Journal of Modern Physics, 2013, 4, 459-462
http://dx.doi.org/10.4236/jmp.2013.44064 Published Online April 2013 (http://www.scirp.org/journal/jmp)
Energy Levels of Helium Nucleus
Cvavb Chandra Raju
Department of Physics, Osmania University, Hyderabad, India
Received January 9, 2013; revised February 10, 2013; accepted February 22, 2013
Copyright © 2013 Cvavb Chandra Raju. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
ABSTRACT
The Helium-4 nucleus is more similar to the Hydrogen atom of atomic physics. In the case of hydrogen atom, there are
many energy levels which were experimentally seen and theoretically explained using non-relativistic quantum me-
chanics. In this note, we use a central potential to derive the energy levels of Helium-4 nucleus. The ground state and
the first few energy levels agree pretty well with experiment. The same potential can be used with nuclei like Oxy-
gen-17 and many more nuclei.
Keywords: Helium-4; Nuclear Energy Levels; Deuteron; Morphed Gravitational Potential
1. Introduction and Formulation of the
Problem
The Deuteron nucleus has no excited states. The ground
state energy of the Deuteron is experimentally found to
be −2.225 MeV and the measured radius of this nucleus
is 2.1 F. This is the distance between the center of mass
and either of the nucleon in the Deuteron nucleus [1].
There are no known solutions of the Schrödinger equa-
tion of this nucleus with Yukawa potential.
Are there any central potentials with which we can
solve the Schrödinger equation for many nuclei such that
their ground state wave functions and excited states can
be obtained? This question led us to a central potential
which is closely related to the gravitational potential en-
ergy.
There is no reason or experimental support to believe
that the universal constant of gravitation G is same for all
values of interacting masses. For interacting masses of
the order of nucleon masses G may not retain its univer-
sality. This led us to the following expression for the
gravitational potential energy of two particles whose
masses are and ,
1
m2 m

2
2
012
gc
GM mm
r






0
1eVr G , (1.1)
where in place of the usual constant of universal constant
G we have a modulating factor.
The constant g2 is a dimensionless real number whereas
M
has dimensions of mass. We believe that the expo-
nential goes to zero when the interacting masses are large
and the Universal law of Gravitation is restored. It is the
parameter M0 that causes the Universal Law of Gravita-
tion restored. An approximation to Equation (1.1) is giv-
en by,

2
12
2
0
11 mm
gc
Vr Gr
GM



 





, (1.2)
where “r” is the distance between the interacting parti-
cles. Simplifying Equation (1.2), we have,
2
12
2
0
mm gc
Vr r
M
 
2
0
. (1.3)
The above potential energy is obtained from the gravi-
tational potential energy. It may be called “the morphed
gravitational potential energy”. There are two constants
g2 and
M
which we will obtain below.
2. Deuteron
The Deuteron is a bound system of a neutron and a pro-
ton with an orbital angular momentum of zero. The total
spin of the two nucleons is one. The deuteron nucleus has
no excited states. The experimentally measured ground
state energy of the Deuteron nucleus is −2.225 MeV and
its orbital angular quantum number . There is no
stable diproton. It is also known that the nuclear potential
depends on the spin orientation [2] of the nucleons inside
a nucleus. If the nuclear force is independent of their spin
orientation then the singlet (total spin = 0) state and the
triplet state (total spin = 1) will have the same energy.
0
C
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C. C. RAJU 460






But this is not observed. This means that the singlet ten-
sor potential is weaker than the triplet potential. We con-
sider an extreme situation wherein the singlet potential of
the diproton is quite negligible. In that event the total
potential energy operating for a diproton nucleus is,
,, ,
nm
rRrY

  

1,2,3, ,0,1, ,1and
,1,2,,
nn
m
222
2
0
p
gcm ec
r Mr

, (2.1)
where, the first term is the morphed gravitational poten-
tial energy and the second term is the Coulomb potential
energy between the two protons. Here 21
137
e
2
is the
fine structure constant. If, 0
2
p
M
m and 22
g
e

, the
total potential energy in the case of a diproton will be
zero and there will be no stable diproton nucleus! But
with these values for g2 and 2
0
M
the theoretically com-
puted binding energy for the deuteron nucleus turns out
to be quite small. Hence we chose the following values
for these parameters which are quite close to the values
mentioned above.
2
2
0.2254
e
g0.0323 84

2
48
10 gm

2
, (2.2)
and,
2
00.931826M . (2.3)
There are a few important points to note about the above
values of the parameters g2 and 0
M
.
 The parameter g2 is the weak interaction constant of
the electro-weak standard model [3] because it con-
tains e2 and the Weinberg mixing parameter sin2θW =
0.2254 as in the Standard model [4]. The other pa-
rameter is nearly equal to the square of the proton
mass and this is required to avoid a stable diproton
nucleus.
 The interaction parameter contains the square of mass
in the numerator and also in the denominator; Be-
cause of this reason the interaction constant of the
morphed gravitational potential has the same dimen-
sions as the electroweak constant. Hence it is renor-
malizable. This is an important point in favor of the
morphed gravitational potential.
The Schrödinger equation for the deuteron is given by,

2
2,,
2Vr r

,,E r








 


Vr 1
, (2.4)
where, is given by Equation (1.3) with
p
mm

and . The reduced mass
2n
mm

is given by,
pn
p
n
mm
mm
 

Vr
. (2.5)
Since is a central potential, from the methods
of quantum mechanics, it just follows that [5,6],
. (2.6)
As in the case of the Hydrogenatom, the quantum
numbers, n, and m have their values,


