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JKAU: Sci., Vol. 23 No. 2, pp: 65-78 (2011 A.D./1432 A.H.)
DOI: 10.4197 / Sci. 23-2.5
65
Polarization Dependence of the Bremsstrahlung Cross-
Section in the Scattering of Electrons by Nuclei with
Electromagnetic Multipole Moments
S. A. Al-Khateeb
Department of Mathematics, Faculty of Science for Girls,
King Abdulaziz University, Jeddah, Saaudi Arabi
Abstract. The present work deals with one of the most important
problem in the quantum electrodynamics, 1.e the problem of
bremsstrahlung process due to the interaction between lepton
(electron) and the field of nuclei having electromagnetic multipole
distributions. The detaild calculations for the cross-sections and the
angular and energy distributions of the process are given in the first
Born approximation of the quantum perturbation theory in two cases:
With neglecting the polarizations of the photon and the electron, and
with studying these polarizations. Applying the obtained formulae to
some nuclei, the light nucleus ( )94Be and the heavier one ( )27
13Al , to
showing the effect of both circular photon polarization and
longitudinal electron polarization on the bremsstahlung cross-section,
and comparing the results obtained for the two nuclei graphically.
I-Introduction
The Study of the bremsstrahlung process was firstly done by Bethe and
Heitler[1]
in 1934, and many works about this process could have
appeared in the period (1934-1954)[2-4]
. In these works, the polarizations
of the photons of bremsstrahlung and of the electrons were not studied
because of the complicity of the calculations in this case.The circular
polarization of bremsstrahlung photons due to the collision of
longitudinally polarized electrons with nuclei, was firstly investigated by
Grodzins[5]
in 1959. At the same time, Olsen and Maximon[6]
studied the
circular polarization of the bremsstrahlung photons experimentally. Motz
66 S. A. Al-Khateeb
and Placious[7]
in 1960, studied the linear polarization of the
bremsstrahlung photons, and in 1962, Bisi and others[8]
made some
measurements about the dependence of the degree of circular polarization
of bremsstrahlung on the electric and magnetic field and on the atomic
number of the target nuclei. The intensity of bremsstrahlung due to the
interaction of electric and magnetic interactions between electrons and
target nucleus, had been calculated by Ginsberg and Pratt[9]
in 1964, and
in the next year, Mork and Olsen[10]
investigated the polarization effect
on bremsstrahlung, and showed that the study of polarization caused to
increase the cross-section of the process. The study of bremsstrahlung
due to the scattering of longitudinal polarized electrons by different
nuclei was done by Neumcke[11]
in 1966, and after many years, Behncke
and Nakel[12]
in1974 calculated the linear polarization of photons in the
scattering of electrons with carbon nucleus. In 1989, Caffo and others[13]
,
investigated the polarization of the scattered electrons and of the emitted
photons in the bremsstrahlung at high energies. The asymmetry of photon
polarization in the bremsstrahlung due to polarized electrons were
investigated in 1992 by Mergl and others[14]
and in 1996 by Haug[15]
. In
2002, Likhachev and others[16]
, calculated experimentally the degree of
linear and circular polarization of bremsstrahlung photons, and in 2005,
Lee and others[17]
investigated the production of bremsstrahlung in the
field of heavy atoms at intermediate range of energies. Haug[18]
and
Nukel[19]
, in 2006, both studied the photon polarization and the spin
asymmetry in the bremsstrahlung process in details. In 2009,
S.A.Alkhateeb[20]
, obtained the relation between the circular polarization
of photons and the longitudinal polarization of the scattered electrons in
the electromagnetic field of light nuclei.
II-Formulation of the Problem
The bremsstrahlung photons γ are produced in the process of the
interaction of electrons with the field of nuclei:
γ++→+−−
fieNNe (1)
Polarization Dependence of the Bremsstrahlung Cross-Section in the … 67
Feynman diagrams of this process are shown in Fig. a.
