1. Bohr's model of the atom describes electrons orbiting the nucleus in fixed, quantized energy levels. 2. Light is emitted when an electron moves from a higher to lower energy level, with frequency determined by Planck's equation. 3. Hydrogen emits a line spectrum due to its quantized energy levels. The Rydberg equation calculates wavelengths from transitions between levels in Lyman, Balmer, Paschen, Brackett, and Pfund series.

1. Atomic Structure
Chapter 3

2. AT THE END OF THE LECTUREAT THE END OF THE LECTURE
STUDENTS SHOULD BE ABLE TO:STUDENTS SHOULD BE ABLE TO:
(a)(a) Describe Bohr’s Atomic postulatesDescribe Bohr’s Atomic postulates
(b)(b) Explain the existence of electron energy levelsExplain the existence of electron energy levels
in an atom.in an atom.
(c)(c) Describe the formation of line spectrum ofDescribe the formation of line spectrum of
hydrogen atom and calculate the energy ofhydrogen atom and calculate the energy of
electron.electron.

3. History of atom’s theoryHistory of atom’s theory
 John Dalton- reintroduced the atom’s concept inJohn Dalton- reintroduced the atom’s concept in
early 19early 19thth
century.century.
 J.J. Thomson & Earnest Rutherford- developedJ.J. Thomson & Earnest Rutherford- developed
the theory with the discovery of electrons.the theory with the discovery of electrons.
 1897- Thompson found that cathode rays1897- Thompson found that cathode rays
consists of –vely charged electrons.consists of –vely charged electrons.
 Goldstein- discover the protons.Goldstein- discover the protons.
 1911, Rutherford came out with his1911, Rutherford came out with his
Rutherford’s atomic modelRutherford’s atomic model

4. ELECTROMAGNETICELECTROMAGNETIC
RADIATIONRADIATION
 Is the emission and transmission of energy inIs the emission and transmission of energy in
the form of electromagnetic wavesthe form of electromagnetic waves
 It has a particles calledIt has a particles called PHOTONS.PHOTONS.
 It’s exhibit wavelikeIt’s exhibit wavelike behaviorbehavior
 They can be characterized byThey can be characterized by wavelength,wavelength, λλ andand
aa frequency,frequency, νν
 It has differentIt has different λλ andand νν but it move with samebut it move with same
constantconstant speed of light, c, (3.00 x 10speed of light, c, (3.00 x 1088
msms-1-1
)) in ain a
vacuum.vacuum.

6.  Frequency of electromagnetic waves is directly
proportional to its energy.
 The light emitted when electron moves from an
orbit with a larger n value to one with smaller n
value is photon

7. EE ∝∝ νν
E = hE = h νν
Where h = Plank’s constant, 6.6256 x
10-34
Js
The frequency is inversely proportional to its wavelength.
ν = С
λ
E = hС
λ
The energy of a photon may be determined from the following
expression :

8. 2.1

9. Example:Example:
 The wavelength of the green light from a trafficThe wavelength of the green light from a traffic
signal is centered at 522 nm. What is thesignal is centered at 522 nm. What is the
frequency of this radiation?frequency of this radiation?
ν = С 1 nm = 1 x 10-9
m,
λ c = 3.00 x 108
ms-1
ν = 3.00 x 108 m/s
522 nm x 1 x 10-9
m
1 nm
= 5.75 x 1014
s-1
or 5.75 x 1014
Hz

10. Example 2:Example 2:
 The critical wavelength of a quantum needed toThe critical wavelength of a quantum needed to
produce the photoelectric effect in Tungsten isproduce the photoelectric effect in Tungsten is
260 nm. Calculate the energy of the photons260 nm. Calculate the energy of the photons
needed.needed.
E = h ν = h С
λ
= (6.6256 x 10-34
Js) x (3.0 x 108
ms-1
)
260 x 10-9
m
= 7.645 x 10-19
J

11. ♦The lone electron in hydrogen atom revolvesThe lone electron in hydrogen atom revolves
round the nucleus in fixedround the nucleus in fixed CIRCULAR PATHCIRCULAR PATH
known asknown as ORBIT.ORBIT.
BOHR’S POSTULATESBOHR’S POSTULATES
♦Each orbit is associated with fixed amount of
energy. The electron CAN RESIDE in the
orbit but CANNOT EXIST anywhere
BETWEEN the orbits.

