Question

Back-of-the-envelope CO: It is well-known (certainly among astronomers) that the Einstein A-coefficient for the Lyman alpha (n = 2 to n = 1) transition in atomic hydrogen is of order 10^9 s^−1 (actually, 5 × 10^8 s^−1). As we briefly discussed in class, this result could be estimated to order-of-magnitude by considering an (accelerating) electron on a spring that displaces a Bohr radius, a0, and has natural angular frequency ω and then taking the inverse lifetime of the excited state as A ∼ P/ω, where P is the power radiated by an accelerating charge. The ubiquitous carbon monoxide molecule, 12CO, is used by astronomers to trace the presence and measure the temperature of molecular gas in various astrophysical environments. We would rather try to detect H2 directly, but sadly H2 has no permanent electric dipole moment because of its symmetry, while Carbon monoxide does have a permanent dipole moment, so is easier to detect.

a)Estimate the wavelength of the lowest energy, rotational transition in CO (J = 1 to J = 0, where J is the rotational quantum number). Do this by considering a barbell spinning about its axis of greatest moment of inertia and recognizing that angular momentum comes quantized in units of ?h. Compare your estimate
to the true answer of 2.6 mm.

b)Use scaling relations to estimate the Einstein A coefficient of this transition, i.e., the inverse lifetime of the excited J = 1 state. You can use that the dipole moment (that is difficult to guess from first principles) of the CO molecule is 0.1 Debyes, not ∼ 1 Debye, as one might have guessed naively (1 Debye= 10^−18cgs. Note that ea0 = 2.5 Debyes, where e is the electron charge). Compare your estimate to the true answer A10 = 7.4 × 10^-8 s^-1.

Note: This smaller-than-expected dipole moment of CO is a consequence of the strong double bond connecting C to O. Most other molecules common in astrophysics – e.g., H2O, CS, SiS, SiO, HCN, OCS, HC3N – have dipole moments that are all of order 1 Debye.

Answer 1

(a)

We know from quantum mechanics that angular momentum is quantized as L^2=h^2J(J + 1), and from classical mechanics we know the rotational kinetic energy of a
barbell to be E = L^2/2I. Here, the moment of inertia I can be approximated as
I =SUM(miri^2) ∼ 30mp ao^2
, where mp is the mass of a proton and we assume that the
width of the barbell is roughly two times the Bohr radius. For transition J = 1 to J = 0
we find:
∆E1→0 ∼¯h^2/(30mp ao^2)∼hc/λCO
which upon plugging in the appropriate values of the constants gives:
λCO ∼ 2.6mm
which just happens to be exactly the same as the real value.

(b)

We can estimate the Einstein A coefficient using the approximate equation derived
in class:
A21 ∼2d^2ω0^3/hc^3
.
Note we can just plug in our value of λCO = 2πc/ω0 and standard constants for every-
thing except for the dipole moment, which has a difficult to guess value of 0.1 Debye
∼ 10^−19[cgs]. Plugging in these values gives:
A21 ∼ 8.7 × 10^−8s^−1
which is pretty close to the actual value of 6 × 10^−8s^−1.