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Basics of UV/Visble spectroscopy

By Mallappa. Shalavadi,Lecturer,Department of Pharmacology,HSK College of Pharmacy,Bagalkot.

Contents:1. Radiation.2. Characteristics of electromagnetic radiation.3. Electromagnetic Spectrum.4. Visible light.5. Interaction of radiation with matter.6. Basic principle of UV/visible spectroscopy.

Radiation:• Radiation is the energy travelling trough space as a

series of waves or a stream of particles.• Visible light can be explained by two theories

Corpuscular theory and Wave theory.• Corpuscular theory- light travels in the form of

particles called photons.• Wave theory- light travels in the form of wave.• Radiant energy has wave nature and being associated

with electric as well as magnetic fields, these radiations are called electromag- netic radiations.

• Ex – Visible light, UV light, Infra-red, X-rays, Radio-waves ect.

Characterization of electromagnetic radiation:

1. These are produced by the oscillation of electric charge and magnetic field residing on the atom and are perpendicular to each other.

2. These are characterized by their wavelengths or frequencies or wavenumbers.

3. The energy carried by an electromagetic radiation is directly proportional to its frequency.

4. When visible light is passed through a prism, it split up into 7 coloures which correspond to definite wavelengths. This phenomenon is called dispersion.

WHAT IS Wavelength ? • It is the distance between the two adjacent

crests or troughs in a perticular wave.• Denoted by λ and expressed in Angsrtom, nm

or milli micrones.

Frequency-• Defined as number of waves which can pass

though a point in one second.• Denoted by ν (nu) and expressed in cycles per

second or in Hertz (Hz). 1Frequency α Wavelength

Wave number-• Defined as the total number of waves can pass

trough a space of one cm.• It is resiprocal to the wave length. Denoted by

ν and expressed in per cm or cm-1. 1wave number = wavelength in cm.

Energy-• Energy of a wave of the perticular radiation

can also be calculated by applying relation: E = hv = h . c/λWher, h = Plank’s constant 6.626 X 10-34 Joules sec

v = Frequency of radiation in cycles per sec.

c = Velocity of light 2.98 X 108 m/sec

λ = Wavelength in mtr.

• Calculated in joules/mole which can also be converted into kcal/mole.

• HOW TO CALCULATE ENERGY FOR PERTICULAR WAVE LENGTH?

• Ex- calculation of energy associated with radiation having wave length 200 nm.

E = hv = h . c/λ h= 6.626 X 10-34

c= 2.98 X 108

Avogadro number N= 6.02 X 1023

E= Nhc/ λ in mtr.

6.626 X 10-34 X 6.02 X 1023 X 2.98 X 108

E = 200 X 10-9

= 6,00,000 J/mole = 600 KJ/mole Since 4.1855= 1 K cal, 600 4.1855 = 143 K cal/mole for 200 nm.

Electromagnetic spectrum:• The arrangement of all radiations in order of

their increasing wavelength or decreasing frequencies is known as complementary spectrum.

• The portion above visible region is called Infra-red while that below it is called ultra-violet.

Ultra violet -------- 200-400nm.Visible--------------- 400-800nm.IR--------------------- 667-4000/cm or 2.5-15 µ.

Relationship between energy, wavelength and frequency:

Visible light:• The visible spectrum is the electromagnetic

spectrum that is visible to the human eye.• The longest wavelength is red and the shortest

is violet.

• Violet: 400 - 420 nm • Indigo: 420 - 440 nm • Blue: 440 - 490 nm • Green: 490 - 570 nm • Yellow: 570 - 585 nm • Orange: 585 - 620 nm • Red: 620 - 780 nm

Relationship between Absorption of radiation and colors:

Newton’s wheal-

• When white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed.

• Here, complementary colors are diametrically opposite each other. • Thus, absorption of 420-430 nm light renders a substance yellow, and

absorption of 500-520 nm light makes it red. • Green is unique in that it can be created by absoption close to 400 nm as

well as absorption near 800 nm.

Matter:• All organic compounds are capable of

absorbing ECM radiation because all contain valency electrons that can be excited to higher energy levels.

• The electrons that leads to absorption are a. Sigma electrons (σ): These associated with

the saturated bonds. located in sigma bond ex- C-C, C-H, O-H, C-N,N-N ect.

b. Pi electrons (π): These electrons are involved in unsaturated compounds

Ex- Alkenes, Alkynes and Aromatic compounds C=C, C=N, C=O, C=S, ect.c. Non bonding electrons (n): n electrons are

less firmly held or non bonding electrons and found on nitrogen, oxygen, sulphur and halogens.

ex- :O: N: :S: ect.

Interaction of matter with radiation:• electromagnetic radiation interacts with

materials because electrons and molecules in materials are polarizable

• Types of interactions • Absorption • Reflection • Transmission • Scattering • Refraction

UV/Visible spectroscopy• The alternate name for this technique is

Electronic Spectroscopy since it involves the promotion of electrons from the ground state to higher energy state.

• This involves the radiations range from 200nm to 800nm.

• It is absorption spectroscopy.

Principle-• Any molecule has either n, π and σ or a

combination of these electrons .• These bonding (π and σ ) and non bonding

electrons absorb the characteristic radiation and undergoes transition from ground state to excited state.

• By the characteristic absorption peaks, the nature of the electrons present.

TYPES OF ELECTRONIC TRANSITIONS

σ → σ* Transitions• The energy required is large because σ- electrones. • The transition occurs in mainly in saturated

compounds. Examples: Methane -122nm Ethane -135nm Propane -135nm Cyclopropane- 190nm• Below 200nm O2 and N2 from air absorb thus whol

path is evacuated thus called Vacum UV region.

• Why Hydrocarbons are called UV transparent?• Because they require high energy for

excitation i.e below 200nm.• Ex- Propane – 135 nm.

n → σ* Transitions• Saturated compounds containing atoms with lone

pairs (non-bonding electrons) are capable of show n → σ* transitions.

• These transitions usually need less energy than σ → σ* transitions.

• Examples: methanol-203 ethanol -204 ccl4 -257 methyl iodide -258 methyl chloride -172-175

• Why methyl iodide has loger wave length compare to methyl chloride?

π → π* Transitions• This type of transition occurs in unsaturated

compounds contain double bonds or triple bonds and also in aromatics.

• The excitation of π electron requires smaller energy hence transition occurs at longer wavelength.

• Mainly in alkenes, alkynes, carbonyl compounds, cyanides, azo compounds, etc.

• Unconjugated or isolated alkenes-below 200 nm. and conjugated compounds-above 200 nm.

• Examples-

n → π*• In this type of transition, an

electron of unshared electron pair on hetero atom gets excited to pi * anti bonding orbital.

• This requires least energy hence occurs at longer wavelength.

• Examples: aldehydes and ketones

References:Instrumental analysis by B. K. SharmaElementary organic spectroscopy by Y. R.

Sharmawww.Google.com