 
 
  (2.7)
The radial function is given by,


21
2
e
nn
RA L






, (2.8)
where,
0
2r
na
 , (2.9)
and,

2 2
0
02
p
n
M
a
g
cmm


. (2.10)
These results are obtained by a simple transcription of
the Hydrogenatom calculations. The energy of the deu-
teron is given by,

2
2
2
22 2
0
2
pn
n
mm
gc
EMn


 



. (2.11)
The normalization constant A in Equation (2.8) is
given by,


3
3
0
1!
21
2!
n
Anan n

  

 
 




10 2.2251 MeV.E
. (2.12)
Using the above results we can now estimate the
ground state energy, ground state wave function and the
radius of the Deuteron nucleus.
(2.13)


0
100 3
0
1e
π
r
a
a



13
04.31734 10cma

. (2.14)
. (2.15)
here a0 is the distance between the two nucleons. The
radius of the Deuteron is the distance of either of the nu-
cleon from the center of mass. The center of mass lies
halfway between the line joining the two nucleons be-
cause the two nucleons have almost equal mass. We can-
not avoid the excited states of the deuteron nucleus
however small these may be.
3. The Helium-4 Nucleus
The Helium-4 nucleus is a strongly bound system of two
neutrons and two protons. It used to be known as an

-
particle. In fact using this

-particle itself the very ex-

C. C. RAJU 461
istence of a nucleus was inferred. Our morphed potential
energy can be put to test using the estimated energy lev-
els for this nucleus. A rough picture of this nucleus is
like this. Any one nucleon experiences a total morphed
gravitational potential due to the remaining three nucle-
ons. This outer nucleon must be a proton because two
protons of this nucleus cannot be together because of
their Coulomb repulsion. This rough picture indicates
that,


22
npp
m
2
0
2mm
g
cec
rr


Vr M

 , (3.1)
where the second term is due to the Coulomb repulsion,
and the first term is the morphed gravitational potential
energy. We will now apply the methods of quantum me-
chanics in an effort to obtain a theoretical description of
Helium-4 nucleus or alpha-particle.
The Schrödinger equation with the above potential is
given by

2
2,,
2Vr r

,,E r








 
, (3.2)
where the reduced mass is now given by,


24
749 10gm



21.254
22
npp
np
mmm
mm



. (3.3)
The central potential

Vr
enables us to find a closed
solution for this nucleus. As in the case of the H-atom the
radial function is given by,


21
2
n
L

e
n
RA








, (3.4)
where,
0
2r
na
 , (3.5)
here,

2
02
2
0
1
2np
a
gcm m
M





2
p
mec

 



. (3.6)
In Equations (3.4) and (3.5), n is the principal quantum
number as in Equation (2.7). The energy Eigen values for
the Helium nucleus are given by,

2
2
2
p
mec

 


2
2
0
2
2
2
np
gcm m
M
En







. (3.7)
The normalization constant A in Equation (3.4) is giv-
en by,


3
3
0
1!
21
2!
n
Anan n


 

 
 




. (3.8)
For the Helium nucleus the ground state orbital angu-
lar momentum quantum number is zero. Hence the
ground state wave function of the Helium-4 nucleus is
given by,
0
100 3
0
1e
π
r
a
a



10 28.5176 MeVE
. (3.9)
The ground state energy of Helium-4 nucleus is given
by,
. (3.10)


This value should be compared with the binding en-
ergy of this nucleus which is about 28.3 MeV; we ob-
tained this value through ordinary quantum mechanics
and by using masses and the charge as shown above. In
general the principal energy levels of the tightly bound
Helium-4 nucleus are given by,
2
28.5176MeV
n
En 
12
34
56
28.5176 MeV,7.1294 MeV,
3.168622 MeV,1.78235 MeV,
1.140704 MeV,0.792156 MeV.
EE
EE
EE
. (3.11)
The possible energies for this nucleus are listed below:


 

(3.12)
These energy levels are obtained ignoring the tensor
forces. These energy levels can be put to test to find out
the viability of the morphed gravitational potential. Our
efforts to find the principal energy levels were not very
successful. In reference [7], only we could obtain the
measured energy spectrum and this contains only two
energy levels. The energy diagram is arranged by setting
the ground state energy zero, this amount to adding
+28.5176 MeV to all the energies above. Our results
agree pretty well with the two principal levels given in
Ref [7]. The general wave function for the Helium-4 nu-
cleus for any orbital angular momentum is given by,






,, ,
nm
rRrY

   . (3.13)
The wave function given here for the ground state of
the Helium-4 nucleus can be used in alpha-decay or in
reactions involving the alpha particle.
4. Conclusion
In this note we assumed that the universal constant G is
not universal for all values of the interacting masses.
Motivated by this idea we changed G and used an ap-
proximated constant for interacting nucleons. When this
potential is applied along with the methods of ordinary

C. C. RAJU 462
2
quantum mechanics the estimated results are all in agree-
ment with experiment. This very procedure can be appli-
ed to such nuclei Oxygen-17, F-17 and many more nuclei.
The constant 0
M
is not universal. It is an adjustable
parameter. The functional dependence of this parameter
on the product of the interacting masses is unclear as of
now. But the idea cannot be dismissed just like that. But
certainly for larger masses the exponential in Equation
(1.1) goes to zero.
5. Acknowledgements
The author is very grateful to Prudhvi Rchintalapati. But
for his help the work could not have seen the light of the
day.
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