2.1 Atomic Structure

J 10 x 2.18R ),n

1(RE 18-

H2Hn

J 10 x 2.18R where),n

1

n

1(RΔE 18-

H2

f

2

i

H -

At the end of this lesson, students should be able to:

describe Bohr’s atomic model.

OBJECTIVES

explain the existence of energy levels in an atom.

calculate the energy of an electron using:

calculate the energy change of an electron during transition.

calculate the photon emitted by an electron that produces a

particular wavelength during transition.

DALTON ------ ATOMS

THOMSON ------ ELECTRONS

RUTHERFORD ------- POSITIVELY CHARGED NUCLEUS

Four Bohr’s atomic postulate :

electron moves in circular orbit about the nucleus.

Each orbit has a different energy level/energy state. 1

In the specific energy level, the energy of electron is

fixed in value or is quantised.

The electron does not release or absorb any energy

when it is in the ground state or at any particular

energy level/orbit.

This means that an electron moving in an orbit can

have a certain amount of energy.

(quantised = fixed/specific/definite)

2

2

1

nRE Hn

electron

nucleus

(proton)

n=1

n=2

energy of electron at n=1 ≠ energy of electron at n=2

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

2

1

nRE Hn

where,

RH (Rydberg constant)= 2.18 10-18J. n (orbit/ energy level) = 1, 2, 3,…, (integer)

NOTES:

Energy is zero (E = 0) if electron is located infinitely

(n = )far from nucleus.

Energy associated with forces of attraction are taken to

be negative (thus, negative sign)

n=2 (higher energy level)

n=3 (higher energy level)

nucleus (proton)

n=1

3

energy absorbed

At ordinary condition, the electron is at the ground

state (lowest energy state).

If energy is supplied, electron absorbed the energy

and is promoted from lower energy level to a higher

energy level (electron is excited)

Energy at excited state is unstable.

It will fall back to lower energy level and released a

specific amount of energy in the form of light or

photon.

The light have specific wavelength. The line

spectrum is formed.

4

n=2 (higher energy level)

n=3 (higher energy level)

nucleus (proton)

n=1

energy released

The energy absorb / released during transition can be

calculated by:

22

11

fi

Hnn

RΔE

Where,

RH = Rydberg constant (2.18 x 10-18 J)

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

ni = position(n) of electron initially

nf = position(n) of electron finally

The amount of energy released by the electron during

transition is called a photon of energy.

A photon of energy is released in the form of light with

appropriate frequency and wavelength.

Therefore, the energy change (released/absorbed) can

be related as:

Photon = a packet of light energy equals to h

hυΔE

where, h (Planck’s constant) =6.63 10-34 Js

= frequency; c (speed of light) = 3.00 108 ms-1

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

2

1

nRE Hn

The energy absorb / released during transition can be

calculated by:

22

11

fi

Hnn

RΔE

hυΔE

A photon of energy is emitted in the form of radiation with

appropriate frequency and wavelength.

c

υWhere:

Thus:

hcΔE

Where:

v = frequency

λ = wavelength

c = speed of light (3.00x108 ms-1)

h = Plank’s constant (6.63 x 10-34 Js)

Energy of electron at particular orbit

2

1

nRE Hn

Calculation 1

where,

RH (Rydberg constant)= 2.18 10-18J.

n (orbit/ energy level) = 1, 2, 3,…, (integer)

Calculate the energy of electron that occupies n=2

Example 1

Answer:

The energy of an electron at its orbit (n) is -1.36x10-19J.

Determine the value of n.

Example 2

Answer:

1. Calculate the energy of electron that

occupies n=6

2. The energy of an electron at its orbit (n) is

-8.72x10-18J. Determine the value of n.

Energy released/absorbed by electron during transition

Calculation 2

Calculate the energy released when an electron falls from n=3

to n=2.

Example 1

Answer:

22

11

fi

Hnn

RΔE

RH = Rydberg constant (2.18 x 10-18 J)

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

ni = position(n) of electron initially

nf = position(n) of electron finally

Calculate the energy required to promote an electron from the first

energy level to the third energy level of a hydrogen atom.

Example 2

J 10 x 1.94

3

1

1

11018.2

n

1

n

1RΔE

18

22

18

2

f

2

i

H

-

Jx

Answer:

1. Calculate the energy released when an electron excited from n=1 to n=4.

Energy released/absorbed by electron during transition

Calculation 3

Calculate the energy absorbed by an electron that produces a wavelength

of 9.38 x 10-8 m.

Example 1

Answer:

hυΔE

h (Planck’s constant) =6.63 10-34 Js

= frequency;

c (speed of light) = 3.00 108 ms-1

An electron absorbed 4.58x10-19 J of energy when falls from n=5 to n=2.

