CHAPTER 3CHAPTER 3

ATOMIC STRUCTUREATOMIC STRUCTURE

3.1 Bohr’s Atomic Model3.1 Bohr’s Atomic Model

3.2 Quantum Mechanical Model3.2 Quantum Mechanical Model

3.3 Electronic Configuration3.3 Electronic Configuration

Bohr’s Atomic ModelBohr’s Atomic Model

At the end of this topic students should be able to:-At the end of this topic students should be able to:-

1)1) DescribeDescribe the Bohr’s atomic postulates. the Bohr’s atomic postulates.

2)2) ExplainExplain the existence of electron energy levels in an atom and the existence of electron energy levels in an atom and calculate the energy of electroncalculate the energy of electron

3)3) DifferentiateDifferentiate between line spectrum and continuous spectrum. between line spectrum and continuous spectrum.

4)4) Perform calculation Perform calculation involving the Rydberg equation for Lyman, involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund seriesBalmer, Paschen, Brackett and Pfund series

5)5) Calculate the Calculate the ionisationionisation energy energy from Lyman seriesfrom Lyman series

6)6) Outline the weaknesses Outline the weaknesses of Bohr’s atomic model.of Bohr’s atomic model.

7)7) StateState the de Broglie’s postulate and Heisenberg’s uncertainty the de Broglie’s postulate and Heisenberg’s uncertainty principleprinciple

2n

1

In 1913, a young Dutch physicist, In 1913, a young Dutch physicist, Niels Böhr proposed a theory of Niels Böhr proposed a theory of atom that shook the scientific world.atom that shook the scientific world.

The atomic model he described had The atomic model he described had electrons circling a central electrons circling a central nucleusnucleus that contains positively that contains positively charged protons. charged protons.

Böhr also proposed that these orbits can only occur at specifically “permitted” levels only according to the energy levels of the electron and explain successfully the lines in the hydrogen spectrum.

BOHR’S ATOMIC MODELSBOHR’S ATOMIC MODELS

1. Electron moves in circular orbits about the nucleus. In moving in the orbit, the electron does not radiate any energy and does not absorb any energy.

Postulates

H Nucleus (proton) H1

1

BOHR’S ATOMIC MODELSBOHR’S ATOMIC MODELS

ii) The energy of an electron in a hydrogen atom is quantised, that is, the electron has only a fixed set of allowed orbits, called stationary states.

n=1

n=2

n=3

H Nucleus (proton)

[ orbit = stationary state = energy level = shell ]

BOHR’S ATOMIC MODELSBOHR’S ATOMIC MODELS

Postulates

1. At ordinary conditions the electron is at the ground state (lowest level). If energy is supplied, electron absorbed the energy and is promoted from a lower energy level to a higher ones. (Electron is excited)

4. Electron at its excited states is unstable. It will fall back to lower energy level and released a specific amount of energy in the form of light. The energy of the photon equals the energy difference between levels.

BOHR’S ATOMIC MODELSBOHR’S ATOMIC MODELS

Postulates

Ground state

the state in which the electrons have their lowest energy

Excited statethe state in which the electrons have shifted from a lower energy level to a higher energy level

Energy levelenergy associated with a specific orbit or state

Points to Remember

The energy of an electron in its level is given by:

RH (Rydberg constant) or A = 2.1810-18J.

n (principal quantum number) = 1, 2, 3 …. (integer)

Note: n identifies the orbit of electron Energy is zero if electron is located infinitely far from nucleus Energy associated with forces of attraction are taken to be

negative (thus, negative sign)

THE BOHR ATOM

2Hn n

1RE

Radiant energy emitted when the electron moves from higher-energy state to lower-energy state is given by the difference in energy between energy levels:

THE BOHR ATOM

E = Ef - Ei

2

i

Hin

1RE

2

f

Hfn

1RE

2

i

H2f

Hn

1R

n

1RE

2

f2

i

Hn

1

n

1RE

where

Thus,

The amount of energy released by the electron is called The amount of energy released by the electron is called a photon of energy.a photon of energy.

