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UNIT 12: ATOMIC STRUCTURE
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12.1 Early Models of the Atom
12.1.1 Thomson’s model of the atom
{
In 1898, Joseph John Thomson suggested a model of an atom that
consists of homogenous positively charged spheres with tiny negatively
charged electrons embedded throughout the sphere as shown in figure
12.1a.
{
The electrons much likes currants in a
plum pudding.
{
This model of the atom is called ‘plum
pudding’ model of the atom.
positively charged
electron
sphere
Fig. 12.1a
12.1.2 Rutherford’s model of the atom
{
In 1911, Ernest Rutherford performed a critical experiment that showed
the Thomson’s model is not correct and proposed his new atomic
model known as Rutherford’s planetary model of the atom as shown in
figure 12.1b.
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{
{
nucleus
electron
{
Fig. 12.1b
{
+Ze
‘plop’
plop’
-e
{
{
According to this model, the atom was
pictured as electrons orbiting around a
central nucleus which concentrated of
positive charge.
The electrons are accelerating because
their directions are constantly changing as
they circle the nucleus.
Based on Maxwell’s electromagnetic
theory, an accelerating charge emits
energy.
Hence the electrons must emit the e.m.
radiation as they revolve around the
nucleus.
As a result of the continuous loss of
energy, the radii of the electron orbits will
be decreased steadily.
This would lead the electrons spiral and
falls into the nucleus, hence the atom
would collapse as shown in figure 12.1c.
energy loss
Fig. 12.1c
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12.2 Bohr’s Model of Hydrogen Atom
{
{
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In 1913, Neils Bohr proposed a new atomic model based on hydrogen
atom.
According to Bohr’s Model, he assumes that each electron moves in a
circular orbit which is centred on the nucleus, the necessary centripetal
force being provided by the electrostatic force of attraction between the
positively charged nucleus and the negatively charged electron as
shown in figure 12.2a.
{
On this basis he was able to show that
the energy of an orbiting electron
-e
depends on the radius of its orbit.
{
This model has several features which
r
r
are described by the postulates
v
FE
(assumptions) stated below :
+e
1. The electrons move only in
certain circular orbits, called
r
STATIONARY STATES or
ENERGY LEVELS.
LEVELS When it is in
one of these orbits, it does not
radiate energy.
energy
Fig. 12.2a
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2. The only permissible orbits are those in the discrete set for which
the angular momentum of the electron L equals an integer times
h/2π . Mathematically,
nh
and L = mvr
2π
nh or
mvr =
mvr = nh
2π
where r : radius of the orbit
m : mass of the electron
L=
(12.2a)
n : principal quantum number = 1,2 ,3 ,...
h : planck constant
h
h =
2π
3. Emission or absorption of radiation occurs only when an electron
makes a transition from one orbit to another.
The frequency f of the emitted (absorbed) radiation is given by
where
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∆E = hf = E f − Ei
E f : final energy state
Ei : initial energy state
(12.2b)
5
Note :
z
If
∆E ⇒ negative value
∆E ⇒ positive value
z If
12.2.1 Bohr’s Radius in Hydrogen atom
{
Emission of e.m.
e.m. radiation.
Absorption of e.m.
e.m. radiation.
Consider one electron of charge –e and mass m moves in a circular
orbit of radius r around a positively charged nucleus with a velocity v
(Figure 12.2a).
{
The electrostatic force between electron and nucleus contributes the
centripetal force as write in relation below :
FE = Fc centripetal force
electrostatic force
{
kq1q2 mv 2
where q1 = e ; q2 = e
=
r2
r2
e
mv 2 =
(12.2c)
4 πε0 r
nh
From the Bohr’s second postulate : mvr =
2π
and
k=
1
4 πε0
By taking square of both side of the equation, we get
m2v 2 r 2 =
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n2h2
4π 2
(12.2d)
6
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{
By dividing eq. (12.2d) and (12.2c), thus
n 2 h 2ε 0
e 2π
n 2 h 2ε 0
r=
; n = 1,2,3,...
me 2π
mr =
{
(12.2e)
which r are radii of the permissible orbits for the Bohr atom.
Eq. (12.2e) can be written as
2
r = a0 n 2 ; n = 1,2,3,...with a0 =
h ε0
me 2 π
Where a0 is called the Bohr’
Bohr’s radius of hydrogen atom.