Calculate the frequency and wavelength produced.

Example 2

Answer:

1 14

34- 19-

10 x 6.91

Js)10 x (6.63 10 x 4.58

hυΔE

s

J

ms

smxc 7-

1 14

18

4.34x10 10 x 6.91

1000.3

cυ

released

1. Calculate the energy released by an electron that produces a wavelength of 9.49 x 10-8 m.

2. An electron released 2.09x10-18 J of energy when excited from n=1 to n=5. Calculate the frequency and wavelength produced.

<9.52 x 10-8m>

absorbed

Calculation 4

Calculate:

(a) The energy (in J) of an electron has when it occupies a

level equivalent to the quantum number of n = 3 and n = 4.

(b) The energy (in kJ/mol) of photon emitted when one mole of

electron drops from the 4th energy level to the 3rd energy

level.

(c) The frequency (in s-1) and wavelength (in nm) of this

photon.

Answer:

Calculate the energy of an electron in the hydrogen atom

when n = 2, and when n = 6.

Calculate the wavelength released when an electron

moves from n = 6 to n = 2.

If this line in the visible region of electromagnetic

spectrum?

(Note: visible light: 400 ~ 700 nm)

At the end of this lesson, students should be able to:

describe the formation of line spectrum of hydrogen atom.

OBJECTIVES

perform calculations involving the Rydberg equation for

Lyman, Balmer, Paschen, Brackett and Pfund series:

calculate the ionisation energy of hydrogen atom from

Lyman series.

n2 n1 and m 10 x 1.079R where

)n

1

n

1(R

1

1-7H

22

21

H -

State the weakness of Bohr’s atomic model.

State the dual nature of electron using de Broglie’s

postulate and Heisenberg’s uncertainty principle.

(calculation is not required)

Formation of line spectrum of hydrogen atom

Spectrum is a series of lines or a set of colours. 1

Line spectrum Continuous spectrum

A spectrum that contains a series of

discrete lines. Each line corresponds

to a specific wavelength.

A spectrum that contains continuous

band of light radiation of all

wavelengths.

Each line separated by blank area Continuous. No blank area in between.

Source: emission spectrum of atom

(from gas discharge tube of

hydrogen, etc.)

Source: white light, sunlight

2 They are two types of spectrum: line spectrum, and

continuous spectrum.

nucleus n=1

n=2

n=3

n=4

When the electron of a hydrogen

atom at its ground state absorbs

sufficient amount of energy, it will

move to a higher energy level

At higher energy evel, electron is

unstable. It will falls back to a lower

energy level.

During the transition (falls form

higher to lower energy level),

energy will be released in a form of

light at certain wavelength and

frequency.

Since energy is quantised (fix in

value), line spectrum is produced.

The electromagnetic spectrum: 3

There is several emission series of lines obtained during

the transition of electrons in hydrogen spectrum 4

The series in the hydrogen line spectrum are found in the

visible region, infrared region and ultraviolet region. 5

series Spectrum region

Lyman

Balmer

Paschen

Brackett

Pfund

Ultraviolet

Visible

Infrared

Infrared

Infrared

These spectral lines are the result of the transition of

excited electron to the lower energy level. 6

Balmer series is the only one visible to the unaided eye. It

contains four coloured line against a black background.

7

Line

number

Colour in

Balmer series

Wavelength

(nm)