A photon of energy is emitted in the form of radiation A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength.with appropriate frequency and wavelength.

where;where; hh (Planck’s constant) =6.63 (Planck’s constant) =6.63 10 10-34-34 J s J s = frequency= frequency

THE BOHR ATOM

E = h

cWhere; c (speed of light) = 3.00108 ms-1

Thus,

hc

ΔE

n =1 n = 2 n = 3 n = 4

Electron is excited from lower to higher energy level. A specific amount of energy is absorbedE = h = E1-E3 (+ve)

Electron falls from higher to lower energy level .A photon of energy is released.E = h = E3-E1 (-ve)

Energy level diagram for the hydrogen atom

Pot

entia

l ene

rgy

n = 1

n = 2

n = 3

n = 4

n =

Energy released

Energy absorbed

Exercises:Exercises:

1)1) Calculate the energy of an electron in the second Calculate the energy of an electron in the second energy level of a hydrogen atom. (-5.448 x 10energy level of a hydrogen atom. (-5.448 x 10-19-19 J) J)

1)1) Calculate the energy of an electron in the energy level Calculate the energy of an electron in the energy level n = 6 of an hydrogen atom.n = 6 of an hydrogen atom.

3) Calculate the energy change (J), that occurs when an 3) Calculate the energy change (J), that occurs when an electron falls from n = 5 to n = 3 energy level in a electron falls from n = 5 to n = 3 energy level in a hydrogen atom.hydrogen atom.

(answer: 1.55 x 10(answer: 1.55 x 10-19-19J)J)

1)1) Calculate the frequency and wavelength (nm) of the Calculate the frequency and wavelength (nm) of the radiation emitted in question 3.radiation emitted in question 3.

Emission SpectraEmission Spectra

Emission Spectra

Continuous Spectra

LineSpectra

Continuous SpectrumContinuous Spectrum

A spectrum consists A spectrum consists all wavelengthall wavelength components components (containing an (containing an unbroken sequence ofunbroken sequence of frequencies) of the frequencies) of the visible portion of the electromagnetic spectrum are visible portion of the electromagnetic spectrum are present. present.

It is produced by incandescent solids, liquids, and It is produced by incandescent solids, liquids, and compressed gases.compressed gases.

Regions of the Electromagnetic Spectrum

When white light from When white light from incandescent lamp is passed incandescent lamp is passed through a slit then a prism, it through a slit then a prism, it separates into a spectrum.separates into a spectrum.

The white light spread out into The white light spread out into a rainbow of colours produces a rainbow of colours produces a continuous spectrum.a continuous spectrum.

The spectrum is continuous in The spectrum is continuous in that all wavelengths are that all wavelengths are presents and each colour presents and each colour merges into the next without a merges into the next without a break.break.

FORMATION OF CONTINUOUS SPECTRUM

Line Spectrum Line Spectrum (atomic spectrum)(atomic spectrum)

A spectrum consists of discontinuous & discrete lines produced by excited atoms and ions as the electrons fall back to a lower energy level. The radiation emitted is only at a specific wavelength or frequency. It means each line corresponds to a specific wavelength or frequency.Line spectrum are composed of only a few wavelengths giving a Line spectrum are composed of only a few wavelengths giving a series of discrete line separated by blank areasseries of discrete line separated by blank areas

prismfilm

The emitted light (photons) is then separated into its components by a prism. Each component is focused at a definite position, according to its wavelength and forms as an image on the photographic plate. The images are called spectral lines.

FORMATION OF ATOMIC / LINE SPECTRUM

FORMATION OF ATOMIC / LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4n = 5n =

En

erg

y

When an electrical discharge is passed through a sample of hydrogen gas at low pressure, hydrogen molecules decompose to form hydrogen atoms.

Radiant energy (a quantum of energy) absorbed by the atom (or electron) causes the electron to move from a lower-energy state to a higher-energy state. Hydrogen atom is said to be at excited state (very unstable).

