Question

2) The vibrational transition from the v = 0 state to the v = 1 state of the CO molecule is found to absorb at 1 = 4.6um. In which region of the spectrum this transition will be taken place? Calculate the vibrational frequency and the force constant for this molecule.

Answer 1

This transition will take place in the Infrared(IR) region of the spectrum since wavelength of 4.6 micrometres falls into Infrared region.

To calculate frequency of vibration, we use the following relation:

Frequency(v) =

Where

λ = wavelength

c = speed of light = 3*108 m/s

Given:

λ = 4.6 micrometers = 4.6 x 10-6 m

Hence we have

$Frequency (\nu )=\frac{c}{\lambda } = \frac{3\times 10^8}{4.6\times 10^{-6}}=6.5\times 10^{13}\: \: Hz$

Hence, vibrational frequency = 6.5 x 1013 Hertz

Now to find the Force constant(k) , we use the following relation:

$Force\: constant (k )=4\pi ^2\nu ^2\mu$

Where the μ is the reduced mass of the molecule

for a molecule the expression for reduced mass(μ) is given by :

$Reduced\: mass(\mu)=\frac{m_1\times m_2}{m_1+m_2}$

Where m1 and m2 is the atomic mass of atoms in the molecule

Here

m1 = atomic mass of carbon = 12 amu

m2 = atomic mass of oxygen = 16 amu

Also the conversion factor for amu to Kg is:

$1\: \: amu=1.660\times 10^^^{-27}\: \: \: kg$

Hence the reduced mass(μ) for Carbon Monoxide is :

$\mu=\frac{m_1\times m_2}{m_1+m_2}=\frac{12\times 16}{12+16}\times 1.660\times 10^^^{-27}=11.38\times 10^^^{-27}kg$

Now we have all the values required to calculate force constant . So we calculate force constant(k) using the equation:

$Force\: constant (k )=4\pi ^2\nu ^2\mu$

Substituting the values for frequency and reduced mass, we get

$k=4\times 3.14^2\times (6.5\times 10^1^3)^2\times 11.38\times 10^^^{-27}=1.90\times 10^3 \: kg\: s^-^2$

Hence we get

Force constant(k) = 1.90 x 103 kg s-2