Research article-2
Utilization of Bohr’s Theory of Hydrogen
to Helium atom
Dr.M.Meyyappan*
[Professor of Physics (Retd), Alagappa Government Arts
College, Affiliated to Alagappa University, Karaikudi- 630003, Tamilnadu,
India, *email id: meydhanam@gmail.com
Orcid No:0000 0002 5194 2992]
Abstract
[An attempt is made to optimize Bohr's atom model—originally developed for
the hydrogen — to the helium atom, a simplest many electron system as a first
step. Correction due
to electron-electron interactions are taken into account. The possibility of
hyper excitation and hyper de-excitation is proposed where two electrons in the
same orbit jump simultaneously into another orbit. Normal excited states and
its de-excitation are also studied by proposing half-half mixing of two allowed
states. With the help of this ingenious approach, various energy levels of
helium atom are calculated and its spectral feature is studied by calculating
the energy radiated out in certain selected allowed transitions. The result is
compared with the observed data. The
branched de-excitation of excited states is possible when two or more orbital
electrons are present in the atom. The branched de-excitation in helium atom is
studied with and without an electron in the innermost orbit. The correction due
to the relativistic variation of mass was applied to helium ion and helium
atom. It accounts for the non-existence of helium negative ion and explains how
far the radius and orbital velocity of the orbital electron change in helium
atom when the orbital electrons exist in different orbits.]
Introduction
Bohr's theory introduced the concept of quantized energy levels in atoms
and successfully explained the hydrogen spectrum. The model was successfully
applied to hydrogen like ions. Due to some limitations [1) the non-relativistic
Bohr’s theory of hydrogen atom failed to explain the spectral feature of atoms
with two or more electrons. However, it
was later replaced by more advanced quantum mechanical models that provided a
more complete understanding of atomic structure and behavior. When Bohr's atomic
theory was reviewed with an idea of extending Bohr's atomic theory to atoms with
higher atomic number, the first thing that struck the mind was the relativistic
variation of mass of the orbital electrons moving rapidly in circular orbits. The
Bohr’s theory of hydrogen atom is redone [2] with and without considering the
relativistic variation of mass of orbital electron. When a free proton and an
electron are allowed to form a hydrogen atom, half of the loss of its potential
energy is converted into its kinetic energy required to move in a stable orbit.
surrounding the proton and the remaining half is deposited as relativistic
increase of mass. The mass-energy conversion is accomplished involuntarily
within the space of nuclear field. The orbital electron binds with the proton
where the atomic binding energy comes from the energy equivalent of the added
mass of the system. This is the energy that is radiated out during the
electronic transitions i.e., ΔE = Δm c2
= (mn1 - mn2)c2 = En2 - En1
In this article, optimization of
Bohr’s theory of hydrogen atom is described with helium atom .The results
obtained with non-relativistic Bohr model of helium atom are compared with that
of relativistic model. The hyper excitation and de-excitation is proposed where
two electrons in one orbit jump simultaneously into another orbit. To estimate
the total energy associated with helium atom in any one of its electronic
states, a new method of computation called half-half mixing of two different
states is proposed. The normal excited
states and its de-excitation is studied elaborately. The energy radiated out in
allowed electronic transition and the corresponding wavelength of
electromagnetic radiation emitted out are predicted. Yet another possibility of
electronic transition called branched de-excitation is pointed out, where both
the electrons which are initially in different orbits other than the innermost
orbit and the electronic transition takes place from the outer orbit to the
innermost orbit. The spectral feature of helium atom is compared with Relativistic
Bohr model.
Non-relativistic model of helium atom
Fig.1.Helium atom
In the
helium atom the two electrons in the 1s orbit are stable with positive binding
energy. They have both kinetic energy due to its motion in the circular orbit
and potential energy due its position in the electrostatic field within the
atom. The kinetic energy of an electron is (1/2) mv2 and for both
the electrons which are identical in the system K.E is mv2. The
electron is stable in the orbit by balancing out the resultant electrostatic
force with the centrifugal force due to circular motion. Let us suppose that
the two electrons are in the 1s orbit itself having radius r1.The
centripetal force acting on an orbital electron due to nuclear attraction is 2e2
/Kr12 . The other electron in the same orbit is
repelled by the electron already existing there and both of them are positioned
at opposite end of a diameter of the orbit. Since all helium atoms have
identical spectral feature the electrons in an orbits take fixed relative
position that is invariant with respect to time.
The
electrostatic force of repulsion between the two electrons which are
diametrically opposite is e2/4Kr12.The
resultant force is balanced by centrifugal force i.e., 2e2/Kr12
- e2/4Kr12 = (7/4)e2/Kr12
= mv12/r1 . The radius of the orbital
electron in helium atom can be determined from the condition insisting
quantized angular momentum. It gives rn = n2(4/7) ao. = 0.5714 n2 ao. If
we ignore the electron-electron interaction rn = ao/2 .The
radius of the 1s orbit in He+ ion is increased by 0.0714 ao due
to mutual electron-electron interaction in helium atom.
Total
kinetic energy of the electrons in the helium atom = mv12
= (7/4) e2/Kr1.Potential energy of electron 1 = - 2 e2
/Kr1 and the potential energy of electron 2 in the presence of electron 1 = - 2 e2
/Kr1 + e2/2Kr1 . It gives its total energy as - (7/4) e2/Kr1 = -
(49/8) [e2/2Kao]= -83.269 eV which provides enough binding
energy to the system. It gives the first ionization energy 83.269 - 54.368 =
28.901 eV. The experimental value of first ionization energy of helium is
24.481 eV which is 4.41 eV smaller than the theoretically predicted value.
Alternatively very same
result can be derived instead of moving the electrons towards the nucleus to
place them in the prescribed electronic orbits, the nucleus with two units of
positive charge is taken from infinity towards the center of the orbit with two
electrons whose circular motion is induced as the nucleus moves closer towards
the coupled electrons or the coupled two electrons move together vertically
towards the nucleus until all of them come to a plane with
nucleus at the center of the coupled electrons.
(a)
(b)
(a).Bring proton from infinity to the center
of a smallest orbit
with two electrons diametrically
opposite
(b).Bring an orbit with two electrons
diametrically opposite from
infinity towards a proton until all of them are in a plane
Fig.2. Constructing Helium atom
As
the two electrons are kept with a distance of separation 2r1 initially
the given system of coupled electrons repel each other with a force e2/4Kr12
and have an initial potential + e2/2Kr1. To find
out the potential energy of the system, the nucleus with charge 2e+ is taken vertically from infinity to the center of the coupled
electrons and the total work done is determined . When the nucleus is at a
distance x from the center, the force
experienced by it due to
electron-1 is 2 e2/K (x 2+r12 )
and its component along the direction of displacement is 2e2
cosφ /K(x 2 + r12 ) = 2 e2
x /K(x 2 + r12 )3/2 . The
resultant component due to both the electrons is 4 e2 x/K (x2 +
r12)3/2. The parallel component of electron-1
is nullified by the equivalent component of electron-2. Since F = - dU/dx , the
change in potential energy U = - ∞∫o
Edx = ∞∫o Fdx
=[4e2/K] ∞ ∫o x dx /(x2 + r12)3/2 = -[4e2 /K (x2 + r12)1/2]∞o = - 4 e2/Kr1,
where E is the electric intensity at x. Adding the initial potential energy
associated with the coupled electrons the total potential energy of the system
then becomes - 4 e2/Kr1 + e2/2Kr1 = -
(7/2) e2/Kr1.
