photochemistry & spectroscopy 3.1 unit iii · photochemistry & spectroscopy 3.3 3.3 laws of...

PHOTOCHEMISTRY & SPECTROSCOPY 3.1

UNIT III

PHOTOCHEMISTRY

AND SPECTROSCOPY

PHOTOCHEMISTRY

3.1 INTRODUCTION

Photochemistry is the study of chemical reactions resulting from the exposure

of light radiations. The radiations of visible and ultraviolet portion spectrum ranging

from 2,000 - 8,000 Å. Wavelength are mainly of concern in photochemical reaction.

3.2 PHOTOCHEMICAL REACTIONS

All photochemical reactions take place in two stages. In the primary process,

the reacting molecules undergo activation by absorption of light. This is followed

by t he seco ndary pr o cess, in which t he act ivat ed mo lecules undergo a

photochemical charge.

Example: Combination of hydrogen and chlorine, the primary process is

Cl2 h 2Cl*

The activated chlorine atoms Cl*

undergoes chemical reaction as

*

This is a secondary process.

H2

+ Cl HCl + H

3.2 Engineering Chemistry-I

6 14 2 6 11

2 2

28 20

(i) Dark reactions

The chemical reactions, which take place in the absence of light are called

dark reactions.

(ii) Thermal reactions

Chemical reactions, which takes place by the absorption of heat are called

thermal reactions.

3.2.1 Classification of photochemical reactions

(i) Dissociation reactions

Example:

2HI

h H I

(g) 2( g) 2(g)

h

2HBr(g)

H2(g)

+ Br2(g)

(ii) Double decomposition reactions

Example : C H Br h C H Br HBr

(iii) Rearrangement reactions

2053Å

Maleic acid

(iv) Addition reactions

Example : CO Cl h COCl

(v) Polymerisation reactions

Example :

2(C14 H10 ) h C H

Anthracene Dianthracene

(vi) Chain reactions

Example : h

H2 Cl2 2HCl


3.3 LAWS OF PHOTOCHEMISTRY

3.3.1 Grotthus-Draper law (or) The principle of

photochemical activation

It also referred to as the first law of photochemistry. This law states that “only

light which is absorbed can be effective in producing a chemical change.”

This law has a certain limitation. It has been found experimentally that a

light of particular wavelength is responsible for the photochemical change. This

light is called the effective light of the photochemical change.

Example: When white light is passed through, alkaline solution of cupric

tartarate, the dissolved substance undergoes photo chemical reduction to cuprous

oxide.

Absorption spectrum measurements show that infrared and ultraviolet are also

absorbed by the solution, but ultraviolet light only covering a small range of

wavelengths brings about the photochemical reduction. Hence, the ultraviolet light

is the effective light for this photochemical change.

3.3.2 Stark-Einstein law

It states that in the primary process, each reacting molecule is activated by

one quantum of effective light. Hence the law implies a 1:1 relationship between

the number of reacting molecules and the number of quanta of light absorbed. It

is applicable only to the primary process.

Explanation

A A B C

Reactant

molecule Excited

molecule

Product

molecule-1

Product

molecule-2

Fig. 3.1 Illustration of law of photochemical equivalence


Fig.3.1 shows that a molecule ‘A’ absorbs a photon of light and gets

excited (Primary step). The excited molecule (A*) then decomposes to yield B

(secondary step).

A h A * (Pr imary step)

A * B (Secondary step)

Overall reaction A h B

3.4 PHOTOPHYSICAL LAW

The absorption of light in the near UV and visible region by a solution is

governed by a photophysical law is known as Beer-Lambert’s law.

3.4.1 Lambert’s law

It states that when a beam of monochromatic radiation passes through a

homogeneous absorbing medium, the rate of decrease of intensity of radiation dI

with the thickness of the absorbing medium dx is proportional to the intensity of

incident radiation I.

Mathematically it is expressed as

dI dx

kI

.....(1)

Where k = absorption coefficient or proportionality constant

On integrating the equation (1) between limits I = I0

at x = 0 and I = I

at x = x, we get

I dI

I Io

xx

x 0

kdx

ln I

Io

kx


I

Io

ekx

I Ioe kx .....(2)

This equation is known as Lambert’s law.

3.4.2 Beer’s law (or) Beer Lambert’s law

This law states that when a beam of monochromatic radiation passes

through a solution of an absorbing substance, the rate of decrease of intensity of

radiation dI with the thickness of the absorbing solution dx is proportional to

the intensity of the incident radiation I as well as the concentration of the

solution C

Mathematically, it is expressed as

dI dx

kI c .....(3)

Where k = Molar absorption coefficient.

On integrating the equation (3) between limits I = I0 at x = 0 and I = I at

x = x, we get

I dI

I Io

x

= k C d x 0

ln I

Io

ln I

Io

= kCx

= – kCx

By taking natural logarithm

2.303 log I

Io

= kCx


log I

Io =

k Cx

2.303

A = Cx

.....(4)

Where, = k / 2.303 = molar absorptivity (or) molar extinction coefficient.

log I 0/I = A= Absorbance (or) Optical density.

The equation (4) is called Beer-Lambert’s law.

This equation is known as Beer’s law or Beer-Lambert’s law.

Thus, the absorbance A is directly proportional to the molar concentration

C and path length (x).

3.4.3 Application of Beer-Lambert’s law

Determination of unknown concentration

First absorbance ‘As’ of a standard solution of known concentration ‘C

s’ is

measured, then according to Beer-Lambert’s law.

AS

= Cs x

AS x CS

....(5)

Now, we measure the absorbance Au, of a solution of unknown concentration Cu, Since and x are constants. We have

Au =

Cu x

A u

Cu

x

From equations (5) and (6)

As

Cs

x

=

A u

Cu

A u

Cu

.....(6)


C

u =

Au Cs

As

.....(7)

Since the values of Au

and As

are experimentally determined values and Cs

is known, the value of unknown concentration Cu

can be calculated from the

equation (7).

3.4.4 Limitations of Beer-Lambert’s law

If the solution contains impurities, deviation may occur.

If the solution undergoes polymerization, deviation also occurs

It is applicable for dilute solutions

It is not obeyed if the radiation used is not monochromatic.

It is not applied to suspensions.

