I missing a factor of 2 when dividing the Binding Energy by the radius?


Affirmative.

This confusion is in the fact that the Bohr Hydrogen model is not a 2 dimensional static model, but instead is a 3 dimensional electrostatic model.

Total Electron Energy:
$$E = K + U = \frac{mv^2}{2} - k \frac{q^2}{r}$$

Newtons Second Law:
$$F_q = F_c$$
F_q - Coulomb's Force
F_c - Centripetal Force
F_q = F_c = 8.238*10^-8 N

$$k \frac{q^2}{r^2} = \frac{mv^2}{r}$$

Electron Kinetic Energy integral:
$$K = \int \frac{mv^2}{2} = k \frac{q^2}{2r}$$

The Coulomb potential half-integral results from the integral of the Electron average kinetic energy, and therefore is non-relativistic.

Total Electron Energy:
$$E_h = \int k \frac{q^2}{2r_h} - k \frac{q^2}{r_h} = k \frac{q^2}{r_h} \left( \frac{1}{2} - 1 \right) = k \frac{q^2}{r_h} \left( -\frac{1}{2} \right)$$

Total Electron Energy:
$$E_h = -k \frac{q^2}{2r_h}$$


Is there a relationship between the value of the Binding Energy of the Deuteron (2.22 MeV) and the actual value of the Force that exists between the Proton and the Neutron at a distance of approximately 2.13 x 10^-15 m ?


The classical strong nuclear force is:
$$F_s = \frac{ \hbar c}{r_n^2}$$

Given:
r_n = 2.13 x 10^-15 m
Then:
F_s = 6968.265 N

Classical Deuterium Binding Energy: (ev)
$$E_b = \frac{F_s \Delta r_b}{2q}$$
r_b = 1.603*10^-16 m - binding radius

Classical Deuterium Binding Energy: (ev)
$$E_b = \frac{\hbar c \Delta r_b}{2qr_n^2}$$

Classical Binding Energy Force:
$$F_b = \frac{E_b}{r_n} = \frac{\hbar c \Delta r_b}{2r_n^2}$$
F_b = 167.721 N

For more predictive results, reference the semi-classical binding energy formula, however note that the mass defect formula is currently still more accurate for this result.

 

[joecoss]

O1
Physics Forum is a great thing, and one of the great things about it is that Orion 1 is involved with it! Thanks for the Binding Energy, and thanks for the Magnetic Moment of the Tau Lepton.
JC