From these diagrams we see that the momentum transfer to the
nucleus in the process is ifpkpq −+=
γ. The matrix element (s-matrix) for
the process could be written as the sum of two amplitudes corresponding
to the above diagrams (1) and (2) and has the form[3]:
2
4
1
4
2211
2 )(),()( xdxdxUxxxUeSifif
Γ= ∫ (2)
),()()()()(),(),(221241421121 γγ
γγ kxAxxGxAxAxxGkxAxxKfNNfK
−+−=Γ (3)
Where K
A is the electromagnetic potential of the emitted photon with
momentum γk ,
NA is the potential of the nuclear field, )(
21xxG
f− is the
Feynman propagator of the intermediate state between the two points 2x
and 1x .The potentials
KA and
NA are given in the form[5]:
)(2
4),( xkixki
Kee
VkxA
⋅−⋅
+=
ω
π
εμ
γ (4)
xqi
Ne
q
qdeZxA ⋅
∫−=
23
3
21
)2(4)(
π
π (5)
Where ),0( εεµ
�
= is the polarization vector of the photon, ω is the photon
energy, V is the normalization volume, Z is the atomic number of the
target nucleus, the potential )(xAN
given in (4) represent the coulomb
field of nucleus. The differential cross-section of the bremsstrahlung with
respect to photon energy is given by [4]:
γγ
ω
πω
σ
kefi
i
f
ddkppFqp
pmeZ
d
dΩΩ= ),,(
42
2
0
62
(6)
N
qf
pki
p −=−
γ
fp
ip
q
2x
1x
)1(
qi
pkf
p +=+γ
ip
q
2x
1x
γk
aFig .
γk
N
fp
)2(
68 S. A. Al-Khateeb
In(6): 0
m is the rest mass of electron, e is the electronic charge.
edΩ and
γkdΩ are the solid angles of the scattered electron and the
emmited photon respectively, ),,(γkppF
fi is a function depends on the
initial )(ip and final )(
fp momenta of electron and the photon
momentum )(γk . The final form of this function is given by [20]:
[{ ] [ ]⋅++−
−
=2222
0222
0
)()()()(4)()(
1
4
1γγγγ
γγ
kpkpEkpEkpmkpkpm
Fifiiff
fi
[ ] [ ]})()()(2))((2 2222
0
2
fifififippEEkpqkpmqkpkp −++−
γγγγ (7)
here:
[ ] [ ]ffffiiii
pEkppEkp θωθωγγ
cos)(,cos)( −=−= (8)
[ ]),cos(2cos2cos22222
fifiiifffippppppppq −−+++−= θωθωω (9)
φθθθθ cossinsincoscos),cos(fififi
pp += (10)
fiθθ , are the polar angles of
fipp , with respect to the photon direction,
),cos(fi
pp is the angle between ip and
fp , and φ is the angle between the
planes ),(γkp
iand ),(
γkp
f.
III-Analytical Results for the Problem
1- The basic Formula for the Differential Cross-Section
The differential cross-section of bremsstrahlung in the coulomb
field of the nucleus can be obtained from the above relations, and the
final form of this cross - section is given by:
γθθη
ω
σkefie
ddFqd
dΩΩ⎟⎟
⎠
⎞⎜⎜⎝
⎛= ),(
104
(11)
Which is known as Bethe - Hietler relation for the bremsstrahlung,
where:
( )i
f
p
pZ⎟⎠
⎞⎜⎝
⎛=
ωπ
αη
1
22
32
(12)
Polarization Dependence of the Bremsstrahlung Cross-Section in the … 69
⋅ΔΔ
+−Δ
+−Δ
=
fi
i
f
ff
f
i
ii
fieqE
pqE
pF
2
22
2
22
22
2
22
0
2)4(
sin)4(
sin),(
ωθθθθ
−+ )sinsin( 2222
ffiipp θθ
)22(),(2
222 qEEf
fi
fi
fi−−
ΔΔ
θθ
(13)
ffffiiiipEpE θθ cos,cos −=Δ−=Δ (14)
φθθθθ cossinsin),(fififi
ppf = (15)
fiEE , and ω are the energies of initial and final electron and of the
emitted photon, where:
2
0
22
0
2
,, mpEmpEEEffiifi+=+=−=ω (16)
2- Study of the Electron and Photon Polarization and the Electro-
magnetic Structure of Target Nucleus
The studying of the interaction between the longitudinally
polarized electrons and the electromagnetic field of target nucleus
corresponding to the charge distribution (Ze) and the magnetic dipole
moment ( )I
µ of the nucleus, taking into consideration the circular
polarization of the bremsstrahlung photons, gives the following formula
for the differential cross-section of the process:
),,(),,(),,,(Ifi
m
essfi
e
essIfiessdZedZed μθθσθθσμθθσ
γγγ+= (17)
),(),(2
1),,(
10 fieefiefi
e
essdssdZed θθσθθσθθσ
γγ+= (18)
),(),(2
1),,(
10 fimefimIfi
m
essdssdd θθσθθσμθθσ
γγ+= (19)
Where BHe
dd σσ =0
is given by (11) and represent the cross-section due to
the charge distribution without studying the polarization of particles and
photons.
