Raman Spectroscopy: Understanding Stokes and Anti-Stokes Lines

[tanaygupta2000]

Homework Statement

In the vibrational Raman spectrum of HF molecule, the Raman lines are observed at wavelengths 2670 Å and 3430 Å respectively. What is the fundamental vibrational frequency of the molecule?

Relevant Equations

E(n) = ℏω(n+1/2) - χℏω(n+1/2)^2
Fundamental band = E(1)-E(0)
Energy, E = hν

I know that for the vibrational Raman spectrum, the energy levels are given by-
E(n) = ℏω(n+1/2) - χℏω(n+1/2)^2

But I'm not getting what does it meant by the Raman lines are observed at given values of wavelengths and the fundamental vibrational frequency.

I equated the E(0) with the energy corresponding to 3430 Å and E(1) with that corresponding to 2670 Å and found out the harmonic frequency (ν) and the anharmonicity constant (χ).

Then for the fundamental band, I calculated the difference E(1)-E(0) and finally divided the result so obtained with Planck's constant h to get the frequency.
The final value of the frequency I'm getting is exactly double as that of given in the answer. Are my steps correct?
Please help! Thank You

 

[Steve4Physics]

Homework Statement: In the vibrational Raman spectrum of HF molecule, the Raman lines are observed at wavelengths 2670 Å and 3430 Å respectively. What is the fundamental vibrational frequency of the molecule?
Relevant Equations: E(n) = ℏω(n+1/2) - χℏω(n+1/2)^2
Fundamental band = E(1)-E(0)
Energy, E = hν

I know that for the vibrational Raman spectrum, the energy levels are given by-
E(n) = ℏω(n+1/2) - χℏω(n+1/2)^2

But I'm not getting what does it meant by the Raman lines are observed at given values of wavelengths and the fundamental vibrational frequency.

I equated the E(0) with the energy corresponding to 3430 Å and E(1) with that corresponding to 2670 Å and found out the harmonic frequency (ν) and the anharmonicity constant (χ).

Then for the fundamental band, I calculated the difference E(1)-E(0) and finally divided the result so obtained with Planck's constant h to get the frequency.
The final value of the frequency I'm getting is exactly double as that of given in the answer. Are my steps correct?

Well, no replies yet so see if this helps...

The question is not clear. Have you stated it completely? Are the Raman lines adjacent? Are they both Stokes (as opposed to anti-Stokes) lines?

But ignoring that, there seems to be some confusion in your approach, even though it might give the correct answer.

Take a step back and forget Raman spectra for the moment. The equation $E(n) = ℏω(n+\frac 12) - χℏω(n+\frac 12)^2$ gives the vibrational energy levels for a diatomic molecule. I’d guess you are allowed to ignore the last term (enhamonicity correction) and just use $E(n) = ℏω(n+\frac 12)$, i.e. treat a molecule as a simple quantum harmonic oscillator.

The molecule’s vibrational energy levels are then $E_0 = \frac 12 ℏω$ (the ground state), $E_1= \frac 32 ℏω, E_2 =\frac 52 ℏω$ …

You are being asked to find the ‘fundamental’ frequency $ω$, but probably expressed in Hz (remember $f$ (or $\nu$ if preferred) $= \frac ω{2\pi}$).

To produce Raman scattering we fire photons (say of photon energy $P$) at the molecules.

The first question to address is “what energies are the photons produced by Raman scattering?”. You need to express these energies in terms of (some of) $P, E_0, E_1, E_2$. Can you do that?.

Typo's (due to LaTeX not rendering) fixed.

 

[tanaygupta2000]

Well, no replies yet so see if this helps...

The question is not clear. Have you stated it completely? Are the Raman lines adjacent? Are they both Stokes (as opposed to anti-Stokes) lines?

But ignoring that, there seems to be some confusion in your approach, even though it might give the correct answer.

Take a step back and forget Raman spectra for the moment. The equation $E(n) = ℏω(n+\frac 12) - χℏω(n+\frac 12)^2$ gives the vibrational energy levels for a diatomic molecule. I’d guess you are allowed to ignore the last term (enhamonicity correction) and just use $E(n) = ℏω(n+\frac 12)$, i.e. treat a molecule as a simple quantum harmonic oscillator.