It is defined as the radius of the lowest orbit or ground state (n=1)
and is given by
2
r = a0 (1) =
2
(6.63 × 10 ) (8.85 × 10 )
(9.11 × 10 )(1.60 × 10 ) π
− 34
− 31
−12
−19 2
a0 = 5.31 × 10 −11 m
{
The radii of the orbits associated with allowed orbits or states n
2,3,… are 4a0,9a0,…, hence the atomic radius is quantized.
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=
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12.2.2 Discrete Energy in Hydrogen atom
{
The total energy E of the system is given by
Kinetic energy of the electron
E = K +U
(12.2f)
Potential energy of the electron
The potential energy, U of the electron is
1
kq1q2
where q1 = e ; q2 = −e and k =
4πε0
r 2
e
(12.2g)
U =−
4 πε0 r
U=
and the kinetic energy, K of the electron is
K=
1 2
mv
2
(12.2h)
By substituting eq. (12.2c) into eq. (12.2h), thus
1  e2 


2  4 πε0 r 
e2
K=
8 πε0 r
K=
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(12.2i)
8
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By substituting eq. (12.2g) and (12.2i) into eq. (12.2f), therefore the
total energy E of the system is

e2 

+  −
8πε 0 r  4πε 0 r 
e2
(12.2j)
E=−
8 πε0 r
Ze 2
In general,
En = −
8 πε0 r
Z
:
atomic
number
where
E=
{
{
e2
n=2
λ = 2πr1
Fig. 12.2b
λ
{
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n=3
2λ = 2πr2
Fig. 12.2c
3λ = 2πr3
Fig. 12.2d
9
λ
n=4
n=5
4λ = 2πr4
5λ = 2πr5
Fig. 12.2e
{
(12.2k)
From de Broglie’s relation, the electron is to be regarded as a wave,
then its stable orbits in an atom are those satisfy the conditions of a
standing (stationary) wave as shown in figures 12.2b, 12.2c, 12.2d ,
12.2e and 12.2f.
λ
λ
λ
orbital
n=1
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Equation of discrete energy
in hydrogen atom only
Fig. 12.2f
If there are n waves in the orbital and λ is wavelength of wave
properties of electron therefore
nλ = 2πrn
(12.2l)
th
where rn : radius of the n orbit
n = 1,2 ,3,...
h
Since λ =
then eq. (12.2l) can be written as
mv
 h 
n
 = 2πrn
 mv 
nh
mvrn =
Bohr’
Bohr’s second postulates
2π
10
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{
The energy level of hydrogen atom is given by
n 2 h 2 ε0
Ze 2
where Z = 1 and r =
me 2 π
8 πε0 r (hydrogen atom)
(1)e 2
En = −
 n 2 h 2 ε0 

8 πε0 
2
 me π 
me 4
me 4
and
E
=
= 13.6 eV
En = − 2 2 2
1
8 ε02 h 2
8 ε0 h n
13.6 eV
E
(12.2m)
En = − 21 or En = −
n2
n
th
where En : energy level of the n orbit(state)
E1 : energy level of ground state
n = 1,2 ,3,...
En = −
The negative sign means that work has to be done to remove the
electron to infinity, where it is considered to have zero energy, i.e. the
electron is bound to the atom.
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12.3 Energy Level of Hydrogen Atom
{
{
The energies of the electron in an atom can have only certain values.
This values are called the energy level of the atom.
The energy level of hydrogen atom can be calculated by using eq.
(12.2m) which is
13.6 eV
En = −
When n
n2
; n = 1,2,3,...
= 1, the ground state (the state of the lowest energy level)
level
13.6 eV
E1 = −
= −13.6 eV
(1)2
13.6 eV
= -3.39 eV
When n = 2, the first excited state, E2 = −
2
2
13.6 eV
n = 3, the second excited state,E3 = −
= −1.51 eV
32
13.6 eV
n = 4, the third excited state, E4 = −
= -0 .85 eV
42
13.6 eV
electron is completely
n = ∞, E ∞ = −
=0
2
removed from the atom.
(∞ )
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{
Figure 12.3a shows diagrammatically the various energy levels in the
hydrogen atom.
n
∞
En (eV )
0.0 Free electron
5
4
3
− 0.54
− 0.85
− 1.51
4th excited state
3rd excited state
2nd excited state
− 3.39
1st excited state
Ionization energy
is defined as the
energy required
by an electron in2
the ground state
to escape
completely from
the attraction of
the nucleus.
An atom
becomes
ion.
1
Excitation energy
is defined as the energy
required by an electron
that raises it to an excited
state from its ground state.