γωηθθσ
keefiedddF
qd ΩΩ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
141
1),( (20)
γωμηθθσ
kemIfimdddF
qad ΩΩ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
020
1)(4),( (21)
70 S. A. Al-Khateeb
γωμηθθσ
kemIfimdddF
qad ΩΩ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
121
1)(2),( (22)
Where I is the spin of target nucleon, epe
dd σσ =1
is the charged cross-
section with studying the polarization, m
d0
σ the magnetic cross-section
without studying polarization, and mpm
dd σσ =1
is the magnetic cross-
section with studying the polarization. The functionsme
FF01
, and m
F1
are
given by [20]:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Δ−
Δ++
ΔΔ−
−++Δ
+
+−+Δ
=
22
2
0
2
0
2
2
22
2
0
2
2
22
1
11),()coscos(
),(2
)coscos(sin
)coscos(sin
),(
if
fiffiiii
fi
fi
fffiiii
f
ff
iffiiii
i
ii
fie
fmpEpEf
mEpEpEp
mEpEpEp
F
θθωθωθωθθ
ωωθωθωθ
ωωθωθωθ
θθ
⎟⎟⎠
⎞⎜⎜⎝
⎛−
Δ+−
ΔΔ+
2
0
2222
22
011
)sinsin(m
Epp
mi
i
iiff
fiω
θθω
(23)
( ) ( )2
22
22
2
22
22
0
4
sin4
4
sin4),(
f
ff
i
i
ii
ffim
ppq
ppqF
Δ++
Δ+=
θθθθ
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+
Δ
Δ+
Δ
Δ+
ΔΔ+⎟⎟
⎠
⎞⎜⎜⎝
⎛−+
ΔΔ− 2
24
),(2 22
2
0
2
i
f
f
i
fi
fi
fi
fi qmEE
qf ωθθ (24)
( )⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−+ΔΔ
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Δ+
Δ−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Δ+
Δ⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛= 2
0
2
22
2
0
2
13
211112
4),( mEEEmE
q
pF
fii
fififi
ifim
ωθθ
( )⎟⎟
⎠
⎞
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
Δ
Δ−
Δ
Δ−
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
ΔΔ
−−
Δ
Δ−+
Δ
Δ+−
i
f
f
ii
fi
fi
f
fi
i
if EmEEppm
2
2222
0
2
2
2
2
2
0
ωωω
(25)
The quantity 1±=e
S represent the helicity of the initial electron.
1+=e
S for the right-longitudinal electron. 1−=e
S for the left-longitudinal
electron. The quantity 1±=γ
S represent the circular polarization of
bremsstrahlung photon. 1+=γ
S for the right circular photon polarization.
1−=γ
S for the left circular photon polarization.
