Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

This question concerns the following exercise from an old exam:

The vibrational motion of a linear diatomic molecule can be approximated as simple harmonic motion.

  • A CO molecule has a bond with force constant $k = 1900 \, Nm^{-1}$. What frequency of radiation would excite transitions between the different vibrational enery levels? (C and O have masses of $12 \, m_u$ and $16\, m_u$ respecitvely, where $m_u$ is the atomic mass unit.)

  • Explain why electromagnetic radiation can not excite vibrational transitions in an $O_2$ or $N_2$ molecule. How is this important for the heating of the earth's atmosphere by the sun (the greenhouse effect)?

For the first part I know the energy levels of a harmonic oscillator (in center of mass reference frame)

$$E_n = \hbar\omega(n+1/2),\quad \text{where }\omega = \sqrt{\frac{k}{\mu}}, \quad \mu = \frac{m_Om_C}{m_O + m_C}$$

and the energy of a photon of frequency $\nu$ is given by $E_{ph} = h\nu$. The photon energy must be the energy difference between two states: $E_{ph} = E_n - E_m = \hbar \omega (n-m)$. This gives me the frequency

$$\nu = \frac{(n-m)}{2\pi}\sqrt{\frac{k}{\mu}}$$

Or minimum frequency $\nu_0 = \frac1{2\pi}\sqrt{\frac{k}{\mu}}$, right?

However, now to my actual question: I don't know why we cannot do the exact same thing for the other two molecules? What is different for $O_2$ and $N_2$?

I'd appreciate some help, thanks! =)

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The molecules $O_2$ and $N_2$ are symmetric and have no dipole momentum. That's why they can not interact with EMW (at least within dipole approximation).

One can say that transitions between the oscillator levels in these molecules are forbidden by symmetry in electrodipole approximation.

The molecule $CO$ consists of two different atoms. The average positions of positive and negative charges are not the same. This molecule is polar.

share|improve this answer
I think for similar reasons the non-adjacent transitions in CO are also not excited. There is no dipole moment coupling the ground state to the third excited state, or the fourth state to the eleventh. –  Marty Green Feb 3 '12 at 18:11
Thanks a lot for your answer! –  Sam Feb 4 '12 at 0:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.