UNIT 1-STRUCTURE OF ATOMS

ATOMIC PHYSICS AND SPECTROSCOPY

UNIT 1-STRUCTURE OF ATOMS

1

•Atom- Smallest particle of chemical element possessing chemical

properties of the element

J.J.THOMSON’S PLUM PUDDING MODEL

•Atom- positive sphere with electrons embedded in it like plum in a

pudding

•Could not explain optical spectra of hydrogen and elements

2

3

RUTHERFORD ATOM MODEL

•Atom- nucleus (positively charged)

•Electrons revolves round the nucleus in circular orbits

•An accelerated charge (electron) must radiate energy and spirals

down into the nucleus

•Atom – not stable

BOHR ATOM MODEL

I Postulate:

•Electron cannot revolve in all possible orbits

•Can revolve in allowed or permissible orbits

•Angular momentum is integral multiple of ħ=h/2ᴨ

•Electrons revolving in these orbits does not radiate energy

•Ang. Momentum=L=(mv)r=m(rω)r=mr2ω=nh/2ᴨ

•n= principal quantum number=1,2,3....

II Postulate

•Atom radiates energy when electron jumps from higher energy to

lower energy

•Difference in energy is given out as a Photon

•Photon Frequency υ=(Ei-Ef)/h 4

Bohr Formulae(i) Radii of stationary orbit

(ii) Total energy of electron in orbit

Consider an atom whose nucleus has a positive charge Ze and mass M.

For hydrogen, Z = 1

Let an electron of charge (–e) and mass m move round the nucleus in an

orbit of radius r

Since M >> m, the nucleus is stationary. Hence the mass of the nucleus

does not come into the calculations.

The electrostatic force of attraction between the nucleus and the electron

)1())((

4

12

0

1 −−−−−=r

eZeF

5

(i) Radii of stationary orbit

The centrifugal force on the electron

The system will be stable if F1 = F2

)3())((

4

12

20

−−−−=r

eZe

r

mv

According to Bohr’s first postulate, 2

nhmvr =

mr

nhv

2=

222

222

4 mr

hnv

=

)2(2

2 −−−−−−=

r

mvF

Sub V2 in (3) )4(

))((

4

1

4 20

222

22

−−−−=r

eZe

mrr

hmn

6

)4())((

4

1

4 20

222

22

−−−−−=r

eZe

mrr

hmn

( )

)5( orbit nth theof Radius

1

2

022

2

0

22

−−−==

=

mez

hnr

zemr

hn

n

For hydrogen atom z=1,

Radius of the nth orbit = )5(2

022

ame

hnrn −−−=

7

From equation (5) we find that rn ∝ n2

The radii of the orbits are in the ratio of 1 : 4 : 9 : 16 : 25 etc.

The radius of the first orbit for hydrogen atom , put n=1 in (5a)

me

hr

2

02

1

= 1

2rnrn =

8

mez

hr

2

02

1

=

rdr

r

r11

2−=

r

Zemv

r

eZe

r

mveqn

)(

4

12

;))((

4

12)3(

2

0

20

=

=→

P.E.= W.D. in bringing an electron from infinity to the orbit

= Integral of electrostatic force between nucleus and electron

(ii) Total energy of electron in orbit

9

)8(8

)(8

)(.

)5(

220

42

2200

22

2

022

2−−−−

−=

−=

−−−=

hn

mezE

hn

mezzeET

mez

hnr

n

n

As n increases En increases

Hence outer orbits have greater energies than inner orbits

For hydrogen atom z=1

)8(1

8 220

4

2a

nh

meEn −−−−

−=

10

Bohr’s interpretation of hydrogen spectrum

If an electron jumps from an outer orbit n2 of higher energy to an inner orbit n1 of

lower energy, the frequency of the radiation emitted is given by

−

−=

−=

−=

−−−−

−=

−=

−=

22

2

2

2

12

23

0

4

22

20

4

2

21

20

4

1

220

4

12

12

11

8

1

8

1

8

)8(1

8

nnh

me

nh

meE

nh

meE

anh

meE

h

EE

EEh

n

n

n

nn

nn

11

The wavenumber of a radiation is defined as the reciprocal of its wavelength λ in

vacuum and gives the number of waves contained in unit length in vacuum.