1st line red 656

2nd line Blue-green 486

3rd line indigo 434

4th line violet 410

The possible energy levels in the hydrogen atom and the transitions

of electrons that produce the lines in the hydrogen spectrum: 7

Lines produced caused by the transition of electrons from

higher energy level to lower energy level. 7

Series

Line

Lyman

Transition of e-

1st

2nd

3rd

4th

5th

n=

n=1

n=2

n=3

n=4

n=5

n=6 n=7

n=2 to n=1

ch

hΔE

E , ,

Continuum

limit

e-

n=3 to n=1

e- n=4 to n=1

e-

n=5 to n=1

e-

n=6 to n=1

e-

Series

Line

Balmer

Transition of e-

1st

2nd

3rd

4th

n=

n=1

n=2

n=3

n=4

n=5

n=6 n=7

Continuum

limit

n=3 to n=2

e-

n=4 to n=2 e-

n=5 to n=2

e-

n=6 to n=2

e-

ch

hΔE

E , ,

Series

Line

Paschen

Transition of e-

1st

2nd

3rd

4th

n=

n=1

n=2

n=3

n=4

n=5

n=6 n=7

n=4 to n=3

Continuum

limit

e- n=5 to n=3

e-

n=6 to n=3

e- e-

n=7 to n=3

ch

hΔE

E , ,

Series

Line

Brackett

Transition of e-

1st

2nd

3rd

n=

n=1

n=2

n=3

n=4

n=5

n=6 n=7

n=5 to n=4

Continuum

limit

e-

n=6 to n=4

e-

n=7 to n=4

e-

ch

hΔE

E , ,

Series

Line

Pfund

Transition of e-

1st

2nd

n=

n=1

n=2

n=3

n=4

n=5

n=6 n=7

Continuum

limit

n=6 to n=5

e-

n=7 to n=5

e-

ch

hΔE

E , ,

Each line in the hydrogen spectrum has a specific

wavelength. 8

9 The wavelength can be calculated by using the

Rydberg equation:

2

2

2

1

1

11

nnRH

Where = wavelength

RH = Rydberg constant = 1.097 x 107 m-1

n1 and n2 = 1, 2, 3, …, (principal quantum numbers)

Since should have a positive value thus n1 < n2

nm441.3

m 10 x 4.41

m106 x 2.27

m 10 x 1.097

7-

1-

1-7

22 5

1

2

1

Example

21 n n ; 2

22

1

111

nnRH

Calculate the wavelength of the indigo line in the

Balmer series.

Solution:

Indigo line 3rd line in Balmer series n=5 to n=2

m nm 1 x 10-9

Try!!!

1. Calculate the wavelength of the continuum limit in the

Paschen series.

2. For the Lyman series, calculate the wavelength when

electron falls from n=4 to n=1.

• The further they fall, more energy is released and the higher the frequency.

• The orbitals also have different energies inside energy levels.

Ultraviolet Visible Infrared

hcΔE hvΔE

Changing the Energy

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

47

Line spectrum

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

Which of the line that corresponds to

i) the shortest wavelength?

ii) the lowest frequency?

A B C D E E

∆

Example

Line spectrum

A B C D E

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.

Example

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

Solution

Line spectrum

W Y

Balmer series

Example

Compare the difference:

i f

E = RH 1

n2

1

n2

1 2

RH

1

n2

1

n2

1 =

n1 can be ni or nf as long as n1 < n2 to get positive value for

Both equations can be used to calculate wavelength ( ) and frequency ( ) of any line of H atom emission series

Remember! RH values and units are different in both equations !

RH = 2.18 x 10–18 J

RH = 1.097 x 107 m-1

Use the Rydberg equation to calculate the wavelength (in nm) of the forth line in the Balmer series of Hydrogen spectrum .

EXAMPLE

ANS: 410.2 nm

ANS:n = 3

An electron in the n = 5 level of an H atom emits a photon of wavelength 1281 nm. To what energy level does the electron move?

EXAMPLE

Calculate what is;

i ) Wavelength

ii ) Frequency

iii ) Wave number

of the last line of

hydrogen spectrum

in Lyman series

Wave number = 1/wavelength

For Lyman series; n1 = 1

& n2 = ∞

EXAMPLE

The minimum energy required to remove

one mole of electrons from one mole of

the gaseous atoms or ions.

M (g) → M+ (g) + e- ΔH1= +ve kJ

Calculate the ionization energy of H atom

in kJmol-1.

EXAMPLE

Ans: EXAMPLE

• Unable to explain the line spectrum of atoms or ions

containing more than one electron (such as helium).

• Electron is restricted to move in a certain distance around

the nucleus of an atom.

• Unable to explain the extra lines formed in the hydrogen

spectrum.

• Unable to explain the dual nature of electrons.

Bohr’s theory has a number of weaknesses. Bohr’s atomic

model

(A) de Broglie’s postulate

• in 1924, Louis de Broglie, a French physicist, suggested

that light and matter appear to have dual natures, that is,

both light and matter are wave-like as well as particle-

like.

• de Broglie suggested an equation that allows the

calculation of the wavelength of an electron or a particle

with mass (m), moving at velocity ( ):

m

h λ

where, h = Planck’s constant (J s)

m = particle mass (kg)

= velocity (m/s)

= wavelength of a matter wave

(B) Heisenberg’s uncertainty principle

• Bohr’s theory:

the electron was thought of as orbiting the nucleus.

This would mean that at any moment, we would

know both the precise position and precise

speed of the electron.

This is not allowed in quantum mechanics.

• In 1927, Werner Heisenberg states in his principle that:

This principle is known as Heisenberg’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”.

where

x= uncertainty in measuring the position

p=uncertainty in measuring the momentum

= mv

h = Planck’s constant

• Mathematically, his principle stated as:

4

h px