Question

-26 Problem # 3 (15 points) Consider a CO molecule. The reduced mass is 1.14 x 10 kg. a) In CO the l = 0 to l = 1 rotational absorption line occurs at a wavelength of 2.6 mm (or frequency f= 1.15 x 104 Hz). What is the bond length R (or equilibrium distance between the 2 atoms) of the CO molecule? b) When CO is dissolved in liquid carbon tetrachloride, infrared radiation of wavelength 4.67 um (or frequency f= 6.42 x 10-3 Hz) is absorbed. This absorption is due solely to the co molecules being excited to the next higher vibrational state (disregard the carbon tetrachloride which is transparent to energy of this wavelength). What is the force constant k (or spring constant) of the bond in the CO molecules?

Answer 1

cal energy are The stational given by J (J+1) 21 ? E (1 1 = ur? 2 ħ >) J (J+1) E = BJ (JH) alhe And E J JH = 2B (J+1) 2 B qlhc Here, J=1 => Eelt - 2B ((+1) >> ودمع 2 B 2. B(0+1) E B à 2B a

2 z 2 2 The =) -K to? c The =) lu t? a Mr2 hc given => 22 - a ħ2 Mhe a=2.6X10²3 = ) 8= a ħ2 u he bond U=1,14 x 10-26 length kg 2.6610-3 x (T-054 X10-34) 1.14x10-26 x 6.62681034 X 3x108 10 >> r= 1.129 X 100 m or, r= 1. 129 A or Ir- Or 1129

For vibrational spectroscopy we know that ارد 1 k 2тс ܥܙ 1 2 .. Ric k ДА >) 2 тус k д = ) k lunch - 4 spring а coustant 2) Чx - Tч (4.47 vie) given А = 4. ст лоят 4.62xtor k - S90. 89 м/с = 1,14 х 10-26