Download Morphed Gravitational Potential Energy, Nuclear Energy Levels

Journal of Nuclear and Particle Physics 2013, 3(2): 25-28
DOI: 10.5923/j.jnpp.20130302.01
Nuclear Binding with Gravitational Interaction
Cvavb Chandra Raju
Department of physics, Osmania University,Hyderabad-500 007 India
Abstract An altered gravitational potential energy is used to calculate the energy levels of nuclei. The universal constant
of gravitation G is changed through an exponential factor. Th is exponential factor is independent of distance and it is quite
small. The negligible exponential factor does not alter the law of universal gravitation for large interacting masses. For
interacting masses of the order of nucleon masses the morphed gravitational potential p lays crucial role. Using the methods of
ordinary quantum mechanics along with the morphed gravitational potential energy the binding energies of (1) Deuteron (2)
Heliu m-4(3) Oxygen-17 (4) Flourine-17 and Alu miniu m-27 areestimated. Allthese estimates agree pretty well with
experiment. The ground state wave functions are also found. The general wave functions of Deuteron and Heliu m-4 are found
in terms of their masses and charge. The energy spectrums of all the above nuclei are estimated.
Keywords Morphed Gravitat ional Potential Energy, Nuclear Energy Levels
𝑔𝑔
𝑉𝑉 (𝑟𝑟) = −𝐺𝐺 � 1 − �1 −
2 ℏ𝑐𝑐
𝐺𝐺𝑀𝑀02
1. Introduction
An understanding of the various physical properties of
nuclei requires a detailed knowledge of the nuclear forces
operating between the nucleons of a nucleus. The
Yu kawa1 potential is very successful in accounting for the
short range strong nuclear interaction. But no specific and
exact solution of the Schrödinger equation is available for
any nucleus. Are there any general potentials for which the
Schrödinger equation can provide an order of magnitude of
some gross properties of anucleus? This question led us to a
potential energy among nucleons wh ich is closely related to
the gravitational potential energy.
The constancy of the universal constant of gravitation G is
well established for larger interacting masses. No
experimental ev idence is availab le to confirm that G retains
its universality for interacting masses of the order of a
nucleon. We believe that the gravitational potential energy is
in general g iven by,
g c 

−
mm
GM 02
 1 2 ,
−G 1 − e
V (r ) =

 r


2
(1.1)
Where, 𝑔𝑔2 is a dimensionless positive number and
𝑀𝑀0 has the dimensions of mass. If 𝑀𝑀0 goes to zero, the
exponential factor tends to zero,and the usual expression for
the gravitational potential energy for part icles of masses
𝑚𝑚1 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚2 at a distance r results. An approximat ion to
Eq.(1.1) is given by,
* Corresponding author:
[email protected] (Cvavb Chandra Raju)
Published online at http://journal.sapub.org/jnpp
Copyright © 2013 Scientific & Academic Publishing. All Rights Reserved
��
𝑚𝑚 1 𝑚𝑚 2
𝑟𝑟
.
(1.2)
Our thesis is that as and when 𝑚𝑚1 𝑎𝑎𝑎𝑎𝑎𝑎/𝑜𝑜𝑜𝑜 𝑚𝑚2 are close
to 𝑀𝑀0 , it is Eq.(1.2) that is more correct. The gravitational
potential energy for particles whose masses are small is then
given by,
𝑉𝑉(𝑟𝑟) = −
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 1 𝑚𝑚 2
.
𝑟𝑟
𝑀𝑀02
(1.3)
The above potential energy is obtained from the
gravitational potential energy. We may call it
the“morphedgravitational potentialenergy (M GPE)”
We will show that Eq.(1.3) can be used to explain many
properties of nuclei.
The factor 𝑀𝑀0 is not a universal constant. It is an
adjustable parameter for each nucleus whereas 𝑔𝑔2 is same
for all nuclei. The M GPE g iven by Eq.(1.3) may still have
an exponentially decreasing Yu kawa factor 𝑒𝑒 −𝜇𝜇𝜇𝜇 wh ich is
set equal to unity in all the cases considered here.
The paper is organized in the fo llo wing way. In Sec.2 we
apply the MGPE for the case of Deuteronnucleus. InSec.3
the binding energy and the principal energy levels of the
Heliu m-4 nucleus are determined using the methods of
ordinary quantum mechanics. Sections 4, 5 and 6 are used
for the nuclei, Oxygen-17, Flourine-17 and Alu min iu m-27
respectively. Our conclusions are presented in the last
Sec.7.
2. The Deuteron Nucleus
The simplest known bound system is a Deuteron nucleus.