12.  The closer the electron to the nucleus, the lowerThe closer the electron to the nucleus, the lower
its energy.its energy.
 The orbits called eitherThe orbits called either energy levelsenergy levels,,
electronic shellselectronic shells oror principle shells.principle shells.
 An integer known asAn integer known as the principle quantumthe principle quantum
number, nnumber, n represented each orbital.represented each orbital.
 Eg:Eg: n = 1, 2, 3, ….n = 1, 2, 3, ….

13.  In the transition of the electron between higherIn the transition of the electron between higher
energy level to lower energy, the excess energyenergy level to lower energy, the excess energy
∆E∆E, will be released., will be released.
 And the frequency will be determined by Planck’sAnd the frequency will be determined by Planck’s
equation:equation:
∆∆E = hE = hνν
h= Planck constant (6.63 x 10h= Planck constant (6.63 x 10-34-34
Js or 3.99 x 10Js or 3.99 x 10-10-10
Js molJs mol-1-1
))

14.  According to Bohr’s theory, the electron in theAccording to Bohr’s theory, the electron in the
hydrogen atom can have those discrete energy,hydrogen atom can have those discrete energy,
EEnn..
EEnn = -R= -RHH 11
nn22
Where, RWhere, RHH = 2.18 x 10= 2.18 x 10-18-18
JJ
n = 1, 2, 3, ………..n = 1, 2, 3, ………..

15. 1. e-
can only have specific
(quantized) energy
values
2. light is emitted as e-
moves from one energy
level to a lower energy
level
Bohr’s Model of
the Atom (1913)
En = -RH ( )
1
n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18
J
2.1

16. Example 1:Example 1:
 Calculate the energy that an electron canCalculate the energy that an electron can
posses at the third orbit.posses at the third orbit.
En = -RH 1 where n = 3
n2
= -2.18 x 10-18
J x 1
32
= -2.422 x 10-19
J

17. Example 2:Example 2:
 Calculate the energy that an electronCalculate the energy that an electron
possesses when it is completely separatedpossesses when it is completely separated
from the nucleus.from the nucleus.
En = -RH 1 where n = ∞
n2
= -2.18 x 10-18
J x 1
∞2
= 0 J

18. ObjectiveObjective
 Differentiate between line spectrum andDifferentiate between line spectrum and
continuous spectrumcontinuous spectrum
 Perform calculations involving the RydbergPerform calculations involving the Rydberg
equation for Lyman, Balmer, Paschen, Brackettequation for Lyman, Balmer, Paschen, Brackett
and Pfund seriesand Pfund series

19. Continuous SpectrumContinuous Spectrum
andand
Line SpectrumLine Spectrum

20. Continuous SpectrumContinuous Spectrum
 When white light from an incandescent lamp isWhen white light from an incandescent lamp is
passed through a prism, it produces apassed through a prism, it produces a continuouscontinuous
spectrumspectrum, or rainbow colours (Fig. 1)., or rainbow colours (Fig. 1).
 The different colours of light represent differentThe different colours of light represent different
wavelengths.wavelengths.
 All wavelengths are present in aAll wavelengths are present in a continuouscontinuous
spectrumspectrum..
 White light is simply a combination of all theWhite light is simply a combination of all the
various colours.various colours.

21. Figure 1Figure 1

22. Line SpectrumLine Spectrum
 If the light from a gas discharge tube containingIf the light from a gas discharge tube containing
a particular element is passed through a prism,a particular element is passed through a prism,
only narrow coloured line are observed (Fig 2).only narrow coloured line are observed (Fig 2).
 Each line corresponds to light of a particularEach line corresponds to light of a particular
wavelength.wavelength.
 The pattern of emitted by an element is called itsThe pattern of emitted by an element is called its
line spectrumline spectrum

23. Figure 2
Line Emission Spectrum of Hydrogen Atoms

24. Hydrogen SpectrumHydrogen Spectrum
 The light emitted from the electron transition inThe light emitted from the electron transition in
hydrogen atom is called hydrogen spectrum,hydrogen atom is called hydrogen spectrum,
with certain wavelengthwith certain wavelength
 Wavelength emitted / absorbed can beWavelength emitted / absorbed can be
calculated by Rydberg equationcalculated by Rydberg equation
 Emission spectrum of hydrogen is classified inEmission spectrum of hydrogen is classified in
five series.five series.