FORMATION OF ATOMIC / LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5n = 6n =

En

erg

y

When the electrons fall back to lower energy levels, radiant energies (photons) are emitted in the form of light (electromagnetic radiation of a particular frequency or wavelength)

FORMATION OF ATOMIC / LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4n = 5n =

Lyman Series

Emission of photon

Linespectrum

E

Energy

FORMATION OF ATOMIC / LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4n = 5n =

Lyman Series

Emission of photon

Linespectrum

Balmer Series

E

Energy

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n =

22 2

AE

23

AE 3

24 4

AE

21 1

AE

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

Exercise: Exercise: Complete the following tableComplete the following table

Series n1 n2Spectrum

region

Lyman 2,3,4,…

2 3,4,5,…

Paschen 4,5,6,… Infrared

4 5,6,7,… Infrared

5 6,7,8,… Infrared

ultraviolet

VisibleBalmer

Brackett

Pfund

1

3

The following diagram depicts the line spectrum of The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the hydrogen atom. Line A is the first line of the Lyman series.Lyman series.

ExerciseExercise

Linespectrum

E

Specify the increasing order of the radiant energy, Specify the increasing order of the radiant energy, frequency and wavelength of the emitted photon. frequency and wavelength of the emitted photon.

Which of the line that corresponds to Which of the line that corresponds to

i) the shortest wavelength?i) the shortest wavelength?

ii) the lowest frequency?ii) the lowest frequency?

A B C D E

Line E

Line A

Describe the transitions of electrons that lead to Describe the transitions of electrons that lead to the lines W, and Y, respectively.the lines W, and Y, respectively.

Solution

Linespectrum

W Y

ExerciseExercise

Balmer series

For W: transition of electron is from n=4 to n=2

For Y: electron shifts from n=7 to n=2

HomeworkHomework

Calculate Calculate EEnn for for nn = 1, 2, 3, and 4. Make a one- = 1, 2, 3, and 4. Make a one-

dimensional graph showing energy, at different dimensional graph showing energy, at different values of values of nn, increasing vertically. On this graph, , increasing vertically. On this graph, indicate by vertical arrows transitions that lead to indicate by vertical arrows transitions that lead to lines inlines in

a)a) Lyman seriesLyman series

b)b) Paschen seriesPaschen series

In Lyman series, the frequency of the convergence of In Lyman series, the frequency of the convergence of spectral lines can be used to find the ionisation energy of spectral lines can be used to find the ionisation energy of hydrogen atom: hydrogen atom:

IE = IE = hh

The frequency of the first line of the Lyman series > the The frequency of the first line of the Lyman series > the frequency of the first line of the Balmer series.frequency of the first line of the Balmer series.

Significance of Atomic SpectraSignificance of Atomic Spectra

Lyman Series

Linespectrum

E

Balmer Series

Linespectrum

ABCDE

ExerciseExercise

Paschen series

Solution

Which of the line in the Paschen series corresponds to the longest wavelength of photon?

Describe the transition that gives rise to the line.

Line A.

The electron moves from n=4 to n=3.

Calculate Calculate a)a) the wavelength in nm the wavelength in nm b)b) the frequency the frequency c)c) the energy the energy

that associated with the second line in the Balmer series that associated with the second line in the Balmer series of the hydrogen spectrum.of the hydrogen spectrum.

ExerciseExercise

Solution (a)

Second line of Balmer series: the transition of electron is from n2=4 to n1=2

== RRHH

11 11

nn1122 nn22

22

11

== (1.097x10(1.097x1077 m m11))11 11

2222 4422

11

= x1

109

mnm

= 486 nm

4.86x107 m

Wavelength emitted by the transition of electron Wavelength emitted by the transition of electron between two energy levels is calculated using between two energy levels is calculated using Rydberg equation:Rydberg equation:

Rydberg Equation

RH = 1.097 107 m-1

= wavelengthSince should have a positive value thus n1 < n2

where

22

21

Hn

1

n

1R

λ

1

Calculate the wavelength, in nanometers of Calculate the wavelength, in nanometers of

the spectrum of hydrogen corresponding to the spectrum of hydrogen corresponding to

nnii = 2 and n = 2 and nff = 4 in the Rydberg equation. = 4 in the Rydberg equation.