It is a bound system of a proton and neutron with an orbital
angular mo mentu m quantum number ℓ = 0 and has a total
spin of one. The binding energy of this nucleus is found to
be 2.2251 MeV and its radius is found tobe 2.1F.The
quadrupole
26
Cvavb Chandra Raju: Nuclear Binding with Gravitational Interaction
Moment is experimentally found to be 2.82 ×
10 −27 𝑐𝑐𝑐𝑐2 .The magnetic mo ment of the Deuteron is
about0.85 nuclear magneton.2, 3.If the nuclear potential is
independent of nuclear spin, then the triplet state (total
spin=1) and the singlet state (total spin=0) of the dinucleon
system will have the same energy. But this is not
observed.This means that the singlet potential is weaker
than the triplet potential. Using this informat ion we can
guess the parameters 𝑔𝑔2 and 𝑀𝑀02 .We consider an extreme
situation wherein the singlet tensor potential of aDiproton
nucleus iszero. Then the potential energy expressions
operative for a dip roton nucleus are,
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑝𝑝 𝑚𝑚 𝑝𝑝
𝑒𝑒 2 ℏ𝑐𝑐
1
+
𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒 2 =
.(2.1)
𝑉𝑉 (𝑟𝑟) = − 2
𝑟𝑟
𝑀𝑀0
𝑟𝑟
137
This total potential energy will be zero if 𝑀𝑀02 = 𝑚𝑚2𝑝𝑝
and 𝑔𝑔2 = 𝑒𝑒 2 and a stable Diproton nucleus will not exist!
Similarly a Dineutron nucleus will have neglig ible binding
energy if 𝑀𝑀02 = 𝑚𝑚2𝑛𝑛 ! For the Deuteron and for all nuclei
we found that,
𝑒𝑒 2
𝑔𝑔2 =
0.2254
= 0.032384 .
(2.2)
For each nucleus the parameter 𝑀𝑀02 is to be chosenso
that the ground state energy agrees with experiment. For
Deuteron nucleus 𝑀𝑀02 is given by,
(2.3)
𝑀𝑀02 = 0.931826 × 10−48 𝑔𝑔𝑔𝑔2
It should be noted that Eq. (2.2) shows that 𝑔𝑔2 is
electroweak interaction parameter4 . The Schrödinger
Equation for the Deuteron nucleus is given by
ℏ2
�− ∇2 + 𝑉𝑉(𝑟𝑟)� Ψ (𝑟𝑟, 𝜗𝜗, 𝜙𝜙) = 𝐸𝐸Ψ(𝑟𝑟, 𝜗𝜗. 𝜙𝜙) , where, (2.4)
2𝜇𝜇
𝜇𝜇Is the reduced mass and V(r) is given by Eq. (1.3) with
𝑚𝑚1 = 𝑚𝑚𝑝𝑝 and 𝑚𝑚2 = 𝑚𝑚𝑛𝑛 .The reduced mass is given by,
𝑚𝑚 𝑝𝑝 𝑚𝑚 𝑛𝑛
.
(2.5)
𝜇𝜇 =
𝑚𝑚 𝑝𝑝 +𝑚𝑚 𝑛𝑛
The spherically symmetric potential5, 6 leads to the
solution
(2.6)
Ψ(𝑟𝑟, 𝜗𝜗, 𝜙𝜙) = 𝑅𝑅𝑛𝑛ℓ 𝑌𝑌ℓ𝑚𝑚 (𝜗𝜗, 𝜙𝜙) .
Here
the
quantum
numbers
n, ℓ𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚 ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 as in the case of
Hydrogen atom.
n= 1, 2, 3…..ℓ = 0,1,2 … . (𝑛𝑛 − 1) , 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚 = −ℓ, −ℓ +
1, … + ℓ
(2.7)
The radial wave function is given by,
(2.8)
𝑅𝑅𝑛𝑛ℓ = 𝐴𝐴𝑒𝑒 −𝜌𝜌 /2 𝜌𝜌 ℓ 𝐿𝐿2ℓ+1
𝑛𝑛+ℓ (𝜌𝜌) , where,
𝜌𝜌 =
𝑎𝑎0 =
2𝑟𝑟
𝑛𝑛𝑎𝑎 0
ℏ2
, and
𝑀𝑀02
𝜇𝜇 𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑝𝑝 𝑚𝑚 𝑛𝑛
(2.9)
.
(2.10)
The energy Eigen-values of the Deuteron nucleus are
given by,
𝐸𝐸𝑛𝑛ℓ = −
𝜇𝜇
1
2 ℏ2 𝑛𝑛 2
�
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑝𝑝 𝑚𝑚 𝑛𝑛
𝑀𝑀02
2
� .