− 13.6
Ground state
Fig. 12.3a
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{
excited state
is defined as the
energy levels that
higher than the
ground state.
is defined as the
lowest stable
energy state of an
13
atom.
Example 1 :
The electron in the hydrogen atom makes a transition from E2=-3.40 eV
energy state to the ground state with E1= -13.6 eV. Calculate
a. the energy in Joule, and
b. the wavelength of the emitted photon.
(Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1)
Solution:
a. The energy of the emitted photon is
∆E = E f − Ei
∆E = E1 − E2
∆E = (− 13.6 ) − (− 3.40 )
∆E = −10.2 (1.60 × 10 −19 )
∆E = 1.63 × 10 −18 J
b. The wavelength of the photon is
hc
λ
hc
λ=
∆E
λ = 1.22 × 10 −7 m
Negative sign
means the energy
is emitted.
∆E =
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{
Example 2 :
The lowest energy state for hydrogen atom is E1= -13.6 eV. Calculate
the frequency of the photon required to ionize the atom.
(Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1)
Solution:
An atom where its electron at ground state is raised to the zero energy
level said to be ionized. The ionization energy is
∆E = E f − Ei
∆E = E∞ − E1
∆E = (0 ) − (− 13.6 )
∆E = 13.6 (1.60 × 10 −19 )
∆E = 2.18 × 10 −18 J
The frequency of the photon required to ionize the atom is
∆E = hf
∆E
f =
h
f = 3.29 × 10 15 Hz
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{
Example 3 :
A hydrogen atom emits radiation of wavelengths 121.5 nm and 102.4 nm
when the electrons make transitions from the 1st excited state and 2nd
excited state respectively to the ground state.
Calculate:
a. the energy of a photon for each of the wavelengths above;
b. the wavelength emitted by the photon when the electron makes a
transition from the 2nd excited state to the 1st excited state.
(Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1)
Solution: λ1=121.5x10-9 m, λ2=102.4x10-9 m
a. The energy of the photon due to transition from 1st excited state to the
hc
ground state is
∆E1 =
λ1
∆E1 = 1.64 × 10 −18 J
The energy of the photon due to transition from 2nd excited state to the
hc
ground state is
∆E2 =
λ2
∆E2 = 1.94 × 10 −18 J
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b.
∆E3
∆E1
∆E2
2nd excited state
1st excited state
Ground state
∆E3 = ∆E2 − ∆E1
∆E3 = 3.00 × 10 −19 J
Therefore the wavelength of the emitted photon due to the transition
from 2nd excited state to the 1st excited state is
hc
λ3
λ3 = 6 .63 × 10 −7 m
∆E3 =
{
Example 4 :
The electron of an excited hydrogen atom make a transition from the
ground state to the 4th excited state. Determine the energy absorbs by
the atom.
(Given h = 6.63 x 10-34 J s and c = 3.00 x 108 m s-1)
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Solution:
a. By applying the equation of energy level in hydrogen atom,
13.6 eV
n2
13.6 eV
Ground state, n = 1
E1 = −
12
E1 = 13.6 (1.60 × 10 −19 ) J
E1 = 2.18 × 10 −18 J
13.6 eV
E5 = −
4th excited state, n = 5
52
E5 = 0.54 (1.60 × 10 −19 ) J
E5 = 8.64 × 10 −20 J
En = −
Hence the energy absorbs by the atom is
∆E = E f − Ei
∆E = E5 − E1
∆E = 2.09 × 10 −18 J
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12.4 Emission Line Spectrum
{
{
The emission lines correspond to photons of discrete energies that are
emitted when excited atomic states in the gas make transitions back to
lower energy levels.
Figure 12.4a shows line spectra produced by emission in the visible
range for hydrogen (H), mercury (Hg) and neon (Ne).
Fig. 12.4a
12.4.1 Hydrogen Emission Line Spectrum
{
Emission processes in hydrogen give rise to series, which are
sequences of lines corresponding to atomic transitions as shown in
figure 12.4b.
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n
∞
5
4
3
2
19
E
n ( eV )
0. 0 Free electron
− 0. 54 4th excited state
*Pfund series
− 0. 85 3rd excited state
*Brackett series
− 1. 51 2nd excited state
Paschen series
involves transitions that end with the Infrared
range
2nd excited state.
−
3.
39
1st excited state
Balmer series
involves transitions ending with the Visible
light range
1st excited state of hydrogen.