PROBLEM 1 :

Find out the absorbance (or) optical density of a solution, if the

transmittance of a solution is 16.4%

Given :

Percentage (%) of transmittance (% T) = 16.4% (or)

16.4 0.164

Transmittance (T) =

100

Solution :

Absorbance (or) Optical Density A =

log 1

(or) log T T

= – log 0.164

= –(– 0.785)

Absorbance (or) Optical density A = 0.785


Absorbance A = – log T

= – log 0.186

= – (– 0.730)

A = 0.730

PROBLEM 2 :

The percentage transmittance of 5×10–4

M solution of disodium fumarate

in a 1 cm c ell is 1 8. 6% . Ca lcu l ate ( i) th e abso rban ce ( A)

(ii) the molar absorption coefficient ( )

Given :

% T = 18.6% (or)

T

C

=

=

0.186

5×10–4

M

x = 1 cm

Solution :

(i)

(ii)The molar absorption coefficient

We know that A =

Cx

A

Cx

0 .7 3 0 =

5 1 0 4

1

= 1.46 × 103

dm3

mol–1

cm–1

PROBLEM 3 :

The percentage transmittance of 5×10–4

m solution of disodium fumarate

in a 1 cm cell is 17.8%. Calculate

(i) the absorbance (A)

(ii) the molar absorption coefficient ( )


Given :

Solution :

% T = 17.8%

(or) T = 0.178

C = 5×10–4

M

x = 1 cm


= – log 0.178

= – (– 0.750)

A = 0.750

(ii)The molar absorption coefficient

A

We know that Cx

A Cx

0 .7 5 0 =

5 1 0 4

1

= 1.5 × 103

dm3

mol–1

cm–1

PROBLEM 4 :

Calculate the molar absorption of a solution 1×10–4

M. The absorbance

of which is 0.24, when the path length is 2.0 cm.

Given : A = 0.24

C = 1 × 10– 4

M

x = 2.0 cm

Solution :

We know that

A

Cx

A Cx

0 .2 4 =

1 1 0 4

2

= 1.2 × 103

dm3

mol–1

cm–1


% T = 40%

(or) T = 0.40

x

=

=

2.0 cm

6,000 dm3

mol–1

PROBLEM 5 :

A solution shows a transm ittance of 20% when taken in a cell of

2.5 cm thickness. Calculate its concentration of the molar absorption coefficient

is 12,000 dm3

mol–1

cm–1

. (AU. Jan.2005)

Given :

Solution :

% T = 20%

(or) T = 0.20

x = 2.5 cm

= 12,000 dm3

mol–1

cm–1


= – log 0.20

A = 0.699

We know that

A

Cx

0.699

A Cx

= 12000 2.5

= 2.33 × 10–5

mol dm–3

PROBLEM 6 :

A solution of thickness 2cm transmits 40% incident light. Calculate the

con cen t rati on o f th e so lu ti on g iven th e = 60 00 d m3

mo l–1

cm – 1

(TAU. June 2010)

Given :

cm–1


Solution :


= – log 0.40

= – (–0.398)

A = 0.398

C A

We know that x A Cx

0.398 =

6000 2

C = 3.316 × 10–5

mol dm–3

PROBLEM 7 :

A m on och rom atic radiation is in ciden t on a solution of 0.05 M

concentration of an absorbing substance. The intensity of the radiation is reduced

to one-forth of the initial value after passing through 10 cm length of the

solu tion. Calculate the m olar extinction coefficient of the substance.

(AU, B.Tech, May 2003)

Given :

C = 0.05 M

x = 10 cm

I = 1/4

I = 1

Solution :

According to Beer-Lambert’s law

log I

Io

Cx

This equation may be written as

=

log I

Cx I o


= log 1 / 4

1

= log 0.25

0.5

0.05 10

= (0.6020)

0.5

= 1.204

Molar extinction co-efficient = 1.204 dm3

mol–1

cm–1

3.5 QUANTUM EFFICIENCY ()

The number of moles of the substance undergoing photochemical change per

Einstein of radiation absorbed is known as quantum efficiency.

Alternatively, it is the number of molecules of the substance undergoing

photochemical change per quantum of radiation absorbed.

Thus, No. of molecules reacting in a given time

No. of quanta of light absorbed in the same time

If we measure the rate of formation of product, then quantum yield can be

written as,

d[Pr oduct] / dt

Number of quanta absorbed

d[Pr oduct] / dt

Iabs

Where, Iabs

= Number of quanta absorbed.


3.5.1 Low or high quantum yield

The quantum efficiency varies from zero to 106. If a reaction obeys the

Einstein law, one molecule is decomposed per photon, the quantum yield

1 .

(i) Low quantum yield

When the number of molecules decomposed is less than one per photon, the

quantum yield 1 and the reaction has a low quantum yield.

(ii) High quantum yield

1

When two or more molecules are decomposed per photon, the quantum yield

and the reaction has a high quantum yield.

3.5.2 Conditions for high and low quantum yield

All the reactant molecules should be initially in the same energy state and

hence equally reactive.

Molecules in the activated state should be largely unstable and decompose

to form the products.

The reactivity of the molecules should be temperature independent.

3.5.3 Causes (or) Reasons for low quantum yield

Excited molecules may get deactivated before they form products.

Excited molecules may lose their energy by collisions with non-excited

molecules.

Molecules may receive insufficient energy thereby reaction does not occur.

Reversion of the primary photochemical reaction.

Recombination of fragments (obtain by dissociation) will give low

quantum yield.


Example:

(i) Dimerization of anthracene to dianthracene

2C14 H 10 h C28H20

Quantum yield = 2, Actually it is found to be 0.5.

Because, the above reaction is reversible

2C14 H 10 h

thermal

C28H20

3.5.4 Causes (or) Reason for high quantum yield

Absorption of radiation produces atoms or free radicals, which initiate a series

of chain reactions.

Formation of an intermediate product, which acts as a catalyst.

If the reaction is exothermic the heat evolved may activate other molecules,

without absorption of additional quanta of radiation.

The active molecules produced by primary absorption may collide with other

molecules and activate them which inturn activate other reacting molecules.

Example:

Decomposition of HI

In the primary reaction, HI molecule absorbs a photon and dissociated to produce

are H and I.

HI h H I .....(1) Primary process

H + HI H2

+ I .....(2) Secondary process

I + I I2

2HI + h H2

+ I2

.....(3) Overall reaction


In the secondary reaction, step(1) H react with HI to produce H2

and I. This I

react with step(1) I to produce I2.

T he o ve r all r eact io n sho ws t hat t he t wo HI a r e de co mpo sed fo r

one photon ( h ). so the quantum yield () =2.