(2.11)
All these results are a simple transcription of Hydrogen
atom co mputations with an appropriate change of the
constants. The normalization constant A in Eq. (2.8) is
given by,
23
1 (𝑛𝑛−ℓ−1)!
[(𝑛𝑛+ℓ )! ]
𝐴𝐴 = � 3 3
𝑛𝑛 𝑎𝑎 2𝑛𝑛
0
.
(2.12)
The ground state wave function of the Deuteron nucleus
is given by,
Ψ10 =
1
�𝜋𝜋𝑎𝑎 30
𝑒𝑒 −
𝑟𝑟 �
𝑎𝑎 0
.
(2.13)
Fro m Eq. (2.10) it just follows that,
(2.14)
𝑎𝑎0 = 4.31734 × 10 −13 cm.
Here 𝑎𝑎0 is the distance between the two nucleons. The
center of mass lies half-way between the two nucleons as
their masses are nearly equal. The radius is half of Eq.
(2.14).This agrees pretty well with experiment. The ground
state energy of the Deuteron nucleus is given by,
(2.15)
𝐸𝐸10 = −2.2251𝑀𝑀𝑀𝑀𝑀𝑀 .
This also agrees pretty well with experiment. The
quadrupole mo ment calculations are presented to another
journal.
3. The Helium-4 Nucleus
The Heliu m-4 nucleus is a strongly bound system of two
protons and two neutrons. Our M GPE can be put to test by
comparing the estimated energy
Spectrum with experiment. A rough picture of this
nucleus is like this. Any one nucleon experiences a total
MGPE due to the remaining three nucleons.
The outer nucleon must be a proton as the two protons
repel each other due to the Coulomb fo rce. This rough
picture indicates that,
𝑉𝑉 (𝑟𝑟) = −
𝑔𝑔 2 ℏ𝑐𝑐 (2𝑚𝑚 𝑛𝑛 +𝑚𝑚 𝑝𝑝 )𝑚𝑚 𝑝𝑝
𝑀𝑀02
𝑟𝑟
+
𝑒𝑒 2 ℏ𝑐𝑐
𝑟𝑟
.
(3.1)
We will now apply the methods of quantum mechanics in
an effort to obtain a theoretical description of Heliu m-4
nucleus. In Eq.(3.1) the parameter 𝑀𝑀02 is given by,
(3.2)
𝑀𝑀02 = 0.935301 × 10 −48 𝑔𝑔𝑔𝑔2
This is slightly d ifferent fro m Eq. (2.3).So lving the
Schrödinger equation
With the potential energy given by Eq.(3.1) leads to the
following Energy spectrum for the Heliu m-4 nucleus:
𝐸𝐸𝑛𝑛ℓ = −
𝜇𝜇
1
2ℏ2 𝑛𝑛
�
2
𝑔𝑔 2 ℏ𝑐𝑐 �2 𝑚𝑚 𝑛𝑛 +𝑚𝑚 𝑝𝑝 �𝑚𝑚 𝑝𝑝
𝑀𝑀02
2
− 𝑒𝑒 2 ℏ𝑐𝑐� .
(3.3)
The constant 𝑎𝑎0 for the Heliu m-4 nucleus is given by,
𝑎𝑎0 =
ℏ2
1
.
2
𝜇𝜇 �𝑔𝑔 ℏ𝑐𝑐�2𝑚𝑚 𝑛𝑛 +𝑚𝑚 𝑝𝑝 �𝑚𝑚 𝑝𝑝 −𝑒𝑒 2 ℏ𝑐𝑐 �
2
𝑀𝑀
(3.4)
0
All other expressionslike ,Ψ, 𝜌𝜌 𝐸𝐸𝐸𝐸𝐸𝐸 have similar form as
in the case of the Deuteron nucleus. For the Heliu m -4
nucleus, we have,
(3.5)
𝜇𝜇 = 1.254749 × 10 −24 𝑔𝑔𝑔𝑔 ,
And the quantum nu mber
n= 1, 2, 3….
(3.6)
The energy Eigen values for the Heliu m-4 nucleus are
given by.
28 .3
(3.7)
𝐸𝐸𝑛𝑛 = − 2 MeV.
𝑛𝑛
Some energy levels are listed below. These energy levels
agree very well
With experiment.
