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CSJMU M Pharm Modern Pharmaceutical Analytical Techniques sessional question paper UV visible IR spectrofluorimetry atomic absorption
Image ke Unit 1 syllabus ke according topics hain: UV-Visible, Derivative UV, IR including FTIR, Spectrofluorimetry, Flame Emission Spectroscopy, aur Atomic Absorption Spectroscopy. CSJMU ke official M.Pharm course outline mein bhi MPH 101T ko Modern Pharmaceutical Analytical Techniques ke roop mein listed kiya gaya hai, see official course document.
Exact paper koi guarantee nahi kar sakta, even teacher ka designation dekhkar bhi exact questions predict nahi hote. Lekin neeche ke MCQs syllabus-line based, conceptual, aur sessional ke liye highest-yield rakhe hain. Faltu advanced topics include nahi kiye gaye hain.
Most Expected MCQs: Unit 1
A. UV-Visible Spectroscopy and Derivative Spectroscopy
1. Beer-Lambert law is expressed as:
A. A = εbc
B. A = ε/bc
C. A = bc/ε
D. A = ε + b + c
Answer: A
A. A = εbc
B. A = ε/bc
C. A = bc/ε
D. A = ε + b + c
Answer: A
2. In Beer-Lambert law, absorbance is directly proportional to:
A. Transmittance only
B. Concentration and path length
C. Wavelength only
D. Refractive index only
Answer: B
A. Transmittance only
B. Concentration and path length
C. Wavelength only
D. Refractive index only
Answer: B
3. The unit of molar absorptivity (ε) is:
A. mol L⁻¹ cm⁻¹
B. L mol⁻¹ cm⁻¹
C. cm mol L⁻¹
D. L cm mol⁻¹
Answer: B
A. mol L⁻¹ cm⁻¹
B. L mol⁻¹ cm⁻¹
C. cm mol L⁻¹
D. L cm mol⁻¹
Answer: B
4. The relation between absorbance and transmittance is:
A. A = log T
B. A = 1/T
C. A = -log T
D. A = T/100
Answer: C
A. A = log T
B. A = 1/T
C. A = -log T
D. A = T/100
Answer: C
5. Percentage transmittance (%T) of a solution having absorbance 1 is:
A. 100%
B. 10%
C. 1%
D. 90%
Answer: B
A. 100%
B. 10%
C. 1%
D. 90%
Answer: B
6. If transmittance decreases, absorbance will:
A. Decrease
B. Increase
C. Remain constant
D. Become zero
Answer: B
A. Decrease
B. Increase
C. Remain constant
D. Become zero
Answer: B
7. Which is the preferred cuvette material for UV-region measurements?
A. Ordinary glass
B. Plastic
C. Quartz
D. Stainless steel
Answer: C
A. Ordinary glass
B. Plastic
C. Quartz
D. Stainless steel
Answer: C
8. Glass cells are generally unsuitable for UV spectroscopy because glass:
A. Reflects visible radiation
B. Absorbs UV radiation
C. Does not hold solution
D. Increases fluorescence
Answer: B
A. Reflects visible radiation
B. Absorbs UV radiation
C. Does not hold solution
D. Increases fluorescence
Answer: B
9. The usual UV range used in UV-Visible spectroscopy is approximately:
A. 10-100 nm
B. 200-400 nm
C. 400-800 nm
D. 800-2500 nm
Answer: B
A. 10-100 nm
B. 200-400 nm
C. 400-800 nm
D. 800-2500 nm
Answer: B
10. Visible region approximately lies between:
A. 100-200 nm
B. 200-400 nm
C. 400-800 nm
D. 800-1000 nm
Answer: C
A. 100-200 nm
B. 200-400 nm
C. 400-800 nm
D. 800-1000 nm
Answer: C
11. The most common source of UV radiation in a UV-Visible spectrophotometer is:
A. Tungsten lamp
B. Deuterium lamp
C. Hollow cathode lamp
D. Mercury lamp
Answer: B
A. Tungsten lamp
B. Deuterium lamp
C. Hollow cathode lamp
D. Mercury lamp
Answer: B
12. Tungsten filament lamp is mainly used for:
A. UV region
B. Visible region
C. IR region
D. X-ray region
Answer: B
A. UV region
B. Visible region
C. IR region
D. X-ray region
Answer: B
13. A chromophore is a group which:
A. Causes emission of sound waves
B. Absorbs UV-Visible radiation
C. Always produces fluorescence
D. Increases molecular weight
Answer: B
A. Causes emission of sound waves
B. Absorbs UV-Visible radiation
C. Always produces fluorescence
D. Increases molecular weight
Answer: B
14. An auxochrome generally causes:
A. No change in absorption
B. Shift in wavelength and/or increase in intensity
C. Only decrease in intensity
D. Complete disappearance of peak
Answer: B
A. No change in absorption
B. Shift in wavelength and/or increase in intensity
C. Only decrease in intensity
D. Complete disappearance of peak
Answer: B
15. Shift of absorption maximum to a longer wavelength is called:
A. Hypsochromic shift
B. Bathochromic shift
C. Hyperchromic shift
D. Hypochromic shift
Answer: B
A. Hypsochromic shift
B. Bathochromic shift
C. Hyperchromic shift
D. Hypochromic shift
Answer: B
16. Shift of absorption maximum to a shorter wavelength is known as:
A. Bathochromic shift
B. Hyperchromic effect
C. Hypsochromic shift
D. Red shift
Answer: C
A. Bathochromic shift
B. Hyperchromic effect
C. Hypsochromic shift
D. Red shift
Answer: C
17. Increase in intensity of an absorption band is called:
A. Hypochromic effect
B. Hyperchromic effect
C. Hypsochromic shift
D. Bathochromic shift
Answer: B
A. Hypochromic effect
B. Hyperchromic effect
C. Hypsochromic shift
D. Bathochromic shift