The parallel
component of electrostatic force acting on electron-1 towards the center of the
orbit is 2e2r1/K(x2 + r12)3/2
.At the end of the displacement of the nucleus this force becomes maximum
and is equal to 2e2/Kr12. Taking into account
the initial electron-electron repulsion the total centripetal force 2e2/Kr12
- e2/4Kr1 = (7/4)e2/Kr12
which induces the circular motion with centrifugal force mv12
/r1 . It gives the kinetic energy of the system as (7/4)e2/Kr1.
By adding the kinetic energy of the electro ns with its potential energy, the
total energy becomes - (7/4)e2/Kr1. Very same result can be obtained by keeping
the helium nucleus at a point and the coupled electrons separated by a distance
2r1 is moved from infinity towards the nucleus vertically.
When the condition -1 of
Bohr's theory is applied the circumference of the n th orbit must be equal to n
times the wavelength of matter-waves associated with the orbiting electron.
(2πrn)2
= 4π2 rn2 =
n2 λ2 = n2
h2/m2 vn2 = (4/7) n2 h2 4 πεo
rn /m e2
rn
= (4/7)n2h2εo/mπe2
= (4/7) n2 ao ...... (1)
Substituting the value for rn the total
energy becomes - (7/4)2 e2/n2
K ao = - (49/8) (1/n2) 13.595 eV. It gives the total energy associated with
the helium atom as - 83.269 eV. The sum of first and second ionization energies
of helium is 24.481 + 54.403 = 78.884 eV.
Total energy of electrons in the helium atom assembled
The same result can be arrived
by computing the component of energy in assembling the helium atom. Let us
suppose a helium atom is assembled with a nucleus having 2 units of positive
charge and two separate electrons.
Fig.3.Helium ion + electron → Helium
atom
There are two stages in the assembling. (1)
e-1 is placed at r1 (= ao/2) to form a helium ion and (2)
when e-2 is brought from infinity to the orbit having radius r2 (= 4
ao/7) the e-1 at r1 is shifted to take up a new orbit
having the same radius r2
When electron (e-1) is brought closer to
the nucleus, it gains acceleration due to nuclear force of attraction and it
starts making a circular motion around the nucleus. In the first stage helium
ion is formed. Let r1 be the
radius of its orbit. The ionized helium atom with single electron has both
kinetic and potential energies. Its kinetic energy can be determined by
equating the electrostatic force and centrifugal force 2 e2/Kr1 = mv12
/r1 or mv12
= 2e2/K r1..Kinetic energy = (1/2) m v12 = e2/K r1 where r1 = h2εo/2mπe2 =
ao/2 .In terms of Bohr radius the kinetic energy becomes 4[e2/2Kao].
Potential energy of the electron is determined by evaluating the work done in
taking the electron from infinity to the assigned orbit. It is worked out as -
2 e2 /Kr1 = - 8 e2/2Kao .Total energy of the system is 4[e2/2 Kao] - 8e2/2K
ao = - 4e2/2Kao
= 54.38 eV This is the second ionization energy of helium. This is in
good agreement with the practical value 54.403 eV
It gives a formula for the Zth ionization energy of
hydrogen-like ions of all elements. If Z is the atomic number, then Z e2 /Kr12 = m v12 / r1
or m v12 = Z e2 / K r1.Kinetic
energy = (1/2) m v12 = Z e2/2K r1 =
Z2 e2/2K ao where r1 = h2εo /Zmπe2 = ao /Z .Potential energy = - Z e2 /Kr1 = - Z2
e2/Kao and the total
energy of the system is -Z2
e2/2 K ao = - Z2[e2/2Kao]
= - Z2 x 13.595 eV .
Stage.1. Helium ion + electron = helium atom
Now the
second electron is placed in an orbit of radius r2 . In the final
assembly both the electrons are in the same orbit having radius r2 and
both of them have same kinetic and potential energies as they are identical in
all respect. When e-2 is brought from infinity to r2 , e-1 is
shifted from r1 to r2 .The potential energy of e-2 in the
presence of nucleus is ∞∫r2 [2e2/Kx2] dx = [2e2/Kx]
∞r2 = 2e2 /Kr2 = - 7[e2/2K ao] .
The increase in the potential energy of e-2 due to e-1 occurs when e-1 is
displacing from r1 to r2 . In calculating the increase in
potential energy of e-2 due to e-1 the electron e-1 is supposed to be at its mean
position (1/2) (r1 + r2) = (15/28) ao . When
e-2 is at an intermediate distance x away from
the central nucleus , the
repulsive force F experienced by
it due to e-1 is -e2 /K [x+
(15/28)ao]2 . For
an infinitesimal small displacement dx towards the nucleus, the workdone dw = F
dx. Total work done is stored as its potential energy.
Potential energy of the e-2 = ∞∫r2 - e2 /K (x+ (15/28)ao)2]
dx.
[e2 /K (x+ (15/28)ao)]∞r2 =
e2 /K [r2+ (15/28)ao] = (e2 /K) {1/[(4/7) ao+
(15/28)ao]} = 28 e2 /
31K ao = (56/31) [e2 /2Kao]
This induced potential
is shared by both the electrons, each electron has an increase of potential by
(56/62)[e2 /2Kao]
Stage.2.shifting of e-1
from r1to r2
When e-2 reaches its assigned orbit of
radius r2, the electron e-1 is shifted back from r1 to r2
.In the final position, the
condition of electrostatic force is equal to centrifugal force requires
2e2/Kr22 - e2/4Kr22
= (7/4) e2/Kr22 = mv22 /r2
The kinetic energy gained
by e-2 is (1/2) m v22 = (7/8)e2/Kr2 = (49/16) [e2/2Kao].
Due to induction, the kinetic energy of e-1 is decreased. It is dropped from 4
[e2/2Kao] to (49/16)
[e2/2Kao].
It is equal to - (15/16) [e2/2Kao]. The decrease in
kinetic energy of e-1 is (1/2) m v12
-
(1/2)m v22 = e2/ Kr1 - (7/8)e2 /Kr2
= 4e2 /2Kao -
(49/16) [e2/2Kao]
=(15/16) [e2/2Kao]
When e-2 is at infinity, the
potential energy of e-1 is increased due to its displacement from r1 to
r2.
A change in potential energy of e-1
occurs in the presence of nucleus only -
[2e2 / K x]r1r2
= - 2e2/K[1/r2 -
1/r1] = - 2e2/K[ (r1 - r2)/ r1
r2 ]= - 2e2/K(
- 1/4ao) = [e2/2Kao]
When e-2 is r2 ,
the change in the potential energy of e-1
due to nucleus is [e2/2Kao]
and due to the presence of e-2, [e2/K][1/(x+r2
)]r1r2 = [e2/K][ 1/2r2 - 1/(r1+
r2 )] = [e2/K] [ 7/8ao - 14/15 ao]
= [e2/2Kao][-7/60] .