−

−=

221

22

30

4 11

8 nnh

me

Wavenumber =ῡ = 1/λ=υ/c

−

−==

221

22

30

4 11

8 nnch

me

c

30

4

28Rconstant sRydberg'

ch

me

==

−=

22

21

11

nnR

12

Spectral series of hydrogen atom

(1)Lyman series

When an electron jumps from second, third, ... etc., orbits to the first orbit, we get

the Lyman series which lies in the ultraviolet region.

Here, n1 = 1 and n2 = 2, 3, 4, 5 ....

2,3,4...n 1

1

122

=

−=

nR

13

(2) Balmer Series

When an electron jumps from third, fourth... etc., orbits to the second orbit, we

get the Balmer series which lies in the visible region of the spectrum.

Here, n1 = 2 and n2 = 3, 4, 5 ....

3,4,5...n 1

2

122

=

−=

nR

The first line in the series (n = 3) is called the Hα line,

the second (n = 4) the Hβ line and so on.

14

(3) Paschen series

When an electron jumps from fourth, fifth ... etc., orbits to the third orbit,

we get the Paschen series in the infrared region

Here, n1 = 3 and n2 = 4, 5, 6 .....

4,5,6...n 1

3

122

=

−=

nR

5,6,7...n 1

4

122

=

−=

nR

(4) Brackett series

If n1 = 4 and n2 = 5, 6, 7 ..... etc., we get the Brackett series.

15

(5) Pfund series

If n1 = 5 and n2 = 6, 7, 8, ..... we get Pfund series

6,7,8...n 1

5

122

=

−=

nR

Brackett and Pfund series lie in the very far infrared region of the hydrogen

spectrum.

By putting n = ∞ in each one of the series, we get the wavenumber of the

series limit, i.e., the last line in the series.

The electron jumps giving rise to the different series in hydrogen spectrum

16

17

18

The energy-level diagramFor hydrogen atom,

eVn

E

nh

emzE

n

n

−=

−−−−

−=

2

220

42

16.13

)8(1

8 2

The lowest energy E1 is the ground state

higher energy E2, E3, E4...are called excited states

19

•As n increases, En increases.

•As n increases, the energy levels crowd and tend to form a continuum

•Discrete energy states are represented by horizontal lines

•Electronic jumps between these states are represented by vertical lines

• Energy Level diagram shows how spectral lines are related to atomic energy

levels

20

21

22

In Bohr theory, nucleus (infinite mass) is fixed

at the centre of the circular orbit and the electron

revolves round it.

If the nuclear a mass M is not infinite:

Both the nucleus and electrons revolve around a

common centre of mass with same angular velocity ω.

Effect of Nuclear Motion on Atomic Spectra

N- Nucleus and e-electron

M-mass of nucleus

m-mass of electron

Nucleus and electron are rotating about their common centre of mass C

Nucleus -moving in a circle of radius r1

Electron -moving in a circle of radius r2

23

)4(mM

Mr

similarly,

)3(mM

mr

mr

m

M1r

r

eqn(2)in sub and r find eqn(1) from

)2(

2

1

1

1

11

2

21

−−−−−

+=

−−−−

+=

=

+

=

+

−=

−−−−=

r

r

rm

M

r

m

Mrr

rrr

According to centre of mass theory

Mr1 = mr2 ...(1)

Let r represent the distance between the nucleus and electron. Then,

r = r1 + r2

24

( )