Example

Solution:

Rydberg equation: Rydberg equation:

1/1/λλ = RH (1/ni2 – 1/nf2) = RH (1/ni2 – 1/nf2) nini = 2 = 2 nfnf = 4 = 4RH = 1.097 x 10m7 RH = 1.097 x 10m7

1/λ = RH (1/22 – 1/42) = RH(1/4-1/16) λ = 4.86m x 102 m = 486nm

Use the rydberg equation to calculate the wavelength of the spectral line of hydrogen atom that would result when an electron drops from the fourth orbit to the second orbit, then identified the series the line would be found.

Example

Solution:

1/λ = RH (1/n1 2 – 1/n2 2)n1 = 2 n2 = 4

1/λ = 1.097 x 107 (1/22 – 1/42) λ = 4.86 x 10-7 m

= 486 nm *e dropped to the second orbit (n=2), >>> Balmer series

EXAMPLE 3

Calculate the wavelengths of the fourth line in the Balmer series of hydrogen.

n1 = 2 n2 = 6RH = 1.097 x 107m-1

λ = 4.10 x 10-7 m

RH

22 62

1 11=

λ

Different values of RDifferent values of RHH and its usage and its usage

1. RH = 1.097 107 m-1

RH

n21 n2

2

1 11=

λ

RH = 2.18 x 10-18 J

2

f2

i

Hn

1

n

1RE

n1 < n2

EXAMPLE 4

Calculate the energy liberated when an electron Calculate the energy liberated when an electron from the from the fifth energy levelfifth energy level falls to the falls to the second second energy levelenergy level in the hydrogen atom. in the hydrogen atom.

ΔE = 4.58 x 10-19 J

hc

ΔE

ΔE = (6.63 10-34Js)X(3.00108 ms-1)

RH

n21 n2

2

1 11=

λ

521.097 x 107

22

1 11=

λ

1

λ= 0.2303 X 107 m-1

X (0.2303 X 107 m-1)

Calculate what is;Calculate what is;

i ) Wavelengthi ) Wavelength

ii ) Frequencyii ) Frequency

iii ) Wave number iii ) Wave number

of the last line of of the last line of

hydrogen spectrum hydrogen spectrum

in Lyman series in Lyman series

Wave number = 1/wavelengthWave number = 1/wavelength

EXERCISE

For Lyman series; n1 = 1

& n2 = ∞Ans:i. 9.116 x10-8m ii. 3.29 x1015 s-1

iii. 1.0970 X 107 m-1

i) 1.097 x 107 12

1 11=

λ ∞2 = 1.097 X 107 m-1

λ = 9.116 X 10-8 m

ii) V = cλ

=3.00 X 108 ms-1

9.116 X 10-8 m= 3.29 X 1015 s-1

iii) Wave number =1

= 1.097 X 107 m-1

λ

SOLUTION

Defination :Defination : Ionization energy is the Ionization energy is the minimum energy required to remove one mole of electron from one mole of gaseous atom/ion.

M (g) M (g) M M++ (g) + e (g) + e H = +veH = +ve

The hydrogen atom is said to be ionised when electron The hydrogen atom is said to be ionised when electron is removed from its ground state (n = 1) to n = is removed from its ground state (n = 1) to n = ..

At n = At n = , the potential energy of electron is zero, here , the potential energy of electron is zero, here the nucleus attractive force has no effect on the electron the nucleus attractive force has no effect on the electron (electron is free from nucleus). (electron is free from nucleus).

Ionization Energy

nn11 = 1, n = 1, n22 = ∞ = ∞

∆∆E E = R = RHH (1/n (1/n1122 – 1/n – 1/n22

22))

= 2.18 X 10 = 2.18 X 10 -18-18 (1/1 (1/122 – 1/ ∞ – 1/ ∞ 22))

= 2.18 X 10 = 2.18 X 10 -18-18 (1 – 0) (1 – 0) = 2.18 X 10 = 2.18 X 10 -18 -18 JJ

Ionisation energy Ionisation energy = 2.18 X 10 = 2.18 X 10 -18-18x 6.02 X 10x 6.02 X 102323J molJ mol-1-1

=1.312 x 10 =1.312 x 106 6 J molJ mol-1-1

= 1312 kJ mol= 1312 kJ mol-1-1

Example

1

1 st lineConvergent limit

Finding ionisation energy experimentally:

Ionisation energy is determined by detecting the wavelength of the convergence point

10.97 10.66 10.52 10.27 9.74 8.2210.97 10.66 10.52 10.27 9.74 8.22

wave number (x10wave number (x1066 m m-1-1))

The Lyman series of the spectrum of hydrogen is shown The Lyman series of the spectrum of hydrogen is shown above. Calculate the ionisation energy of hydrogen from above. Calculate the ionisation energy of hydrogen from the spectrum.the spectrum.

