Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Chapter One Atomic Structure and Energy Levels
1-1 Atomic Model
In order to explain many phenomena associated with material we must know
the atom structure . Atomic models were proposed to explain the distributions of the
charged particles in an atom.
1-2 Early Models for Atom
The model of the atom in the days of Newton was a tiny, hard, indestructible
sphere. Although this model was a good basis for the kinetic theory of gases, new
models had to be devised when later experiments revealed the electronic nature of
atoms.
J. J. Thomson, in 1898,
proposed that an atom possesses a
spherical shape (radius
approximately 10–10 m) in which
the positive charge is uniformly
distributed. The electrons are
embedded into it in such a manner
as to give the most stable
electrostatic arrangement .Many
different names are given to this
model, for example, plum pudding, raisin pudding
or watermelon. This model can be visualized as a
pudding or watermelon of positive charge with
plums or seeds (electrons) embedded into it. An
important feature of this model is that the mass of
the atom is assumed to be uniformly distributed
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
over the atom. Although this model was able to explain the overall neutrality of the
atom, but was not consistent with the results of photoelectric effect , spectrum of
hydrogen atom , X-ray , …… .
In 1911 Ernest Rutherford and his students Hans Geiger and Ernest Marsden
performed a critical experiment showing that Thomson’s model couldn’t be correct.
In this experiment he found that: the atom consists of a nucleus of positive charge that
contains nearly all the mass of the atom. Surrounding this central nucleus negatively
charged electrons .
There are two basic difficulties with Rutherford’s planetary model. First, an
atom emits certain discrete characteristic frequencies of electromagnetic radiation and
no others; the Rutherford model is unable to explain this phenomenon. Second, the
electrons in Rutherford’s model undergo a centripetal acceleration. According to
Maxwell’s theory of electromagnetism, centripetally accelerated charges revolving
with frequency f should radiate electromagnetic
waves of the same frequency. Unfortunately,
this classical model leads to disaster when
applied to the atom. As the electron radiates
energy, the radius of its orbit steadily decreases
and its frequency of revolution increases. This
leads to an ever-increasing frequency of emitted
radiation and a rapid collapse of the atom as the electron spirals into the nucleus.
1-3 Bohr's Models for Hydrogen Atom
At the beginning of the 20th century, scientists were perplexed by the failure
of classical physics to explain the characteristics of spectra. Why did atoms of a given
element emit only certain lines? Further, why did the atoms absorb only those wave-
lengths that they emitted?
Neils Bohr (1913) was the first to explain quantitatively the general features of
Hydrogen atom structure and its spectrum. Bohr’s model for Hydrogen atom is based
on the following postulates:
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
1. The atom has a massive positively-charged nucleus.
2. The electron in the hydrogen atom can move around the nucleus in a circular
path of fixed radius and energy. These paths are called orbits, energy level or
energy states. These orbits are arranged concentrically around the nucleus.
3. The electrons revolve in these orbits, influence of the Coulomb force of
attraction (between the nucleus and electron) which it's balanced by the
centrifugal force for revolving electron.
4. An electron cannot revolve round the nucleus in any arbitrary orbit but in just
certain definite quantization
orbits. Only those orbits are
possible (or permitted) for
which the Hydrogen atom
doesn’t emit energy in the
form of electromagnetic
radiation. Hence, the total
energy of the atom remains
constant. Such orbits are also
known as stationary orbits
(also called Bohr's stationary
orbits). those orbits are possible (or permitted) for which the orbital angular
momentum of the electron is equal to an integral multiple of h/2 i.e.
Angular momentum of electron in the nth orbit is
m vn rn=nh2 π
where n is an integer (n= 1, 2, 3 etc. for the first, second and third
orbits respectively.
5. Let the different permitted orbits have energies of E1, E2, E3 etc. The electron
can be raised from n = 1 orbit to any other higher orbit if it is given proper
amount of energy equal to energy difference between these two orbits.
