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Atomic Absorption: Energy Levels / transitions

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.

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Atomic Absorption: Energy Levels / transitions

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  1. 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

  2. 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

  3. Molecular Absorption Spectra: Dr. David Johnson: CH0004 Lecture Notes - Term 1

  4. 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

  5. Molecular Absorption: E2 Vibrational State E1 Electronic State Rotational State E0 IR VIS UV Dr. David Johnson: CH0004 Lecture Notes - Term 1

  6. 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

  7. Dr. David Johnson: CH0004 Lecture Notes - Term 1

  8. 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

  9. Dr. David Johnson: CH0004 Lecture Notes - Term 1

  10. Dr. David Johnson: CH0004 Lecture Notes - Term 1

  11. 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

  12. 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

  13. 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

  14. Dr. David Johnson: CH0004 Lecture Notes - Term 1

  15. Transmittance v Absorbance: Dr. David Johnson: CH0004 Lecture Notes - Term 1

  16. 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

  17. 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

  18. 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

  19. Transmission and Perceived Colour: Dr. David Johnson: CH0004 Lecture Notes - Term 1

  20. Complimentary Colours: Dr. David Johnson: CH0004 Lecture Notes - Term 1

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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

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