1. (a) For T = 273 K and pressure 1 atmosphere, that is 106 dyne cm−2 (760 mm of Hg), find the density, n, of an ideal gas in cm−3. Repeat for conditions in a molecular cloud, that is T = 10 K, pressure 10−12 mm of Hg.

(b) For both sets of conditions, find the mean free path, λ, which is defined as 1∕(σ n), and the mean time between collisions, τ, which is 1∕(σ n v), where v is the average velocity. In both cases, take σ = 10−16 cm−3. For the laboratory, take the average velocity to be 300 m s−1; for the molecular cloud, take the average velocity of H2 as 0.2 km s−1.

(c) Suppose that the population of the upper level of a molecule decays in 105 s. How many collisions in both cases occur before a decay?

(d) For extinction we define the penetration depth, λ v , in analogy with the mean free path. When λ v  = 1 the light from a background star is reduced by a factor 0.3678. For a density of atoms n, λ v , in cm, is 2 × 1021/n. Calculate the value of λ v for a molecular cloud and for standard laboratory conditions. The parameters for both are given in part (a) of this problem.

2. (a) The result of problem 2(c) of chapter 13 gives \(T_{\mathrm {k}}=21.2\left ({m/m}_{\mathrm {H}}\right )\, \left (\varDelta {V}_{\mathrm {t}}\right )^2\) where ΔV t is the FWHP thermal width, i.e. there is no turbulence and the gas has a Maxwell–Boltzmann distribution. Apply this formula to the CO molecule (mass 28 m H) for a gas of temperature T. What is ΔV t for T = 10 K, T = 100 K, T = 200 K?

(b) The observed linewidth is 3 km s−1 in a dark cloud for which T = 10 K. What is the turbulent velocity width in such a cloud if the relation between the observed FWHP linewidth, ΔV 1∕2, the thermal linewidth, ΔV t and the turbulent linewidth ΔV turb is

$$\displaystyle \begin{aligned} \varDelta {V}_{1/2}^2=\varDelta {V}_{\mathrm{t}}^2+\varDelta {V}_{\mathrm{turb}}^2\,? \end{aligned}$$

3. The following expression is appropriate for the spontaneous decay between two rotational levels, (u, l) of a linear molecule: \(A_{\mathrm {ul}}=1.165\times 10^{-11}\, \mu _0^2 \nu ^3\, {(J+1)}/ {(2J+3)}\) where ν is in GHz, μ 0 is in Debyes and J is the lower level in the transition from J + 1 → J. Use this to estimate the Einstein A coefficient for a system with a dipole moment of 0.1 Debye for a transition from the J = 1 level to the J = 0 level at 115.271 GHz.

4. To determine whether a given level is populated, one frequently makes use of the concept of the “critical density”, n ∗, defined as:

$$\displaystyle \begin{aligned}A_{\mathrm{ul}}=n^*\,\left\langle{}\sigma v\right\rangle\,\,.\end{aligned}$$

where u is the quantum number of the upper rotational level, and l is that for the lower level . If we take \(\left \langle {}\sigma v\right \rangle \) to be 10−10 cm3 s−1, determine n ∗ from the following A ul coefficients

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathrm{CS}: A_{10}&\displaystyle =&\displaystyle 1.8 \times 10^{-6}\,\mathrm{s}^{-1} \\ \mathrm{CS}: A_{21}&\displaystyle =&\displaystyle 2.2 \times 10^{-5}\,\mathrm{s}^{-1} \\ \mathrm{CO}: A_{10}&\displaystyle =&\displaystyle 7.4 \times 10^{-8}\,\mathrm{s}^{-1}. \end{array} \end{aligned} $$

5. Suppose the effective radius r e = 1.1 × 10−8 cm and the reduced mass, m r, of a perfectly rigid molecule is 10 atomic mass units, AMU (an AMU is 1/16 of the mass of a 16-oxygen atom; 1 AMU= 1.660 × 10−24 g), where \(\varTheta =m_{\mathrm {r}} r_{\mathrm {e}}^2\).

(a) Calculate the lowest four rotational frequencies and energies of the levels above the ground state. One needs a simplified version of Eq. 15.11 to 15.13 from ‘Tools’ for the rotational constant is

$$\displaystyle \begin{aligned} B_{\mathrm{e}} = \frac{\hbar}{4 \pi \,\varTheta_{\mathrm{e}}} \end{aligned} $$
(15.1)

The energy of level J is:

$$\displaystyle \begin{aligned} E_{\mathrm{rot}} = W(J) = \frac{\hbar^2}{2 \,\varTheta_{\mathrm{e}}} \,J ( J + 1 ) - h D \left[ J ( J + 1 ) \right]^2 \,. \end{aligned} $$
(15.2)

and the frequency is the difference between the energy of level J + 1 and J divided by the Planck constant:

$$\displaystyle \begin{aligned} \nu(J) = \frac{1}{h} \left[ W(J + 1) - W(J) \right] = 2 \,B_{\mathrm{e}} \,\left[( J + 1 ) - 4 D(J+1)^3 \right] \end{aligned} $$
(15.3)

(b) Repeat if the reduced mass is (2/3) AMU with a separation of 0.75 × 10−8 cm; this is appropriate for the HD molecule. The HD molecule has a dipole moment μ 0 = 10−4 Debye, caused by the fact that the center of mass is not coincident with the center of charge. Take the expression for A(ul) from Problem 3 and apply to the J = 1 − 0 and J = 2 − 1 transitions of the HD molecule.