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
6. Radiation is emitted by the hydrogen
atom when the electron “jumps” from a
more energetic initial state to a less
energetic state. The “jump” can’t be
visualized or treated classically. In
particular, the frequency f of the
radiation emitted in the jump is related
to the change in the atom’s energy and
is independent of the frequency of the
electron’s orbital motion . The
frequency and wavelength of the
emitted radiation is given by:
f =Ei−E f
hλ= hc
Ei−E f
where Ei is the energy of the initial state, E f is the energy of the final state, h is
Planck’s constant, and Ei>E f .
1-4 De-Broglie's Hypothesis for Bohr's Stationary Orbits
For more than a decade following Bohr’s publication, no one was able to
explain why the angular momentum of the electron was restricted to these discrete
values. Finally, de Broglie (1924) gave a direct physical way of interpreting this
condition. In his doctoral dissertation in 1924, Louis de Broglie postulated that,
because photons have wave and particle characteristics, perhaps all forms of matter
have both properties . This was a highly revolutionary idea with no experimental
confirmation at that time. According to de Broglie, all moving particles, such as
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
electrons, atoms, molecules,….., having an associates wave called De-Broglie wave
with wavelength:
λd=h
mv
where v is the velocity of the particle. And the momentum
P=mv= hλd
De Broglie's idea was that, if light can have both a particle and a wave nature,
perhaps electrons can too! Perhaps the quantization of the angular momentum of an
electron in the hydrogen atom was due to the wave nature of the electron. An electron
with a linear momentum p=mv would have a wavelength λd =h/p. This is now called
the de Broglie wavelength . This relationship applies not only to photons and
electrons, but as far as we know, to all particles!
With a formula for the electron wavelength, de Broglie was able to construct a
simple model explaining the quantization of angular momentum in the hydrogen
atom. In de Broglie's model, one pictures an electron wave chasing itself around a
circle in the hydrogen atom. If the circumference of the circle, 2π r did not have an
exact integral number of wavelengths, then the wave, after going around many times,
would eventually cancel itself out as illustrated in Figure (1-A). But if the
circumference of the circle were an exact integral number of wavelengths as
illustrated in Figure (1-B), there would be no cancellation. This would therefore be
one of Bohr's allowed orbits.
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Figure (1) de-Broglie wave length of electron in Hydrogen atom
Suppose (n) wavelengths fit around a particular circle of radius rn as shown in
Figure(2),then we have:
n λd=2 π rn
Using the de Broglie formula λd =h/mv for the electron wavelength, we get:
nhm vn
=2 π rn m vn rn=nh2 π
Figure(2) de-Broglie wavelength of electron in each orbit
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Normal, Excited and Ionized Atom
When electron in its inner most orbit (n = 1), then the atom is said to be in its
normal (or unexcited) state. Generally, if atom possesses proper energy (by heating
or electrical voltage or ….)Then electron jump from unexcited orbit (n=1) to higher
(excited) permitted orbits having n = 2, 3, 4 etc. This atom called excited atom. When
the electron is completely removed from the atom (due to high energy), the atom
is said to be ionized.
1-5 Atomic Energy Level Diagram
Instead of drawing various electron orbits to the scale of their radii as
in Fig. (3 a), it is customary to draw horizontal lines to an energy scale as shown in
Fig. (3 b) and such a diagram is called energy level diagram of an atom. In this array
of energies, the higher energies are at the top while the lower energies are at the
bottom. The various electron jumps between allowed orbits now become vertical
arrows between energy levels.
Notes:
1-Usually energy in atom is measured in eV instate of Joule where:1 eV=1 . 6×10−19 Joules
2- The first level (n=1) called ground state or unexcited state.
3- The above states (n>1) called excited state where first excited state at n=2 and second exited state at n=3 and so on.
4- Ionized state at n=∞ where energy level =0eV.
Therefore ionized energy (absorbed energy that made atom ion) will be E∞−E 1=−E 1
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Example 1-1 : The six lowest energy levels of Sodium vapor are:{ -15, -12.9, -
11.81, -11, -10.9 and -10.74eV}. When a photon of wavelength 3300Ao is observed, a
three other photons are emitted. Find wave length of these three emitted photons.