formulaeBohr allin by replaced be tois m nucleus of mass finite

by replaced is m

)7()2/(mr

is (6)equation motion thenuclear of absence In the

)6()2/(

postulate 1 Bohrs toAccording

electron theof mass reduced thecalled is

1mM

Mm

)5(mM

Mm)(

mM

Mm

mM

Mr

mM

mr

rMrL momentumAngular

2

2

st

222

2

22

2

2

2

1

For

nh

nhr

M

m

mwhere

rrrMm

mM

m

−−−−=

−−−−=

+

=+

=

−−−=

+=+

+=

++

+=

+=

25

32

0

4

32

0

4

32

0

4

32

0

4

2

2

2

1

32

0

24

2

2

2

1

32

0

24

n1n212

222

0

24

n

8

e

)10(

1

1

M

m1

1

mM

Mm

8

me

8

e

8

e

)9(11

8

e

11

8

e emittedradiation theoffrequency The

E-E level, n ton from jumpselectron an When

)8(1

8

eEbecomes atomhydrogen of levelEnergy

ch

mR

M

mRR

where

mM

M

chmM

mM

chchRelementanyforconsantRydberg

nnch

z

C

nnh

z

h

nh

z

z

z

z

z

z

zz

=

−−−−−

+

=

+

=+

=

+

=

+

==

−−−−−−

−==

−==

=

−−−−

−==

26

)11(11

;10097770R

M

m1

1RR

1840/1/

10973740mpically spectrosco estimated is R

number Z atomic of nucleus of mass Mz

rest at is nucleus when the,Mhen constant w Rydberg of value

2

2

2

1

2

1

H

H

H

1-

z

−−−−

−=

=

+

=

=

=

=

=

−

nnRz

m

Mm

The

z

z

27

Evidences for Bohr’s theory

(1) The ratio of mass of electron to the mass of proton

The Rydberg constants for hydrogen and helium are

HHe

He

H

H

He

He

He

H

H

MM

M

m

M

m

R

R

M

m

RR

M

m

RR

4

1

1

1

1

=

+

+

=

+

=

+

=

28

4

4

4

14

1

41

1

HeH

HHe

H

HeH

H

HHe

H

He

H

HHHe

H

H

H

He

H

H

H

He

RR

RR

M

m

RR

M

mRR

M

mR

M

mRRR

M

mR

M

mR

M

m

M

m

R

R

−

−=

−=−

−

=−

+=

+

+

+

=

From spectroscopic data, RHe = 10972240 m–1 and RH = 10967770 m–1

∴ m/MH ≈ 1 /1837

This value is in excellent agreement with the value obtained by other methods

29

(2) Spectrum of singly ionised helium

Singly ionised helium He+ (a helium atom which has lost a single electron)

resembles a hydrogen atom, except that Z = 2

The nucleus is nearly four times as heavy.

Putting Z = 2 ,

Singly-ionised helium has the same type of atomic spectrum as hydrogen,

except that all wavenumbers are four times larger.

This conclusion from the theory agrees with observation except for a slight

numerical discrepancy

The reduced mass µ is slightly greater for He than for H.

−=

22

21

114

nnRH

30

(3) The discovery of deuterium

Deuterium is an isotope of hydrogen whose atomic mass is double that of

ordinary hydrogen (1 neutron and 1 proton) in the nucleus.

According to theory, for atoms with same Z, but different nuclear mass

there should be lines of slightly different wavenumbers

Because of the greater nuclear mass, the spectral lines of deuterium are all

shifted slightly to shorter wavelengths

For example, Hα line of deuterium has a wavelength of 6561 Å, while that

of hydrogen is 6563 Å.

It was also observed that the intensity of lines which are slightly shifted

towards the short wavelength side is extremely less than the corresponding

hydrogen lines.

This is because the concentration of deuterium in natural hydrogen is only

one atom in 5000.

−=

2

2

2

1

2 11

nnRz z

31

Ritz combination principle

Statement:By a combination of the terms that occur in the Rydberg or Balmer

formula, other relations can be obtained holding good for new lines and new

series.

By this principle, Ritz predicted new series of lines in the hydrogen

spectrum before they were actually discovered

−=−

−=

−=

2

2

2

42

3

42

2

32

2

11

formulae, two theseCombining

11

11

R

R

R

It is the first line of a new series in the infrared, discovered by Paschen

32

Similarly, the second line of the Paschen series can be obtained by forming

the difference of Hγ and Hα and so on.

Ritz combination principle may also be stated as follows :

If lines at frequencies νij and νjk exist in a spectrum with j > i and k > j,

then there will usually be a line at νik where ν ik = ν ij + ν jk.

