Mixing
Dominant contribution of v = 0 → v = 1 transition in vibrational spectroscopy
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Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.
After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($Delta v = pm 1$), my professor's notes state that:
Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $v = 0 to v = 1$ transition and is called fundamental transition.
I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $v = 0 to v = 1$ will take place?
physical-chemistry quantum-chemistry ir-spectroscopy
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
$endgroup$
add a comment |
$begingroup$
Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.
After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($Delta v = pm 1$), my professor's notes state that:
Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $v = 0 to v = 1$ transition and is called fundamental transition.
I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $v = 0 to v = 1$ will take place?
physical-chemistry quantum-chemistry ir-spectroscopy
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
$endgroup$
2
$begingroup$
A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $nu$ (nu). The Greek letter $nu$ is used to refer to a frequency; so we can have a transition for which $Delta E = hnu$, for example.
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– orthocresol♦
Feb 3 at 21:23
$begingroup$
@orthocresol, did you mean wavelength? In $Delta E=hnu$ , $nu$ is a frequency in 1/second.
$endgroup$
– porphyrin
Mar 6 at 20:00
add a comment |
$begingroup$
Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.
After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($Delta v = pm 1$), my professor's notes state that:
Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $v = 0 to v = 1$ transition and is called fundamental transition.
I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $v = 0 to v = 1$ will take place?
physical-chemistry quantum-chemistry ir-spectroscopy
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
$endgroup$
Recently I had studied vibrational spectroscopy using the quantum harmonic oscillator model.
After stating the gross selection rule [that vibrational oscillation demands a change in dipole moment for polar molecule] and showing that transitions are possible only between adjacent energy states ($Delta v = pm 1$), my professor's notes state that:
Most of the molecules will populate the lowest energy states, following Maxwell–Boltzmann distribution, the dominant contribution to IR absorption spectra comes from $v = 0 to v = 1$ transition and is called fundamental transition.
I do not understand the point that 'Most of the molecules will populate the lowest energy states...'. What does it mean and how does it lead to the conclusion that most of the times a transition of $v = 0 to v = 1$ will take place?
physical-chemistry quantum-chemistry ir-spectroscopy
physical-chemistry quantum-chemistry ir-spectroscopy
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
edited Feb 3 at 21:26
orthocresol♦
40.3k7117247
40.3k7117247
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
asked Feb 2 at 20:32
Onkar SinghOnkar Singh
514
514
2
$begingroup$
A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $nu$ (nu). The Greek letter $nu$ is used to refer to a frequency; so we can have a transition for which $Delta E = hnu$, for example.
$endgroup$
– orthocresol♦
Feb 3 at 21:23
$begingroup$
@orthocresol, did you mean wavelength? In $Delta E=hnu$ , $nu$ is a frequency in 1/second.
$endgroup$
– porphyrin
Mar 6 at 20:00
add a comment |
2
$begingroup$
A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $nu$ (nu). The Greek letter $nu$ is used to refer to a frequency; so we can have a transition for which $Delta E = hnu$, for example.
$endgroup$
– orthocresol♦
Feb 3 at 21:23
$begingroup$
@orthocresol, did you mean wavelength? In $Delta E=hnu$ , $nu$ is a frequency in 1/second.
$endgroup$
– porphyrin
Mar 6 at 20:00
2
2
$begingroup$
A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $nu$ (nu). The Greek letter $nu$ is used to refer to a frequency; so we can have a transition for which $Delta E = hnu$, for example.
$endgroup$
– orthocresol♦
Feb 3 at 21:23
$begingroup$
A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $nu$ (nu). The Greek letter $nu$ is used to refer to a frequency; so we can have a transition for which $Delta E = hnu$, for example.
$endgroup$
– orthocresol♦
Feb 3 at 21:23
$begingroup$
@orthocresol, did you mean wavelength? In $Delta E=hnu$ , $nu$ is a frequency in 1/second.
$endgroup$
– porphyrin
Mar 6 at 20:00
$begingroup$
@orthocresol, did you mean wavelength? In $Delta E=hnu$ , $nu$ is a frequency in 1/second.
$endgroup$
– porphyrin
Mar 6 at 20:00
add a comment |
1 Answer
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The reference should really be to the Boltzmann distribution where the population is proportional to $exp(-E_n/(k_mathrm BT))$ where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_mathrm BT$ at room temperature $approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $ngt 0$.
edited Feb 3 at 10:12
Loong♦
34.3k884179
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The reference should really be to the Boltzmann distribution where the population is proportional to $exp(-E_n/(k_mathrm BT))$ where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_mathrm BT$ at room temperature $approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $ngt 0$.
edited Feb 3 at 10:12
Loong♦
34.3k884179
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
$endgroup$
add a comment |
$begingroup$
The reference should really be to the Boltzmann distribution where the population is proportional to $exp(-E_n/(k_mathrm BT))$ where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_mathrm BT$ at room temperature $approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $ngt 0$.
edited Feb 3 at 10:12
Loong♦
34.3k884179
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
$endgroup$
add a comment |
$begingroup$
The reference should really be to the Boltzmann distribution where the population is proportional to $exp(-E_n/(k_mathrm BT))$ where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_mathrm BT$ at room temperature $approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $ngt 0$.
edited Feb 3 at 10:12
Loong♦
34.3k884179
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
$endgroup$
The reference should really be to the Boltzmann distribution where the population is proportional to $exp(-E_n/(k_mathrm BT))$ where $E_n$ is the energy of the $n^{th}$ energy level. This shows that the population is greatest when $n$ is small. Also you need to know that absorption is proportional to the population difference between any two levels. As the vibrational frequency is generally a few hundred wavenumbers, or higher, and $k_mathrm BT$ at room temperature $approx 210$ wavenumbers you can see that the population of $n= 0$ levels can be very large compared to that when $ngt 0$.
edited Feb 3 at 10:12
Loong♦
34.3k884179
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
edited Feb 3 at 10:12
Loong♦
34.3k884179
edited Feb 3 at 10:12
Loong♦
34.3k884179
edited Feb 3 at 10:12
Loong♦
34.3k884179
34.3k884179
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
answered Feb 2 at 21:50
porphyrinporphyrin
18.4k3157
18.4k3157
add a comment |
add a comment |
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$begingroup$
A tangential note: the vibrational quantum number is commonly denoted with the English letter $v$, not the Greek $nu$ (nu). The Greek letter $nu$ is used to refer to a frequency; so we can have a transition for which $Delta E = hnu$, for example.
$endgroup$
– orthocresol♦
Feb 3 at 21:23
$begingroup$
@orthocresol, did you mean wavelength? In $Delta E=hnu$ , $nu$ is a frequency in 1/second.
$endgroup$
– porphyrin
Mar 6 at 20:00