Lyman series
Ultraviolet
involves transitions that end with the
range
ground state of hydrogen.
− 13 . 6 Ground state
1
Fig. 12.4b
*Brackett series involves transitions that end with the
3rd excited state.
Infrared
*Pfund series involves transitions that end with the range
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Simulation
4th excited state.
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{
Figure 12.4c shows “permitted” orbits of an electron in the Bohr model
of a hydrogen atom.
Fig. 12.4c : not to
scale
{
Figure 12.4d illustrates the lines spectrum of Balmer series for
hydrogen atom.
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Fig. 12.4d
12.4.2 Wavelength of Emission Line Spectrum for Hydrogen atom
{
{
If an electron makes a transition from an outer orbit (ni) to an inner
orbit (nf), energy is radiated.
The energy radiated can be calculated by using equation below:
∆E = E f − Ei = En f − Eni
Emission of e.m.
e.m.
radiation (you
can ignore it)
{
∆E = −
me4  1
1 
− 2
2
2 2 
8 εo h  n f
ni 
(12.4a)
Hence the wavelength of the photon emitted (energy radiated in form of
e.m. radiation) is shown below.
Since
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
me4  
me4 


∆E = − 2 2 2 −  − 2 2 2 
 8ε h n 
8 εo h ni 
o
f  

∆E =
hc
then eq. (12.4a) can be written as
λ
hc
me4  1
1 
= 2 2 2− 2
λ 8 εo h  n f
ni 
22
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1
λ
=
me4  1
1 
− 2
2
2
3 
8 εo ch  n f
ni 
 1
1 
= RH  2 − 2 
n
λ
ni 
 f
1
where
{
RH : Rydberd constant
n f : final value of n
ni : initial value of n
Note:
z
For Lyman series (nf=1) :
z
For Balmer series (nf=2) :
z
For Paschen series (nf=3) :
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z
For Brackett series (nf=4) :
z
For Pfund series (nf=5) :
(12.4b)
RH =
me4
= 1.0974 × 107 m −1
2
3
8ε 0 ch
1
1 
= RH  2 − 2 
λ
 1 ni 
 1
1
1 
= RH  2 − 2 
λ
 2 ni 
1
 1
1 
= RH  2 − 2 
λ
 3 ni 
1
23
 1
1 
= RH  2 − 2 
λ
ni 
4
 1
1
1 
= RH  2 − 2 
λ
 5 ni 
1
To calculate the shortest wavelength in any line series, ni=∞.
Example 5 :
Determine the wavelength for a line spectrum in Lyman series when
the electron makes a transition from n=4 level.
(Given Rydberg constant ,RH = 1.0974 x 107 m s-1)
z
{
Solution:
ni=4
By applying the equation of wavelength for Lyman series (nf=1), thus
1
1 
= RH  2 − 2 
λ
 1 ni 
1
1 1
= 1.0974 × 107  2 − 2 
λ
1 4 
−8
λ = 9.72 × 10 m
1
(
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)
24
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{
Example 6 :
The Balmer series for the hydrogen atom corresponds to electronic
transitions that terminate at energy level n=2 as shown in the diagram
below.
E (eV )
n
n
∞
6
5
4
3
0. 0
− 0. 38
− 0. 54
− 0. 85
− 1. 51
2
− 3. 40
Calculate
a. the longest-wavelength, and
b. the shortest-wavelength photon emitted in this series.
(Given Rydberg constant ,RH = 1.0974 x 107 m s-1)
Solution: nf=2
The equation of wavelength for Balmer series is
 1
1 
= RH  2 − 2 
λ
 2 ni 
1
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a. The longest-wavelength photon results from the transition from
n = 3 to n = 2 (Balmer series). Thus
1
λmax
 1 1
= RH  2 − 2 
2 3 
∆Emin = E f − Ei = E2 − E3
∆Emin = (− 3.40 ) − (− 1.51)
∆Emin = −1.89 × 1.60 × 10 −19
∆Emin = 3.02 × 10 −19 J
hc
= 3.02 × 10 −19 J
or
λmax = 6.56 × 10 −7 m
λmax
b. The shortest-wavelength photon results from the transition from
n = ∞ to n = 2 (Balmer series). Thus
1
λmin
1 
 1
= RH  2 − 2 
2 ∞ 
λmin = 3.65 × 10 −7 m
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or
∆Emax
∆Emax
∆Emax
∆Emax
hc
λmin
= E f − Ei = E2 − E∞
= (− 3.40 ) − (0 )
= −3.40 × 1.60 × 10 −19
= 5.44 × 10 −19 J
= 5.44 × 10 −19 J
26
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12.5 Many Electrons Atomic Model
12.5.1 Limitations of Bohr’s Model
{
The Bohr’s theory
z
predicts successfully the energy levels of the hydrogen atom but
fails to explain the energy levels of more complex atoms.
z
can explain the spectrum for hydrogen atom but some details of the
spectrum cannot be explained especially when the atom is placed
in a magnetic field.
z
cannot explain the Zeeman effect.