3.6 DETERMINATION OF QUANTUM EFFICIENCY (OR)

QUANTUM YIELD

Quantum yield, of a photochemical reaction is expressed as

Number of molecules reacted (or) formed

Number of photons (or) Quanta of radiation absorbed

(or)

Number of moles reacted (or) formed

Number of Einsteins of radiation absorbed

We can calculate the quantum yield from the determination of the following:

(i) Determination of the number of molecules reacted in a given time.

(ii) Determination of the amount of photons absorbed in the same time.

3.6.1 Experimental determination of number of

molecules reacted

The number of molecules reacted in a given time can be determined by the

usual analytical techniques used in chemical kinetics.

The rate of reaction is measured by the usual methods. Small quantities of

the samples are pipetted out from the reaction mixture from time to time and the

concentration of the reactants are continuously measured by the volumetric methods

(or) optical rotation (or) absorption.

From the data, the amount and number of molecules can be calculated.


Fig. 3.2 Apparatus used for measurement of light intensity

3.6.2 Experimental determination of amount of

photons absorbed

Photochemical reaction takes place by the absorption of photons of the light

by the reactant molecules. Hence, it is essential to determine the intensity of the

light absorbed.

Radiation emitted from a source of light (sunlight, mercury vapour lamp,

tungsten filament) is passed through the lens, which produces parallel beams.

(Fig.3.2). Then the parallel beams are passed through a monochromator (or) filter,

which yield s a beam o f t he o ne wave leng t h o nly. T he lig ht fr o m t he

monochromator is allowed to enter into the reaction cell immersed in a thermostat,

containing the reaction mixture.

The part of the light that is not absorbed fall on a detector, which measures

the intensity of radiation.

Among the so many detector, the most frequently employed one is the

chemical actinometer.

The chemical actinometer

It is a device used to measure the amount of radiation absorbed by the system

in a photochemical reaction. The rate of a chemical reaction can be measured by

using chemical actinometer.


2 2

2

Uranyl oxalate actinometer

The most commonly used chemical actinometer is uranyl oxalate actinometer.

It contains 0.05 M oxalic acid and 0.01 M Uranyl sulphate in water.

When it is exposed to radiation, oxalic acid undergoes decomposition to give

H2O, CO and CO

2

UO2

h UO2 *

UO2 *

+ COOH

COOH

2+

UO + CO + CO + H O

2 2 2

The residual concentration of oxalic acid can be calculated by titrating with

standard KMnO4. The amount of oxalic acid consumed is a measure of the

intensity of radiation.

Calculation of the amount of radiation absorbed.

Step-1 : The empty cell (or) the cell filled with solvent is exposed to raidation

and reading is noted = Total incident energy.

Step-2 : The cell is filled with the reactants and again the reading is noted

= Residual energy.

Total Residual

Total energy absorbed by the reacting mixture = incident energy energy transmitted

3.7 PHOTOPHYSICAL PROCESS

Generally atoms or molecules goes to excited state by the absorption of

suitable radiation. When the absorbed radiation is not used to cause chemical

reaction, it will be re-emitted as light of longer wavelength.


3.7.1 Types of photophysical process

There are two types of process:

(i) Fluorescence

(ii) Phosphorescence

(i) Fluorescence

When a molecule or atom absorbs radiation of short wavelength (high

frequency) it gets excited. Then the excited atom (or) molecule re-emits the

radiation of the same frequency (or) lower frequency within short time about 10–8

sec. Fluorescence stops as soon as the incident radiation is cut off. This process

is called fluorescence.

The substance which exhibits fluorescence is called fluorescent substance.

Examples: Calcium fluoride (CaF2), Uranium, Petroleum, iodine, mercury,

organic dyestaff (fluorescein), etc.

Uses

Fluorescent dyes are used in paints and fabrics to make them glow, when

bombarded with ultraviolet photons in sunlight.

(ii) Phosphorescence

When a molecule or atom absorbs radiation of higher frequency, the emission

of radiation is continuous for sometime, even when the incident radiation is cut

off. This process is called phosphorescence (or) delayed fluorescence.

The substance which exhibits phosphorescence is called phosphorescent

substance.

Examples: Zinc sulphide, alkaline earth sulphides (CaS, BaS and SrS).

Applications

Phosphorescent materials are used in luminous clock dials and in other

objects that are made to glow in the dark.


3.7.2 Difference between phosphorescence and

fluorescence

Photophorescence Fluorescence

Decay period is very short.

i.e., 10–9

to 10–4

S.

Decay period is much longer.

i.e., 10–4

to 100 s.

It can be observed in solution

at room temperature.

It is not observed in solution

at room temperature.

It is the radiation emitted in a

transition between states at

same multiplicity.

It is the radiation emitted in a

transition between states of different

multiplicity.

Its spectrum is not mirror image

of the absorption spectrum.

Its spectrum is mirror image of

the absorption spectrum.

Example: Zinc sulphide,

Sulphides of alkaline earth-metals.

Example: CaF2, Petroleum, Uranium.

3.7.3 Mechanism of photophysical process (or)

Jablonski diagram

Most molecules I possess on even number of electrons and in ground state,

all electrons are paired.

(a) (b)

( c)

Fig. 3.3 Spin orientation due to absorption of light


(i) Spin multiplicity

The term 2S+1 is known as spin multiplicity.

Where, S is the total electronic spin.

(ii) Singlet ground state

When the spins are paired ( ) [Fig. 3.3(a)], the clockwise orientation of

one electron is cancelled by the anticlockwise orientation of other electron.

Thus, S = s1

+ s2

1

1 0

=

2 2

2S + 1 = 2 (0) + 1

= 0 + 1 = 1

i.e., spin multiplicity of the molecule = 1

The molecule is in the singlet ground state.

(iii) Triplet excited state

On absorption of a photon of energy, h , one of the paired electrons goes

to a higher energy level (i.e. excited state), the spin orientations of the two electrons

may be either

(a) Parallel () [Fig.3.3(b)]

Then, S = s1

+ s2

= 1

1 1

2 2

2S + 1 = 2 (1) + 1

= 2 + 1 = 3


The molecule is in the triplet excited state.


(iv) Singlet excited state

(b) Anti parallel () Fig. 3.3. (c)

Then, S = s1

+ s2

1

1 0

=

2 2

2S + 1 = 2 (0) + 1

= 0 + 1 = 1


The molecule is in the singlet excited state.

Mechanism on explanation

Depending upon the energy h of the photon, electron can jump to any

higher electronic state. So we get the series of

(i) Singlet excited states S1, S

2, S

3.... etc.