The molecule’s vibrational energy levels are then $E_0 = \frac 12 ℏω$ (the ground state), $E_1= \frac 32 ℏω, E_2 =\frac 52 ℏω$ …

You are being asked to find the ‘fundamental’ frequency $ω$, but probably expressed in Hz (remember $f$ (or $\nu$ if preferred) $= \frac ω{2\pi}$).

To produce Raman scattering we fire photons (say of photon energy $P$) at the molecules.

The first question to address is “what energies are the photons produced by Raman scattering?”. You need to express these energies in terms of (some of) $P, E_0, E_1, E_2$. Can you do that?.

Typo's (due to LaTeX not rendering) fixed.

Thanks for the response, sir.
The question might be incomplete or incorrect, as I too felt I require additional information to proceed.

I consider to approach the first question as, since we bombarded photons of energy $P$ at the molecule and given that we obtain the Raman lines at these two wavelengths, then taking them to be corresponding to $E_0$ and $E_1$ respectively,
then $P = E_0 + E_1$

Solving this might help, if P had been provided.
Thanks

 

[Steve4Physics]

I consider to approach the first question as, since we bombarded photons of energy $P$ at the molecule and given that we obtain the Raman lines at these two wavelengths, then taking them to be corresponding to $E_0$ and $E_1$ respectively,
then $P = E_0 + E_1$

I think you have misunderstood how the Raman effect works.

Take a careful look at this diagram: https://ptal.eu/sites/ptal.eu/files/Raman laser.png. This shows various possibilties, but look at the transitions for Stokes Raman Scattering.

Suppose the molecule is in its ground vibrational state (‘level 0’). Its energy is $E_0$. When a photon of energy $P$ is absorbed, the molecule is raised into a ‘virtual’ state of energy $E_0 + P$.

Case 1: A Raman line is produced if the molecule decays to level 1. A photon of energy $(E_0 + P) – E_1$ is emitted. This is shown in the above diagram.

Case 2: A Raman line is produced if the molecule decays to level 2. A photon of energy $(E_0 + P) – E_2$ is emitted. (Not shown in the above diagram.)

If we assume that 2670 Å and 3430 Å are the wavelengths for the above 2 cases, we have two equations and can eliminate$P$ (so the value of $P$ is not required).

 

[tanaygupta2000]

I think you have misunderstood how the Raman effect works.

Take a careful look at this diagram: https://ptal.eu/sites/ptal.eu/files/Raman laser.png. This shows various possibilties, but look at the transitions for Stokes Raman Scattering.

Suppose the molecule is in its ground vibrational state (‘level 0’). Its energy is $E_0$. When a photon of energy $P$ is absorbed, the molecule is raised into a ‘virtual’ state of energy $E_0 + P$.

Case 1: A Raman line is produced if the molecule decays to level 1. A photon of energy $(E_0 + P) – E_1$ is emitted. This is shown in the above diagram.

Case 2: A Raman line is produced if the molecule decays to level 2. A photon of energy $(E_0 + P) – E_2$ is emitted. (Not shown in the above diagram.)

If we assume that 2670 Å and 3430 Å are the wavelengths for the above 2 cases, we have two equations and can eliminate$P$ (so the value of $P$ is not required).

Thanks a lot for clearing my doubt sir.
Can we suppose that the first line be Rayleigh and second be Stokes?
In that case we can write
$(E_0 + P) - E_0$ = Energy corresponding to 2670 $Å$
$(E_0 + P) - E_1$ = Energy corresponding to 3430 $Å$

Solving these two equations, giving me the frequency as 2.5×10^15 Hz, about 20 times greater than what is given in the answer.

Even using
$(E_0 + P) - E_1$ = Energy corresponding to 2670$Å$
$(E_0 + P) - E_2$ = Energy corresponding to 3430$Å$
makes no difference to the answer.

 

[Steve4Physics]

Can we suppose that the first line be Rayleigh and second be Stokes?

No because the question explicitly says "the Raman lines are observed at wavelengths 2670 Å and 3430 Å"

In that case we can write
$(E_0 + P) - E_0$ = Energy corresponding to 2670 $Å$
$(E_0 + P) - E_1$ = Energy corresponding to 3430 $Å$

Solving these two equations, giving me the frequency as 2.5×10^15 Hz, about 20 times greater than what is given in the answer.

