Question

-26 Problem #3 (15 points) Consider a CO molecule. The reduced mass is 1.14 x 10kg. 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 10^1 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

Answer:

Given, reduced mass of CO molecule is μ = 1.14 x 10^-26 kg

The expression for the rotational energy is E_rot = L²/2I

where L is angular momentum and I is the moment of inertia of the molecule and it is μR², R is the bond length.

and L² = l(l+1) h²/4$\pi$²

Thus, E_rot = [l(l+1) h²/4$\pi$² ] / 2I = l(l+1) h² / 8I$\pi$²

E_rot is in units of joules (J). Now, in terms of wave number (cm^-1). Wave number is $\bar{\nu }$ = 1/$\lambda$

In spectroscopy, $\bar{\nu }$ in written as $\varepsilon$ _rot

$\varepsilon$_rot = E_rot/hc = [l(l+1) h² / 8I$\pi$²] / hc

Therefore, $\varepsilon$ _rot = l(l+1) h / 8$\pi$²Ic = B l(l+1) cm^-1      (l = 0,1,2 ....)   ------------- (1)

where B, the rotational constant, is given by

B = h/8$\pi$²Ic cm^-1    --------------------- (1A)

The rotational transition between two state l and l+1 for emission spectra is

= 2B(l+1) cm^-1       --------------------------------- (2)

VI+1+1

For absorption spectra, = - 2B(l+1) cm^-1 ------------------- (3)

VI+1+1

where '-ve' sign indicates the absoprtion.

(a) Given, wavelength of the absorption line is $\lambda$ = 2.6 mm = 0.26 cm

Wave number $\bar{\nu }$ = 1/$\lambda$ = 1/0.26cm = 3.846 cm^-1

The trasnition takes place from l = 0 to l = 1 (absorption line)

Therefore, = 2B(0+1) = 2B

10-1

Then, 3.84 cm^-1 = 2B

or B = (3.84 cm^-1) / 2 = 1.923 cm^-1

Now, using eq(1A), B = h/8$\pi$²Ic cm^-1

Thus, I = h/8$\pi$²cB

or I = (6.626x 10^-34 J.s) / 8$\pi$²(3 x 10¹⁰ cm/s) (1.923 cm^-1) = 14.56 x 10^-47 kg.m²

and I = μR²

Therefore, bond length R = (I/μ)^1/2 = [ (14.56 x 10^-47 kg.m²) / (1.14 x 10^-26 kg) ]^1/2 = 1.130 x 10^-10 m = 1.130 Å

Hence, this is the bond length of the CO molecule.

(b) Given, wavelength $\lambda$ = 4.67 x 10^-6 m = 4.67 x 10^-4 cm

Then, wavenumber $\bar{\nu }$ = 1/$\lambda$ = 1/ 4.67 x 10^-4cm = 2.14 x 10³ cm^-1

The expression for the force constant is k = 4$\pi$²$\bar{\nu }$²c²μ

Therefore, k = 4$\pi$²(2.14 x 10³ cm^-1)²(3 x 10¹⁰ cm/s)²(1.14 x 10^-26 kg) = 1853.080 N/m.

Hence, this is the force constant in the CO molecule.