Polarization Dependence of the Bremsstrahlung Cross-Section in the … 71
3- The Angular Distribution of the Circular Polarized Bremsstrahlung
Photon
The integration of (17), (18) and (19) with respect to the solid angle
of the scattered electron e
dΩ gives the angular distribution of the circular
polarized bremsstrahlung photon into the form:
),(),(),,(Ii
m
essi
e
essIiessdZedZed μθσθσμθσ
γγγ+= (26)
)()(2
1)(
1 ieeioei
e
essdssdd θσθσθσ
γγ+= (25)
)()(2
1)(
1 imeiomi
m
essdssdd θσθσθσ
γγ+= (28)
Where:
γθηπθσ
kieioedd ΩΦ= )(4)(
0 (29)
γθηπθσ
kieiedd ΩΦ= )(2)(
11 (30)
γθμηπθσ
kimIiomdad ΩΦ= )()(4)(
0 (31)
γθμηπθσ
kimIimdad ΩΦ= )()(2)(
11 (32)
In the ultra relatives case ( )20
,, cmEEkfi>>
γ, the functions
mmee 1010,,, ΦΦΦΦ are given by:
⎩⎨⎧
Δ
Δ−+−+
Δ
−−
Δ
−−−
Δ=Φ
0
0
0
2
0
2
2
00
22
0
0
)2()22(127)1(2
8
1)( ε
γγ
β
γγγθ
amie
⎭⎬⎫
⎟⎟⎠
⎞⎜⎜⎝
⎛ −+
−−
−⎟⎟⎠
⎞⎜⎜⎝
⎛
Δ
+−+−+
β
ε
β
γ
γ
γ
γ
εγγγγ T
2
2
0
2 13
11
)22(2)2( (33)
⎩⎨⎧
Δ−
Δ−−+
Δ
−−
Δ
−+
Δ=Φ L
amie
0
0
0
2
2
00
22
0
1
)1(
)2()2(1614
8
1)(
γ
γγ
β
γγγθ
⎭⎬⎫
⎟⎟⎠
⎞⎜⎜⎝
⎛ −+
−−
β
ε
β
γ
γ
γ0
2
2
13
1
(34)
Timε
βγ
γγγε
γ
γ
β
γγθ
3
0
22
0
2
0
2
0
)1(2
)1(
)1(42
)1(
2
3)(
−
Δ+−+
−−
Δ
Δ−−+
−=Φ
0
0
0
2
)1(4
)2()22(ε
γ
γγ
Δ−
Δ++−+ (35)
Timε
βγ
γγγε
γ
γ
β
γγγθ
3
0
0
2
0
1
)1(
)1(
)1(2
)1(1
2)(
−
Δ+−+
−−
⎩⎨⎧
Δ
Δ−−+−=Φ
⎭⎬⎫
Δ−
Δ+−+
0
0
0
)1(2
)2()2(ε
γ
γ (36)
72 S. A. Al-Khateeb
where:
i
i
icm
Ea
Eθ
εγ
γ
cos1,,02
0
−=Δ== (37)
γ
γγγβ
)1(2ln2,2)1(
0
2−
=Δ+−=a
L (38)
[ ] γεγβ
γβεγε ln2,
1
1ln,)1(2ln2
0−=
−+
+−
=−=T
a (39)
The second term in (27) and (28) represents the effect of the spin
correlation between the electron and photon spins on the charge and
magnetic distributions of the target nucleus respectively.
4- Calculation of the Degree of Circular Polarization of the
Bremsstrahlung Photons
The degree of circular polarization in the case of the
bremsstrahlung emitted from the collision of electron with nuclei having
charge distribution and magnetic dipole moment, is give by:
[ ] [ ][ ] [ ]
11
11
),,(),,(
),,(),,(),,(
−==
−==
+
−
=
γγ
γγ
γγ
γγ
μθσμθσ
μθσμθσμθ
sIiess
sIiess
sIiess
sIiess
Ii
ZedZed
ZedZed
ZeP (40)
Using the relations (26) - (36) we get the following relation for (P):
⎥⎦
⎤⎢⎣
⎡
+
+=
)()(1
)()(1)(),,(
0
1
iI
iI
ieeIi
Ra
RapsZeP
θμ
θμθμθ (41)
Where: e
m
i
e
m
i
e
e
ieRRp
1
1
1
0
0
0
0
1 )(,)(,)(Φ
Φ=
Φ
Φ=
Φ
Φ= θθθ
)(ie
p θ represents the degree of charged circular polarization of
bremsstrahlung produced in the collision of longitudinally polarized
electrons )1,1( −=es with the field of nuclei having zero magnetic dipole
moment. If we take the averge on the spin state of electron )1( ±=es and
the summation over the photon spin )1( ±=γs , we have the relation:
[ ]0000
)(1)()(),,( RadddZedIeimieIi
μσθσθσμθσ +=+= (42)
Which represents the angular distribution of the bremsstrahlung
produced by un-polarized electron in the field of a nucleus with spin
Polarization Dependence of the Bremsstrahlung Cross-Section in the … 73
I.The second term in (42) gives the magnetic correction to Bethe-Heitler
relation for bremsstrahlung. This correction is more effective for the light
nuclei, which has a large magnetic moment at large angles of the photon
emission.
