PS2_2011Ans - Name Section Rec TA Ch 1b, Solution Set for...

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Name Ch 1b, Solution Set for Problem Set Two Section Side 1 of 14 Due Friday, Jan. 14, 2011 Rec TA 1. Intro to Spectroscopy: Rotational Spectroscopy (9 points) Solution To the first order, only molecules with permanent dipoles exhibit peaks during rotational spectroscopy. (“To the first order” means “to a rough approximation.” There are other interactions involving more subtle properties of molecules which can result in weak absorption of radiation. Diatomic oxygen does not have a permanent dipole ( 3 points ), and, to first order, does not exhibit a peak during rotational spectroscopy. (Side note: Oxygen is paramagnetic, allowing it to undergo a magnetic dipole transition (higher order, weaker), which makes it a candidate for magnetic rotation spectroscopy.) Both carbon monoxide and hydrogen cyanide have permanent dipoles, so they will exhibit peaks. ( 6 points total; 3 points each ) (Side note: in the gas phase, carbon monoxide and hydrogen cyanide have permanent dipoles of 0.112 and 2.98 Debye units, respectively.)
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Name Ch 1b, Solution Set for Problem Set Two Section Side 2 of 14 Due Friday, Jan. 14, 2011 Rec TA 2. Rotations of Molecules (29 points) When solving problems, it is good practice to use symbolic algebra and variables/parameters as often as possible to develop a governing equation before substituting in numerical values. By doing so, you can solve the problem more quickly, you can see the relationships between different variables more easily, and you are more likely to avoid rounding errors. This practice will be carried out below. Solution for 2a Allowable absorption frequencies are defined as (OGC, p. 831): ( 2 points ) [1] The absorption of light is allowed only between states that differ by for a heteronuclear diatomic molecule like HCl. We don’t know which states (values of the rotational quantum number) are measured, but we do know that . Therefore, we’ll keep with as an unknown and associate it with the smallest measured frequency (2.50 x 10 12 s -1 ). We’ll also let . ( 2 points ) [2a] [2b] Solving both Equations [2a] and [2b] for : [3a] [3b] Subtract Equation [3a] from Equation [3b] and solve for I : [4]
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Name Ch 1b, Solution Set for Problem Set Two Section Side 3 of 14 Due Friday, Jan. 14, 2011 Rec TA ( 2 points ) (Any two adjacent states may be used to calculate difference.) ( yields 2 sig. figs.) ( ) NOTE: full credit as long as student used adjacent frequencies in the calculation: ( 3 sig. figs. ) ( 3 sig. figs. ) ( student should receive full credit if they just use the relation instead ) Solution for 2b Rotational Energies take on the values (OGC, pg. 830): ( 1.5 points for this equation or equation with value for moment of inertia already substituted) [5] Substitute Equation [4] into Equation [5] to develop a governing equation. Substitute
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PS2_2011Ans - Name Section Rec TA Ch 1b, Solution Set for...

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