𝐸𝐸1 = −28.3 𝑀𝑀𝑀𝑀𝑀𝑀 , 𝐸𝐸2 = −7.075 𝑀𝑀𝑀𝑀𝑀𝑀
(3.8)
𝐸𝐸3 = −3.14 𝑀𝑀𝑀𝑀𝑀𝑀 , 𝐸𝐸4 = −1.768 𝑀𝑀𝑀𝑀𝑀𝑀 .
Journal of Nuclear and Particle Physics 2013, 3(2): 25-28
The ground state wave function of the Heliu m-4 nucleus
(or 𝛼𝛼particle) is given by,
𝑟𝑟
1
𝑒𝑒 − �𝑎𝑎 0 .
(3.9)
Ψ100 =
�𝜋𝜋𝑎𝑎 30
4. Oxygen- 17 Nucleus
There are eight protons and nine neutrons in this nucleus.
The core for this nucleus is Oxygen-16.The single particle
shell model shows that the total J for this nucleus is 5/ 2 due
to a neutron outside the core part.The spin of the neutron is
½ and hence the orbital angular mo mentu m quantum
number ℓ = 2.So the principal quantum nu mber n=3 for
this nucleus in its ground state.The potential energy for
Oxygen-17 nucleus is given by,
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑚𝑚
(4.1)
𝑉𝑉 (𝑟𝑟) = − 2 𝑜𝑜 𝑛𝑛 , where,
𝑟𝑟
𝑀𝑀0
𝑚𝑚0 = �8𝑚𝑚𝑝𝑝 + 8𝑚𝑚𝑛𝑛 � = 26.780272 × 10 −24 𝑔𝑔𝑔𝑔 . (4.2)
The parameter 𝑀𝑀02 for this nucleus is given by,
(4.3)
𝑀𝑀02 = 0.88696 × 10 −48 𝑔𝑔𝑔𝑔2 .
The Schrödinger equation with the above potential
energy shows that,
(4.4)
Ψ(𝑟𝑟 , 𝜗𝜗, 𝜙𝜙) = 𝑅𝑅𝑛𝑛ℓ (𝑟𝑟)𝑌𝑌ℓ𝑚𝑚 (𝜗𝜗, 𝜙𝜙) ,
where n=3,4,5 …etc. The energy Eigen values are given by,
𝐸𝐸𝑛𝑛 = −
𝜇𝜇
1
2ℏ2 𝑛𝑛 2
𝑔𝑔 2 ℏ𝑐𝑐𝑚𝑚 𝑜𝑜 𝑚𝑚 𝑛𝑛
�
𝑀𝑀02
�
2
= −
1185 .88
MeV. Where, (4.5)
𝑛𝑛 2
n=3, 4,5 ….
(4.6)
The reduced mass in this case is given by,
(4.7)
𝜇𝜇 = 1.576331 × 10 −24 𝑔𝑔𝑔𝑔.
Some energy levels are g iven below:
𝐸𝐸3 = −131.7646𝑀𝑀𝑀𝑀𝑀𝑀 , 𝐸𝐸4 = −74.12 𝑀𝑀𝑀𝑀𝑀𝑀
𝐸𝐸5 = −47.44𝑀𝑀𝑀𝑀𝑀𝑀 , 𝐸𝐸6 = −36.94𝑀𝑀𝑀𝑀𝑀𝑀 . (4.8)
The binding energy of Oxygen-17 nucleus is 137.763
MeV.This should be compared with the ground state energy
obtained above.
5. The Fluorine-17 Nucleus
This nucleus is an isobar of Oxygen -17 nucleus. The
core part of this nucleus contains 8protons and 8 neutrons.
The single particle shell model shows that there is a proton
outside the core with a total J= 5/2 in the ground state. The
orbital angular mo mentum quantum nu mber of the outer
proton is ℓ = 2 as in the case of Oxygen-17 nucleus.
Therefore the principal quantum number n of this nucleus
must be 3 in the ground state. The total potential energy of
the outer proton is given by,
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑜𝑜 𝑚𝑚 𝑝𝑝
8𝑒𝑒 2 ℏ𝑐𝑐
+
, where,
(5.1)
𝑉𝑉 (𝑟𝑟) = − 2
𝑀𝑀0
𝑟𝑟
𝑟𝑟
𝑀𝑀02 = 0.866433 × 10 −48 𝑔𝑔𝑔𝑔2 .
(5.2)
Solving the nonrelativistic Schrödinger equation we note
that,
𝐸𝐸𝑛𝑛ℓ = −
𝜇𝜇
1
2ℏ2 𝑛𝑛
�
2
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑜𝑜 𝑚𝑚 𝑝𝑝
𝑀𝑀02
2
− 8𝑒𝑒 2 ℏ𝑐𝑐� = −
1153 .602
𝑛𝑛 2
MeV.