The actual change of potential energy of e-1 is taken as the mean of
change of potential energy of e-1 when e-2 is at infinity and e-2 is at r2,
where the change of potential due to electron-electron interaction is
equally shared by the participants. The
change of potential energy of e-1 when e-2 is at infinity is [e2/2Kao]
and when e-2 is at r2 is [e2/2Kao] - [7/60][e2/2Kao]
which give a mean as [1 - 7/120][e2/2Kao].= (113/120)[e2/2Kao]
Kinetic energy of e-1
when e-2 is present [e2/2Kao][4 - 15/16] = (49/16) [e2/2Kao]
Potential energy of e-1
when e-2 is present [e2/2Kao][ -8 +(56/62) +1 -7/240] = -
6.126 [e2/2Kao]
Kinetic energy of e-2
[e2/2Kao](49/16)
Potential energy of e-2
[e2/2Kao][ -7+ (56/62) - 7/240 ] = - 6.126[e2/2Kao]
Total energy of the
helium atom is the sum of all the four components and is equal to 2[3.0625 - 6.125][e2/2Kao] = -6.125 x 13.595 = -83.269 eV
The relativistic variation of mass where
energy can be exchanged with the mass of the moving electron, spin-spin
interaction between electrons and with nucleus
may be responsible for this small difference between practical and
theoretical values of total energy of the system. By studying the total energy
of helium like ions, one can find the cause of deviation. Let Ze be the nuclear
charge with two 1s electrons in the helium like ions. The radius of the innermost
orbit is given by
Ze2/Kr12 - e2/4Kr12
= (Z-1/4) e2/Kr12
[(4Z-1)/4]e2/Kr1 =
mv12 which gives
r1 = [4/(4Z-1)] ao
Kinetic energy of the electrons = mv12
= [(4Z-1)/4]e2/ Kr1 = [(4Z-1)2 /8]e2/2Kao
Potential energy of the electrons =
-2Ze2/Kr +e2/2Kr =(e2 /Kr)[-2Z + 1/2]=- (e2/2Kr) [4Z-1]
= - [(4Z-1)2 /4] e2/2Kao
Total energy of the system = - [(4Z-1)2 /8]e2/2Kao
Using this relation the theoretical
value of total energy is worked out and compared with its experimental value
[the sum of z th and (z-1) th ionization energy]
Table.1. z th and (z-1) th
ionization energies of first few helium like ions
.............................................................................................................................................. Z Symbol x 13.595 eV Total energy Ionization experimental Difference
Z
(Z-1) value D
D/Z
..............................................................................................................................................
2
He
-(49/8) = 6.125 -
83.269 54.403 24.481
-78.884 4.385 2.19
3
Li+ - (121/8)= 15.125 -205.624
122.419 75.619 -198.038 7.586 2.52
4
Be2+ -(225/8) = 28.125 -382.359 217.657
153.85 - 371.507 10.852
2.71
5
B3+ - (361/8) = 45.125
-663.474 340.127 259.298
-599.425 64.049 12.81
6
C4+ -(529/8) = 66.125 -848.969 489.84 391.986
-881.826 32.857 5.476
7
N5+ -(729/8) = 91.125 -1238.844 666.83 551.925
-1218.755 20.089 -2.87
.............................................................................................................................................
Hyper excited states and de-excitation
Fig.4. Helium atom - normal, hyper excited and normal excited states
When
sufficient energy is available both the electrons in the ground state of helium
may be excited. Such excitation is called hyper excited. When a helium atom is hyper excited, the
process of de-excitation takes place in two ways. In the first way one of the
electrons in the hyper excited state jumps to the inner orbits either directly
or step by step with intermediate energy levels. Then the second electron
follows its own de-excitation. In the second way called hyper de-excitation
both the electrons jumps to the lower energy level. The hyper excitation of
helium is possible in dense stars enriched with helium at high temperature
which provide more probability for hyper excitation to happen. Since the
instability of the atomic system is increased many-fold, the hyper
de-excitation will be faster than the normal de-excitation. The hyper
excitation and de-excitation are not possible in hydrogen atom, a single
electron system.
Fig.5. Hyper excitation and Hyper de-excitation
Total
energy of helium in its nth hyper excited state is given by - (1/n2)(49/8)
(e2/2Kao). In two different hyper excited states with n =
n1 and n2, the total energies are - (1/n12)(49/8)
(e2/2K ao) and -
(1/n22)(49/8) (e2/2Kao)
respectively. During hyper de-excitation, both the electrons simultaneously
jump into to any inner or innermost orbit.
In hyper de-excitation between any two states, the transition energy ΔE
is given by (49/8) (e2/2Kao)
[(1/n12) - (1/n22)] = 83.269 [(1/n12)
- (1/n22)] eV.
Since λ = hc/ΔE, where ΔE is in joules, the corresponding wavelength of
radiation emitted in any transition is given by λn2* → n1* is
12.4 x 10-7/(49/8) (e2/2K ao) [(1/n12)
(1/n22)]m. The star (*) is used to represent the hyper
excited state. Using this formula, the transition energy and the corresponding
wavelength of radiation emitted can be predicted. The energy of transition and
the wavelength of radiation emitted in various hyper de-excitation are given in
Table.2
Table.2. Hyper de-excitations and wavelengths in helium
.....................................................................
transition energy wavelength . eV nm
....................................................................
n2* → n1* 62.452 19.85 n3*→ n1* 74.017 16.75 n4*
→ n1* 78.065 15.88 n3* → n2* 11.565 117.22 n4*
→ n2* 15.613 79.42 n4*
→ n3* 4.048 306.32 n5*
→ n3* 5.921 209.42
n6* → n3* 6.939 178.70
n5* → n4* 1.874 661.69 n6* → n4* 2.891 428.91
..............................................................................
The
ultraviolet wavelength in the helium spectrum is most notably characterized by
the strong atomic line at 58.4 nm. The strongest UV lines for astrophysical
observation are at 30.38 nm and 58.43 nm. The spectroscopical study in UV
region of helium atom shows that hyper excitation and hyper de-excitation have
very little probability to happen under normal situations. In hot stars
enriched with helium, hyper de-excitation may be one of the causes for the
emission of UV radiation. Helium has several spectral lines in the ultraviolet
(UV) region, which are defined as having wavelengths shorter than approximately
400 nm. Key UV lines for neutral helium include lines around 396.5 nm and 388.9
nm. It shows that the cause of transition in helium atom liberating energy in
the UV radiation is not hyper de-excitation
Normal excited states and de-excitation
Fig.6. Normal de-excitation in helium atom
Method-I : Half - half mixing of two states
Hydrogen has
only one electron, so its emission spectrum is relatively simple with few
lines. Helium, on the other hand, has two electrons, which means there are more
possible transitions and therefore more emission lines. When the electron
jumps from the n (n>1) to the
first orbit of neutral helium atom, the
transition energy and the corresponding wavelength can be worked out by
half-half mixing of two states involved in the transition. At first let us try
to understand this technique. The spectral feature of helium can be studied if
we are able to calculate the energy associated with an excited/ionized state of
helium, where the two electrons are in two different orbits. In general the
most probable transition is with the de-excitation of helium atom with one of
the electrons in the innermost orbit and the other electron is in the nth orbit
(n ≥ 2). When electrons are in different
orbits and making jumping between the
permitted energy levels, its relative position varies which makes changes in its velocity which in turn
alters the radius of the electronic
orbits as they have to obey the condition-1 of Bohr's theory of hydrogen. Such perturbation in the system makes the
problem of estimating the total energy of the system cumbersome. However one
can solve by a technique called half-half mixing of two hyper-states. Let E11 and Enn be
the energy of the helium atom in its ground state and in its nth hyper-excited
state where both the electrons
are in the first and nth orbit respectively.