Answer: B
18. The main reason for deviation from Beer-Lambert law at high concentration is:
A. Change in refractive index and solute-solute interactions
B. Use of quartz cell
C. Low path length
D. Use of monochromatic radiation
Answer: A
A. Change in refractive index and solute-solute interactions
B. Use of quartz cell
C. Low path length
D. Use of monochromatic radiation
Answer: A
19. Stray light generally produces which type of deviation from Beer-Lambert law?
A. Positive deviation
B. Negative deviation
C. No deviation
D. Random deviation only
Answer: B
A. Positive deviation
B. Negative deviation
C. No deviation
D. Random deviation only
Answer: B
20. The best absorbance range for quantitative UV analysis is generally:
A. 0-0.05
B. 0.2-0.8
C. 2-4
D. More than 5
Answer: B
A. 0-0.05
B. 0.2-0.8
C. 2-4
D. More than 5
Answer: B
21. A solvent used in UV analysis should ideally have:
A. High UV absorption at analytical wavelength
B. Absorption cutoff below the analytical wavelength
C. High fluorescence
D. High color intensity
Answer: B
A. High UV absorption at analytical wavelength
B. Absorption cutoff below the analytical wavelength
C. High fluorescence
D. High color intensity
Answer: B
22. Solvent effect on UV spectrum may occur due to:
A. Hydrogen bonding and polarity
B. Only path length
C. Only sample volume
D. Only cuvette shape
Answer: A
A. Hydrogen bonding and polarity
B. Only path length
C. Only sample volume
D. Only cuvette shape
Answer: A
23. In derivative spectroscopy, the first derivative spectrum represents:
A. A versus λ
B. dA/dλ versus λ
C. d²A/dλ² versus λ
D. T versus λ
Answer: B
A. A versus λ
B. dA/dλ versus λ
C. d²A/dλ² versus λ
D. T versus λ
Answer: B
24. In the first derivative spectrum, the signal generally crosses zero at:
A. λmax of the original spectrum
B. λmin of the original spectrum only
C. Solvent cutoff
D. Detector wavelength
Answer: A
A. λmax of the original spectrum
B. λmin of the original spectrum only
C. Solvent cutoff
D. Detector wavelength
Answer: A
25. A major application of derivative UV spectroscopy is:
A. Measuring melting point
B. Resolution of overlapping spectra
C. Determining flame temperature
D. Separating ions by mass
Answer: B
A. Measuring melting point
B. Resolution of overlapping spectra
C. Determining flame temperature
D. Separating ions by mass
Answer: B
26. Derivative spectroscopy is especially useful for:
A. Removing spectral interference from overlapping bands
B. Replacing all chromatographic methods
C. Measuring particle size directly
D. Determining atomic mass
Answer: A
A. Removing spectral interference from overlapping bands
B. Replacing all chromatographic methods
C. Measuring particle size directly
D. Determining atomic mass
Answer: A
B. IR Spectroscopy and FTIR
27. IR spectroscopy mainly involves transitions between:
A. Electronic energy levels only
B. Vibrational energy levels
C. Nuclear energy levels
D. Atomic orbitals only
Answer: B
A. Electronic energy levels only
B. Vibrational energy levels
C. Nuclear energy levels
D. Atomic orbitals only
Answer: B
28. The commonly used mid-IR region is:
A. 4000-400 cm⁻¹
B. 200-400 nm
C. 400-800 nm
D. 800-2500 nm
Answer: A
A. 4000-400 cm⁻¹
B. 200-400 nm
C. 400-800 nm
D. 800-2500 nm
Answer: A
29. The unit commonly used for IR absorption frequency is:
A. Hertz
B. Nanometre
C. Wavenumber (cm⁻¹)
D. Joule
Answer: C
A. Hertz
B. Nanometre
C. Wavenumber (cm⁻¹)
D. Joule
Answer: C
30. For a molecule to show IR absorption, there must be a change in:
A. Molecular weight
B. Dipole moment
C. Boiling point
D. Refractive index only
Answer: B
A. Molecular weight
B. Dipole moment
C. Boiling point
D. Refractive index only
Answer: B
31. The functional-group region in IR spectrum is approximately:
A. 4000-1500 cm⁻¹
B. 1500-400 cm⁻¹
C. 400-100 cm⁻¹
D. 200-400 nm
Answer: A
A. 4000-1500 cm⁻¹
B. 1500-400 cm⁻¹
C. 400-100 cm⁻¹
D. 200-400 nm
Answer: A
32. The fingerprint region in IR spectrum is approximately:
A. 4000-2500 cm⁻¹
B. 2500-2000 cm⁻¹
C. 1500-400 cm⁻¹
D. 400-800 nm
Answer: C
A. 4000-2500 cm⁻¹
B. 2500-2000 cm⁻¹
C. 1500-400 cm⁻¹
D. 400-800 nm
Answer: C
33. The fingerprint region is highly useful for:
A. Confirming identity of compounds
B. Measuring solution pH
C. Determining atomic number
D. Measuring fluorescence lifetime
Answer: A
A. Confirming identity of compounds
B. Measuring solution pH
C. Determining atomic number
D. Measuring fluorescence lifetime
Answer: A
34. The approximate IR absorption range for carbonyl, C=O, stretching vibration is:
A. 3300 cm⁻¹
B. 1700 cm⁻¹
C. 2250 cm⁻¹
D. 1000 cm⁻¹
Answer: B
A. 3300 cm⁻¹
B. 1700 cm⁻¹
C. 2250 cm⁻¹
D. 1000 cm⁻¹
Answer: B
35. A broad absorption around 3200-3600 cm⁻¹ generally indicates:
A. C=O stretching
B. O-H stretching
C. C≡N stretching
D. C-Cl stretching
Answer: B