Solution
Energy of observed photon is Ephoton=hf =hc
λ
=6 .62×10−34×3×108
3300×10−10=6.02×10-19J=3.76eV
This energy observed by an electron in ground state (whose energy =-15eV) therefore
the new energy of electron =-15+3.76=-11.24eV. This new energy made electron
jump up to new position lie between third and fourth level, therefore the electron must
jump down to third level, that made the photon will emit the energy difference where:
ΔE1=(-11.24)-(-11.81)=0.57eV=9.12×10-20 J
Wave length of first photon is
8
E2=-12.9
E1=-15
E3=-11.81
E4=-11Enew=-11.24 ΔE1
ΔE2
ΔE3
First photon
Third photon
Secondphoton
Incident photon
Illustration for example 1-1
Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
λ 1= hcΔE 1
=6 . 62×10−34×3×108
9 .12×10−20=2.177×10-6m
The second photon will emit when electron jump from third level to second level
where:
ΔE2=(-11.81)-( -12.9)=1.09 eV=1.744×10-19 J
Wave length of second photon is:
λ 2= hc
ΔE 2=6 . 62×10−34×3×108
1.744×10−19=1.138×10-6m
The third photon will emit when electron jump from second level to ground level
where:
ΔE3=( -12.9)-(-15)=2.1 eV=3.36×10-19 J
Wave length of third photon is
λ 3= hcΔE3
=6 .62×10−34×3×108
3. 36×10−19=5.91×10-7m
1-6 Hydrogen Atom Energy Level Diagram
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
The above postulates concerning Bohr's atomic model can be utilized to
calculate not only the radii of different electron orbits but also the velocity and orbital
energy possessed by different electrons. We will use Hydrogen atom as example to
illustrate the energy levels because it contain a single electron in its orbit.
1-6-1 Radius of Orbits
Now, the stability of the atom requires that the centrifugal force acting on the
revolving electron be balanced by the electrostatic pull exerted by the positively
charged nucleus on the electron.
Centrifugal force = pull force (Coulomb's force)
m v2
r= e2
4 π ϵ0 r2
Where:
ϵ 0: permittivity of free space and e: electron charge
Also, according to Bohr's postulates, mvr = nh/2 .
Now we will substitute v= nh
2 π mr in above equation we will obtain :
rn=( h2ϵ 0
πme2 )× n2
From this equation we can see that value of possible radii is discrete then the above
equation can be rewrite:
rn=r1× n2
Where rn is the radius of nth orbit and r1 is the radius of first orbital =0.52×10-10 m .
1-6-2 Velocity of Electron in the Orbits
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
In the same way by substituting equation of radius in the Bohr's equation we
will find:
vn=e2
2 h ϵ 0× 1
n=v1×( 1
n )Where vn is the velocity of revolving electron in the nth orbit and v1 is the velocity of
electron in the first orbital = 2 .1818×106 m /sec
1-6-3 Electron Energy for Each Orbit
There two type of energy for electron in Hydrogen atom:
1- Kinetic Energy (K.E=12
mv2
) since electron is a moving mass.
By substituting vnin the K.E equation we will have
K . E= me4
8 h2ε 02×
1n2
Joules
2- Potential Energy (P.E) since electron is a negative charge move in electrical
field of positive nucleus.
P .E= −e2
4 πε0 r by substituting rn in this equation we will obtain
P . E=−me4
4 h2ε02 × 1
n2 Joules
Therefore the total energy will be:
En=P. E+K . E=−m e4
8 h2 ϵ02 × 1
n2
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En=−21 .7×10−19
n2 =E1
n2 Joule
Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Where En is the total energy of electron in the nth orbital or energy of the nth
level Energy can found in term of e.V:
Now we can found energy for each orbital
E1 is the energy of first orbit (ground or stable state)= =-13.6eV.
E2 is first excited level is =-3.4eV,
.
.
E∞=0eV
That made ionized energy for Hydrogen atom = E∞- E1=13.6eV
The energy level diagram of Hydrogen atom is shown in the Figure below .
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En=−13.6
n2 eV
Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
What does the negative electronic energy (En) for hydrogen atom mean?