In any state the system has three different components of energy- 1.kinetic energy of the electrons due to its orbital motion (KE), 2.negative potential energy of the orbital electrons by virtue of its position in the nuclear field (NPE) and 3.positive potential energy due to electron-electron interaction. (e-e).This energy is equally shared by the interacting participants.
Fig.7. Half-half mixing of two allowed states
When both the
electrons are in the innermost orbit the total energy of the system E11 =
NPE11
+ (e-e)11 + KE11 =
- 4 e2/Kr11 + e2/2Kr11 + (7/4) e2/Kr11
= -(7/4) e2/Kr11= (7/4)2 e2/Kao
= 83.269 eV. When both the electrons are in the same orbit labelled by n,
its total energy is Enn = NPEnn + (e-e ) nn + KEnn = - (7/4) e2/Krnn
= - (7/4)2 (1/n2)e2/Kao. When
one of the electrons is in the innermost orbit and other electron is in the nth
orbit, its total energy E1n = NPE1n + (e-e)1n +
KE1n. NPE1n and KE1n
are half of the sum of total negative potential energies and kinetic
energies respectively with electrons in the innermost and n th orbit
NPE1n = -[2e2/K][1/r11
+ 1/rnn] = - [2e2 /Kr11][1 +1/n2]
- [2e2 /Kr11][(n2 +1)/n2]
KE1n =
[(7/8)e2 /K][ 1/r11 + 1/rnn] = [(7/8)e2 /K][(n2
+1)/n2]
(e-e)1n = e2/Kd1n
where d1n is the distance between the two electrons when they
are in two different orbits. But d1n = r11 + rnn and
e2/K = 2(e-e)11 r11
=
2[(e-e)11x r11]/(r11 + rnn) =
2(e-e)11/ (1+n2) =[e2 /Kr11][1/(n2+1)]
E1n = [e2/Kr11]{[(n2 +1)/n2][(7/8)-2]
+ [1/(n2 + 1)]}
This expression can be verified by computing the total energy
associated with the system with one electron in the innermost orbit and other
electron is removed.
E1∞ = [e2/Kr11][
1 x (-9/8)] = -[9x7/16] e2/2Kao = 53.53 eV. The observed
second ionization energy of the helium atom is 54.4 eV.
When an excited helium atom exists
with one electron in the innermost orbit and the other electron in the nth
orbit , then the state E1n gets
1/2 share of kinetic and negative
potential energies from each contributing states (E11 and
Enn), which are due to the
interaction with stable central nucleus. The sharing of electron-electron
interactional energy is different. When an electron in lower energy state (m)
donates its energy to higher energy state (n) it carries an amount of energy
(e-e)mm [m2/(n2+m2)]. When an
electron from higher energy state (n) donates its energy to lower energy state
(m) it carries an amount of energy + (e-e)nn [n2/(n2+m2)].
This energy contributed by E11 is utilized by E1∞ to have more potential
and kinetic energy with higher binding energy. E1∞.
= (1/2) [KE11+ NPE11] + (e-e)11 [l/(1+∞)]
=. (1/2) [ KE11+ NPE11 ] .In fact E1∞
refers the second ionization of helium atom and is equal to (1/2) [E11
- (e- e)11] = (1/2)[(-83.269) - (7/4)13.595] = -53.53 eV
and the energy contributed by E11
to E1∞ is the first
ionization energy i.e., (-83.269) + (53.53) = 29.739 eV. .Enn is formed by mixing of two
identical states Enn each state contributes 1/2
of its energy and keep the energy same. When a hyper state Enn is
formed by half-half mixing of two hyper-states Enn and Enn , then its energy by this method becomes
2 (1/2)[ KEnn + NPEnn] + 2(e-e)nn n2/2n2] = Enn = [KEnn + N
PEnn + (e-e ) nn] .
The above expression for Enm can
be derived by yet another way. When an excited
helium atom exists with one electron in
the inner orbit labelled m and the other
electron in the outer orbit n , then the
state Emn gets 1/2 share of
kinetic and negative potential
energies from each contributing
states (Emm and Enn)
, which are due to the interaction with
stable central nucleus. The sharing of electron-electron interactional energy
is different. When an electron in lower energy state (m) donates its energy to
higher energy state (n) it carries an amount of energy (e-e) mm [m2/(n2+m2)]. When an electron from higher energy state (n)
donates its energy to lower energy state (m) it carries an amount of energy +
(e-e)nn [n2/(n2+m2)].
In helium atom (e-e)mm = e2/2krm, (e-e)mm
= e2/2krn and
(e-e)mn = e2/k(rn + rm). Let
us suppose that the fraction of contribution by the hyper excited is x. Then
x[e2/2krm] + (1-x) [e2/2krn] = e2/k(rm+
rn) .By solving, we can determine x, x
[1/rm - 1/rn] = 2/(rm+ rn) - 1/rn
; x = rm
/(rm + rn)
and (1-x) = rn / (rm + rn)
Since r is proportional to square of its orbital quantum number x = m2/(m2
+ n2) and (1-x) = n2/(m2 + n2).
Emn =(1/2)[KEmm + NPEmm + KEnn +
NPEnn] + (e-e)mm [m2/(m2 + n2)]
+ (e-e)nn[n2/(m2 + n2)]
Since (e-e)mm = (e-e)11/ m2 and (e-e)nn = (e-e)11/n2 , (e-e)mm
= (m2/n2)(e-e)nn. Substituting this
value in the above relation, we get,
Emn =(1/2)[KEmm +
NPEmm + KEnn + NPEnn] +2(e-e)mm [m2/(m2
+ n2)]
Lyman series for Helium atom
Fig.8.De-excitation from n=2 to n=1 (Type-I)
On the basis of
this description let us make an attempt to find the total energy associated
with an ionized helium where one electron is in its innermost orbit and the
other in next higher orbit. The energy associated with the system E12 is
derived from the calculated energies of the systems E11 and E22.
The radius of the inner most electronic orbit is (4/7)ao and
the radius of the next higher orbit is (16/7)ao The three components
of energy pertaining to the ground state of helium atom is kinetic energy KE11= (7/4) e2/Kr11 ; negative potential energy NPE11 =
- 4e2 /Kr11 and
the positive electron-electron interaction energy (e-e)11 = e2/2Kr11.
The corresponding components for next higher energy state are KE22 =
(7/4) e2 /Kr22, PE22 = - 4e2 /Kr22
and (e-e)22 = e2/2Kr22
. The energy of excited helium in its state E12 is {(1/2) [KE11
+ KE22 +NPE11 + NPE22] + 2(e-e)11 (1/5) eV. Substituting the values of each component
we get [e2/2Kr11][ -(9/4) (5/4)+(2/5)]
= 57.396 eV. When electron jumps from n=2 to n=1, the energy
liberated ΔE(12→11) is 83.269 57.396 = 25.873
eV . The wavelength of this radiation corresponds to 12.4 x 10-7 /25.873
= 47.926 nm
Knowing this technique, one can derive a formula suitable for any
electronic transition in helium atom. Consider an excited state of helium atom where
one electron is in the nth orbit and other electron in the innermost
orbit. The energy associated with the
system E1n can be computed as before from its contributors E11 and Enn .The energy contributed by E11 to
E1n is (1/2)[ KE11
+NPE11], the energy contributed by Enn to E1n is
(1/2)[ KEnn + NPEnn], and the electron-electron interactional
energy in the resultant assembly is (e-e)1n = 2(e-e)11[1/(n2 +1)] =
2(e-e)nn [n2/(n2+1)] ,. The energy associated with E1n is
the sum of these three contributions and is equal to (1/2)[(KE11 +
KEnn) + (PE11 + PEnn)] + [2(e-e)11 [(1/(n2+1)] .
