Raman Spectroscopy Questions and Answers

Q: The selection rules for vibrational Raman spectra and rotational Raman spectra are :

The rules are Δν = ±1 and ΔJ = 0, ±2 respectively.Raman scattering involves two photons, so it follows the polarisability rather than the dipole moment. For a harmonic vibration the polarisability changes linearly with displacement, giving Δv = ±1 (Stokes and anti-Stokes lines). For rotation of a linear molecule the polarisability repeats twice per turn, giving ΔJ = 0, ±2; ΔJ = 0 is the unshifted Rayleigh line.ΔJ = 0, ±1 is the rule for infrared rotation-vibration spectra, and including Δv = 0 in the vibrational rule only adds the Rayleigh line, not a vibrational Raman line.

Q: When excited with mercury line at 435.8 nm, Benzene shows first Raman shift at 606 cm\(^{-1}\). What will be the Raman shift if excited by He-Ne Laser (632.8 nm).(Given : n = 6.6 × 10\(^{-34}\) Js) :

The Raman shift is still 606 cm\(^{-1}\).A Raman shift equals the energy of a molecular vibration (or rotation) expressed in wavenumbers. It is a property of the molecule, not of the exciting light, so changing the source from the mercury line to the He-Ne laser changes the absolute wavelengths of the Raman lines but not their shift from the exciting line.The other values come from wrongly scaling the shift with the excitation wavelength or frequency. The constant given in the question (printed as n, meaning Planck's constant) is not needed.

Q: Which of the following molecules give no infrared or microwave spectra but Rotational Raman spectra ?

The molecules are \(O_{2}\) and \(H_{2}\).A pure rotational (microwave) spectrum needs a permanent dipole moment, and an infrared vibrational band needs a vibration that changes the dipole moment. Homonuclear diatomics such as \(O_{2}\) and \(H_{2}\) have no dipole and their stretching cannot create one, so they are inactive in both. Their polarizability is anisotropic, however, so they show rotational Raman spectra.HCl, HCN, CO and \(H_{2}O\) are polar and show microwave and IR spectra. \(CO_{2}\) has no microwave spectrum but its bending and asymmetric stretch are IR active, and \(CH_{4}\), a spherical top, is IR active and shows no rotational Raman spectrum.

Q: Relative intensities of successive lines in the observed Raman spectra of nitrogen molecule is :

Successive lines alternate in intensity in the ratio 1 : 2.The \({}^{14}N\) nucleus has spin I = 1, so \({}^{14}N_{2}\) is a homonuclear molecule of bosons. The nuclear spin states split into symmetric and antisymmetric sets with statistical weights in the ratio\[ \frac{I + 1}{I} = \frac{2}{1} \]and these combine with even and odd rotational levels respectively. Rotational Raman lines from even-J levels are therefore twice as intense as those from odd-J levels, and successive lines alternate weak and strong in the ratio 1 : 2.The ratio 1 : 3 is the hydrogen case (I = 1/2, ortho to para 3 : 1). The ratios 3 : 4 and 2 : 3 do not correspond to I = 1.

Q: The energy separation between two consecutive Stokes lines in Raman spectrum depends on :

The separation depends on the energy separation between vibrational levels in the excited state, option A.In a Stokes line the molecule gains energy from the photon and is left in a higher vibrational level (v = 1, 2, ...) of the electronic ground state. The first Stokes line is the shift \(E_{1} - E_{0}\) and the next is \(E_{2} - E_{0}\), so their separation is \(E_{2} - E_{1}\), the spacing between the vibrational levels the molecule is left in. For an anharmonic oscillator this spacing shrinks, which is why overtone lines are not equally spaced.Options B and D are wrong because the lines are shifts from the incident line: the position of the incident light changes the absolute wavelengths but not their separation, and the intensity affects only how strong the lines are. Option C is the weaker reading: the upper vibrational levels reached are themselves levels of the ground electronic state, so the wording is loose, but a single level v = 0 has no spacing, and the key reads the stem as referring to the excited levels.

Q: Light of wavelength 1.5 μm incident on a material with a characteristic Raman frequency of 20 × 1012 Hz results in a Stokes-shifted line of wavelength :

The Stokes line is at 1.67 μm.Work in frequency. The incident frequency is \(c/\lambda = 3 \times 10^{8}/1.5 \times 10^{-6} = 2.0 \times 10^{14}\) Hz. A Stokes line is shifted down by the Raman frequency:\[ \begin{aligned} \nu_{s} &= 2.0 \times 10^{14} - 0.2 \times 10^{14} \\ &= 1.8 \times 10^{14}\ \text{Hz} \\ \lambda_{s} &= \frac{3 \times 10^{8}}{1.8 \times 10^{14}} \\ &= 1.67\ \mu\text{m} \end{aligned} \]1.47 μm would be on the anti-Stokes side (shorter wavelength). 1.57 and 1.77 μm do not correspond to a 20 THz shift.

The selection rules for vibrational Raman spectra and rotational Raman spectra are :

A.Δν = 0, ±1 and ΔI = 0, ±1 respectively
B.Δν = ±1 and ΔJ = 0, ±1 respectively
C.Δν = ±1 and ΔJ = 0, ±2 respectively
D.Δν = 0, ±1 and ΔJ = 0, ±2 respectively

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The rules are Δν = ±1 and ΔJ = 0, ±2 respectively.

