Lecture 1 Spectrophotmetry Dr. Niveen.pptx

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1. INSTRUMENTAL ANALYSIS Prof. Niveen A. Mohamed Pharmaceutical Analytical Chemistry Department Faculty of Pharmacy Assiut University 2025-2026 Lecture No. 1 Pharmaceutical Analytical Chemistry Department

2. Spectrophotometry Spectroscopy : is the study of interaction between electromagnetic radiation and matter . Spectrophotometry : deals with the specific interaction of light energy and solutions. Molecules absorb, reflect ( انعكاس ) or transmit light energy. The color of a solid object is determined by what wavelengths are reflected by the object (the other wavelengths being absorbed ). The color of a solution is determined by those wavelengths that are absorbed or transmitted by the molecules in that solution. Absorbance is a characteristic of a substance just like melting point, boiling point, density, and solubility. Absorbance can be related to the amount of the substance in solution; thus, it can be used to quantitatively determine the amount of substance that is present. 3

3. How Light Interacts with Matter? By three ways: 1- Absorption 2- Reflection 3- Transmission 4

4. Light and radiation Light is an electromagnetic radiation (EM), composed of both electric and magnetic components. It displays the property of continuous waves and can be described by the characteristics of wave motion. Light also behaves like a particle . Today, we envision ( نتصور الضوء ) light as a self-contained packet of energy, a photon , which has both wave and particle-like properties. 5

5. 6 Light waves , propagate at the highest known velocity of 300.000 Km/Sec. Such wave motion may be classified according to: 1- Wavelength (  , lambda) which is the linear distance measured along the line of propagation, between crest of one wave to that of the next wave 2- Amplitude which is the vertical distance from midline of a wave to the peak or trough.

6. 7 Different units of length are used to express wavelengths and their amplitude in different parts of the EM-spectrum. For example; In UV-Visible region (190-800 nm) , the units used are A (Angstrom) and nm (nanometer), while in infra red region the units are the  m (micron). m = meter Thus 1 m = 10 2 cm = 10 3 mm = 10 6 m = 10 9 nm = 10 10  o . or  = 10 -1 nm = 10 -4 m = 10 -7 mm = 10 -8 cm = 10 ‑10 m. 3- Frequency (  , nu ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s) 4- Wavenumber (  - , nu par), which is expressed in cm -1 . When the wavelength (  ) is expressed in cm, 1/  gives the number of waves per cm or wavenumber (  - ) = 1/  , cm -1 .

7. Fundamentals of Spectrophotometry Summary of Properties of Light 1.) Particles and Waves Light waves consist of perpendicular, oscillating electric and magnetic fields Parameters used to describe light amplitude (A) : height of wave’s electric vector Wavelength (l) : distance (nm, cm, m) from peak to peak Frequency ( n ): number of complete oscillations that the waves makes each second Hertz (Hz): unit of frequency, second -1 (s -1 ) 1 megahertz (MHz) = 10 6 s -1 = 10 6 Hz - Wavenumber (  - , nu par), which is expressed in cm -1 . When the wavelength (  ) is expressed in cm, 1/  gives the number of waves per cm or wavenumber (  - ) = 1/  , cm -1 .

8. 9 Relations between l , n and n are given by the following equations: C =  x  Since  = 1/  Then  = 1/  =  /C Or C =  /  Where C is the velocity of light in vacuum = 3 x 10 10 cm/Sec. Note that, the longer the wavelength, the lower the frequency and the smaller the wavenumber and vice versa. Relations between l , n and n Example: If we have a visible radiation of 500 nm, then: l in cm = 500 x 10 -7 = 5 x 10 -5 cm  = 1/  = 1/5 x 10 -5 = 0.2 x 10 5 = 2 x 10 4 cm -1 and n = C X n = 3 X 10 10 . 2 X 10 4 = 6 X 10 14 Hz

9. Light as energy Energy in matter occurs in the form of packets referred to as quanta . Light like any other matter consists of energy packets called photons . The absorption and emission of light by compounds occur in these packets (photons). The energy (E) of a photon is directly proportional to the frequency and inversely proportional to the wavelength. It can be related to C, l and n by the following equation: E = h  = h C/  Where h is a constant called Planck’s constant , which equal to 6.625 x 10 -27 erg. sec. or 6.625 X 10 -34 J. sec Example: What is the energy of a 500 nm photon? C =  x  n = c/ l = (3 x 10 10 cm s -1 )/(5.0 x 10 -5 cm) nm = 10 9cm n = 6 x 10 14 s -1 E = h n =(6.626 x 10 -27 erg.s )(6 x 10 14 s -1 ) = 4.0 x 10 -12 erg Photons composed of either; 1- Monochromatic beam (one single wavelength), or 2- Polychromatic beam (of several wavelengths). Dr.Noha M. Hosny 10

10. Fundamentals of Spectrophotometry Summary Properties of Light 1.) Particles and Waves Relationship between Frequency and Wavelength Relationship between Energy and Wavelength where: c = speed of light (3.0x10 8 m/s in vacuum)) n = frequency (sec -1 ) l = wavelength (m) where: = (1/ l ) = wavenumber As frequency ( l ) decreases, energy (E) of light increases Photons composed of either; 1- Monochromatic beam (one single wavelength), or 2- Polychromatic beam (of several wavelengths).

