slide1
Download
Skip this Video
Download Presentation
Atomic Absorption: Energy Levels / transitions

Loading in 2 Seconds...

play fullscreen
1 / 26

Atomic Absorption: Energy Levels / transitions - PowerPoint PPT Presentation


  • 406 Views
  • Uploaded on

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. Atomic Absorption Spectra:. 285 nm. 5p. 4p. 330 nm. 3p. 590 nm. 3s. Absorption Spectrum for Sodium Vapour:. 200. 300. 400. 500.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' Atomic Absorption: Energy Levels / transitions' - quinta


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

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

slide2

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

slide3

Molecular Absorption Spectra:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide4

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

slide5

Molecular Absorption:

E2

Vibrational State

E1

Electronic State

Rotational State

E0

IR

VIS

UV

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide6

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

slide8

X

V

M

G

UV

IR

R

 / nm

10-2

102

106

1010

The Regions of the Electromagnetic Spectrum:

Energy: Transition: Spectroscopy:

G Gamma Nuclear Gamma

X X-ray Inner (core) XRF XPS

Electrons

UV-Vis Bonding UV-Vis

Electrons AAS / AES

IR Infrared Rotation IR

Vibration

M Microwave

R Radiowave Spin orientation NMR / ESR

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide11

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

slide12

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

transmittance absorbance the beer lambert law
Transmittance, Absorbance & The Beer-Lambert Law:

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

slide15

Transmittance v Absorbance:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide16

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

So Io / I = 3.68

A = log [3.68] = 0.57

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide17

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

slide18

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

 Colour Comp. Colour

< 380 UV Colourless

380-435 Violet Green/Yellow

435-480 Blue Yellow

480-500 Blue/Green Orange

500-560 Green Red

560-580 Green/Yellow Violet

580-595 Yellow Blue

595-650 Orange Blue/Green

650-770 Red Green

> 770 IR Colourless

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide19

Transmission and Perceived Colour:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide20

Complimentary Colours:

Dr. David Johnson: CH0004 Lecture Notes - Term 1

slide21

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

slide22

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

slide23

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

slide24

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

So A = 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

slide25

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

slide26

i. (I / Io) = 0.42  (I / Io) = 2.38

A = 0.377

ii.  = c /  = 300 x 106 s-1

 = 1.0 m

iii E = h = 300 x 106 s-1

E = 2.0 x 10-25 J

iv E = 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

ad