However, not all combinations of frequencies are observed because certain

selection rules operate

Example:If lines of frequencies ν12 and ν23 can be represented as

ν12 = T1 – T2

ν23 = T2 – T3

then a line of frequency ν13 will exist,

where ν13 = (T1 – T2) + (T2 – T3) = T1 – T3

33

Alkali atomic spectra

Hydrogen and the alkali metals (Li, Na, K, Rb, and Cs form group 1 of the

periodic table)

All these atoms have one valence electron.

Valence electron determines the chemical characteristics of the atom.

In all these atoms except the hydrogen, the valence electron moves in a net field

of the nucleus of positive charge + Ze and the core of electrons with negative

charge – (Z–1) e surrounding the nucleus.

These electrons act as a shield between the valence electron and the nucleus.

Owing to this shielding, the effective nuclear charge is not Ze but a lesser value

Zeff. The energy is then given by

2

22

2

)( −−=−= Zn

Rhc

n

RhcZE

effn

34

Here, σ is a screening constant which is different for the different l-states of a given

value of n.

The value of Zeff is largest for 3s. Hence it is lowered much more than the 3p state.

For a given n, the S state has less energy than the P state, the P state less energy

than the D state, the D state less than the F state, and so on.

With increasing n, the energy difference between states becomes less and less

For sodium the 5d and 5f levels almost coincide

The spin angular momentum of the valence electron combines with its orbital

angular momentum and gives the total angular momentum

If l ≠ 0, the total angular momentum quantumnumber j can have the values

j = l + (1/2) or j = l– (1/2). That is, each of the l levels (l ≠ 0) splits into a doublet.

If l = 0, j takes the only value 1/2. Therefore, the S states remain as singlet. The

selection rules for the transitions are Δn = any value, Δl = ± 1, Δj = 0, ± 1

)1( +jj

35

Any new theory in Physics must reduce to well-established corresponding

classical theory when the new theory is applied to the special situation in

which the less general theory is known to be valid.

Bohr’s theory gives only the frequencies of the spectral lines and says nothing

about the nature (whether polarised or not) and intensity of lines, whereas

classical theory is very successful in this respect.

Also, according to classical theory, the frequency of the spectral line is the same

as the orbital frequency of the electron (ν = ω/2π).

But in Bohr’s theory, the frequency of the spectral line is determined by the

difference in energy between two orbital states: ν = (Ei – Ef)/h.

But it can be shown that, for transitions between states whose quantum

numbers are relatively high, the frequency of the spectral line coincides very

nearly with the orbital frequency.

Bohr’s Correspondence Principle

36

Let us consider an atom of effectively infinite mass. Then

)2(rm32

)mr2 (4)(4

mr2 =nh

or mr = nh/2 postulate,first sBohr’ toAccording

1

4

E

2

8E

large; isn whereand n'' w.r.t eqn(1) tingifferentia

)1(1

8

3 6330

4

3 20

4

30

4

2

2

330

4n

320

4

n

220

4

2

22

2

2

2

−−−−=

==

=

=

−−=

−−−−−

−=

me

nme

nnh

me

nnh

me

h

nnh

me

D

nh

meEn

37

)3(4161

241

)(

4

1

))((

4

12

4

2

0

2

6

2

0

3

2

2

0

2

2

0

2

−−−−=

=

=

=

e

m

r

squarring

e

m

r

r

emr

r

ee

r

mv

For the orbit to be in equilibrium,

2

1

2

416

m32 4

2

0

2

333

0

4 2

2

=

=

=

=

n

n

ne

mem

Sub (3) in (2)

)2(rm32 3 633

0

4

2−−−−=

me

38

Frequency given by the quantum theory becomes identical with the orbital

frequency (classical frequency) provided

n = large and Δn=1

Therefore, we may conclude that the behaviour of the atom tends

asymptotically to that expected from the classical theory in the region of

large quantumnumbers.

This correspondence principle has proved to be of great value in the

computation of :

intensity

polarisation

coherence of spectral radiation

formulation of selection rules.

39

According to Bohr, the lines in the hydrogen spectrum should each have a well-

defined wavelength.

Spectrographs of high resolving power showed that the Hα, Hβ, and Hγ lines in the

hydrogen spectrum are not single.