5- Numerical Application to some Nuclei (Results and Discussion)
We shall now consider some applications of the obtained formulas
to particular nuclei. The electric and magnetic scattering cross-sections
for the two nuclei ( )27
13Al and ( )9
4Be at different values of energy
)5000,3000,1000( Mev=ε and angles )60,30,10( 000
=θ , and for the polarized
and unpolarized cases are considered, and we have the following results:
1- the polarized electric cross-sections for the two nuclei are
decreased with increasing both energies and scattering angles, such that
edσ is more effective at energy Mev1000=ε and angle 0
10=θ [Fig.1+3],
for ex:
)10,1000(107982.6 031
==×=−
θεσ Mevatdep
,
)60,5000(1072937.4 036
==×=−
θεσ Mevatdep
2- the polarized magnetic cross-sections for the two nuclei are also
decreased with increasing both energies and scattering angles, such that
edσ is more effective at energy Mev1000=ε and angle 0
10=θ [Fig(2) +
(4)], for ex:
)10,1000(107122.5 019
==×=−
θεσ Mevatdmp
,
)60,5000(1005474.1 022
==×=−
θεσ Mevatdmp
10 20 30 40 50 60
Angle Degrees q
- 1´ 10 - 23
0
1´ 10 - 23
2´ 10- 23
3´ 10- 23
4´ 10- 23
ssor
Cnoit
ceSHds
peL-g
=05.
Polarization Bremsstrahlung Radiation- AL2 7
1 3
He=0001
,0003
,0005
ve
ML
10 20 30 40 50 60
Angle Degrees q
- 1´ 10 - 24
0
1´ 10 - 24
2´ 10 - 24
3´ 10 - 24
4´ 10 - 24
ssor
Cnoit
ceSHds
peL
Polarization Bremsstrahlung Radiation- Be94
He=0001
,0003
,0005
ve
ML
Fig.3 Fig.1
74 S. A. Al-Khateeb
3- comparing the polarized ep
dσ and unpolarized e
dσ electric cross-
sections for the two nuclei ( )27
13Al and ( )9
4Be [Fig(5) + (6)], we see that:
i) the values of cross-sections for ( )27
13Al are larger than the value for
( )94
Be .
ii) the value of polarized ep
dσ electric cross-section for the two
nuclei is larger than the value for the unpolarized e
dσ cross-section, for
ex:
)10,1000(107982.6 031
==×=−
θεσ Mevatde
,
)10,1000(1075822.4 023
==×=−
θεσ Mevatdep
10 20 30 40 50 60
Angle Degrees q
- 1´ 10 - 19
0
1´ 10 - 19
2´ 10- 19
3´ 10- 19
4´ 10- 19
5´ 10- 19
ssor
Cnoit
ceSHds
pmL-g
=05.
Polarization Bremsstrahlung Radiation - AL2 7
1 3
He=0001
,0003
,0005
ve
ML
10 20 30 40 50 60
Angle Degrees q
- 2´ 10 - 20
0
2´ 10 - 20
4´ 10 - 20
6´ 10 - 20
8´ 10 - 20
1´ 10 - 19
ssor
Cnoit
ceSHds
pmL-g
=05.
Polarization Bremsstrahlung Radiation - Be94
He=0001
,0003
,0005
ve
ML
10 20 30 40 50 60
Angle Degrees q
0
2 ´ 10 -31
4 ´ 10 -31
6 ´ 10 -31
ssor
Cnoit
ceSH-
LA
72
31
,-e
B94L
Bremsstrahlung Radiation ds e
He=0001
ve
ML
10 20 30 40 50 60
Angle Degrees q
0
5 ´ 10 -24
1 ´ 10 -23
1.5 ´ 10 -23
ssor
Cnoit
ceSH-
LA
72
31
,-e
B94L
Polarization Bremsstrahlung Radiation ds ep
He=0001
ve
ML
Fig.4 Fig.2
Fig.6 Fig.5
Polarization Dependence of the Bremsstrahlung Cross-Section in the … 75
4- comparing the polarized mp
dσ and un polarized m
dσ magnetic
cross-sections for the two nuclei ( )27
13Al and ( )9
4Be [Fig(7) + (8)],we see
that:
i) the value of cross-sections for ( )27
13Al are larger than the value for
( )94
Be .