(5.3)
In the above expression the principal quantum nu mber n
will have,
27
n = 3 ,4 , 5 ….
(5.4)
The reduced mass in this case is given by,
(5.5)
𝜇𝜇 = 1.574289 × 10 −24 𝑔𝑔𝑔𝑔 .
The principal energy levels of this nucleus are listed
below:
𝐸𝐸32 = −128.178 𝑀𝑀𝑀𝑀𝑀𝑀 , 𝐸𝐸4 = −72.1 𝑀𝑀𝑀𝑀𝑀𝑀 ,
𝐸𝐸6 = −32.04 𝑀𝑀𝑀𝑀𝑀𝑀 .
(5.6)
𝐸𝐸5 = −46.14 𝑀𝑀𝑀𝑀𝑀𝑀 ,
The very first ground state energy should be compared
with the binding energy of this nucleus which is 128
MeV.The above energy levels are very close to the principal
energy levels of Oxygen-17 nucleus. These two are
mirrornuclei. Moreover the parameter 𝑀𝑀02 for these two
nuclei is nearly equal.
6. The Aluminium-27 Nucleus
The Alu min iu m-27 nucleus also has a total J= 5/2 .In this
case there are 13 protons and 13 neutrons in the core part of
this nucleus. There is a neutron outside the core. The mass of
the core part is given by,
(6.1)
𝑚𝑚𝑜𝑜 = 43.517942 × 10 −24 𝑔𝑔𝑔𝑔 .
The reduced mass for this system is given by,
(6.2)
𝜇𝜇 = 1.612845 × 10 −24 gm.
The potential energy of the outer neutron in this case is
given by,
𝑉𝑉(𝑟𝑟) = −
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑜𝑜 𝑚𝑚 𝑛𝑛
𝑀𝑀02
𝑟𝑟
.
(6.3)
The parameter 𝑀𝑀02 for this nucleus is given by,
(6.4)
𝑀𝑀02 = 1.094011 × 10 −48 𝑔𝑔𝑔𝑔2 .
The energy spectrum for the Alumin iu m-27 nucleus is
given by,
𝐸𝐸𝑛𝑛 = −
𝜇𝜇
1
2ℏ2 𝑛𝑛
𝑔𝑔 2 ℏ𝑐𝑐 𝑚𝑚 𝑜𝑜 𝑚𝑚 𝑛𝑛
�
2
𝑀𝑀02
�
2
.
(6.5)
In the above expression the principal quantum nu mber n
has the follo wing values:
n= 3, 4 , 5 etc,
(6.6)
The ground state energy and some principal levels are
listed below:
𝐸𝐸4 = −131.6𝑀𝑀𝑀𝑀𝑀𝑀 ,
𝐸𝐸32 = −233.99 𝑀𝑀𝑀𝑀𝑀𝑀 ,
𝐸𝐸6 = −58.5 𝑀𝑀𝑀𝑀𝑀𝑀etc. (6.7)
𝐸𝐸5 = −84.24 𝑀𝑀𝑀𝑀𝑀𝑀 ,
The ground state energy 𝐸𝐸32 should be co mpared with the
binding energy of this nucleus which is 234 MeV.
7. Discussion
The M GPE is used here to exp lain all the properties 7 of
Deuteron nucleus except its magnetic mo ment. The ground
state wave functions ofDeuteron, Heliu m, Oxygen-17,
Flourine-17 and Alu miniu m-27 are obtained here in terms of
their masses and charges. The energy levels o f Heliu m-4
nuclei agree pretty well with experiment8 .For the First time it
is clear why mirror nuclei have almost identical spectrum
(Oxygen-17 and Flourine-17).
The M GPE can be used to explain Alpha-Decay, Fusion
and Fission Reactions and radioactiv ity. In princip le the
MGPE can be applied to any nucleus in a way that one
28
Cvavb Chandra Raju: Nuclear Binding with Gravitational Interaction
applies the Coulo mb potential energy to many electron
systems.
Even in the case of Elementary particles also the M GPE
plays an important role. For examp le ,let the Standard Higgs
Boson be a bound system of ZZ Bosons with zero angular
mo mentu m (as in REF.4).Then the estimated binding energy
of this bound system will be 54.43 GeV. if the parameter
𝑀𝑀02 = 5.357 × 10−46 𝑔𝑔𝑔𝑔2 in the M GPE. This reproduces a
mass of 125.5 GeV for the Standard Higgs Boson.
Finally this paper should be seen as the beginning of
Quantum Gravity.
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