(ii) Triplet excited state T1, T

2, T

3 ... etc.

(a) S1, S

2, S

3, ... etc. are called first singlet excited state, second singlet

excited state, third singlet excited state, etc. respectively.

(b) T1, T

2, T

3, ... etc. are called first triplet excited state, second triplet

excited state, third triplet excited state, etc. respectively.

Generally singlet excited state has higher energy than the corresponding

triplet excited state. Thus,

E(S1) > E (T

1) ; E(S

2) > E (T

2) ; E(S

3) > E (T

3) and so on.

When a molecule absorbs light radiation, the electron may jump from So

to S1,

S2

(or) S3

singlet excited state depending upon energy of the light radiation as shown

in Jablonski diagram (Fig. 3.4). For each singlet excited state there is a corresponding

triplet excited state.


S3

Internal conversion T3

S2 Inter-system

crossing

Internal conversion

S1

T2

Inter-system crossing

T1

Absorption Fluorescence Phosphorescence

S0

Fig. 3.4 Jablonski diagram of various photophysical processes

Thus, Mo h M*

Where, Mo

= Molecule in the ground state

M*

= Molecule in the excited state

3.7.4 Types of transitions

The activated molecules returns to the ground state by emitting its energy

through the following general types of processes.

(i) Non-radiative transitions

These transition do not involve the emission of any radiation, so these are

referred to as radiationless (or) non-radiative transitions.

(a) Internal conversion (IC)

It involves the return of the activated molecule from the higher excited states

to the first excited states.

The energy of the activated molecule is emitted in the form of heat, through

molecular collisions. This process occurs in less than about 10–11

s.


S3 S1

S2 S1

(b) Intersystem crossing (ISc)

The energy of the activated molecule is lost through transition between states

of different spins (i.e. different multiplicity). For example, S2

to T2

or S1

to T1.

Such transitions are forbidden occurs relatively at slow rates.

(ii) Radiative transition

It involves the return of the activated molecule from the singlet excited state

S1

and triplet excited state T1

to the singlet ground state. So, these transitions are

accompanied by the emission of radiation.

(a) Fluorescence

The emission of radiation due to the transition from singlet excited state S1

to ground state So

is called fluorescence.

S

1 So

This transition is allowed transition and occurs in about 10–8

second.

(b) Phosphorescence

The emission of radiation due to the transition from the triplet excited state

T1

to the ground state So

is called phosphorescence.

T

1 So

This transition is forbidden transition and occurs in slow rate.

(iii) Quenching of Fluorescence

The fluorescence may be quenched (stopped), when the excited molecule collides

with a normal molecule before it fluoresces. During quenching, the energy of the excited

molecules get transferred to the molecule with which it collides.


3.8 CHEMILUMINESCENCE

If light is emitted at ordinary temperature, as a result of chemical reaction

the process is called as chemiluminescence. It is the reverse of a photochemical reaction.

Explanation:

In a chemiluminiscence reaction, the energy released during the chemical

reaction makes the product molecule electronically excited. The excited molecule

then emits radiation, as it returns to the ground state.

Examples:

(i) When pyragallol is oxidised by hydrogen, peroxide, chemiluminescence

is produced.

(ii) Oxidation of decaying wood and containing certain bacteria.

(iii) Grignard reagent produces greenish blue light on oxidation by air.

3.8.1 Mechanism of chemiluminescence

It can be explained by considering anion-cation reactions.

Example:

Interaction between the aromatic anions (Ar–) and cations (Ar

+). (Fig.3.5)

AMBO

+ +

BMO

Ar – Ar + Ar * Ar

Fig. 3.5 Mechanism of chemiluminescence

Ar Ar

1

Ar* Ar

1 Ar

*

Ar h


The aromatic anion (Ar–) contains one unpaired electron in the antibonding

molecular orbital (AMBO) and two paired electrons in the bonding molecular

orbital (BMO). The AMBO of the aromatic cation Ar+

is empty. When the electron

is transferred from the AMBO of the anion (Ar–) to the AMBO of the cation (Ar

+).

So, the singlet excited state 1Ar

* is formed. The excited state can be deactivated by

the emission of photon h .

3.9 PHOTOSENSITIZATION

The foreign substance (called sensitizer) which absorbs the radiation and

transfers the absorbed energy to the reactants, is called a photosensitizer.

This process is called photosensitized reaction(or) photosensitization.

Examples:

(i) Molecular photosensitizers: Benzophenone, sulphurdioxide

(ii) Atomic photosensitizers: Cadmium, Mercury, Zinc, etc.

3.9.1 Mechanism of photosensitization

Consider a general Donor (D) – Acceptor (A) system in which D (the

sensitizer) absorbs the incident photon. The triplet state of D is higher in energy

than that of A (the reactant).

(i) On absorption of the photon, the singlet excited state of donor (i.e., 1D)

is produced. 1D in-turn gives the triplet excited state

3D via inter-system

crossing (ISC).

(ii) The triplet excited state 3D then collides with A (i.e., acceptor or

reactant), thereby producing the triplet excited state of acceptor, 3A and

the ground state of donor (So).

(iii) If 3A gives the desired reaction products, the mechanism is known as

photosensitization. However, if the desired reaction products results from 3D state, then A is known as quencher and the process is called

quenching. Thus, the reactions depicting photosensitization and quenching

may be described as below:


D h 1D

1 D ISC 3D

3 D A D

3 A

3 A products (photosensitization)

3 D products (Quenching)

1 1

D or S )

ISC

A ( 1

2D 2A

(or T1) (or T1)

S0

Donor, D

(Sensitizer)

S0

Acceptor, A

(Reactant)

Fig. 3.6 Mechanism of photosensitization

The triplet excited state of the sensitizer or donor (1D) must be higher in

energy level than the triplet excited state of the reactant or acceptor (3A) so that

enough energy is available to raise the reactant molecule to its triplet excited

state.

Examples for photosensitized reactions:

1. Dissociation of H2

molecule.

2. Photosynthesis in plants.


SPECTROSCOPY

3.10 INTRODUCTION

Analytical technique is a powerful tool used to study the structure of atoms and

molecules. It is also used in the analysis of various compounds.

It deals with the study of interaction of electromagnetic radiation with the matter.

During the interaction, the energy is absorbed or emitted by the matter. The

measurement of the absorbed or emitted radiation frequency is made using spectroscopy.

3.11 TYPES OF SPECTROSCOPY

The study of spectroscopy can be carried out by the following title.

