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Atomic Absorption: Energy Levels / transitionsPowerPoint Presentation

Atomic Absorption: Energy Levels / transitions

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Atomic Absorption: Energy Levels / transitions

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3d

4d

5p

5d

3p

4p

6s

5s

4s

3s

7s

Atomic Absorption:

Energy Levels / transitions

E

3s transitions

e.g. Na Al

3p transitions

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Atomic Absorption Spectra:

285 nm

5p

4p

330 nm

3p

590 nm

3s

Absorption Spectrum for Sodium Vapour:

200

300

400

500

600

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Molecular Absorption Spectra:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Molecular Absorption:

E = Eelec + Evib + Erot (+ Etrans)

- see next diagram

Where…..

Eelec > Evib > Erot

x10 x10 approx

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Molecular Absorption:

E2

Vibrational State

E1

Electronic State

Rotational State

E0

IR

VIS

UV

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Properties of Electromagnetic Radiation:

To account for the phenomena associated with spectroscopy, radiation must be viewed as having both WAVE and PARTICLE properties.

(a) Wave:

Continuous sinusoidal oscillations: c =

(b) Particle:

To fit QUANTUM THEORY, radiation must be considered as “packets of energy” (PHOTONS) such that:

E = h

h (Planck constant): 6.63 x 10-34

Hence:

E = h c /

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Dr. David Johnson: CH0004 Lecture Notes - Term 1

X

V

M

G

UV

IR

R

/ nm

10-2

102

106

1010

The Regions of the Electromagnetic Spectrum:

Energy:Transition:Spectroscopy:

G GammaNuclearGamma

X X-rayInner (core)XRF XPS

Electrons

UV-VisBondingUV-Vis

ElectronsAAS / AES

IR InfraredRotationIR

Vibration

M Microwave

R RadiowaveSpin orientationNMR / ESR

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Dr. David Johnson: CH0004 Lecture Notes - Term 1

UNITS:E:Joules (J)

:Metres (m)

:Hertz (Hz or s-1)

Example Calculation:

(a)Calculate the frequency (in Hz) of an X-ray of wavelength 2.65 A (Angstrom Units)

(b)Calculate the corresponding energy (in electron volts, eV)

h (Planck's constant):6.63 x 10-34 Js

c (velocity of light):3 x 108 ms-1

1 A (Angstrom unit):1 x 10-10 m

1 eV (electron volt):1.6 x 10-19 J

(a) = c / = 3 x 108 / 2.65 x 10-10

= 1.13 x 1018 Hz

(b)E = h = 1.134 x 1018 x 6.63 x 10-34 J

= 7.49 x 10-16 J

= 4682.4 eV

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Absorption Spectra:-

In either molecular spectroscopy or atomic spectroscopy, the absorbance characteristics of a species are described in terms of an ABSORBANCE SPECTRUM, which is a plot of some function of beam attenuation against , or .

The y-axis is usually either transmittance (T), Absorbance (A) or log(Absorbance).

Dr. David Johnson: CH0004 Lecture Notes - Term 1

I0

I

Path length “l”

Cell containing solution of

concentration “c”

TRANSMITTANCE:

T = (I / I0) x 100%

ABSORBANCE:

A = log (I0 / I)

BEER-LAMBERT LAW:

A = c l

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Transmittance v Absorbance:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Example calculations:

Convert the following transmittances to absorbance:

(a) 0. 1

(b) 27.2 %

(a)I / Io = 0.1

So:Io / 1 = 10

A = log [10] = 1

(b)I / Io = 27.2 / 100

SoIo / I = 3.68

A = log [3.68]= 0.57

Dr. David Johnson: CH0004 Lecture Notes - Term 1

The Beer-Lambert Law:

Absorption Concentration

More specifically:

A = log (Io / I) = abc

a = constant

b = path length

(of radiation through absorbing medium)

In particular,

If c is in mols dm-3

b is in cm

“a” is given a specific name (MOLAR ABSORPTIVITY) and is given the symbol ““

i.e. A = c l

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Colour in Solution:

e.g. why is copper sulphate solution blue???

The Cu(H2O)62+ complex ion (which is classed as a molecular ion) absorbs the RED light at 650-800 nm, leaving the resultant solution green/blue: - the COMPLIMENTARY COLOUR

ColourComp. Colour

< 380UVColourless

380-435VioletGreen/Yellow

435-480BlueYellow

480-500Blue/GreenOrange

500-560GreenRed

560-580Green/YellowViolet

580-595YellowBlue

595-650OrangeBlue/Green

650-770RedGreen

> 770IRColourless

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Transmission and Perceived Colour:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Complimentary Colours:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

1.

2.

- Atomic vs Molecular Spectra:
- Copper Nitrate – Molecular Spectrum (UV-VIS)
- 1. Nitrate anion (in the UV)
- 2. Cu(H2O)62+ (in the visible)
- (blue / green colour)

Abs

200

/ nm

800

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Atomic vs Molecular Spectra:

Copper Nitrate – Atomic Spectrum (UV-VIS)

Line Spectrum

Main line at 327.4 nm

327.4 nm

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Example Calculation - Beer Lambert Law:

Compound X (RMM = 167.0) exhibits a molar absorptivity of 1850 dm3 mol-1 cm-1 at its corresponding max (420 nm). A solution was prepared by dissolving 0.1106 g of a sample that contained 14.4% by mass of X in water and diluting to 500 cm3. Calculate a value for the transmittance (expressed as a %) of the solution measured at 420 nm using a 2 cm cell.

What colour would you expect the solution to be? - assuming that compound X was the only substance present in the sample to absorb in the visible region.

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Beers Law:A = c l

l = 2 cm

= 1850

c:0.1106 x (14.4 / 100) g in 500 mL

= 0.0159 g in 500 mL

= 0.0319 g L-1= 1.907 x 10-4 mol dm-3

SoA = 0.706(Io/I) = 5.08

T = 19.7 %(I/Io) = 0.197

Violet light absorbed (420 nm)

Yellow/Green = complimentary colour

Dr. David Johnson: CH0004 Lecture Notes - Term 1

Example Calculations:

Calculate:

i. the absorbance associated with a transmittance of 42%

ii. the wavelength associated with a frequency of 300 MHz

iii. the energy associated with a frequency of 300 MHz

iv. the energy associated with a wavenumber of 2800 cm-1

v. the frequency of X-Rays with a wavelength of 0.4 nm

Dr. David Johnson: CH0004 Lecture Notes - Term 1

i.(I / Io) = 0.42 (I / Io) = 2.38

A = 0.377

ii. = c / = 300 x 106 s-1

= 1.0 m

iiiE = h = 300 x 106 s-1

E = 2.0 x 10-25 J

ivE = h c = 2800 x 100 m-1

E = 5.57 x 10-20 J

v. = c / = 0.4 x 10-9 m

= 7.5 x 1017 Hz

Dr. David Johnson: CH0004 Lecture Notes - Term 1