When the electron jumps from n th orbit to the innermost orbit , the transition energy is the
energy difference between these two states. It is E1n - E11 = (1/2)[(KE11 +NPE11)
+ (NPEnn + KEnn)] +
2(e-e)11 [(1/(n2+1)] - [KE11 +NPE11
+ (e-
e)11] = (1/2)
[KEnn - KE11 + NPEnn - NPE11] + 2(e-e)11 [(1/(n2+1)]-
(e-e)11 = (1/2)[ KEnn -
KE11 + NPEnn - NPE11] + [(1-n2)/(1
+ n2)][(e-e)11]
Substituting the values of various components of energy we get (E1n
- E11) ={(1/2)[49/8n2
- 49/8 - 14/n2 + 14]+ [(1-n2)/(1 + n2)](7/4)}
(e2/2Kao). On simplification, ΔE(1n→
12) = [(1-n2)/2n2]{-63/8 +(7/2)[n2/(1+n2)]} (e2/2Kao).
When an electron jumps from the orbit n=2 to the innermost orbit, the
transition energy (E12 - E11)
is derived by using the above formula ΔE(2→
1) =(13.595) (-3/8)[
-(63/8)+(14/5)] = 25.873 eV and
λ2 →1 = 12.4 x 10-7 /25.873= 47.93 nm
For any transition E1n
→ E11, the transition energy is given by [(1-n2)/2n2]{-63/8
+(7/2)[n2/(1 +n2)]} (e2/2Kao). Using this
formula the Lyman series for helium atom can be predicted. Table.2 given below
gives transition energy and the corresponding wavelength for various possible
transition from orbits with n ≥ 2 to the
innermost orbit with n =1 .
Table.3.Spectral lines in
Lyman series of normal helium atom
...................................................................
transition energy Wavelength
in eV in nm
............................................................
n2→n1 25.87 47.93 n3→n1 28.55 43.43 n4→n1 29.19 42.48 ..........................................................
The radiation components in this region are the resonance
transitions from atomic helium originating from the upper n (n = 2, 3, 4,---) P state to the lower n= 1 ,
ground state S . The observation of atomic resonance emission shows 58.43 nm
and 30.38 nm.(Lyman series in Helium atom)
The helium spectrum in the ultraviolet range includes several prominent
lines, with the strongest being at 58.43 nm and 30.38 nm. Additionally, other
lines can be found between 60-110 nm
There is yet another
way by which the Lyman series may take place, where both the electrons are in
the same orbit with n greater than 1 and one of the electrons jump from the
orbit to the innermost orbit. For
example, in the initial state the helium atom is in its first hyper-excited
state, where both the electrons are in the second permitted orbits. During
transition, one of the electrons jumps to the innermost orbit. Usually this transition
will be followed by another successive transition where the remaining electron
in the second orbit will jump to the innermost orbit with half-filled. When the
hyper de-excitation is hindered by some reasons, the processes of de-excitation
takes place in steps.
The energy of the atom in its initial stat E22 = KE22
+ NPE22 + (e-e)22
The energy of the atom in its final state E12 =
(1/2)[KE22 +NPE22 +
KE11 + NPE11] + 2(e-e)11 (1/5)
The transition energy is given by ΔE(22→
12) = E22
-E12 = (1/2)[KE22 + NPE22 - KE11 - NPE11] +
[(e-
e)22 -(2/5) (e-e)11].Substituting the
values of the components of energy we
get the transition energy as (1/2)[(7/4) e2/Kr22 - 4 e2/Kr22
- (7/4) e2/Kr11 +
4 e2/Kr11] + (1/4)e2/2Kr11 - (2/5)
e2/2Kr11
= (1/2)[e2/Kr11] [(7/16) -1 - 7/4 + 4] +
[e2/2Kr11][(1/4) - (2/5)]= (7/4) [e2/2Kao]
[(27/16) - (3/20)]= 2.69 x 13.595 = 36.57 eV
and wavelength of radiation
emitted is λ22*
→12 = 33.9 nm
The transition energy in jumping of an electron fron
the energy level Enn to E1n
is Enn - E1n =
(1/2)[KEnn + NPEnn - KE11
- NPE11] + [(e-e)nn - 2(e-e)11 /[1/(n2+1)]N
= [(35+63n2)/16(n2+1)] [(n2 -1)/n2].
Using this formula the other possible transition can be studied. For example E33
→ E13 gives 27.27 nm.
The another possibility of this kind of transition is both the electrons
are in different orbits other than the innermost orbit and the electronic
transition take place from the outer orbit to the innermost orbit.
Fig. 9.Branched De-excitation of excited states under Lyman
series
The branched de-excitation may happen among the excited states
of helium atom with and without an electron in the innermost orbit.
Without an electron in the innermost orbit transition may happen between
Emn and E1n or E1m . For example E23 can
undergo transition through either E12 or E13.
E23 = (1/2)[KE33 +NPE33 +KE22 + NPE22] +(4/13)
(e-e)22 +(9/13)(e-e)33]
E12 = (1/2)[KE22 + PE22 +KE11
+ PE11] +(1/5) (e-e)11 +(4/5)(e-e)22]
E13 = (1/2)[KE33
+ PE33 +KE11 + PE11] +(1/10) (e-e)11
+(9/10)(e-e)33]
E23 → E12
= (1/2)[KE33 + PE33 -KE11
- PE11] - (1/5) (e-e)11 - (32/65) (e-e)22 +(9/13)
(e-e)33
= [49/144 - 7/9 - 49/16 +7 - 14/65 + 7/52 - 7/20](13.595)= 41.725
eV
and λ23 →121 = 30 nm.
Similarly
E23 →E13 = (1/2)[KE22
+ PE22 -KE11 - PE11] -(1/10) (e-e)11
+(4/13)(e-e)22-(9/10)(e-e)33 + (9/13)(e-e)33
= (1/2)[ KE22
+ PE22 -KE11 - PE11] -
(1/10) (e-e)11 +(4/13) (e-e)22 -(27/130)(e-e)33
= [49/64 -7/4 - 49/16 +7 + 7/52 -7/40
-21/520]13.595 = 39.05 eV λ23 →13 = 31.75 nm.
Even though
the possibility is very little there is yet another way for the transition
under Lyman series to happen. Two electrons in two different orbits with n >
1 jump simultaneously to the inner most orbit. For example E23 may
undergo to E11
E23 = (1/2)[KE33 +NPE33 +KE22 + NPE22] +(4/13)
(e-e)22 +(9/13)(e-)33]
E11 = KE11 + NPE11 + (e-e)11
E23 →E11 =
(1/2)[KE33 +NPE33 + KE22
+ NPE22 -KE11 - NPE11] -(e-e)11 +(4/13)(e-e)22+(9/13)(e-e)33 = {(1/2)[49/72 -14/9
+49/32 -14/4 - 49/8 + 14] -(7/4) +(4/13)(7/16) + (9/13)7/36}
(e2/2Kao) =
{(1/2) [ 5.0312] - 1.4807 }(e2/2Kao) = 14.069 eV λ23 →11 = 88.13 nm.