1. Atomic spectroscopy

2. Molecular spectroscopy

3.11.1 Atomic spectroscopy

It is obtained due to the interaction of the electromagnetic radiation with atoms.

The atoms absorb radiation and gets excited from the ground state electronic energy

level to another level. The structure and energy levels of atoms can be obtained from

their atomic spectra.

3.11.2 Molecular spectroscopy

It is obtained due to the interaction of the electromagnetic radiation with

molecules. The molecules absorb radiation and get excited from the ground state

energy (electronic, vibrational and rotational) levels to another level. The structure,

stability and other useful informations of molecules can be obtained from molecular

spectra.


E

Differences between atomic spectra and molecular spectra

S.No. Atomic spectra Molecular spectra

1

It is line spectra

It is a complicated spectra

2

It is due to electronic transition

from one lower level to another

higher level in an atom

It is due to electronic,

vibrational and rotational

transitions in a molecule.

3

It is due to the interaction of

atoms with electromagnetic

radiation.

It is due to the interaction of

molecules with electromagnetic

radiation.

3.12 SPECTRUM

(i) Absorption spectrum

If the electromagnetic radiations are passed through a substance, radiations of

particular wavelengths are absorbed by the substance. Appearance of dark pattern of

lines which correspond to the wavelength absorbed is called the absorption spectrum

(Fig. 3.7a). (Excited

E2 state) E2

h h

(Ground 1 state)

E1

(a) Absorption spectrum (b) Emission spectrum

Fig. 3.7 Spectrum

(ii) Emission spectrum

It is produced by the emission of radiant energy by an excited atom

(Fig. 3.7b).


3.13 ELECTROMAGNETIC RADIATION

The electromagnetic radiations or waves possess electrical and magnetic properties.

For example, radio waves, visible light, UV light, infrared rays, rays, etc. They

do not require a supporting medium for their transmission. They can readily pass through

vac uum. T he y t r avel in t he spee d o f lig ht

(3×1010

cm/sec.)

The energy of various rays are compared are given below:

Cosmic ray > rays > X-rays > UV-rays > Visible rays > IR rays >

Microwave > Radiowave

Fig. 3.8 Electromagnetic radiation

An electromagnetic radiation consists of electric and magnetic fields which

oscillates in planes at right angles to each other (perpendicular to each other) and

also right angles to the direction of propagation of the waves.

3.13.1 Some importants definitions

(i) Wavelength ( ) : The linear distance between successive maxima or

minima of a wave is known as wavelength.

(ii) Frequency ( ) : The number of vibrations or oscillations per second

is known as frequency.

(iii) Velocity (C) : The product of wavelength and frequency is equal to the

velocity of light.

(iv) Wave number ( ) : It is defined as “the reciprocal of wavelength.”


Relationship between and C

Relationship between wavelength, frequency, wave number and velocity is given

as

1

C

3.14 ELECTROMAGNETIC SPECTRUM

The electromagnetic spectrum can be defined as “the arrangement of all types

of electromagnetic radiation in order to their wavelengths or decreasing frequencies

is known as complete electromagnetic spectrum.”

100 10

1 10

2 10

3 10

4 10

6 10

8 10

16

X UV Visible Near Far Micro Radio

rays rays rays region IR IR waves waves

V I B G Y O R

3800 4800 4500 4900 5300 5800 6500 7600

Fig. 3.9 Complete magnetic spectrum

3.15 ABSORPTION OF RADIATION

When electromagnetic radiation is passed through a matter, the following

changes takes place.

(i) As the photons of electromagnetic radiations are absorbed by the matter,

vibrational changes or rotational changes, electronic transition may takes

place.

After absorption, molecules get excited from the ground state to excited

state. Then they reemit electromagnetic radiation (or) liberate energy

quickly in the form of heat.


(ii) In some cases, the portion of electromagnetic radiation, which passes into

the matter, instead of being absorbed may be reflected or re-emitted or

scattered.

(iii) When the electromagnetic radiation is absorbed or scattered, it may undergo

changes in orientation or polarisation.

(iv) In some cases the molecules absorbs radiation and get excited.

3.15.1 Factors influencing absorbance

(i) The concentration of the molecular.

(ii) The nature of the absorbing molecules.

(iii) The length of the path of the radiation through the matter.

Absorbing molecules

Light source

Incident light Io

Transmitted light I x

Fig. 3.10 Absorbance of photons by the matter

3.16 MOLECULAR (OR) ABSORPTION SPECTRA

The molecular spectra arises from the following three types of transitions,

viz., (i) vibrational transition (ii) Rotational transition (iii) Electronic transition.

Those transitions are accompanied by the absorbance of electromagnetic

radiation.


Energ

y

Ero

t E

rot

Ex

cite

d

elec

tro

nic

st

ate

G

roun

d e

lect

ron

ic s

tate

3.17 ENERGY LEVEL DIAGRAM

A molecule of a compound have a large number of energy levels. Small

amount of energy is required for transition between some of these while large

amount of energy is required for other levels.

The energy level diagram, showing different transitions in molecules are

shown in the diagram Fig.3.11.

V’ = 1

J’ = 4 J’ = 3

J’ = 2 J’ = 1

Evib

V’ = 0

J’’ = 4 J’’ = 3 J’’ = 2 J’’ = 1

V’’ = 2 V’’ = 1

Evib

V’’ = 0

Zero point energy

Fig. 3.11 Energy level diagram showing the various molecular energies

The various types of spectra, the regions in which these spectra lie and the

energy changes that takes place in the molecules due to absorption of radiation

are listed below.


3.17.1 Vibrational spectra (or) Vibrational transition

It is associated with the to– and –fro motion of the nuclei of molecules such that

the centre of gravity does not change.

Vibrations of the atoms present in the molecule are causing vibrational energy level

spectrum.

The wavelength of infra red region is 8000 – 350000 Å. The frequency

ranges from 3.75×1014

to 8.6×1012

sec–1

. This is the low energy about 10 kJ region.

3.17.2 Rotational spectra (or) Rotational transition

It arises when the molecule rotates about an axis perpendicular to the inter-

nuclear axis and passing through the centre of gravity of the molecule. The rotation

of molecule with atoms causes rotational energy levels.

The rotational energy levels appear in microwave region which is a very low energy

level in the solar energy.

3.17.3 Electronic spectra (or) Electronic transition

Electronic energy arises due to the motion of electrons while considering the

nuclei of atoms of a molecule as fixed points. Electronic spectra results from the

transition of electron from the ground state energy level to an excited state energy

level of a molecule, due to the absorption of radiations in the visible and ultraviolet

regions. Electronic spectra occur in the visible region of 12,500 to 25,000 cm–1

those in the UV region of 25,000 to 70,000 cm–1

.


