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This is my m pharm Modern pharmaceutical analytical techniques unit 1 syllabus mujhe is unit ke predicted MCQ bnao jo ki mere sessional peper me aa skte he or aye ge ye dhayn rkha deep me bhi ho skta he MCQ par ye dhyan rkhna ki teacher csjmu kanpur university me associate professor he toh ye bhi dhyan rkhna or koshis Krna ki jo tum btao usi me se question ane chiye faltu ki mat baata jo expected ho vhi btana
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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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