A. C=O stretching
B. O-H stretching
C. C≡N stretching
D. C-Cl stretching
Answer: B
36. The approximate absorption region for C≡N or C≡C stretching is:
A. 3300-2500 cm⁻¹
B. 2260-2100 cm⁻¹
C. 1800-1600 cm⁻¹
D. 1500-400 cm⁻¹
Answer: B
A. 3300-2500 cm⁻¹
B. 2260-2100 cm⁻¹
C. 1800-1600 cm⁻¹
D. 1500-400 cm⁻¹
Answer: B
37. Hydrogen bonding usually causes O-H stretching frequency to:
A. Shift to a higher wavenumber and sharpen
B. Shift to lower wavenumber and broaden
C. Completely disappear
D. Remain unchanged
Answer: B
A. Shift to a higher wavenumber and sharpen
B. Shift to lower wavenumber and broaden
C. Completely disappear
D. Remain unchanged
Answer: B
38. Conjugation with a carbonyl group usually shifts C=O stretching frequency:
A. To higher wavenumber
B. To lower wavenumber
C. To zero
D. Into UV region
Answer: B
A. To higher wavenumber
B. To lower wavenumber
C. To zero
D. Into UV region
Answer: B
39. Number of fundamental vibrational modes for a non-linear molecule containing N atoms is:
A. 3N
B. 3N-3
C. 3N-5
D. 3N-6
Answer: D
A. 3N
B. 3N-3
C. 3N-5
D. 3N-6
Answer: D
40. Number of fundamental vibrational modes for a linear molecule containing N atoms is:
A. 3N-6
B. 3N-5
C. 3N-3
D. 3N
Answer: B
A. 3N-6
B. 3N-5
C. 3N-3
D. 3N
Answer: B
41. Which is NOT a fundamental molecular vibration?
A. Stretching
B. Bending
C. Rotation of monochromator
D. Scissoring
Answer: C
A. Stretching
B. Bending
C. Rotation of monochromator
D. Scissoring
Answer: C
42. KBr pellet technique is generally used for:
A. Solid samples in IR spectroscopy
B. Gaseous samples in UV spectroscopy
C. Metals in AAS
D. Solutions in flame photometry
Answer: A
A. Solid samples in IR spectroscopy
B. Gaseous samples in UV spectroscopy
C. Metals in AAS
D. Solutions in flame photometry
Answer: A
43. Nujol mull method is used for:
A. Solid samples in IR spectroscopy
B. UV solvent selection
C. Flame emission analysis
D. Atomic absorption source preparation
Answer: A
A. Solid samples in IR spectroscopy
B. UV solvent selection
C. Flame emission analysis
D. Atomic absorption source preparation
Answer: A
44. In an FTIR instrument, the key component is:
A. Michelson interferometer
B. Hollow cathode lamp
C. Flame burner
D. Nebulizer
Answer: A
A. Michelson interferometer
B. Hollow cathode lamp
C. Flame burner
D. Nebulizer
Answer: A
45. FTIR initially records data as an:
A. Absorbance spectrum directly
B. Interferogram
C. Chromatogram
D. Calibration curve
Answer: B
A. Absorbance spectrum directly
B. Interferogram
C. Chromatogram
D. Calibration curve
Answer: B
46. Fourier transformation converts an interferogram into:
A. Mass spectrum
B. IR spectrum
C. Fluorescence spectrum
D. Atomic emission spectrum
Answer: B
A. Mass spectrum
B. IR spectrum
C. Fluorescence spectrum
D. Atomic emission spectrum
Answer: B
47. The advantage of FTIR where all wavelengths are measured simultaneously is called:
A. Beer advantage
B. Fellgett or multiplex advantage
C. Lambert advantage
D. Stokes advantage
Answer: B
A. Beer advantage
B. Fellgett or multiplex advantage
C. Lambert advantage
D. Stokes advantage
Answer: B
48. Higher energy throughput in FTIR due to absence of slits is called:
A. Jacquinot advantage
B. Stokes shift
C. Quenching effect
D. Beer effect
Answer: A
A. Jacquinot advantage
B. Stokes shift
C. Quenching effect
D. Beer effect
Answer: A
49. The major application of IR spectroscopy in pharmaceutical analysis is:
A. Functional-group identification and drug identification
B. Elemental analysis of sodium only
C. Particle counting only
D. Determination of molecular mass by ions
Answer: A
A. Functional-group identification and drug identification
B. Elemental analysis of sodium only
C. Particle counting only
D. Determination of molecular mass by ions
Answer: A
C. Spectrofluorimetry
50. Fluorescence occurs when an excited molecule returns from:
A. Excited singlet state to ground singlet state
B. Excited triplet state to ground singlet state only
C. Ground state to excited state
D. Ionic state to metallic state
Answer: A
A. Excited singlet state to ground singlet state
B. Excited triplet state to ground singlet state only
C. Ground state to excited state
D. Ionic state to metallic state
Answer: A
51. Fluorescence emission generally occurs at:
A. Shorter wavelength than excitation
B. Longer wavelength than excitation
C. Exactly zero wavelength
D. Same wavelength in every case
Answer: B
A. Shorter wavelength than excitation
B. Longer wavelength than excitation
C. Exactly zero wavelength
D. Same wavelength in every case
Answer: B
52. The difference between excitation and emission maxima is called:
A. Beer shift
B. Stokes shift
C. Bathochromic effect
D. Multiplex effect
Answer: B
A. Beer shift
B. Stokes shift
C. Bathochromic effect
D. Multiplex effect
Answer: B
