{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

assn2sol - Chemistry 350 Assignment 2 Solutions 1 The...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Chemistry 350 Assignment 2 Solutions October 13, 2004 1. The energies and degeneracies of the two lowest electronic states of atomic iodine are: e 1 = 0 cm - 1 , g e 1 = 4; e 2 = 7603.2 cm - 1 , g e 2 = 2. What temperature is required so that 10% of the atoms are in the excited state? Disregard any contribution from higher electronic states. Answer: As in any question like this, this requires a simple calculation of the Boltzmann factor, p j , for the desired state. We can assume we can state the percentage to as many significant digits as we desire, so we’ll choose six here. p j = e - e j /kT i e - e i /kT 0 . 100000 = 2 e - 7603 . 2 cm - 1 / (0 . 69509 cm - 1 K - 1 ) T 4 + 2 e - 7603 . 2 cm - 1 / (0 . 69509 cm - 1 K - 1 ) T ln 0 . 400000 1 . 80000 = - 7603 . 2 cm - 1 (0 . 69509 cm - 1 K - 1 ) T T = 7272 . 5 K 2. Show that the fraction of harmonic oscillators in the ground vibrational state is given by f 0 = 1 - e - e /kT Calculate f 0 for 15 N 2 (g) ( k f = 2243 N m - 1 ) at 300, 500 and 1500 K. Answer: Using the harmonic oscillator partition function, we find that for the v ’th vibrational energy level, p v = e - βhν e v e - βhν e / 2 (1 - e - βhν e ) e - βhν e / 2 = e - βhν
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}