ii) the value of polarized mp
dσ magnetic cross-section for the two
nuclei is larger than the value for the un polarized m
dσ cross-section, for
ex:
)10,1000(1073899.4 023
==×=−
θεσ Mevatdm
,
)10,1000(107122.5 019
==×=−
θεσ Mevatdmp
5- comparing the total electric cross-section epe
dddE σσ += and the
total magnetic cross-section mpm
dddM σσ += for the two nuclei [Fig(9) +
(10)],we see that:
The values of E
dσ and M
dσ for the ( )27
13Al nucleus are larger than
that for the ( )94
Be nucleus, and have the largest value at 01000 , 10 .Mevε θ= =
10 20 30 40 50 60
Angle Degrees q
0
1 ´ 10 -23
2 ´ 10 -23
3 ´ 10 -23
4 ´ 10 -23
ssor
Cnoit
ceS
-L
A7
23
1,-
eB
94
Bremsstrahlung Radiation dsm
e=
0001
ve
M
10 20 30 40 50 60
Angle Degrees q
- 1´ 10- 19
0
1´ 10- 19
2´ 10- 19
3´ 10- 19
4´ 10- 19
5´ 10- 19
ssor
Cnoit
ceS
-L
A7
23
1,-
eB
94
Polarization Bremsstrahlung Radiation dsmp
e=
0001
ve
M
Fig.8 Fig.7
76 S. A. Al-Khateeb
6- comparing between the electric and magnetic parts of the cross-
section for the two nuclei [Fig (11) + (12)], where
mpmepedddMdddE σσσσ +=+= , . We see that:
i) the magnetic part is more effective (i.e has larger values) in the
polarized and un polarized cross for the two nuclei.
ii) in the polarized case, the values of cross-sections (electric or
magnetic) are larger than the unpolarized case, which means that the
study of polarization in the bremsstrahlung process gives us best result,
and make the process more evident than that if we neglect the
polarization effect, a result in consistent with the experimental results.
10 20 30 40 50 60
Angle Degrees q
- 1´ 10 - 23
0
1´ 10 - 23
2´ 10 - 23
3´ 10 - 23
4´ 10 - 23
ssor
Cnoit
ceS
EdH-
LA
72
31
,-e
B94L
Polarization Bremsstrahlung Radiation
He=0001
ve
ML
10 20 30 40 50 60
Angle Degrees q
- 1´ 10 - 19
0
1´ 10 - 19
2´ 10 - 19
3´ 10 - 19
4´ 10 - 19
5´ 10 - 19
ssor
Cnoit
ceS
MdH-
LA
72
31
,-
eB
94L
Polarization Bremsstrahlung Radiation
He=0001
ve
ML
10 20 30 40 50 60
Angle Degrees q
0
1 ´ 10 -31
2 ´ 10 -31
3 ´ 10 -31
4 ´ 10 -31
5 ´ 10 -31
6 ´ 10 -31
ssor
Cnoit
ceS
EdH-
LA
72
31
,-e
B94L
Bremsstrahlung Radiation
He=0001
ve
ML
10 20 30 40 50 60
Angle Degrees q
0
1 ´ 10-23
2 ´ 10-23
3 ´ 10-23
4 ´ 10-23
ssor
Cnoit
ceS
MdH-
LA
72
31
,-
eB
94L
Bremsstrahlung Radiation
He=0001
ve
ML
Fig.11 Fig.12
Fig.9Fig.10
Polarization Dependence of the Bremsstrahlung Cross-Section in the … 77
Acknowledgements
The author would Like to thank the Deanship of Scientific
Research at King Abdulaziz University to support this research (no. 21-
h008/429) and encourage researchers to continue the march of scientific
research.
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78 S. A. Al-Khateeb
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