53. Fluorescence is generally more sensitive than UV-Visible absorption spectroscopy because it measures:
A. Emitted radiation against a dark background
B. Only transmitted radiation
C. Only heat produced
D. Atomic weight directly
Answer: A
A. Emitted radiation against a dark background
B. Only transmitted radiation
C. Only heat produced
D. Atomic weight directly
Answer: A
54. The most commonly used radiation source in spectrofluorimetry is:
A. Xenon arc lamp
B. Hollow cathode lamp
C. Deuterium lamp only
D. Tungsten lamp only
Answer: A
A. Xenon arc lamp
B. Hollow cathode lamp
C. Deuterium lamp only
D. Tungsten lamp only
Answer: A
55. The detector commonly used in spectrofluorimetry is:
A. Photomultiplier tube
B. Thermometer
C. Flame ionization detector
D. Conductivity cell
Answer: A
A. Photomultiplier tube
B. Thermometer
C. Flame ionization detector
D. Conductivity cell
Answer: A
56. In fluorescence measurement, detector is usually placed at:
A. 0° to incident beam
B. 90° to incident beam
C. 180° to incident beam
D. Any angle has no effect
Answer: B
A. 0° to incident beam
B. 90° to incident beam
C. 180° to incident beam
D. Any angle has no effect
Answer: B
57. Measurement at 90° in fluorimetry mainly reduces:
A. Scattered excitation radiation reaching detector
B. Sample concentration
C. Fluorescence emission
D. Molecular vibration
Answer: A
A. Scattered excitation radiation reaching detector
B. Sample concentration
C. Fluorescence emission
D. Molecular vibration
Answer: A
58. Quenching means:
A. Increase in fluorescence intensity
B. Decrease in fluorescence intensity
C. Increase in absorbance only
D. Shift to UV region only
Answer: B
A. Increase in fluorescence intensity
B. Decrease in fluorescence intensity
C. Increase in absorbance only
D. Shift to UV region only
Answer: B
59. Dynamic quenching is mainly due to:
A. Collision between fluorophore and quencher
B. Permanent covalent bond formation only
C. Change in path length
D. Use of quartz cuvette
Answer: A
A. Collision between fluorophore and quencher
B. Permanent covalent bond formation only
C. Change in path length
D. Use of quartz cuvette
Answer: A
60. Static quenching occurs because of:
A. Formation of a non-fluorescent complex
B. Increase in excitation intensity only
C. Use of a flame
D. Use of an interferometer
Answer: A
A. Formation of a non-fluorescent complex
B. Increase in excitation intensity only
C. Use of a flame
D. Use of an interferometer
Answer: A
61. Which factor commonly decreases fluorescence intensity?
A. High concentration causing inner-filter effect
B. Suitable dilution
C. Use of pure solvent
D. Proper wavelength selection
Answer: A
A. High concentration causing inner-filter effect
B. Suitable dilution
C. Use of pure solvent
D. Proper wavelength selection
Answer: A
62. Fluorescence intensity at low concentration is proportional to:
A. Concentration of fluorophore
B. Atomic number
C. Molecular weight alone
D. Flame temperature only
Answer: A
A. Concentration of fluorophore
B. Atomic number
C. Molecular weight alone
D. Flame temperature only
Answer: A
63. Rigid planar aromatic compounds commonly show:
A. Higher fluorescence
B. No electronic transition
C. Only atomic absorption
D. No interaction with radiation
Answer: A
A. Higher fluorescence
B. No electronic transition
C. Only atomic absorption
D. No interaction with radiation
Answer: A
64. Fluorescence can be reduced by:
A. Heavy atom substitution and molecular collisions
B. Proper wavelength selection only
C. Increasing detector sensitivity only
D. Use of a blank
Answer: A
A. Heavy atom substitution and molecular collisions
B. Proper wavelength selection only
C. Increasing detector sensitivity only
D. Use of a blank
Answer: A
65. A pharmaceutical application of spectrofluorimetry is:
A. Assay of fluorescent drugs and drugs derivatized to fluorescent products
B. Measuring viscosity only
C. Determining sodium by emission only
D. Measuring IR fingerprint only
Answer: A
A. Assay of fluorescent drugs and drugs derivatized to fluorescent products
B. Measuring viscosity only
C. Determining sodium by emission only
D. Measuring IR fingerprint only
Answer: A
D. Flame Emission Spectroscopy and Atomic Absorption Spectroscopy
66. Flame emission spectroscopy is based on:
A. Absorption of radiation by ground-state atoms
B. Emission of radiation by excited atoms
C. Molecular vibration only
D. Fluorescence from aromatic compounds only
Answer: B
A. Absorption of radiation by ground-state atoms
B. Emission of radiation by excited atoms
C. Molecular vibration only
D. Fluorescence from aromatic compounds only
Answer: B
67. Flame photometry is particularly useful for determination of:
A. Alkali and alkaline-earth metals
B. Carbonyl compounds only
C. Proteins only
D. Non-metals only
Answer: A