3.18 ULTRAVIOLET (UV) AND VISIBLE SPECTROSCOPY

3.18.1 Principle

UV-Visible spectra arises from the transition of valency electrons within a

molecule or ion from a lower electronic energy level (ground state E0) to higher

electronic energy level (excited state E1). This transition takes place due to the

absorption of UV (wavelength 100-400nm) or visible (wavelength 400-750 nm) region

of the electronic spectrum by a molecule (or) ion.

The actual amount of energy required depends on the difference in energy

between E0

and E1.

h E1 Eo

3.18.2 Types of electrons involved in organic molecule

The energy absorbed by an organic molecule involves transition of valency

electrons. The following three types electrons are involved in the transition.

S.No. Type of electron Examples Region Present in

1

-electrons

Saturated

hydrocarbons

CH3–(CH2)n–CH3

–

–

2. electrons Unsaturated hydro-

carbons and

aromatic compounds

UV or

Visible

C=C

or

C C

3 n-electrons Organic compounds

having N,O,X atoms

UV

Unshared (or)

non-bonded

electrons

The above three types of electrons are shown in the simple molecule

formaldehyde (HCHO).


n *

n *

*

*

3.18.3 Electronic transitions and energy level diagram

Energy absorbed in the UV and visible region by a molecule causes transitions

of valence electrons in the molecule. The possible transitions are

* ; n * ; n * and *

The energy level diagram for a molecule is shown in the Fig.3.12.

* (antibonding)

* (antibonding)

E

N

E

R

G

Y

E Increasing

n (antibonding)

(bonding)

(bonding)

Fig. 3.12 Energy level diagram

From the figure, it is clear that the energy levels for different transitions are in

the following order.

* n * * n *

3.18.4 Types of transitions involved in organic molecule

(i) σ σ * Transitions

* Transitions occur in the compounds where all the electrons are involved

in single bonds and there is no lone pair of electrons. These transitions are found in

saturated hydrocarbons.

The energy required for this transition is very large. This absorption bands takes

place in the far UV region of 120-140 nm.

Examples:

(i) CH4

at 122 nm

(ii) CH3 – CH3 at 135 nm


(ii) n * Transitions

n * transitions occur in the saturated compounds having lone pair (non-

bonding) of electrons in addition to * transitions. As the energy required

for n * transitions is less than * transitions. This absorption bands takes

place at longer wavelength in the near UV region (180 - 200 nm).

Example:

Tr imet hylamine ( (CH3 )3 N) : n *

o c cur s

max at 227 nm a nd

* occurs at

max 100nm.

(iii) *

*

Transitions

transitions occur due to the transition of an electron from a bonding

orbital to an antibonding * orbital. Molecules having electron system

(unsaturated hydrocarbons and aromatic compounds) exhibit these transitions. But,

selection rules determine whether transitions to a particular

or forbidden.

* orbital is allowed

Example:

UV spectrum of ethylene :

It shows two bands – one at 174nm, an intense band and at 200nm, a

weak band. Both are due * to transitions. As per the selection

rules, the intense band at 174 nm is due to allowed transition.

UV spectrum of unsaturated ketone

methyl vinyl ketone

: O : ||

CH 2 CH

This compounds shows

C CH3

Low intensity band at 324 nm. This is due to n *

transition.

High intensity band at 219 nm. This is due to

*

transition.


(iv) n *

n *

Transitions

transitions are shown by unsaturated molecules containing hetero atoms

like N,O,S and X. It occurs due to the transition of non-bonding, lone pair of electrons

to the antibonding orbitals ( * ). This transition shows a weak band and takes

place in longer wavelength with low intensity.

Examples:

Aldehydes and ketones having no C=C or C C bonds like acetaldehyde,

acetone.

At 270-300 nm, weak n *

transitions are found in them.

n *

transition takes place in the range of 270-300nm.

Aldehydes and ketones having double bond like

CH3 – CH = CH – CO – CH3

n * transition takes place in the range of 300-350nm.

3.18.5 Important terms used in UV-visible spectroscopy

(i) Chromophores (colour producing groups)

The presence of one or more unsaturated groups in a compound is mainly

r esponsible for the co lour o f t he compound. This gro ups are called as

chromophores.

Examples:

C C ; C C ; C N ; N N ; C O ; etc.

(ii) Chromogen

The compound containing the chromophoric group is called chromogen.

(iii) Auxochromes

Certain groups, which did not cause any colour effects in the absence of

chromophoric groups. It refers to an atom or a group of atoms which does not


give rise to absorption band on its own, but when conjugate to chromophore will cause

a red shift. Such groups are called as auxochromes or colour augmenters or deepeners.

Examples:

–OH, –OR, –NH2, –NHR, –NR

2 , –X

(iv) Some important definitions related to change in wavelength and intensity

(a) Bathochromic shift or shift: A shift of an

is called bathochromic shift.

max towards longer wavelength

(b) Hypsochromic shift or blue shift: A shift of an

wavelength is called hypsochromic shift.

max towards shorter

(c) Hyperchromic effect: It is an effect leading to an increased absorption

intensity.

(d) Hypochromic effect: It is an effect leading to decreased absorption

intensity.

Illustrations :

In bromo ethylene CH2 =CHBr, C=C is a chromo phor e and Br is an

auxochrome.

Substitution of a hydrogen atom in ethylene by a halogen atom causes a

bathochromic shift and a hyperchromic effect.

3.18.6 Instrumentation

I. Components

The most important components of UV-Visible spectrometer are

(i) Radiation source

(ii) Monochromators

(iii) Cells

(iv) Detectors

(v) Recorder


Hydrogen or deuterium lamps are commonly used as radiation source.


De

tec

tor

Requirements of a radiation source : It must be stable and supply continuous

radiation, sufficient intensity.

Source

Monochromator

sample Recorder

Beam

splitter

reference

Fig. 3.13 Block diagram of uv-visible spectrophotometer

(ii) Monochromator:

Selects appropriate wavelength of light. The important elements of a

monochromator are

an entrance slit

a dispersing element

an exit slit

The dispersing element may be a prism or a grating or a filter.

(iii) Cells

It is made out of quartz or fused silica. The cells must contains the following

characteristics

They must be uniform in construction.

The material of cell should be inert to solvents.

They must transmit the light of the wavelength used.