Raman scattering involves two photons, so it follows the polarisability rather than the dipole moment. For a harmonic vibration the polarisability changes linearly with displacement, giving Δv = ±1 (Stokes and anti-Stokes lines). For rotation of a linear molecule the polarisability repeats twice per turn, giving ΔJ = 0, ±2; ΔJ = 0 is the unshifted Rayleigh line.

ΔJ = 0, ±1 is the rule for infrared rotation-vibration spectra, and including Δv = 0 in the vibrational rule only adds the Rayleigh line, not a vibrational Raman line.

Discuss in Forum

When excited with mercury line at 435.8 nm, Benzene shows first Raman shift at 606 cm\(^{-1}\). What will be the Raman shift if excited by He-Ne Laser (632.8 nm).

(Given : n = 6.6 × 10\(^{-34}\) Js) :

A.1200 cm\(^{-1}\)
B.409 cm\(^{-1}\)
C.803 cm\(^{-1}\)
D.606 cm\(^{-1}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The Raman shift is still 606 cm\(^{-1}\).

A Raman shift equals the energy of a molecular vibration (or rotation) expressed in wavenumbers. It is a property of the molecule, not of the exciting light, so changing the source from the mercury line to the He-Ne laser changes the absolute wavelengths of the Raman lines but not their shift from the exciting line.

The other values come from wrongly scaling the shift with the excitation wavelength or frequency. The constant given in the question (printed as n, meaning Planck's constant) is not needed.

Discuss in Forum

Which of the following molecules give no infrared or microwave spectra but Rotational Raman spectra ?

A.HCl, HCN
B.\(\text{CO}_{2}\), \(\text{CH}_{4}\)
C.\(\text{O}_{2}\), \(\text{H}_{2}\)
D.CO, \(\text{H}_{2}\text{O}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The molecules are \(O_{2}\) and \(H_{2}\).

A pure rotational (microwave) spectrum needs a permanent dipole moment, and an infrared vibrational band needs a vibration that changes the dipole moment. Homonuclear diatomics such as \(O_{2}\) and \(H_{2}\) have no dipole and their stretching cannot create one, so they are inactive in both. Their polarizability is anisotropic, however, so they show rotational Raman spectra.

HCl, HCN, CO and \(H_{2}O\) are polar and show microwave and IR spectra. \(CO_{2}\) has no microwave spectrum but its bending and asymmetric stretch are IR active, and \(CH_{4}\), a spherical top, is IR active and shows no rotational Raman spectrum.

Discuss in Forum

Relative intensities of successive lines in the observed Raman spectra of nitrogen molecule is :

A.3 : 4
B.1 : 2
C.2 : 3
D.1 : 3

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

Successive lines alternate in intensity in the ratio 1 : 2.

The \({}^{14}N\) nucleus has spin I = 1, so \({}^{14}N_{2}\) is a homonuclear molecule of bosons. The nuclear spin states split into symmetric and antisymmetric sets with statistical weights in the ratio

\[ \frac{I + 1}{I} = \frac{2}{1} \]

and these combine with even and odd rotational levels respectively. Rotational Raman lines from even-J levels are therefore twice as intense as those from odd-J levels, and successive lines alternate weak and strong in the ratio 1 : 2.

The ratio 1 : 3 is the hydrogen case (I = 1/2, ortho to para 3 : 1). The ratios 3 : 4 and 2 : 3 do not correspond to I = 1.

Discuss in Forum

The energy separation between two consecutive Stokes lines in Raman spectrum depends on :

A.Energy separation between vibrational levels in the excited state
B.Wavelength of the incident light
C.Energy separation between vibrational levels in the ground state
D.Intensity of the incident light

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The separation depends on the energy separation between vibrational levels in the excited state, option A.

In a Stokes line the molecule gains energy from the photon and is left in a higher vibrational level (v = 1, 2, ...) of the electronic ground state. The first Stokes line is the shift \(E_{1} - E_{0}\) and the next is \(E_{2} - E_{0}\), so their separation is \(E_{2} - E_{1}\), the spacing between the vibrational levels the molecule is left in. For an anharmonic oscillator this spacing shrinks, which is why overtone lines are not equally spaced.

Options B and D are wrong because the lines are shifts from the incident line: the position of the incident light changes the absolute wavelengths but not their separation, and the intensity affects only how strong the lines are. Option C is the weaker reading: the upper vibrational levels reached are themselves levels of the ground electronic state, so the wording is loose, but a single level v = 0 has no spacing, and the key reads the stem as referring to the excited levels.

Discuss in Forum

Light of wavelength 1.5 μm incident on a material with a characteristic Raman frequency of 20 × 10¹² Hz results in a Stokes-shifted line of wavelength :

A.1.47 μm
B.1.57 μm
C.1.67 μm
D.1.77 μm

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The Stokes line is at 1.67 μm.

Work in frequency. The incident frequency is \(c/\lambda = 3 \times 10^{8}/1.5 \times 10^{-6} = 2.0 \times 10^{14}\) Hz. A Stokes line is shifted down by the Raman frequency:

\[ \begin{aligned} \nu_{s} &= 2.0 \times 10^{14} - 0.2 \times 10^{14} \\ &= 1.8 \times 10^{14}\ \text{Hz} \\ \lambda_{s} &= \frac{3 \times 10^{8}}{1.8 \times 10^{14}} \\ &= 1.67\ \mu\text{m} \end{aligned} \]

1.47 μm would be on the anti-Stokes side (shorter wavelength). 1.57 and 1.77 μm do not correspond to a 20 THz shift.

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