A. Alkali and alkaline-earth metals
B. Carbonyl compounds only
C. Proteins only
D. Non-metals only
Answer: A
68. In flame emission spectroscopy, the flame acts mainly as:
A. A detector
B. Atomizer and excitation source
C. A monochromator
D. A sample cell
Answer: B
A. A detector
B. Atomizer and excitation source
C. A monochromator
D. A sample cell
Answer: B
69. The characteristic emission of sodium is approximately:
A. 589 nm
B. 254 nm
C. 3400 cm⁻¹
D. 1700 cm⁻¹
Answer: A
A. 589 nm
B. 254 nm
C. 3400 cm⁻¹
D. 1700 cm⁻¹
Answer: A
70. Atomic absorption spectroscopy is based on absorption by:
A. Excited molecules
B. Ground-state free atoms
C. Solid KBr pellet
D. Fluorescent molecules only
Answer: B
A. Excited molecules
B. Ground-state free atoms
C. Solid KBr pellet
D. Fluorescent molecules only
Answer: B
71. The source used in atomic absorption spectroscopy is generally:
A. Hollow cathode lamp
B. Xenon lamp
C. Tungsten lamp
D. Deuterium lamp
Answer: A
A. Hollow cathode lamp
B. Xenon lamp
C. Tungsten lamp
D. Deuterium lamp
Answer: A
72. A hollow cathode lamp is:
A. Element-specific radiation source
B. Universal IR source
C. Fluorescence detector
D. Type of cuvette
Answer: A
A. Element-specific radiation source
B. Universal IR source
C. Fluorescence detector
D. Type of cuvette
Answer: A
73. The function of nebulizer in AAS is to:
A. Convert liquid sample into fine aerosol
B. Detect emitted light
C. Select wavelength
D. Record interferogram
Answer: A
A. Convert liquid sample into fine aerosol
B. Detect emitted light
C. Select wavelength
D. Record interferogram
Answer: A
74. In AAS, the burner/flame is primarily used for:
A. Atomization of analyte
B. Measuring transmittance
C. Generating UV radiation
D. Producing fluorescence
Answer: A
A. Atomization of analyte
B. Measuring transmittance
C. Generating UV radiation
D. Producing fluorescence
Answer: A
75. AAS is most commonly used for determination of:
A. Trace metals
B. Functional groups of organic compounds
C. Protein sequence
D. Melting point
Answer: A
A. Trace metals
B. Functional groups of organic compounds
C. Protein sequence
D. Melting point
Answer: A
76. Chemical interference in AAS occurs due to:
A. Formation of stable compounds that reduce free atoms
B. Wrong wavelength selection only
C. Detector failure only
D. Higher volume of solvent only
Answer: A
A. Formation of stable compounds that reduce free atoms
B. Wrong wavelength selection only
C. Detector failure only
D. Higher volume of solvent only
Answer: A
77. Ionization interference in AAS occurs when:
A. Analyte atoms become ionized in the flame
B. Sample becomes fluorescent
C. Cuvette is dirty
D. KBr is impure
Answer: A
A. Analyte atoms become ionized in the flame
B. Sample becomes fluorescent
C. Cuvette is dirty
D. KBr is impure
Answer: A
78. One method to reduce ionization interference in AAS is use of:
A. Ionization suppressor/buffer
B. Nujol mull
C. Derivative spectrum
D. Quartz cell
Answer: A
A. Ionization suppressor/buffer
B. Nujol mull
C. Derivative spectrum
D. Quartz cell
Answer: A
79. Spectral interference in AAS can result from:
A. Overlapping absorption or emission lines
B. Only low sample concentration
C. Only use of distilled water
D. High path length only
Answer: A
A. Overlapping absorption or emission lines
B. Only low sample concentration
C. Only use of distilled water
D. High path length only
Answer: A
80. Major difference between flame emission spectroscopy and AAS is:
A. Flame emission measures emitted light; AAS measures absorbed light
B. Both measure only absorbed light
C. Both use only KBr pellets
D. Both are IR techniques
Answer: A
A. Flame emission measures emitted light; AAS measures absorbed light
B. Both measure only absorbed light
C. Both use only KBr pellets
D. Both are IR techniques
Answer: A
Teacher ke conceptual/deep MCQ ke liye highest-priority revision list
In topics ko definitions ke saath nahi, logic ke saath prepare karo:
- Beer-Lambert law, absorbance-transmittance relation, deviations.
- Chromophore, auxochrome, bathochromic/hypsochromic and hyperchromic/hypochromic effects.
- UV solvent selection, solvent cutoff, quartz versus glass cell.
- Derivative spectroscopy, first derivative zero crossing and overlapping spectra.
- IR selection rule, vibrational modes, functional group versus fingerprint region.
- Factors affecting IR vibrational frequency: bond strength, reduced mass, conjugation, hydrogen bonding.
- FTIR principle: Michelson interferometer, interferogram, Fourier transform, Fellgett and Jacquinot advantages.
- Fluorescence versus phosphorescence, Stokes shift, 90-degree detector arrangement.
- Fluorescence quenching: static versus dynamic, inner-filter effect.
- Flame emission versus AAS, hollow cathode lamp, atomization, nebulizer, interferences.
