Solved by Expert Tutors
Solved by Expert Tutors
Question

Using statistical mechanics it is straight forward to show the rotational

quantum number of the most populated rotational level is given by J_max=(kT/2hcB)^(1/2)-1/2 Where B is the rotational constant in wave numbers.
a) Calculate the moments of inertia for 12CH and 13CH given a bond length of 1.12 Å.
b) Using the above formula find the value of J closest to J_max for both radicals at 298K. Compute the energy difference between this state and the next highest state. Repeat for T = 1000K.

Please see attachments for detailed notations and lecture notes:)

Background image of page 1
4/6/2015 11 Rotation Summary Linear Molecules: 2 degenerate inertial axes; 3n-5 vibrations g R Eg. Polyatomic Molecules: Asymmetric Top g C O O x z R y z Eg. O H H y x out of plane of molecule Symmetric Top Oblate: Eg. R R ٣ R ٣ Prolate: Eg. pherical Top: I H 3 C R R ٣ R ٣ Spherical Top: Eg. y x z Raman Spectroscopy: For example, the A 1 and E vibrations in CH 4 are not infrared active but they are Raman active due to the fact polarizability components transform as A 1 and E . Spectroscopy : Chapter 12 Here we will only briefly introduce the topic and primarily restrict ourselves to diatomic molecules. Our intent is to explore the reasons why molecules absorb and emit radiation. irst the nergetics r diatomic molecules: First the energetics for diatomic molecules: Rotation:
Background image of page 1
4/6/2015 12 Vibration: Orders of Magnitude: Where there can be significant changes in these values depending on the molecule. etc. Now there can be transitions between these levels that obey certain selection rules. We shall see the infrared selection rules are; So what would the spectrum of a diatomic molecule look like: Degeneracies: We will deal more with populations of energy levels in statistical mechanics. For now realize the rotational degeneracy gives small populations at low J, maximum populations at intermediate J and declining populations as J increases further. The detailed populations depend on the specific molecule since Hence the spectra of a diatomic molecule would start with a P- branch, have a space for and then have an R-branch: R-branch P-branch
Background image of page 2
4/11/15 1 Raman Selection Rules Δ J = 0 : Q – branch Δ J = -2 : O – branch Δ J = +2 : S – branch When you do a Raman experiment, the Raman signal is the difference between the incident radiation and the scattered radiation. Note: Since these “energies” are given in cm -1 you often simply see used instead of . However I feel this can be confusing so I have used the notation. Typical Raman Spectrum: or Raman shift frequency in cm -1 Q-branch S-branch O-branch Origin of Selection Rules Electric Dipole Transitions: In each kind of transition we will consider that there is an interaction of a photon with a molecule. A photon can be thought of in the following way; Transition energies: 10 to 4000 cm -1
Background image of page 1
4/11/15 2 Consider a diatomic molecule; The wavelength is much greater than the bond length and at any one time the molecule sees a constant electric field; Now, the electric field of the photon will interact with the dipole moment of the molecule; A B ~10 -10 m molecule Now for a transition to occur in the molecule from state i state f the following integral must be non-zero Now since the molecular dimensions << λ we can simplify this expression by taking the outside of the integral as it is essentially a constant at any time t Hence the condition for a transition becomes; Born – Oppenheimer Approximation: Electrons are much smaller than nuclei and hence will move much faster than nuclei. Therefore we can decouple nuclear and electronic motion. Harmonic Oscillator – Rigid Rotor Assumption: Vibrations only slightly change bond lengths and hence have only small effects on moments of inertia, hence vibrational and rotational motions are decoupled. Here we will assume we are in the electronic ground state: Pure Rotation: A B Permanent dipole moment in vibrational state υ x y z
Background image of page 2
Show entire document
4/13/2015 1 2. Transition Dipole Moment A B ces e Z A R A e j r j r j is the distance of the j th electron from the center of electrical symmetry the distance of nucleus ith charge om the center of R k is the distance of nucleus k with charge e Z k from the center of electrical symmetry Where we will assume Where the second term is zero since are orthogonal. Franck-Condon factor Rotation selection rules. Franck – Condon Factors: overlap integral of initial vibrational state with all vibrational states in the final electronic state. Note: There are no selection rules since the vibrational states in the integral are located in separate electronic states. Note: Most excited electronic states have more anti-bonding character than the ground state: See the O 2 example. Hence the bond length is usually larger and the minimum ifted to larger values of r es shifted to larger values of r. gs Rotational Sates: but Δ J = 0 also allowed since the electronic transition can satisfy angular oment m conser ation momentum conservation Energetics:
Background image of page 1
4/13/2015 2 Three Possibilities: Case 1: Normal P, Q, R rotational spectra observed Case 2: Recall: and bond length is greater in xcited state (common) excited state (common) as J increases gets more negative. (Normal) at low J, increases with J but at high J, decreases with J. Where J “turns around” is termed a “band head” Case 3: Therefore bond length shorter in the excited state (uncommon) increases as J increases for all J.
Background image of page 2
Show entire document
Sign up to view the entire interaction

Step-by-step answer

, dictum vitae odio. Donec aliquet. Lorem ipsum
ce

sum dolor sit amet, consectetur adipiscing elit. Nam lacinia pulvinar tortor nec facilisis. Pellentesque dapibus efficitur laoreet. Nam risus ante, dapibus a molestie consequat, ultrices ac magna. Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. Donec aliquet. Lor

DSC_0118.JPG
r

Subscribe to view the full answer

Subject: Chemistry, Science

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question