You haven't shown your working so I can't tell what you've done. But it is based on the incorrect assumption that one line is not a Raman line.

Even using
$(E_0 + P) - E_1$ = Energy corresponding to 2670$Å$
$(E_0 + P) - E_2$ = Energy corresponding to 3430$Å$
makes no difference to the answer.

Using this, I get a completly different answer to "2.5x10^15 Hz". Let's call $R_1$ and $R_2$ the energies of the 2670Å and 3430Å photons. Then:
$R_1 = (E_0 + P) - E_1$
$R_2 = (E_0 + P) - E_2$
If we subtract this gives
$R_1 - R_2 = E_2 - E_1$
That’s the equation you need to use.

If gives me the 'fundamental frequency' to be $2.49\times10^{14}$ Hz. From what you have said, this appears to be about double your 'official' answer.

The official answer could be wrong (it happens). Or, if one Raman line is a Stokes line, and the other is an anti-Stokes line, this would give half the above frequency.

You need to show your detailed working or there's no way identify any mistake(s).

 

[tanaygupta2000]

No because the question explicitly says "the Raman lines are observed at wavelengths 2670 Å and 3430 Å"You haven't shown your working so I can't tell what you've done. But it is based on the incorrect assumption that one line is not a Raman line.Using this, I get a completly different answer to "2.5x10^15 Hz". Let's call $R_1$ and $R_2$ the energies of the 2670Å and 3430Å photons. Then:
$R_1 = (E_0 + P) - E_1$
$R_2 = (E_0 + P) - E_2$
If we subtract this gives
$R_1 - R_2 = E_2 - E_1$
That’s the equation you need to use.

If gives me the 'fundamental frequency' to be $2.49\times10^{14}$ Hz. From what you have said, this appears to be about double your 'official' answer.

The official answer could be wrong (it happens). Or, if one Raman line is a Stokes line, and the other is an anti-Stokes line, this would give half the above frequency.

You need to show your detailed working or there's no way identify any mistake(s).

Okay I finally got the mentioned answer by considering one Stokes line and other Anti-Stokes.
I constructed following equations:
$(E_1+P) - E_0$ = hc/(3430 A)
$(E_1+P) - E_2$ = hc/(2670 A)

Solving these two equations (subtracting them and substituting values of $E_0$ and $E_2$) finally gives me the fundamental frequency to be 1.245e14 Hz as mentioned.

I am highly grateful to you sir for two take-home messages, irreleavnt of the authenticity of the problem:
(1.) The diagram of Raman Effect. The distinct energy levels really made it clear to me how to mathematically put conditions of Rayleigh, Stokes and Anti-Stokes lines. Prior to this, I was only visualizing them to be equally-spaced in wavenumbers and that the Rayleigh line is most intense, followed by Stokes and then Anti-Stokes lines.

(2.) That the two (randomly given) Raman lines were actually one Stokes and one Anti-Stokes. In the future while solving similar questions, I will make sure to consider this as my first approach to solve questions.

 

[Steve4Physics]

Okay I finally got the mentioned answer by considering one Stokes line and other Anti-Stokes.
I constructed following equations:
$(E_1+P) - E_0$ = hc/(3430 A)
$(E_1+P) - E_2$ = hc/(2670 A)

I think you have the correct answer but for the wrong reasons!

The first Stokes photon is produced when a molecule (in the ground state, energy $E_0$) is excited to energy $E_0+P$ and decays back to energy $E_1$. The emitted photon has energy $(E_0+P) – E1$.

The first anti-Stokes photon occurs when a molecule (which happens to be in the 1st excited state, energy $E_1$) is excited to energy $E_1+P$ and decays back to energy $E_0$. The emitted photon has energy $(E_1+P) – E0$.

So your two equations should be
$(E_0+P) – E_1 = hc/(3430Å)$ (Stokes)
$(E_1+P) – E0= hc/(2670Å)$ (anti-Stokes)

This is what is shown in the diagram https://ptal.eu/sites/ptal.eu/files/Raman laser.png

By the way, if you do this using symbols, you find Planck's constant cancels leaving only simple arithmetic at the very end.