Sessional strategy: Pehle Q. 1-26, phir 27-49, phir 50-65, aur end mein 66-80 revise karo. Is order mein syllabus ke almost har stated point se question cover ho raha hai.
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Mujhe isme uv visible spectroscopy samjao jo ki mein m pharm paper me likh pao mujhe iss tarike se samajna takki mujhe ek baar me sb yaad ho jye or usse me apne analytical ke paper me likh du jisse mujhe full Mark mile ye bhi dhya rkha techer ki nazar mujhpe he toh mujhe paper me accha Krna jaroori he teacher marks bhi kaat sakti he toh vo koi mauka na dena simple language me smajho koi diagram ho toh vo bhi btana jisse acche marks score ho jye
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UV visible spectroscopy Beer Lambert law pharmaceutical analysis instrument components derivative spectroscopy review
UV-Visible Spectroscopy: M.Pharm Answer Notes
Paper mein heading, diagram, formula, aur applications zaroor likhna. Is format mein likhoge to answer complete lagega aur marks cut hone ka chance kam hoga.
One-line definition:
UV-Visible spectroscopy is an absorption spectroscopic technique in which a substance absorbs ultraviolet or visible radiation, causing electronic transition of its molecules. The amount of absorbed light is used for identification and quantitative estimation of drugs.
UV region approximately 200-400 nm aur visible region 400-800 nm hota hai. Quantitative analysis generally Beer-Lambert law par based hota hai. Beer-Lambert law explanation
1. Introduction / Principle
Jab UV ya visible light kisi solution par pass hoti hai, to drug molecule us light ki kuch energy absorb karta hai. Is energy ko absorb karke molecule ke electrons lower energy level se higher energy level par chale jate hain. Isse electronic transition kehte hain.
Simple rule yaad rakho
Light absorb hui → electrons excite hue → absorbance measure hui → concentration calculate hui.
Electronic transitions
| Transition | Kahan hota hai? | Energy requirement |
|---|---|---|
| σ → σ* | Saturated bonds, such as C-C and C-H | Very high |
| n → σ* | Molecules with O, N, S lone-pair electrons | High |
| π → π* | Double bonds and aromatic compounds | Moderate |
| n → π* | Carbonyl, nitro, azo compounds | Lower |
Pharmaceutical compounds jisme conjugated double bonds, aromatic rings, carbonyl group, nitro group, azo group etc. hote hain, unka UV analysis easily ho sakta hai.
Chromophore
Molecule ka woh atom/group jo UV-Visible radiation absorb karta hai, chromophore kehlata hai.
Examples:
- C=C
- C=O
- N=N
- NO₂
- Benzene ring
Auxochrome
Aisa group jo khud weak absorption show karta hai, lekin chromophore ke saath judkar absorption wavelength ya intensity change kar deta hai, auxochrome kehlata hai.
Examples: -OH, -NH₂, -OR, -SH.
2. Important Terms: Exam mein 2-3 lines likh dena
λmax
Wavelength at which a substance shows maximum absorption is called lambda max (λmax).
Why analysis at λmax?
Because at λmax absorbance maximum hoti hai, sensitivity better hoti hai, aur small wavelength error ka effect comparatively kam hota hai.
Because at λmax absorbance maximum hoti hai, sensitivity better hoti hai, aur small wavelength error ka effect comparatively kam hota hai.
Absorbance
Light ka woh part jo sample absorb kar leta hai.
[
A = \log_{10}\frac{I_0}{I}
]
Where:
- (I_0) = intensity of incident light
- (I) = intensity of transmitted light
- (A) = absorbance
Transmittance
Sample se pass hone wali light ka fraction.
[
T = \frac{I}{I_0}
]
[
%T = \frac{I}{I_0} \times 100
]
Relation:
[
A = -\log T = \log \frac{I_0}{I}
]
One easy numerical fact
- If %T = 100, then A = 0
- If %T = 10, then A = 1
- If %T = 1, then A = 2
3. Beer-Lambert Law
This is the most important part. Isko formula, explanation aur deviations ke saath likhna.
Statement
When monochromatic radiation passes through a homogeneous absorbing solution, the absorbance is directly proportional to concentration of absorbing species and path length of the cell, provided all other conditions remain constant.
[
A = \varepsilon bc
]
Where:
| Symbol | Meaning |
|---|---|
| A | Absorbance |
| ε | Molar absorptivity or molar extinction coefficient |
| b | Path length of cell in cm |
| c | Concentration in mol/L |
Molar absorptivity ki unit:
[
L\ mol^{-1}\ cm^{-1}
]
Law ko simple example se samjho
Agar concentration double kar di jaaye, solution mein light absorb karne wale molecules double ho jaayenge. Therefore absorbance bhi double ho jaayegi.
Similarly, agar cuvette path length 1 cm se 2 cm kar dein, light ko solution ke andar double distance travel karna hoga. Isliye absorbance increase hogi.
Lambert's law
Absorbance is directly proportional to path length.
[
A \propto b
]
Beer's law
Absorbance is directly proportional to concentration.
[
A \propto c
]
Dono ko combine karne par:
[
A \propto bc
]
[
A = \varepsilon bc
]
Calibration curve
Drug ke different known concentrations prepare karte hain aur unki absorbance λmax par note karte hain. Then:
- X-axis: Concentration
- Y-axis: Absorbance
Straight line milti hai.
Diagram to draw in answer sheet
Absorbance
^
| /
| /
| /
| /
| /
|__________/________________> Concentration
Unknown sample ka absorbance measure karke calibration curve se concentration calculate karte hain.