Example

ΔE ΔE = h= hc/c/λλ

==h x ch x c / / λλ = = h x ch x c x wave no. x wave no.

= = 6.626 x 106.626 x 10-34 -34 J s x 3 x 10J s x 3 x 108 8 m sm s-1-1 x 10.97x 10 x 10.97x 1066 m m-1-1 = = 218.06218.06x 10x 10-20 -20 JJ

= = 2.18 2.18 x 10x 10-18-18JJ

Ionisation energyIonisation energy

= 2.18 X 10 = 2.18 X 10 -18-18x 6.02 X 10x 6.02 X 1023 23 J molJ mol-1-1

=1.312 x 10=1.312 x 106 6 J molJ mol-1-1

= 1312 kJ mol= 1312 kJ mol-1-1

Solution

Compute the ionisation energy of hydrogen atom Compute the ionisation energy of hydrogen atom in kJ molin kJ mol11..

Exercise

Solution

n1 = 1n2 =

11== RRHH

11

nn1122 nn22

22EE

== 2.18x102.18x101818 JJ11 11

1122 22

==2.18x102.18x101818 JJ 6.02x106.02x102323

1 atom1 atom molmol JJ

kJkJx x

10001000

11

= = 1312 kJ mol1312 kJ mol11

11

atomatom

The weakness of Bohr’s Theory

Bohr was successful in introducing the idea of quantum Bohr was successful in introducing the idea of quantum energy and in explaining the lines of hydrogen energy and in explaining the lines of hydrogen spectrum.spectrum.

His theory could not be extended to predict the energy His theory could not be extended to predict the energy levels and spectra of atoms and ions with more than one levels and spectra of atoms and ions with more than one electron.electron.

His theory can only explain the hydrogen spectrum or His theory can only explain the hydrogen spectrum or ions contain one electron eg Heions contain one electron eg He++, Li, Li2+.2+.

Modern quantum mechanics retain Bohr’s concept of Modern quantum mechanics retain Bohr’s concept of discrete energy states and energy involved during discrete energy states and energy involved during transition of electrons but totally reject the circular orbits transition of electrons but totally reject the circular orbits he introduced. he introduced.

Davisson & Germer observed the diffraction of Davisson & Germer observed the diffraction of electrons when a beam of electrons was directed at electrons when a beam of electrons was directed at a nickel crystal. Diffraction patterns produced by a nickel crystal. Diffraction patterns produced by scattering electrons from crystals are very similar to scattering electrons from crystals are very similar to those produced by scattering X-rays from crystals. those produced by scattering X-rays from crystals. This experiment demonstrated that electrons do This experiment demonstrated that electrons do indeed possess wavelike properties. indeed possess wavelike properties.

Thus, can the ‘position’ of a wave be specifiedThus, can the ‘position’ of a wave be specified??????

Point to Ponder

de Broglie’s Postulatede Broglie’s Postulate

In 1924 Louis de Broglie proposed that not only light but all matter has a dual nature and possesses both wave and corpuscular properties. De Broglie deduced that the particle and wave properties are related by the expression:

h = Planck constant (J s)h = Planck constant (J s)m = particle mass (kg)m = particle mass (kg) = velocity (m/s)= velocity (m/s) = wavelength of a matter wave= wavelength of a matter wave

=h

m

Heisenberg’s Uncertainty PrincipleHeisenberg’s Uncertainty Principle

It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain.

Stated mathematically,Stated mathematically,

where where x = uncertainty in measuring the positionx = uncertainty in measuring the position

p = uncertainty in measuring the momentump = uncertainty in measuring the momentum

= = mvmv

h = Planck constant h = Planck constant

h4xx p p