So far, we assumed that rotation does not affect the shape of a molecule (rigid rotor). In fact, centrifugal forces that stretch the bonds to some extent will occur. We will confine the treatment of this problem in the example of the linear rotor.

An increased length of the bond corresponds to a higher moment of inertia (I
= µr^{2}). In turn, the rotational constant B (B ∝ I^{-1})
falls, i.e. we would expect a rotational energy with a minor slope than in the
case of _{rot} = B_{rigid} J(J + 1).

This is a case where we have no general and exact solution to Schrödinger's equation, but we are able to deal with the problem and determine the eigenvalues for the rotational energy by using a series expansion. The newly introduced coefficient D is but a small correction, in other words, D is much smaller than B.

E_{rot} = B · J(J + 1) - D · J^{2}(J +
1)^{2} + ... |

From the treatment of the elastic rotor, we would already expect that the
weaker the bond, the more pronounced the deviations from the rigid rotor become.
A weak bond corresponds to a low vibrational energy (E_{vib} =
* h*ω(v + ½); ω = (

We now want to focus on the selection rules for rotational spectra.

Auf diesem Webangebot gilt die Datenschutzerklärung der TU Braunschweig mit Ausnahme der Abschnitte VI, VII und VIII.