4. Deviations from Beer-Lambert Law
Teacher conceptual question pooch sakti hain: "Explain deviations from Beer-Lambert law." Isko headings mein likhna.
A. Real deviations
High concentration par solute molecules ek dusre ke close aa jate hain. Inke beech interaction hota hai aur refractive index change ho sakta hai. Isse ε constant nahi rehta.
Result: Linearity lost ho jati hai.
B. Chemical deviations
Analyte solution mein chemical changes kar sakta hai, such as:
- Association
- Dissociation
- Ionization
- Complex formation
- Tautomerism
Example: Weak acid or weak base ka ionization pH ke saath change hoga, isliye absorbance bhi change ho sakti hai.
C. Instrumental deviations
- Polychromatic radiation: Beer law strictly monochromatic light ke liye valid hai.
- Stray light: Unwanted light detector tak pahunchti hai.
- Wide slit width: Different wavelengths sample tak reach karti hain.
- Dirty or scratched cuvette: Absorbance inaccurate aayegi.
- Detector non-linearity.
Prevention
- Dilute solution use karo
- Monochromatic radiation use karo
- Blank correct use karo
- Proper wavelength select karo
- Clean quartz cuvette use karo
- Stray light ko minimize karo
5. Instrumentation of UV-Visible Spectrophotometer
Exam mein block diagram bana kar har component ka function likhna best hota hai.
Block diagram
Radiation source
↓
Monochromator
↓
Entrance/Exit slit
↓
Sample cell (cuvette)
↓
Detector
↓
Amplifier / Signal processor
↓
Readout / Computer / Recorder
Double-beam instrument diagram
Radiation source
↓
Monochromator
↓
Beam splitter
/ \
↓ ↓
Reference cell Sample cell
\ /
↓ ↓
Detector
↓
Display/Recorder
Diagram ke neeche likho:
In double beam instruments, one beam passes through blank/reference and the other through sample. This reduces error due to source intensity fluctuations.
In double beam instruments, one beam passes through blank/reference and the other through sample. This reduces error due to source intensity fluctuations.
Components and their functions
1. Radiation source
| Region | Source |
|---|---|
| UV region | Deuterium lamp or hydrogen discharge lamp |
| Visible region | Tungsten filament lamp / tungsten-halogen lamp |
Memory trick:
UV = Deuterium
Visible = Tungsten
UV = Deuterium
Visible = Tungsten
2. Monochromator
Monochromator polychromatic light mein se required single wavelength select karta hai.
Types:
- Prism
- Diffraction grating
Modern instruments mein diffraction grating commonly used hota hai.
3. Slits
Slits narrow beam allow karti hain aur spectral resolution improve karti hain.
- Entrance slit: Light ko monochromator mein enter karata hai
- Exit slit: Selected wavelength ko sample cell tak bhejta hai
4. Cuvette or sample cell
Sample ko hold karne ke liye used.
| Region | Cuvette material |
|---|---|
| UV region | Quartz or fused silica |
| Visible region | Glass or plastic can be used |
Very important: Ordinary glass UV radiation absorb karta hai. Isliye UV measurement ke liye quartz cell mandatory hai.
Usual path length: 1 cm.
5. Detector
Detector transmitted light ko electrical signal mein convert karta hai.
Common detectors:
- Phototube
- Photomultiplier tube
- Photodiode
- Diode array detector
Photomultiplier tube highly sensitive hota hai.
6. Readout system
Electrical signal ko absorbance, transmittance, spectrum, ya concentration ke form mein display karta hai.
6. Single Beam and Double Beam Spectrophotometer
| Feature | Single beam | Double beam |
|---|---|---|
| Beam path | One beam | Two beams |
| Blank and sample | Separately measured | Simultaneously/alternately measured |
| Cost | Less expensive | More expensive |
| Accuracy | Comparatively less | Better |
| Correction for source fluctuation | Poor | Better |
| Use | Routine simple analysis | Accurate quantitative analysis |
One line for exam
Double beam spectrophotometer provides better accuracy because it compensates for variations in source intensity and detector response.
7. Selection of Solvent and Solvent Effects
Ideal solvent ki properties
A solvent used for UV analysis should:
- Drug ko completely dissolve kare.
- Analytical wavelength par absorb na kare.
- Chemically inert ho.
- Drug ke saath reaction na kare.
- High purity ka ho.
- Stable ho.
- Low UV cutoff rakhta ho.
UV cutoff
Wavelength below which a solvent itself strongly absorbs radiation is called its UV cutoff.
Rule:
Solvent ka cutoff, drug ke λmax se lower hona chahiye.
Solvent ka cutoff, drug ke λmax se lower hona chahiye.
For example, if drug ka analysis 260 nm par ho raha hai, solvent ko 260 nm par appreciable absorption nahi dikhani chahiye.
Common solvents
| Solvent | General use |
|---|---|
| Water | Polar drugs, if drug soluble and stable |
| Methanol | Very commonly used |
| Ethanol | Common UV solvent |
| Acetonitrile | Low UV absorption and good solvent |
| Hexane | Non-polar compounds |
Solvent effect
Solvent polarity and hydrogen bonding UV spectrum ko affect kar sakte hain. Isse λmax shift ho sakta hai aur absorbance intensity change ho sakti hai.
