
L-11 Rotational Inertia symbol I
... • A figure skater has a rotational inertia I1 when her arms are stretched out, and I2 when her arms are pulled in close to her body. If her angular velocity is 1 when she spins with her arms stretched out, what is her angular velocity when she pulls hers arms in, so that I2 = ½ I1 = 0.5 I1 • Soluti ...
... • A figure skater has a rotational inertia I1 when her arms are stretched out, and I2 when her arms are pulled in close to her body. If her angular velocity is 1 when she spins with her arms stretched out, what is her angular velocity when she pulls hers arms in, so that I2 = ½ I1 = 0.5 I1 • Soluti ...
Review of Definitions
... 11. Hydrophilic Molecules. Molecules which can form hydrogen bonds with water and hence, are miscible with water, i.e."water-loving," (e.g. sugars, alcohols, soluble proteins, DNA, poly(ethylene oxide)). 12. Bond Rupture Strength. Binding energy needed to break a bond. 13.Ionization. Any process by ...
... 11. Hydrophilic Molecules. Molecules which can form hydrogen bonds with water and hence, are miscible with water, i.e."water-loving," (e.g. sugars, alcohols, soluble proteins, DNA, poly(ethylene oxide)). 12. Bond Rupture Strength. Binding energy needed to break a bond. 13.Ionization. Any process by ...
Quantum Chemistry and Spectroscopy
... and one will participate to the C=O bond. The sp2 orbital and O:2pz orbital will form a s-bond. The px orbitals from the C and O will form a p-bond and the electron pair on the O:py orbital does not interact much with the carbon, it is marked with n O. (The O:2s orbital does not interact much with t ...
... and one will participate to the C=O bond. The sp2 orbital and O:2pz orbital will form a s-bond. The px orbitals from the C and O will form a p-bond and the electron pair on the O:py orbital does not interact much with the carbon, it is marked with n O. (The O:2s orbital does not interact much with t ...
Midterm exams I
... 4. Where would be Raman lines (in cm-1 ) for (1) diamond and (2) for helium? 5. Which molecule would have higher vibrational frequency: Fluorine (F2) or iodine (I2)? 6. Why Raman spectra should be measured at very low laser power? 7. Why Raman lines of amorphous phase looks broad in comparison to li ...
... 4. Where would be Raman lines (in cm-1 ) for (1) diamond and (2) for helium? 5. Which molecule would have higher vibrational frequency: Fluorine (F2) or iodine (I2)? 6. Why Raman spectra should be measured at very low laser power? 7. Why Raman lines of amorphous phase looks broad in comparison to li ...
I believe the chemical bond is not so simple as people seem to think
... the reduction of the effective slit width resulting from the finite size of the dimer. From a theoretical analysis of the data which also takes into account the van der Waals interaction with the grating bars, the bond length (mean internuclear distance) and the binding energy are found to be !r" ! ...
... the reduction of the effective slit width resulting from the finite size of the dimer. From a theoretical analysis of the data which also takes into account the van der Waals interaction with the grating bars, the bond length (mean internuclear distance) and the binding energy are found to be !r" ! ...
Rotational spectroscopy

Rotational spectroscopy is concerned with the measurement of the energies of transitions between quantized rotational states of molecules in the gas phase. The spectra of polar molecules can be measured in absorption or emission by microwave spectroscopy or by far infrared spectroscopy. The rotational spectra of non-polar molecules cannot be observed by those methods, but can be observed and measured by Raman spectroscopy. Rotational spectroscopy is sometimes referred to as pure rotational spectroscopy to distinguish it from rotational-vibrational spectroscopy where changes in rotational energy occur together with changes in vibrational energy, and also from ro-vibronic spectroscopy (or just vibronic spectroscopy) where rotational, vibrational and electronic energy changes occur simultaneously.For rotational spectroscopy, molecules are classified according to symmetry into spherical top, linear and symmetric top; analytical expressions can be derived for the rotational energy terms of these molecules. Analytical expressions can be derived for the fourth category, asymmetric top, for rotational levels up to J=3, but higher energy levels need to be determined using numerical methods. The rotational energies are derived theoretically by considering the molecules to be rigid rotors and then applying extra terms to account for centrifugal distortion, fine structure, hyperfine structure and Coriolis coupling. Fitting the spectra to the theoretical expressions gives numerical values of the angular moments of inertia from which very precise values of molecular bond lengths and angles can be derived in favorable cases. In the presence of an electrostatic field there is Stark splitting which allows molecular electric dipole moments to be determined.An important application of rotational spectroscopy is in exploration of the chemical composition of the interstellar medium using radio telescopes.