(iv) Detectors

The most commonly used detectors are

(a) Barrier layer cell


(b) Photomultiplier tube

(c) Photocell

These detectors converts light radiations into electrical signals.

(v) Recorder

The signal from the detector is finally received by the recorder.

II. Working of UV-Visible spectrometer

The radiation from the source is allowed to pass through the monochromator unit.

The monochromator allows a narrow range of wavelength to pass through an exit slit.

The beam of radiation coming out of the monochromator is split into two equal

beams. One half of the beam is directed to pass through a transparent cell containing

a solution of the compound to be analyzed. The another half is directed to pass through

an identical cell which contains only the solvent.

The instrument is designed in such way that it can compare the intensities of these

two beams.

If the sample absorbs light at a particular wavelength, then intensity of the sample

beam (I) will be less than that of the reference beam (I0).

The instruments give output graph, which is a plot of wavelength Vs. absorbance

of the light. The graph obtained is known as absorption spectrum.

3.18.7 Applications of UV-Visible spectroscopy

( i ) Determination of molecular weight

Molecular weight of a compound can be determined if the compound forms a

suitable derivative which gives an absorption band.

( i i ) Determination of calcium in blood serum

Calcium in blood serum can be estimated by using electronic absorption spectra.

( i i i ) Studying kinetics of chemical reactions

UV-Visible spectroscopy can be used to study the kinetics of a reaction.

Generally absorbance is directly proportional to the concentration. Hence,


absorbance is measured at different intervals of time. From the data obtained, the kinetics

of the reaction is studied.

(iv) Determination of inorganic substance

UV spectrometry can be used for the determination of inorganic substances. For

examples

Many metals (Mg, Mo, Nb, Ni, Al, Au, Cu, Cd) non-metals (B, Br2, Cl

2,Si)

– – – - and anions (NO3 , Cl , I , NO2 )

Ammonia

Lead in bone ash

(v) Identification of electron

UV and Visible spectroscopy is generally used for the identification of the

presence of a conjugated electron in a given compound.

(vi) Predicting relationship between different groups

UV Spectroscopy is used in predicting the relationship between different

functional groups.

– between two or more C–C multiple bonds (double or triple bonds)

– between C–C and C–O double bonds

– between C–C double bonds and aromatic benzene ring

(vii) Detection of purities

Ultraviolet absorption spectra have been used for the identification of

degradation products and for testing the purity in biological and pharmaceutical

research.

(viii) Determination of concentration of ozone

The concentration of ozone in urban smog was measured using UV

spectroscopy.


a

(ix) Detection of impurities

The impurities in organic compounds can be detected using electronic spectra since

the bands due to impurities are very intense. Saturated compounds have little absorption

band and unsaturated compounds have strong absorption band.

(x) Determination of vitamin

UV-spectroscopy is the best method for determining the structure of several

vitamins.

(xi) Dissociation constants of Acids and Bases

The dissociation constant (Pka) of an acid HA can be determined by

determining the ratio of [HA]/[A-] spectrophotometrically from the graph plotted

between absorbance Vs wavelength at different pH values. These values are

substituted in the equation

Pk = pH + log [HA] / [A–]

3.19 INFRARED SPECTROSCOPY

3.19.1 Principle

It is an important tool in the structural elucidation of organic compounds.

The region of electromagnetic radiation in between the visible and microwave

region is called infrared region. IR spectra is produced by the absorption of energy

by a molecule in the infrared region and the transitions occur between vibrational

levels.

Range of Infrared Radiation

The range in the electromagnetic spectrum extending from 12500 to

50 cm–1

is commonly referred to as the infrared.

The infrared region may be divided into the following three regions for

convenience.

(i) Near infrared : The region is from 12500 to 4000 cm–1

.

(ii) Infrared (or) Ordinary IR : The region is form 4000 to 667 cm–1

.

(iii) Far infrared : The region is from 667 to 50 cm–1

.


=0.8 2.5 15 20

Near IR

IR

Far IR

V=12500 4000 667 50 cm-1

Fig. 3.14 Range of IR radiation

3.19.2 Molecular Vibrations

The absorption of infrared radiations causes vibration in a molecule. There are

two types of fundamental vibrations in a molecule.

(i) Stretching

(ii) Bending

(i) STRETCHING VIBRATIONS

In this type of vibrations, the bond length increases or decreases but the bond

angle remains unlatered.

Types of stretching vibrations

(a) Symmetric stretching

(b) Asymmetric stretching

(a) Symmetric stretching

In this t ype, at oms moves in a same

direction with respect to the central atom.

(b) Asymmetric stretching

In this type, one atom approaches the central

atom while the other atom departs from it.


(ii) BENDING VIBRATIONS

In this type of vibrations, the bond angle changes but the bond length remains

unaltered.

Types of bending vibrations

(a) Scissoring

(b) Rocking

(c) Wagging

(d) Twisting

(a) Scissoring :

Two atoms approach each other.

(b) Rocking :

Scissoring Rocking

+ + + -

Wagging Twisting

The movement of the atoms takes place in the same direction.

(c) Wagging : Two atoms move up and down the plane with respect to

the central atom.

(d) Twisting : One of the atom moves up the plane while the other moves

down the plane with respect to the central atom.

3.19.3 Types of Stretching and Bending Vibrations

(i) For non-linear molecule

A non-linear molecule containing ‘n’ atoms has (3n-6) fundamental vibrational

modes.


Example :

(a) Methane (CH4)

Fundamental vibrational modes = 3n – 6

= 3 × 5 – 6

(b) Benzene (C6H

6)

= 15 – 6 = 9


= 3 × 12 – 6

(c) Water (H2O)

= 36 – 6 = 30


= 3 × 3 – 6

=

9 – 6 = 3

Stretching vibrational modes =

=

n – 1

3 – 1

= 2

These are (i) Symmetric stretching

(ii) Asymmetric stretching

Hence,

The number of bending vibrations = Fundamental – Stretching vibrational

vibrational modes modes

= 3 – 2 = 1

(i)

Symmetric stretching vibration

1 = 3652 cm-1


(ii)

Asymmetric stretching vibration

2 = 3756 cm-1

(iii)

Bending stretching vibration

3 = 1596 cm-1

Fig. 3.15 Vibrational modes of H2O

All these three vibrations involve change in dipole moment. Hence, they are

infrared active. Thus, for a vibration to be active in IR, the dipole moment of the

molecule must change.