The whole Lyman series of helium atom
falls in UV region. The wavelength of the emitted radiation in this series is
around 30-60 nm
Balmer series of Helium atom
There are few ways by which the Balmer series in helium atom may arise. The first one corresponds to electronic transition in helium atom where one of the electrons is in the inner most orbit, while the other electron jumps from the orbit with n ≥ 3 to n =2 The second one corresponds to electronic transition in helium atom where one of the electrons is in second orbit and the other electron jumps from the orbit with n ≥ 3 to n =2 .The former case is more probable than the other due to its different transient nature.
Fig.10. Normal de-excitation from n=3 to n =2
Let us
calculate the energy associated with systems denoted by E13 and E12 where one of the
electrons is in the innermost orbit and other electron is in the third and
second orbit respectively. By using the half-half mixing the energy content of
the systems can be evaluated. For the system E13, the contributors
are E11 and E33 and for the system E12 they are E11 and E22
E13 = (1/2) [KE11 + NPE11 + KE33 +
NPE33] + (e-e)11 (1/10) + (e-e)33(9/10)
E12 = (1/2) [KE11 +NPE11 +KE22 +
NPE22] + (e-e)11(1/5)+ (e-e)22(4 / 5)
ΔE(13→ 12) =E13 - E12 = (1/2)[KE33
+ NPE33 - KE22 - NPE22 - (e-e)11(1/10)
- (e-e)22(4/5) +(ee)33(9/10)
Substituting the values for all the components of energy, we get
ΔE (13→ 12) = {(1/2) [49/72 -
14/9 - 49/32 + 7/2 ] +[- 7/40 + 7/40 -
7/20]}(e2 /2Kao) = [0.1969]
13.595 = 2.6768 eV and λ13 →12 = 463.24 nm
In the Type II transition, it is E23
→ E22 , where
E23 = (1/2) [KE33 + NPE33 + KE22 +
NPE22] + (4/13) (e-e)22 + (9/13) (e-e)33
E22 = KE22 + NPE22 + (e-e)22
ΔE(23→ 22)
=E23 - E22 = (1/2)[KE33 + NPE33 -
KE22 - NPE22] + (9/13)[(e-e)33 - (e-e)22]
= {(1/2)[ 49/72
-14/9 - 49/32 + 7/2] + (9/13) [7/36 - 7/16]} (e2 /2Kao)
= {(1/2[0.6805 - 1.5555 - 1.5312+ 3.5] +
(63/52)(-5/36)} (e2 /2Kao)
= 0.5469 - 0.1682 = 0.3787 X 13.595 = 5.1484 eV
λ23 →22 =
240.85 nm
By driving a formula, one can determine the
wavelengths of various spectral lines of Balmer series of helium atom.
For Type-I transition,
E1n = (1/2) [KE11 + NPE11 + KEnn +
NPEnn] + [1/(n2 +1)](e-e)11 + [n2/(n2
+ 1)](e-e)nn
E12 = (1/2) [KE11 +
NPE11 + KE22 + PE22] + N[1/(5)](e-e)11 +
(4/5)](e-e)22
ΔE(1n→ 12)
=E1n → E12 = (1/2)[KEnn + NPEnn -
KE22 - NPE22] + [1/(n2 +1)](e-e)11 -
[1/(5)](e-e)11 (4/5)](e-e)22 + [n2/(n2 +
1)](e-e)nn
=(1/2)[KEnn + NPEnn
-KE22 -NPE22]+[(1/n2+1) -1/5](e-e)11
-(4/5) (e-e)22 +[n2/(n2 + 1)](e-e)nn
= {(1/2)[49/8n2
- 49/32 -14/n2 +14/4]- (7/4)[(4-n2)/5(n2+1)]
-7/4 (1/5) +[n2/(n2 + 1)](7/4n2)}( e2 /
2Kao) = (49/16)[(4-n2)/4n2]
- 7 [(4-n2)/4n2] + (7/10) [(4-n2)/(n2 +1)]
= (7/4) (4-n2) [ 7/16 n2 - 1/n2 +
(2/5) [1/(n2+1)] = (7/2) (n2 - 4) [13 n2 +
45]/[16 n2 (n2 +1)]
n = 3 ; ΔE(13→ 12) =
[0.1987] (13.595) = 2.6765 eV and λ13
→12 = 463.3 nm
n=4 ; ΔE(14→ 12) = [0.2442] (13.595) = 3.320 eV and λ14 →12
= 373.5 nm
Fig.11.Helium emission spectrum
As the nuclear charge is twice that of hydrogen, all
the electronic orbits are little closer to the nucleus and as a consequence of
which the electronic transition between n ≥ 3 to
n=2 emits more energy which fall in UV region
Paschen series are due to the transition between n ≥ 4 to n = 3. In the most probable transition one of the
electrons is bound in the innermost orbit and the other electron make
transitions from orbits with n ≥ 4 to n = 3. As the process of de-excitation is
not completed usually it is followed by another successive transition.
E1n =
(1/2) [KE11 + NPE11 + KEnn +
NPEnn] + [1/(n2 +1)](e-e)11 + [n2/(n2
+ 1)](e-e)nn
E13 = (1/2) [KE11 + NPE11
+ KE33 + NPE33] + (1/10)](e-e)11 +
[9/10](e-e)33
ΔE(1n→ 13) =E1n - E13 =(1/2)[KEnn +
NPEnn - KE33 - NPE33]+[(9 - n2)/10(n2 +1)](e-e)11 - [9/10](e-e)33 + [n2/(n2 +
1)](e-e)nn
= (1/2)
[49/8n2 - 14/n2 - 49/72 + 14/9](e2 /2Kao)
+(7/4)[(9-n 2)/10(n2+1)]
-9/10 (7/4) (1/9) + 7/4 [1/(n2+1)][e2 /2Kao]
=
[(9-n2)/9n2 ] (-63/16)
+ (7/20)[(9-n2)/5(n2+1)][e2 /2Kao]
= (7/4) (n2 - 9) [(n2 +5)/20n2
(n2 +1)][e2 /2Kao]
when n = 4, ΔE(14→ 13) =E14 - E13
= (49/4)[21/20x16x17)](13.595)= 0.0473 x 13.595 =0.643 eV and λ14 →13 = 1928.5 nm
n = 5, ΔE(15→ 13) =E15 - E13 =
(7/4)[(16x 30)/(20x25 x26)](13.595)= 0.0646 x 13.595 =0.8784 eV and λ15 →13 = 1411.6 nm
n=6 , ΔE(16→ 13) =E16
- E13 = (7/4)[(27 x 41)/(20x36 x 37)](13.595)= 0.0727 x 13.595
=0.9884 eV and λ16 →13 = 1254 nm
Normal helium
spectral lines in the visible range include wavelengths around 587.6 nm
(yellow), 667.8 nm (red), and 706.5 nm (red). Other visible lines also exist,
such as 447.1 nm (blue-green), 492.2 nm (blue-green), 501.6 nm (green), and
667.8 nm (red). The visible part of the helium spectrum falls roughly between
388.8 nm and 781.3 nm, while the invisible parts include ultraviolet (UV) and
infrared (IR) radiation. Specifically, the visible helium spectrum contains
lines at 388.8 nm, 447.1 nm, 471.3 nm, 492.1 nm, 501.5 nm, 504.7 nm, 587.5 nm,
667.8 nm, 686.7 nm, 706.5 nm, 728.1 nm and 781.3 nm, . UV radiation has
wavelengths shorter than 380 nm, and IR radiation has wavelengths longer than
780 nm
The successive
secondary transition followed after a transition can be identified with the
energy balance relation. .If a transition is split into two successive transitions,
hν1 + hν2 = hν3 1/λ1 + 1/λ2 = 1/λ3 or λ3 = λ1λ2 / (λ1 +λ2)
For example 728.1 nm and 781 .3 nm are two visible radiations in
helium spectrum. When it happens as a single transition its wavelength will
be (781.3 x 728.1)/ (781.3 + 728.1) = 376.88
nm .
