Info

Figure 5.6 Rotational levels of 12C16O within the ground (v = 0) vibrational state. The astro-physically important J = 1 ^ 0 transition at 2.60 mm is shown.

5.3.1 Populating the Rotational Ladder

The rotational levels again have energies given by equation (5.6). These levels are more closely spaced than for H2 because the moment of inertia is greater. (See Figure 5.6.) More importantly, the faster electric dipole transitions can now occur. Here, J changes by ±1. The J = 1 state is elevated above the ground state in 12C16O by AE10 = 4.8 x 10"4 eV, or by an equivalent temperature of only 5.5 K. It is therefore easy to excite this level inside a quiescent cloud and even to populate J = 2, which lies 16 K above the ground state. When the J =1 ^ 0 transition is made radiatively, the emitted photon has a wavelength of 2.60 mm.

Within a molecular cloud, excitation of CO to the J = 1 level occurs primarily through collisions with the ambient H2. In a cloud of relatively low total number density ntot, each upward transition is followed promptly by emission of a photon. Conversely, when ntot is high, the excited CO usually transfers its excess energy to a colliding H2 molecule, with no emission of a photon. The critical density separating the two regimes is given by A10/y10. Using A10 = 7.5 x 10"8 s"1 and y10 = 2.4 x 10"11 cm3 s"1 (the appropriate value for collisions with H2 at a temperature of 10 K), we find that ncrit is 3 x 103 cm"3.

In general, the rate of spontaneous photon emission per unit volume from the J =1 ^ 0 transition is n1A10. Here nj denotes the number density of CO molecules in the level with quantum number J. We may determine these populations by balancing the rates of collisional and radiative excitation and deexcitation. Appendix B presents the simplified but instructive example of a two-level system. The ratio of the densities n1 and n0 is usually expressed through the excitation temperature, Tex. We define this quantity through a generalization of Boltzmann's

Was this article helpful?

0 0

Post a comment