(ii) For Linear Molecules

A linear molecule containing ‘n’ atoms has 3n–5 fundamental vibrational

modes

Example : CO2


= 3 × 3 – 5

= 9 – 5 = 4

Stretching vibrational modes = n – 1

= 3 – 1 = 2

These are (i) Symmetric stretching vibration

(ii) Asymmetric stretching vibration


O C O

symmetric stretching

1340 cm1

O C O

In plane bending

666 cm1

O C O

Asymmetric stretching

2350 cm1

O C O Out of plane bending

666 cm1

Of the four normal modes of vibration of CO2, only the asymmetric stretching

and bending vibrations (i.e., in-plane bending and out-of-plane bending) involve

change in dipolemoment. So all are IR-active. But the symmetric stretching

vibration does not involve any change in the dipolement. So it is IR inactive.

Hence,

The number of bending vibrations = Fundamental – Stretching vibrational

vibrational modes modes

= 4 – 2 = 2

The two bending vibrations have the same energy. Hence, they are called

degenerate.

3.19.4 Instrumentation

I. Components

The important components of IR spectrophotometer are


(ii) Monochromator

(iii) Cell

(iv) Detector

(v) Recorder


The main sources are


De

tec

tor

(a) Nernst glower :

It consists of a rod of the sintered mixture of oxides of Zr, Y and Er. This rod

is electrically heated to 1500oC to produce IR radiation.

(b) Nichrome wire :

Source

Monochromator

sample

cell

Recorder

Beam

splitter

reference

cell

Fig. 3.16 Block diagram of IR spectrophotometer

(ii) Monochromator

It allows the light of the required wavelength to pass through it.

Example : Optical, prism made of NaCl, KBr and grating.

(iii) Sample cell

The cell holding the test sample, must be transparent to IR radiation.

(iv) Detectors

The important IR detectors are

(a) Thermocouple

(b) Pyroelectric detectors

(c) Photoconductivity cell

They convert thermal radiant energy into electrical energy.


(v) Recorder

The recorder records the signal coming out from the detector.

II. Working or IR Spectrophotometer

The radiation emitted by the source is split into two equal, identical beams having

equal intensity.

One of the beams passes through the sample and the other through the reference

sample. When the sample cell contains the sample, the half beam travelling through it

becomes less intense.

When the two half beams (one coming from the reference and the other from

the sample) recombine, they produce an oscillating signal. This signal is measured

by the detecor. The signal from the detector is passed to the recorder unit and

recorded.

3.19.5 Applications of IR Spectroscopy

(i) Hydrogen bonding

It is a powerful and widely used for studying hydrogen bonding as it

decreases the vibrational frequency considerably and the decrease depends on the

strength and nature of hydrogen bonding.

There are two types of hydrogen bonding is there.

(i) Intermolecular hydrogen bonding

(ii) Intramolecular hydrogen bonding

(ii) Testing purity of a sample

Pure sample will give a sharp and well-resolved absorption bands. But impure

sample will give a broad and poorly resolved absorption bands. By comparison

with IR spectra of pure compound, presence of impurity can be detected.

(iii) Progress of reaction

Infrared spectroscopic method can be used follow the progress of a reaction.

The examination of a small portion of the spectrum is sufficient to indicate whether the

desired product is formed or not.


(iv) To study tautomerism

Tautomeric equilibria can be studied with the help of IR spectroscopy.

Example :

The common systems like keto-enol, mercapto-thioamide, lacto-lactam contain

group such as C=O, –OH, –NH (or) C=S. These groups show a characteristic

absorption band in the IR spectrum, which enable us to find out which form

predominates in the equilibrium.

(v) IR spectra is used recently for identifying inorganic materials like ferricyanide,

ferrocyanide, all types of phosphates, nitrate, etc.

(vi) Determination of molecular weight

Mo lecular weight o f a co mpo und can be det er mined by using I R

spectroscopy.

(vii) Crystallinity

The physical property like crystallinity, can be studied through changes in

IR spectra.

(viii) Determination of structure of chemical products

During the polymerisation, the polymer structure can be determined using

IR spectra.

(ix) It can be used to study the structure of coordination compounds.


ANNA UNIVERSITY QUESTIONS

AND EXPECTED QUESTIONS

1. Define Beer-Lambert’s law. (TAU, Jan.2000)

2. Derive Beer-Lambert’s law and write all the limitations observed in the quantitative

analysis. (Chen.AU, June 2009)

3. Derive the expression for the Beer-Lambert’s law. State its disadvantages.

(Coim. AU, July 2009)

4. Explain Stark-Einstein law of photochemical equivalence.

5. State Grotthus-Draper law.

6. Explain the conditions and causes for law and high quantum yield.

7. Explain the determination of quantum yield.

8. Define quantum yield.

9. With a Jablonski diagram, explain radiative and non-radiative pathways for

an electronic transition.

10. Wha t is che milu mine scence? Br ing o u t t he me chanisms o f

chemiluminescence.

11. What is photosensitization? Bring out the mechanisms of photosensitization.

12. What are the types of photophysical process? Explain.

13. Define Interconversion and Inter system crossing.

14. Explain electromagnetic spectrum.

15. Explain the different changes occuring during absorption of radiation? What

are the factors influencing it?

16. Explain in detail about the rotational, vibrational and electronic transitions.


17. What are the limitations of flame photometry.

(Coim.AU, May 2011, July 2009)

18. How is sodium estimated by Flame photometer? (Coim.AU, May 2011,

19. Describe the estimation of nickle by atomic absorption spectroscopy method.

(TAU, Jan.2010)

20. Mention the applications for IR spectroscopy. (Coim.AU, July 2010)

21. Calculate the frequency of radiations having wavelength 5000 Å.

Where C = 2.996×1010 cm/s. (Coim. AU, July 2010)

22. Discuss the application of uv spectroscopy. (Coim.AU, July 2010)

23. Explain the estimation of nickel by atomic absorption spectroscopy.

(Coim.AU, July 2009)

24. Mention few applications of calorimetry. (Coim.AU, Dec.2009)

25. Describe the components and working principle of an IR spectrophotometer

with a block diagram. (Coim.AU, Dec.2009)

26. Discuss how Nickel is estimated using atomic absorption spectroscopy.

(Coim. AU. Dec.2009)

27. Discuss in detail the principle and instrumentation of UV-visible spectroscopy.

(TAU, July 2009)

28. E xplain briefly t he pr inciple and applicat ions o f flame pho to met ry.

(TAU, July 2009)

photochemistry & spectroscopy 3.1 unit iii · photochemistry & spectroscopy 3.3 3.3 laws of...

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