Results
and Discussion
The radius and orbital
velocity of orbital electron change in helium atom when the electrons are in
different orbits. When they are in the same orbit 2e2/Krnn2
- e2/4Krnn2 = (7/4) e2/K = (me)nn
vnn2 rnn and nh/2π = (me)nn
vnn rnn .By solving these two relations we get rnn =(4/7)n2 ao (1- vnn2/c2)1/2
and vnn = (7/4)e2/ 2nhεo
Let us suppose that one of the electrons is in orbit n1 and
the other electron is in orbit n2 . The electron in orbit n1 has radius (rn1)n1-n2 mass (mn1)n1-n2 and velocity
(vn1)n1-n2
2e2/K(rn1)2 - e2/K(rn1 +
rn2)2 =2e2/K- e2/K[1/(1+rn2/rn1)]2
= mn1vn12rn1
[e2/K][2 -
[1/(1+n22/n12]2 = [e2/K](n14
+2n24 +4n12n22)/(n12
+ n22)2 = mn1 vn12rn1
. n1
h/2π = mn1 vn1 rn1
By
solving these two relations, vn1 = [e2/2n1hεo](n14
+2n24 +4n12n22)/(n12
+ n22)2
rn1 = n12 ao [(n12
+ n22)2/(n14 +2n24
+4n12n22)] (1- vn12/c2)1/2
The electron in orbit n2 has radius (rn2)n1-n2 , mass (mn2)n1-n2 and
velocity (vn2)n1-n2
2e2/K(rn2)2 -
e2/K(rn1 + rn2)2 =2e2/K-
e2/K[1/(1+rn1/rn2)]2 = mn2vn22rn2 [e2/K][2
-[1/(1+n12/n22]2=[e2/K](2n14+n24+4n12n22)/(n12+n22)2= mn2 vn22rn2 n2 h/2π = mn2 vn2 rn2
By
solving these two relations, vn2 = [e2/2n1hεo](2n14
+n24 +4n12n22)/(n12
+ n22)2
rn2 = n22 ao [(n12
+ n22)2/(2n14 +n24
+4n12n22)] (1- vn22/c2)1/2
Knowing
the orbital velocity, the binding energy of the orbital electron in orbits n1
and n2 can be estimated from (1/2)mo vn12
, (1/2)mo vn22
respectively. Total binding energy BEn1-n2 = (1/2)mo[(e2/2hεo)2/(n12
+ n22)4][(n14 +2n24
+4n12n22)2/n12+(2n14
+n24 +4n12n22)2/n22]
= [e2/2Kao][1/(n12
+ n22)4][(n14 +2n24
+4n12n22)2/n12+(2n14
+n24 +4n12n22)2/n22]
Using
this general expression, one can find out the binding energy of helium atom in
any of its state.
n1=
n2= 1 ;[e2/2Kao] (1/16)[ 98] =(49/8)[e2/2Kao]
= 83.269 eV
n1
= 2; n2 = 3 ; [e2/2Kao] [1/(13)4]
[(322/2)2 +(257/3)2] = 15.8316 eV
n1=
n2 = 2; [e2/2Kao] [1/(8)4] [(322/2)2
+(257/3)2] = 20.8173 eV
BE12
- BE23 = 20.8173 - 15.8316 = 4.9857 eV and λ23→12 =
248.72 nm It represents the first line
of Balmer series of helium atom .
Relativistic Bohr Model of Helium
Atom
Rejuvenation of Bohr's Theory
of Hydrogen atom with the relativistic
change of mass of the orbital electron predicts that the energy corresponding
to any electronic transition from orbit with quantum number n2 to n1
is equal to the energy equivalent of
the relativistic variation of mass of the electron i.e., ΔE
= Δmc2 = (mn1-
mn2)c2. The transition energy of the electron
in the hydrogen atom is derived in terms of wavelength of matter waves of the
electron in the concerned orbits. It is shown that the atomic binding energy is
exactly equal to the energy equivalent of relativistic increase of mass of the
orbital electron.
Many
literature is available for the application of Bohr's Theory to Helium atom [3-6].The proposed relativistic Bohr model is applied
to helium to verify the observed experimental data and to study its spectral
lines. According to conservation of energy, the loss of total potential energy
of the system is equal to sum of its total kinetic energy and energy equivalent
of relativistic increase of masses of the electrons in the same orbit labelled
n
4e2/Krnn -
e2/2Krnn = (7/2) e2 /Krnn = mnn vnn2 +
dmnn c2
here
dmnn represents the relativistic increase of mass of both the
electrons in the orbit n. Since mnn vnn2 = 2e2/Krnn - e2/4Krnn
= (7/4) e2/Krnn , dmnn c2 =
(7/4) e2/Krnn, where mnn, vnn are the mass and velocity of the
electron in the orbit n,when both the
electrons are in the same orbit having radius rnn. The loss of
potential energy of both the electrons is equally shared by its kinetic energy
and energy required for its relativistic increase of mass.
From
the relations mnn vnn2 rnn = (7/4)
e2/K and mnn vnn rnn = nh/2π , the
radius of the orbit of the electrons and
its orbital velocity in helium atom can be derived as rnn = (4/7) n2
ao [1-vnn2/c2]1/2 ≃ (4/7) n2 ao [1-
vnn2/2c2] and vnn = (7/4) [e2/2εonh]
. In the ground state
of helium atom, substituting the values for r11 and v11 in dm11 c2, that
measures its binding energy
dm11 c2 = (7/4)2
e2/ [Kao (1- v112/c2)1/2]= (49/8) x (e2/2Kao)
x [1/(1- v112/c2)1/2]
= 6.125 x 13.595 [1/(1- v112/c2)1/2]
v112/c2
= (49/64) e4/εo2h2c2
= 0.000163 and (1- v112/c2)1/2 =
0.9999184
dm11c2 =83.269 x [1/ 0.9999184] = 83.275 eV
This can be verified by computing dm11c2 by calculating the relativistic
increase of mass of each electron. For an electron dm = m1 - mo
where m1 = (4/7) h2εo/πe2 r1
and mo = h2εo/πe2 ao. Substituting
the value for r1 we get m1 - mo = h2εo/πe2
[4/7r1 - 1/ao] = h2εo/aoπe2
[(1/[(-v12/c2)1/2 - 1] =
41.651 eV. For both the electrons it is 2 x 41.651 =83.302 eV.
It is
dmc2 that measures the binding energy of the system. When both the
electrons in helium atom are in the same orbit labelled n, dmnnc2
= (7/4) e2/Krnn. i.e., each electron contributes
equally to the binding energy of the system. When one of the electrons is taken
away from the system, the binding energy of the remaining electron will not be
exactly half of the binding energy of the system. This can be verified with the
practical values of first (24.48 eV) and second (54.40 eV) ionization energies
of helium atom. The binding energy of helium atom is sum of its first and
second ionization energies 78.88 eV. It
shows that when one of the electrons in the ground state of helium is taken
away from the system, the binding energy is not halved. The outgoing or
departing electron gives some fraction of its binding energy to the other electron
that remains in the system. This is due to the absence of electron-electron
interaction. That is why (x+ 1)th ionization
energy is greater than xth.