11. Electromagnetic spectrum Electromagnetic radiation can be divided into various regions according to wavelength, wavenumber or frequency. There are many regions useful for analysis such as x-ray, UV-Visible, infrared, radio wave, microwave and more…. Dr.Noha M. Hosny 12 The near ultraviolet region has wavelengths from 200 to 400 nm. The visible spectrum extends from 400 to 800 nm and is further subdivided into colors according to their wavelengths.

12. Electromagnetic Spectrum -ray x-ray Visible IR w Rw 10 -9 10 -7 10 -5 10 -3 10 -1 10 2 Wavelength (  , cm ) Frequency (  , Hz ) 10 8 10 12 10 14 10 16 10 18 10 20 UV Wavelength (nm) 400 500 600 700 800

13. Spectrophotometry Dr.Noha M. Hosny 14

14. The absorbed and perceived ملموس spectral band are diametrically opposite each other in the color wheel . Dr.Noha M. Hosny 15 Note: the absorbed color is different from the actually observed color. The observed color is complementary color which remained after the substance absorbed from white light.

15. How Light Interacts with Matter? 16 An electron will interact with a photon . An electron that absorbs a photon will gain energy. An electron that loses energy must emit a photon. The total energy (electron plus photon) remains constant during this process; involves transitions from ground state energy levels to “excited” states. The reverse process is called emission For absorption to occur, the energy of the photon must exactly match an energy level in the atom (or molecule) it contacts E photon = E electronic transition There are two types of absorption: 1- Atomic 2- Molecular

16. 17 1- ATOMIC ABSORPTION The energy of photon that can interact with a transition jump depends on the energy difference between the electronic levels . Absorption and emission for the sodium atom in the gas phase, Illustrates discrete منفصلة energy transfer 590 nm 330 nm Comment as increase  as E,,,,,,

17. 18 2- MOLECULAR ABSORPTION More complex than atomic absorption because many more potential transitions exist. A molecule may absorb light energy in three ways: By raising an electron to a higher energy level ( electronic ) visible and UV light . By increasing the vibration of constituent nuclei ( vibrational ) when molecule absorb IR irradiation . By increasing the rotation of molecule around its axis ( rotational ) when molecule absorb F-IR irradiation. E total = E electr + E vibrat + E rotat The total energy of a molecule is the sum of its electronic, its vibrational energy and its rotational energy. When a molecule interacts with photons in UV-Vis. Region , the absorption of energy results in displacing an outer electron (valence electron) in the molecule; is given by the equation:  E = Es - Eg = h  = h C/ 

18. SPECTRUM Spectrum is a plot of absorpance (A) versus  or  , it may be: a) Line spectrum: Occur with atoms such as sodium metal which has a sharp line of  at 589 nm. b) Band spectrum: occurs with molecules due to the presence of different vibrational and rotational sub-levels which the molecules may occupy on transition to excited state; combined with electronic transition result in band rather than line spectra. Dr.Noha M. Hosny 19 A l n Band spec. (molecules) Line spec.(atoms) l max

19. b) Band spectrum: 20 There are two parameters which define an absorption band: 1) Its position (  max ) on wavelength scale. 2) Its intensity ( A max ) on the absorbance scale. An excited molecule may be returned to the ground state in about 10 -8 seconds. Energy must be released in the form of: 1- Heat : When excited electron returns directly to the ground state. 2- Light: If it returns via a second excitation state, light is emitted as fluorescence or phosphorescence. 3- Molecular collisions. A l n l max A max

20. Types of electronic transitions - Absorption of radiation in the UV-VIS region depends upon the number and arrangement of electrons in absorbing molecules. - The outer electrons in an organic molecule may occupy one of three different energy levels (  - ,  - or n- energy level) . Accordingly, there are three types of electrons; a)  - electrons ; They are bonding electrons, represent valence bonds and possess the lowest energy level (the most stable) b )  -electrons ; They are bonding electrons, forming the p -bonds (double bonds), and possess higher energy than s -electrons. c ) n -electrons ; They are non-bonding electrons, present in atomic orbitals of hetero atoms (N, O, S or halogens). They usually occupy the highest energy level of the ground state. 21

21. Types of electronic transitions In excited state the s -electrons occupy an anti-bonding energy level ( s *) and the transition is termed s - s * transition. p -electrons occupy an anti-bonding energy level ( p *) and the transition is termed p - p * transition, While the n-electrons may occupy s * or p * levels to give n- s * or n- p * transition. 22

22. Thank you for your attention