Spectroscopy involves the use of laser systems (light amplification by stimulatedemission of emitted radiation) to understand the properties of chemical systems. Saidanother way, spectroscopists direct laser light into a chemical system and then observethe light that comes out of the system. Spectropsocoptics compare the light before itinteracts with the chemical system to the light after the interaction to be able to decipherhow the light interacts with the system.To help with the task of deciphering what happened, spectroscopists havedeveloped what are known as spectroscopic selection rules which are rules dictating theallowable transitions for a system. We will talk more about such selection rules later inthe class. Within the harmonic approximation, a well-known selection rule is thatchemical vibrations can only gain one quanta of vibrational energy at a time. Vibrationalquanta are often noted by the greek letter nu, v, so another way to write the selectionrule would be transitions like v - (absorption of a photon) orv₁ - v₂, and v₁ -(emission of a photon) are allowed but transitions like1 - 1/₂ or 1/₂ - vare not. You will be asked later to connect this idea to raisingand lowering operators. Commonly, the most intense transition (the easiest transition todetect) is the v - 01for the various vibrational modes of the molecule being studied.One such mode is the nitrile stretching mode that was referenced in problem 4 of"2-1 - The Classical and Quantum Harmonic Oscillator. The v-transition for thismode occurs at an energy of 2230 cm (infrared light with a ~4.5μm wavelength). Withthis setup, we will explore vibrational modes and the relationship of raising and loweringoperators to spectroscopic selection rules. -11. Prove the claim that a photon with an energy of 2230 cm corresponds to ~4.5μm light.2. PIPES Problem: Assume the number of states for vand vare the same for thenitrile vibrations. With this assumption, calculate how many benzonitrilemolecules are in the state for every 1 benzonitrile in the v₁ state at roomtemperature (~27C°) using the Boltzmann distribution. (hint: you'll have tomanipulate the population ratio to the obtain population relative to a vpopulation of 1).3. Using the result of problem 2, justify, in plain language, why you can expect (atroom temperature) for the v-transition to be more intense than thev₁ - ₂ transition (hint: a spectroscopist only sees the superimposed,collective signal from the system).4. Explain, in plain language, how the spectroscopic selection rule that vibrationscan only change one quanta of vibrational energy at a time is directly a result ofraising and lowering operators.Important Note: This problem is impossible to solve with your givenknowledge. However, that doesn't mean the problem is impossible. I havestealthily given the answer in the reading before these problems whentalking about the allowed vibrational quanta (see v - v being allowedbut -not being allowed in the harmonic approximation). If you stillcannot answer the question, the internet surely knows the answer. Youjust have to go and find it.

Question

Spectroscopy involves the use of laser systems (light amplification by stimulatedemission of emitted radiation) to understand the properties of chemical systems. Saidanother way, spectroscopists direct laser light into a chemical system and then observethe light that comes out of the system. Spectropsocoptics compare the light before itinteracts with the chemical system to the light after the interaction to be able to decipherhow the light interacts with the system.To help with the task of deciphering what happened, spectroscopists havedeveloped what are known as spectroscopic selection rules which are rules dictating theallowable transitions for a system. We will talk more about such selection rules later inthe class. Within the harmonic approximation, a well-known selection rule is thatchemical vibrations can only gain one quanta of vibrational energy at a time. Vibrationalquanta are often noted by the greek letter nu, v, so another way to write the selectionrule would be transitions like v -> (absorption of a photon) orv₁ -> v₂, and v₁ ->(emission of a photon) are allowed but transitions like1 -> 1/₂ or 1/₂ -> vare not. You will be asked later to connect this idea to raisingand lowering operators. Commonly, the most intense transition (the easiest transition todetect) is the v -> 01for the various vibrational modes of the molecule being studied.One such mode is the nitrile stretching mode that was referenced in problem 4 of&#34;2-1 - The Classical and Quantum Harmonic Oscillator. The v->transition for thismode occurs at an energy of 2230 cm (infrared light with a ~4.5μm wavelength). Withthis setup, we will explore vibrational modes and the relationship of raising and loweringoperators to spectroscopic selection rules. -11. Prove the claim that a photon with an energy of 2230 cm corresponds to ~4.5μm light.2. PIPES Problem: Assume the number of states for vand vare the same for thenitrile vibrations. With this assumption, calculate how many benzonitrilemolecules are in the state for every 1 benzonitrile in the v₁ state at roomtemperature (~27C°) using the Boltzmann distribution. (hint: you'll have tomanipulate the population ratio to the obtain population relative to a vpopulation of 1).3. Using the result of problem 2, justify, in plain language, why you can expect (atroom temperature) for the v->transition to be more intense than thev₁ -> ₂ transition (hint: a spectroscopist only sees the superimposed,collective signal from the system).4. Explain, in plain language, how the spectroscopic selection rule that vibrationscan only change one quanta of vibrational energy at a time is directly a result ofraising and lowering operators.Important Note: This problem is impossible to solve with your givenknowledge. However, that doesn't mean the problem is impossible. I havestealthily given the answer in the reading before these problems whentalking about the allowed vibrational quanta (see v -> v being allowedbut ->not being allowed in the harmonic approximation). If you stillcannot answer the question, the internet surely knows the answer. Youjust have to go and find it.

Accepted Answer

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Prove the claim that a photon with an energy of 2230 cm corresponds to ~4.5 μm light.