By using
the relation BEnn = dmnnc2 =(439/8) [e2/2Kao](1/n2)[1
+ vnn2/2c2 ], one can study the process of hyper de-excitation and
its spectral feature. BE11 - BEnn = (49/8) [e2/2Kao][(1
-1/n2) + (1/2c2) (v112 - vnn2/n2)
]. Substituting the values for vnn = (49/64) [e4/h2 εo2
n2], BE11 - BEnn = (49/8) [e2/2Kao][(n2
-1)/n2] [ 1 + (49/16) (1/moc2) ( e2/2Kao)
= (83.269) [(n2 -1)/n2] [1 .000081]. When n= 2 ΔE(22→
11) =BE11 - BE22 = 62.457 eV and the wavelength of
the emitted radiation is 19.85 nm , when n =3 ,ΔE(33→ 11) =BE11
- BE33 = 74.0232 eV and λ33 →11 = 16.75
nm. (Ref.Table.2)
When the energy level of a
state with the two electrons in the helium atom are in two different orbits
labelled with m and n respectively the binding energy BEmn can be assessed by half-half mixing of two
states with both the electrons are in m and n respectively. The binding energy
of the mixed state with one electron in n = 1 and another electron in n = n is
given by
E11 = [KE11
+NPE11 +(e-e)11] = - dm11c2
Enn = [KEnn +NPEnn +
(e-e)nn] = -dmnnc2
E1n = (1/2)[NPE11 + KE11 +
NPEnn + KEnn] + (e-e)1n = - dm1n c2
(e-e)1n = e2/K(r11
+ rnn) = 2(e-e)11 r11/(r11
+ rnn) = 2(e-e)11 /(1+n2) = (e-e)nn[n2/(1+ n2)]
E1n = (1/2)[NPE11 +
KE11 + NPEnn + KEnn] + 2(e-e)11 /(1+n2) = - dm1n c2
BE1n = (1/2)[dm11c2 + (e-e)11 +
dmnn c2 + (e-e)nn ] -2(e-e)11/(1+n2)
BE11 - BE1n =
(1/2)[dm11c2 - (e-e)11- (e-e)nn -
dmnnc2] + 2(e-e)11/(1+n2)
= (1/2)[dm11c2 - dmnnc2]
+ (e-e)11 {-(1/2) - (1/2n2) + [2/(1+n2)]}
Substituting the values for dm11 c2 , dmnn
c2, and (e-e)11 ,
BE11 - BE1n =
(1/2) (49/8)[e2/2Kao] [(n2 - 1)/n2] + (7/8)[e2/2Kao][(-n4
+2n2-1)/n2(1+n2)]
= (7/8)[e2/2Kao][
5n4 +4n2 - 9]/ 2n2(n2+1)]
Giving
different values (n= 2,3,4,5......) the Lyman series of the helium atom can be determined.(Ref.Table
3)
n BE11
- BE1n
λ1n→
11
n=2;
0.875 x13.595 x 2.175 = 25.873 eV 47.93 nm
n
=3; 0.875 x13.595 x 2.4 =
28.550 eV 43.43 nm
n=4; 0.875 x 13.595 x 2.454 =29.192
eV 42.48 nm
This can be verified by calculating the
relativistic variation of mass of the orbital electrons in normal and excited
helium atom.
The Lyman series of the helium atom can be
studies by deriving a relation for its binding energy in the energy levels when
the two electrons are in the states 11 and 1n
BE11
= dm11c2 ≃
2 [(1/2) mo v112]
where v11 = m11 v112 r11
/m11 v11 r11 = (7/8) [e2/hεo].
Substituting this value in the above relation BE11 =[mo (49/64)e4/h2εo2]
= 2 x 41.681 = 83.374 eV
BE1n is the sum of binding energies of the electrons in two different orbits labelled with 1 and n. BE1n = (dm1)1n c2 + (dmn)1n c2
Fig.12. Excited helium atom labelled 1n
First let us consider the electron in the innermost orbit.
The resultant force acting on the electron is counter balanced by the
centrifugal force 2e2/K - e2/K(1
+rn/r1)2 = [e2/K][(2n4+4n2
+ 1)/(1 +n2)2]
= m1 v12 r1 . Solving
for (v1)1n with the condition for orbital angular
momentum m1 v1 r1 = nh/2π, we get (v1)1n
= [e2/2hεo] [(2n4 +4n2+1)/(1+n2)].
The binding energy contributed by the electron in
the inner most orbit is dm1c2 ≃ (1/2) mo (v1)1n2 = (1/2) mo [e3/h2εo2]
[(2n4 +4n2+1)/2(1 +n2)]2 in terms of eV. When the other electron is in
the orbit labelled with n the resultant force and the centrifugal force
together keep it in stable orbit 2e2/K - e2/K(1+r1/rn)2
= [e2/K][(n4+4n2 +2)/(n4 +2n2
+1)] = mn vn2
rn . Solving for (vn)1n as before, we get (vn)1n
= [e2/2nhεo][
n4+ 4n2 + 2/(1 +n2)2].
The binding energy contributed by the excited electron is dmnc2
= (1/2) mo [e3/h2εo2]
[ (n4+ 4n2 + 2)/2n (1+n2)2]2.
The binding energy of the system dm1n c2 is the sum of
its contribution by both the electrons and is equal to (1/2) mo
[e3/h2εo2]{[(2n4 +4n2+1)/2(1+n2)]2
+ [(n4+ 4n2 +
2)/2n (1 +n2)]2 =
(54.45){[(2n4 +4n2+1)/2(1+n2)]2 +
[(n4+ 4n2 + 2)/2n (1+n2)]2 When the
excited electron jumps to the innermost orbit, the change in the binding energy
dm11c2 - dm1n c2. dm11c2
represents the energy equivalent of the relativistic change of mass of
both the innermost electrons. dm11 c2 = mo v112
, substituting the value of v11
= (7/8)[e2/hεo], dm11c2 =
(1/2)mo [e4/h2εo2]
(49/32) = 83.4378 eV. dm11c2
- dm12 c2 = (1/2)mo [e4/h2εo2]{(49/32)
-[(2n4 +4n2+1)/2(1+n2)]2 - [(n4+
4n2 + 2)/2n (1 +n2)]2}. = (1/2)mo
[e4h2εo2]{(49/32) - [34/100]2
+ [49/50)]2. When excited electron in 2nd orbit
jumps to innermost orbit, the binding energy difference is 83.4378 - 58.5882 =
24.786 eV which corresponds to the electromagnetic radiation 50.03 nm. If the
excited electron is in the 3rd orbit, the change in binding energy
and the wavelength of emitted radiation
become dm11c2 -
dm13 c2 = (1/2)mo [e4h2εo2]{(49/32)
- [119/600]2 + [199/200)]2 = 83.4378 - 56.029 = 27.345 eV
and λ13→ 11 = 45.34 nm.
Acknowledgement
The Author is grateful to RM.Alagappa Chettiar, the philanthropist
and fonder of all Alagappa educational
institutions from where the author is educated.
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