25. Rydberg EquationRydberg Equation
 This equation solve problems based onThis equation solve problems based on
transitions of the electrons in the Lyman totransitions of the electrons in the Lyman to
Pfund Series.Pfund Series.
1 /1 / λλ = R= RHH (1/n(1/nii
22
– 1/n– 1/nff
22
))
where :where :
i)i) ni < nfni < nf
ii)ii) RRHH = 1.097 x 10= 1.097 x 1077
mm-1-1

27. Table 1: The various series in atomicTable 1: The various series in atomic
hydrogen emission spectrum.hydrogen emission spectrum.
Series nf
ni Spectrum Region
Lyman 1 2,3,4, .. Ultraviolet
Balmer 2 3,4,5, ..
Visible and
Ultraviolet
Paschen 3 4,5,6, .. Infrared
Brackett 4 5,6,7, .. Infrared
Pfund 5 6,7,8, .. Infrared

28. Figure 3 : The various emission series for hydrogen atom.Figure 3 : The various emission series for hydrogen atom.

29. Example 1Example 1
Calculate the wavelengths of the first line and the onset of
the continum limit for the Lyman series for hydrogen.
First line : 1/λ = RH (1/ni
2
– 1/nf
2
)
= 1.097 x 107
(1/12
– ½2
)
∴λ = 1.21 x 10-7
m-1
Onset of the continum limit : 1/λ = RH(1/ni
2
– 1/nf
2
)
= 1.097 x 107
(1/12
– 1/∞)
= 1.097 x 107
(1/12
– 0)
∴λ = 9.12 x 10-8
m-1

30. Example 2Example 2
Calculate the wavelength of the third line in theCalculate the wavelength of the third line in the
Brackett series of the hydrogen emission spectrumBrackett series of the hydrogen emission spectrum
SOLUTION
In the Brackett series, nIn the Brackett series, n11 = 4 and the third line (n= 4 and the third line (n22) = 7) = 7
1/1/λλ = 1.097 x 10= 1.097 x 1077
(1/4(1/422
– 1/7– 1/722
) = 4.617 x 10) = 4.617 x 1055
mm-1-1
∴λ∴λ = 2.166 x 10= 2.166 x 10-6-6
m = 2166 nm (1nm = 10m = 2166 nm (1nm = 10-9-9
m)m)

31. Example 3Example 3
Calculate the energy liberated when the electronCalculate the energy liberated when the electron
from the fifth energy level falls to the secondfrom the fifth energy level falls to the second
energy level in a hydrogen atomenergy level in a hydrogen atom
SOLUTION
∆∆E = RE = RHH (1/n(1/nii
22
– 1/n– 1/nff
22
) where RH = 2.18 x 10) where RH = 2.18 x 10-18-18
JJ
= (2.18 x 10= (2.18 x 10-18-18
) x (1/2) x (1/222
– 1/5– 1/522
)J)J
= 4.58 x 10= 4.58 x 10-19-19
JJ

32. At the end of the class, studentsAt the end of the class, students
should be able to:should be able to:
 Differentiate between line spectrum andDifferentiate between line spectrum and
continuous spectrumcontinuous spectrum
 Perform calculations involving the RydbergPerform calculations involving the Rydberg
equation for Lyman, Balmer, Paschen, Brackettequation for Lyman, Balmer, Paschen, Brackett
and Pfund seriesand Pfund series

33. Ionization Energy (IE)Ionization Energy (IE)
 Is the minimum energy required to
removed 1 mole of electron from 1 mole
of gaseous atom or ion.
 E (g)  E+
(g) + e-
(g)
 All IE are positive because energy always
required to remove an electron
 IE for H : ni = 1, nf = ∞
 IE can be calculated by multiplied ΔE with
Avogadro number (NA)

34. Exercise :Exercise :
a)a) Calculate the ionization energy of oneCalculate the ionization energy of one
hydrogen atomhydrogen atom
b)b) Calculate the ionization energy for 2.3 molesCalculate the ionization energy for 2.3 moles
of hydrogen atomsof hydrogen atoms
(Answer : a)2.18 x 10-18J/electron
b) 3.0199x 103kJ