{
Zeeman effect is defined as the splitting of spectral lines
when the radiating atoms are placed in a magnetic field.
{
Figure 12.5a shows the zeeman effect.
2p
Energy Levels
Transitions
1s
No magnetic field
Magnetic field
Fig. 12.5a
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Spectra
27
12.5.2 Utilises 4 quantum numbers:
{
Principal quantum number, n:
Specifies the main energy level and corresponds to the number n in
the Bohr atomic model where n = 1,2,3,...
{
Orbital (azimuthal
(azimuthal)) quantum Number, l :
Specifies the angular momentum of electron, where
l = 0,1,2,..., (n − 1)
For n
= 1, l = 0
For n = 2, l = 0 and 1
For n = 3, l = 0,1 and 2
{
Magnetic quantum number, ml :
Specifies the orientation of the electron orbitals in a magnetic field and
plays an important part in explaining the Zeeman effect
E.g : a. l
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ml = −l ,−(l − 1),...,0,1,..., (l − 1),+l
= 0, ml = 0
b. l = 1, ml = -1,0,+1
c. l = 2, ml = -2,-1,0,+1,+2
28
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Spin quantum number, ms:
Electron spins when put in a magnetic field and have a magnetic
moment because of this spin. Its value is given by
Specify whether the spin aligned
1
ms = ±
with, or counter to, an applied
2
magnetic field.
12.5.3 Pauli Exclusion Principle
{
States “No
No two electrons in an atom can have the same set of four
quantum numbers.”
numbers.
{
It is needed in describing the electron configuration of multi-electron
atoms.
{
Example 7 :
Write down the set of quantum number for electron with principal
quantum number n=2.
Solution:
{
When n=2; l = 0,1; ml=0,-1,0,+1; ms = ±½, ±½, ±½, ±½
The following set of quantum numbers are, therefore, possible :
(2,0,0,+½) ; (2,0,0,-½)
(2,1,-1, +½); (2,1,-1, -½)
(2,1,0, +½); (2,1,0, -½)
(2,1,1, +½); (2,1,1, -½)
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or in table form :
{
Table 12.5a
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There are 8 electron states
with n = 2 and these form
the L shell
29
n
2
l
0
1
ml
0
-1
0
+1
m s +½ - ½ +½ - ½ +½ - ½ +½ - ½
Note :
z
Table 12.5a shows the electron quantum states in the first four shell.
Number of
electron
n
l
ml
Subshell
1
0
0
1s
2
2
2
0
1
0
-1,0,1
2s
2p
2
6
3
0
0
3s
2
3
1
-1,0,1
3p
6
3
2
-2,-1,0,1,2
3d
10
4
0
0
4s
2
4
1
-1,0,1
4p
6
4
2
-2,-1,0,1,2
4d
10
4
3
-3,-2,-1,0,1,2,3
4f
14
Shell
K
8
L
18
M
32
N
30
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z
The total number of electrons in all the shells equal to Z, the atomic
number. Hence the electron configuration for various elements can
be written as shown below in table 12.5b.
Element
Atomic
Number (Z)
He
2
1s2
Be
4
1s22s2
C
6
1s22s22p2
O
8
1s22s22p4
Ne
10
1s22s22p6
Mg
12
1s22s22p63s2
Si
14
1s22s22p63s23p2
S
16
1s22s22p63s23p4
Ar
18
1s22s22p63s23p6
Ca
20
1s22s22p63s23p64s2
Electron Configuration
Table 12.5b
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31
z
Figure 12.5b shows the energy levels for various subshell.
4p
Energy
level
3d
4s
3p
3s
Maximum
electron
in shell
2p
n = 2 2s
n = 1 1s
Fig. 12.5b
{
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Example 8 : (exercise)
Write down the set of quantum number of electron in form of table for
nitrogen atom with atomic number, Z=14.
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THE END…
Next Unit…
UNIT 13 :
X-rays
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