Important spectral shifts
| Effect | Meaning |
|---|---|
| Bathochromic shift / red shift | λmax shifts to longer wavelength |
| Hypsochromic shift / blue shift | λmax shifts to shorter wavelength |
| Hyperchromic effect | Absorption intensity increases |
| Hypochromic effect | Absorption intensity decreases |
Easy memory:
- Batho = bigger wavelength
- Hypso = shorter wavelength
- Hyper = higher intensity
- Hypo = lower intensity
8. Procedure for Quantitative Drug Estimation
Paper mein application ya practical procedure aaye to ye flow likhna:
Prepare standard stock solution
↓
Prepare different dilutions
↓
Scan spectrum and find λmax
↓
Measure absorbance of standard solutions at λmax
↓
Plot calibration curve: Absorbance vs concentration
↓
Prepare and measure unknown sample
↓
Calculate concentration using calibration curve / Beer-Lambert equation
Blank ka role
Blank mein all reagents and solvent hote hain except drug. It corrects absorption due to solvent, reagent, and cuvette.
9. Applications of UV-Visible Spectroscopy in Pharmaceutical Analysis
Is heading mein 6-8 points likhoge to answer strong banega.
- Assay of bulk drugs by direct UV method.
- Estimation of drugs in pharmaceutical dosage forms, such as tablets, capsules, syrups, injections.
- Dissolution studies: Dissolved drug quantity at different time intervals determine karne ke liye.
- Content uniformity testing.
- Stability studies and degradation monitoring.
- Impurity estimation, where the impurity has a different absorption maximum.
- Simultaneous estimation of two drugs using absorbance ratio or multiwavelength methods.
- Determination of pKa and study of ionization behaviour.
- Complex formation studies.
- Detection in HPLC, because many drugs possess UV-absorbing chromophores.
- Protein and nucleic acid estimation in pharmaceutical biotechnology.
- Colorimetric analysis, where a coloured complex is formed before measurement.
UV-Vis is widely applied in pharmaceutical QA/QC and research, as also noted in pharmaceutical UV-Vis use.
10. Derivative Spectroscopy
Ye syllabus ka specific point hai, isko skip mat karna.
Definition
Derivative spectroscopy is a technique in which the rate of change of absorbance with respect to wavelength is plotted against wavelength instead of plotting normal absorbance against wavelength.
Normal spectrum:
[
A \text{ versus } \lambda
]
First derivative spectrum:
[
\frac{dA}{d\lambda} \text{ versus } \lambda
]
Second derivative spectrum:
[
\frac{d^2A}{d\lambda^2} \text{ versus } \lambda
]
Principle
Normal UV spectrum mein do drugs ki absorption peaks overlap kar sakti hain. Derivative spectrum peaks ko sharper bana deta hai aur overlapping bands resolve karne mein help karta hai.
Zero crossing method
First derivative spectrum mein original absorption maximum, λmax, ke around signal often zero cross karta hai.
Agar mixture mein:
- Drug A ka derivative amplitude zero hai at a selected wavelength
- Drug B ka derivative amplitude non-zero hai
To Drug B ko Drug A ki interference ke bina estimate kiya ja sakta hai.
Simple derivative diagram
Normal spectrum:
Absorbance
^
| /\
| / \
|________/____\________> Wavelength
First derivative spectrum:
dA/dλ
^
| /\
---|-------/--\---------> Wavelength
| / \
|_____/ \_____
↑
Zero crossing
Advantages of derivative spectroscopy
- Overlapping spectra resolve karta hai.
- Binary and multicomponent formulations mein useful hai.
- Background interference reduce karta hai.
- Minor spectral differences clearly show karta hai.
- Peak shoulder ko identify karne mein help karta hai.
- No physical separation required in many cases.
Limitations
- Noise amplification ho sakta hai, especially higher derivative order mein.
- Proper wavelength interval and smoothing required hote hain.
- Very dilute samples mein accuracy reduce ho sakti hai.
- Instrument and data-processing settings ka effect aa sakta hai.
11. 10-Mark Answer ka Final Writing Format
Agar question ho: “Explain UV-Visible spectroscopy with instrumentation and applications.”
To exact sequence mein likhna:
To exact sequence mein likhna:
- Definition
- Principle and electronic transitions
- Beer-Lambert law with equation
- Absorbance and transmittance relation
- Instrumentation block diagram
- Function of source, monochromator, cuvette, detector and readout
- Solvent selection and solvent effects
- Applications
- Short conclusion
Conclusion for answer sheet
Thus, UV-Visible spectroscopy is a simple, rapid, sensitive and economical technique used for qualitative identification and quantitative estimation of pharmaceuticals. Its quantitative application is based on Beer-Lambert law, while derivative spectroscopy improves analysis of overlapping spectra.
1-Minute Revision Memory Chain
UV-Visible
↓
Molecule absorbs light
↓
Electron excitation
↓
Absorbance measured
↓
Beer-Lambert law: A = εbc
↓
Instrument: Source → Monochromator → Cuvette → Detector → Display
↓
Use quartz for UV
↓
Measure at λmax
↓
Calibration curve
↓
Drug assay, dissolution, stability, multicomponent analysis
↓
Derivative method for overlapping spectra
Must-remember one liners
- UV range: 200-400 nm
- Visible range: 400-800 nm
- UV source: Deuterium lamp
- Visible source: Tungsten lamp
- UV cuvette: Quartz
- Detector: Photomultiplier tube
- Quantitative law: Beer-Lambert law
- Formula: (A = \varepsilon bc)
- Best analytical wavelength: λmax
- Derivative spectroscopy use: Overlapping spectra ka analysis
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