Each spectral line actually consisted of several very close lines packed together.

Michelson found that under high resolution, the Hα line can be resolved into two

close components, with a wavelength separation of 0.13 Å.

This is called the fine structure of the spectral lines.

Bohr’s theory could not explain this fine structure.

Sommerfeld’s Relativistic Atom Model

40

To explain the observed fine structure of spectral lines, Sommerfeld

introduced two main modifications in Bohr’s theory.

(1) According to Sommerfeld, the path of an electron around the

nucleus, in general, is an ellipse with the nucleus at one of the

foci. The circular orbits of Bohr are a special case of this.

(2) The velocity of the electron moving in an elliptical orbit varies

considerably at different parts of the orbit. This causes relativistic

variation in the mass of the moving electron.

Therefore he took into account the relativistic variation of the mass of

the electron with velocity.

Hence this is known as the relativistic atom-model.

41

Elliptic orbits for hydrogen

Consider the electron moving in an elliptical orbit round the nucleus (N).

In the case of circular orbits, there is only one coordinate that varies periodically,

namely, the angle φ that the radius vector makes with the X-axis.

In the case of elliptic motion, not only does the angle φ vary but the length of the

radius vector r also varies periodically

We have now to quantise the momenta associated with both these coordinates (φ

and r) in accordance with Bohr’s quantum condition. The two quantisation

conditions are

42

Pφ=mr2ω

43

44

45

46

47

48

49

Usually the allowed orbits are described by giving values of n and nφ.

The three orbits for n = 3 are represented by 33, 32 and 31, the subscript being

the azimuthal quantumnumber (nφ)

Another notation:

In this notation, the value of azimuthal quantum number nφ is described by

the letters, s, p, d, f, etc.

The value of nφ corresponding to these letters is 1, 2, 3, 4 etc., respectively

In this notation, the orbit determined by n = 3 and nφ = 1 is represented by

3s.

Similarly 4d will represent the orbit n = 4 and nφ = 3.

50

51

We find that the expression for the total energy is the same as that obtained by

Bohr.

This means that the theory of elliptical orbits introduces no new energy levels,

other than those given by Bohr’s theory of circular orbits.

No new spectral lines, which would explain the fine structure, are to be expected

because of this multiplicity of orbits.

Hence, Sommerfeld proceeded further to find a solution to the problem of fine

structure of spectral lines, on the basis of the variation of the mass of the electron

with velocity.

52

The velocity of the electron in a circular orbit is constant.

But the velocity of the electron in an elliptical orbit varies, being

a maximum at the perihelion and a minimum at the aphelion.

This velocity is quite large (1 / 137) ×C

Sommerfeld modified his theory, taking into account variation of

the mass of the electron with velocity.

He showed that the relativistic equation describing the path of an

electron is

Sommerfeld’s relativistic theory

53

α is a dimensionless quantity and is called the fine structure constant.

The first term in the above equation is the energy of the electron in the orbit with

the principal quantumnumber n.

The second term is the Sommerfeld relativity correction.

This term shows that the energy does depend on the azimuthal quantumnumber nφ.

This results in a splitting of the energy levels of the atom, for a given value of n,

into n components, corresponding to the n permitted values of nφ.

It can be shown that the total energy En in the relativistic theory is,

54

Hα line is due to the transition from n = 3 state to n = 2 state of the hydrogen

atom.

For n = 3, there are three possible energy levels corresponding to the three values

of nφ, 1, 2, and 3.

Similarly there are two possible energy levels for n = 2.

Theoretically, six transitions are possible : 33 → 22; 33 → 21; 32 → 22; 32 → 21 ;

31 → 22 ; 31 → 21 , and they are shown in figure

Fine structure of the Hα line

55

Actually the Hα line has only three components.

To make experiment and theory agree, some of the transitions have to be ruled

out by some selection rule.

The selection rule is that nφ can change only by +1 or –1,

i.e., Δnφ = ±1.

There is no such restriction on Δn.

On the basis of this selection rule, there are only three allowed transitions. I

In the figure, the allowed transitions are shown by continuous lines and the

forbidden lines by dotted lines