The energy of the electron in a hydrogen atom has a negative sign for all possible
orbits . What does this negative sign convey? This negative sign means that the
energy needed to free electron from atom . As example in the hydrogen atom
electron in a ground state energy of has energy -13.6eV. This means that the electron
needs at least 13.6eV of kinetic energy in order to become free of the nucleus.
As the electron gets closer to the nucleus (as n decreases), En becomes larger
in absolute value and more and more negative. The most negative energy value is
given by n=1 which corresponds to the most stable orbit. We call this the ground
state.
1-7 Spectrum Lines of Hydrogen
The spectrum of radiation emitted by a substance that has absorbed energy is
called an emission spectrum. Atoms, molecules or ions that have absorbed radiation
are said to be “excited”. To produce an emission spectrum, energy is supplied to a
sample by heating it or irradiating it and the wavelength (or frequency) of the
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
radiation emitted, as the sample gives up the absorbed energy, is recorded as shown in
Figure below.
The light emitted by a sample of excited atoms (or any other element) can be
passed through a prism and separated into certain discrete wavelengths. Thus an
emission spectrum, which is a photographic recording of the separated wavelengths is
called as line emission spectrum.
Line emission spectra are of great interest in the study of electronic structure.
Each element has a unique line emission spectrum. The characteristic lines in
atomic spectra can be used in chemical analysis to identify unknown atoms in the
same way as finger prints are used to identify people.
For Hydrogen atom line
spectrum can be calculated by using
Bohr's atomic model . If the electron
jumps from initial energy state ni with
energy Ei into final state energy state n f
with energy E f (where ni>n f ), it emits
a photon of wavelength λ given by
λ= hcE i−E f
Wave number can define as reciprocal
of wavelength
1λ=
Ei−E f
hc
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Where Ei=−m e4
8h2 ϵ 02 × 1
n i2 and E f=
−m e4
8h2ϵ 02 × 1
n f2 that made:
1λ= −m e4
8ch3 ϵ 02 ×( 1
ni2 −
1nf
2 )
¿ RH ( 1nf
2−1ni
2 ) Where RH Rydberg constant
RH =m e4
8 ch3 ϵ 02=1.1×107 m-1
The Swedish spectroscopes, Johannes
Rydberg, drive this expression of all
series of lines in the hydrogen
spectrum.
We can use this expression to
evaluate the wavelengths for the
various series in the hydrogen
spectrum as shown in Figure .
For example, in the Balmer
series (Johann Balmer,1884, Swiss mathematics teacher ), n f=2 and ni=2,3,4 , ….
that made emitted wavelength given by:
λ=364.56( n i2
ni2−4 )nm
The Balmer series of lines are the only lines in the hydrogen spectrum which
appear in the visible region of the electromagnetic spectrum. The observed lines have
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
wavelength 656.21nm (Red), 486.07nm (blue/green) , 434.01nm (blue/violet) and
410.12 nm (violet)
In the same way series of lines that correspond to n f=1,3,4,5 are known as
Lyman, Paschen, Bracket and Pfund series, respectively, as give in Table
Example1-2
An electron in Hydrogen atom has Kinetic energy of 2.416×10-19 J, drop to ground
state.
Calculate:
1- Radius and velocity of electron orbit before dropping into ground state.
2- Wavelength of the emitted radiation due to dropping.
3- The ionized energy of Hydrogen.
4- Total energy of electron at fourth excited level.
Solution
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
1-K.E n=
21 .7×10−19
n2 → n=
⌈√21 . 7×10−19
2 .416×10−19 ⌉= 3 electron at 3rd level
Radius is r3=0.529×10-10×(3)2 =4.761×10-10 m
And velocity is v3 =
2. 1818×106
3 =
2- At 3rd level electron energy is E3=
−13 .6 eV32
=-1.511eV
When electron drop to ground state its new energy will be -13.6eV
Then:
λ= hcE3−E1
= 6 . 626×10−34×3×108
(−1 .511−(−13 .6) )×1. 6×10−19=1.027×10-7m
3- Ionized energy =E∞−E1=−E1 =13.6 eV
4- Forth excited state → n=5 →E5=
−13 .6 eV52
×1.6×10−19
=
Example1-3
An electron in Hydrogen atom at ground state absorbs photon with energy of 2.04×10-
18J.
Calculate:
a- The orbital radius and electron energy at excited state.
b- How many spectral lines are emitted if electron made all possible transition in
its return to ground state?
c- The highest and lowest frequencies of these spectral lines.
Solution
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
a- The electron new energy will be
En =E1+E(photon)
En=-13.6eV+
2. 04×10−18
1.6×10−19=-0.85=
−13. 6 eVn2
→n=4, electron at 4th level
Radius is r4=0.529×10-10×(4)2=
b- There are 4 possible ways to jump from 4th state to ground state
First way (4) → (3 ) → (2 ) → (1 )
Second way (4) → (2 ) → (1 )
Third way (4) → (3 ) → (1 )
Fourth way (4) → (1 )
Therefore there are 6 different possible spectra lines can be emitted.
b- Frequency of emitted radiation (
ΔEh ) depend on energy difference between
these level
Maximum radiation frequency =ΔE (max )
h =
E4−E1
h =3.078×1015 Hertz
Minimum radiation frequency =ΔE (min )
h =
E4−E3
h =1.593×1014 Hertz
Tutorial QuestionQ\1: Explain the main weakness points in nuclear atomic model
proposed by Rutherford and how could Bohr able to solve them.
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Q\2: Explain the meaning Bohr's stationary orbits, and how could de
Broglie able to explain why the angular momentum of these orbits
should be discrete values.
Q\3: Explain De-Broglie's Hypothesis for Bohr's Stationary Orbits.
Q\4: Explain the meaning of negative electron energy for Hydrogen
atom and why first level is the most stable level.
Q\5: The six lowest energy levels of Sodium vapor are [-15, -12.9, -11.81, -11, -10.9 and -10.74eV]. When a photon of observed, a two other photons are emitted with wavelength (1.138 and 0.591 )µm, Find :1. Wavelength of observed photon.2. Find velocity of electron in of 3rd exited level If potential energy is
-22eV.3. Lowest and highest frequencies could be emitted when electron in 3 rd
excited state drop into ground state.
Q\6: Cold Mercury vapor is bombarded with radiation and as a result
the fluorescent line 2.537 and 4.078A0 appear. What is the wavelength
have been present in the bombarding photon ?
Q\7: An electron in Hydrogen atom with De-Broglie wavelength=9.690A0 drop into ground state , Find:
1- The orbital radius and electron energy before dropping into ground state.
2- How many spectral lines are emitted if electron made all possible transition in its return to ground state?
3- The highest and lowest frequencies of these spectral lines.4- Ionized energy of Hydrogen atom.
Q\8: An electron in Hydrogen atom has centrifugal force 1.0109×10-9
N, find:
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
1- Number of revolution /sec for the electron
2- The longest photon wavelength would be required to ionize atom.
3- Number of spectral lines if electron made all possible transition in
its return to ground state?
Q\9: For an electron revolving in the Hydrogen atom, find number of
revolution /sec for first three orbits. (Note: number of revolution per
second = velocity of electron in this orbital /circumference of orbit)
Q\10: What is the wavelength of photon that able to ionize a Hydrogen
atom in the normal state and give the ejected electron a kinetic energy of
3.4 eV.
Q\11: Prove that wavelength of spectrum lines in the Balmer series are
given by:
Then calculate the wave-number for the longest emitted wavelength in the Balmer series.Q\12: Prove that for Hydrogen atom kinetic energy for any energy level is given by :
Q\13: Prove that De-Broglie wavelength for electron revolving in
Hydrogen atom is given by :
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Chapter ONE Atomic Structure and Energy Levels First year class Sameer Abdul Kadhim
Q\14: Prove that the time for one revolution of electron in the Hydrogen atom is:
T= √me2
4 √2 ε° (−En )1.5
Where Enis the total energy ( in Joule) of the nth level.
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