(c) Find the “critical density”, n ∗≈ 1010 A(ul).

6. The12C16O molecule has B e = 57.6360 GHz and D e = 0.185 MHz. Calculate the energies for the J = 1, 2, 3, 4, 5 levels and line frequencies for the J = 1 − 0, 2 − 1, 3 − 2, 4 − 3 and 5 − 4 transitions. Use the expression energy E(J)∕h ≈ B e J(J + 1) − D e J 2(J + 1)2 for the energy calculation. Check the results against the relevant parts of Table 16.1 in ‘Tools’, given here as Table 15.1.

Table 15.1 Parameters of the more commonly observed carbon monoxide lines (problem 6)
Full size table

7. Apply for J = 0, 1 the analysis in problem 6 to the linear molecule HC11N, which has B e = 169.06295 MHz and D e = 0.24 Hz. Estimate J for a transition near 20 GHz. What is the error if one neglects the distortion term?

8. In the following, we neglect the distortion term D e and assume that the population is in LTE. The population in a given J level for a linear molecule is given by Eq. (15.33):

$$\displaystyle \begin{aligned}n(J)/n(\mathrm{total})=(2J+1)\mathrm{e}^{B_0 J(J+1)/kT}/Z\end{aligned} $$

where Z, the partition function, does not depend on J. Differentiate n(J) with respect to J to find the state which has the largest population for a fixed value of temperature, T. Calculate this for CO if T = 10 K and T = 100 K. Repeat for CS (B 0 = 24.584 GHz) and HC11N, for T = 10 K.

9. Extend Eq. (15.33 in ‘Tools’), which is:

$$\displaystyle \begin{aligned} \begin{array}{rcl} N(J)/N(\mathrm{total}) = \frac{(2 J + 1)}{Z} \, \exp \left[ - \frac{ h \,B_{\mathrm{e}}J (J+1)}{k T} \right ]\vspace{-3pt} \end{array} \end{aligned} $$

to include the optical depth relation Eq. (15.26), which is:

$$\displaystyle \begin{aligned} \begin{array}{rcl} N_l= 1.95 \times 10^{3} \frac{g_l\, \nu^2 } {g_u\, A_{ul}\,} \int T_{\mathrm{B}} \mathrm{\, d} V \end{array} \end{aligned} $$

to obtain an estimate of which J level has the largest optical depth, τ, in the case of emission for a linear molecule.

(a) Show that when the expression for the A coefficient for a linear molecule is inserted into Eq. (15.26 of ‘Tools’), we have

$$\displaystyle \begin{aligned}N_{\mathrm{l}} = \frac{1.67 \times 10^{14}}{ \mu_0^2 \, \, \nu\mathrm{[GHz]} } \times \, \frac{2J+1}{J+1} \, {T_{\mathrm{ex}}}\, \tau \varDelta\, v\,\,,\end{aligned}$$

where μ is in Debyes and v is in km s−1.

(b) Use the above expression to estimate whether the J for the maximum T MB = T ex τ is larger or smaller than the J obtained in Problem 8.

10. Find the ratio of the intensities of the J = 2 − 1 to J = 1 − 0 transitions for a linear molecule if the excitation temperature of the system, T, is very large compared to the energy of the J = 2 level above the ground state, and both lines are optically thin. What is the ratio if both are optically thick? Use the last equation in the statement of Problem 9 of this Chapter.

11. The ammonia molecule, NH3, is an oblate symmetric top. For ammonia, B = 298 GHz, C = 189 GHz. If T ≫ B, C, the value of Z, the partition function, with C and B in GHz, is \(Z=168.7 \sqrt {{(T^3)}/{(B^2\, A)}}\).

(a) Evaluate Z for NH3 for T = 50 K, 100 K, 200 K, 300 K. For this approximation to be valid, what is a lower limit to the value of T?

(b) The (3,3) levels are 120 K above ground. Use the partition function and

$$\displaystyle \begin{aligned}n(J)/n(\mathrm{total})=(2J+1)\mathrm{e}^{120/T}/Z\end{aligned}$$

to calculate the ratio of the total population to that in the (3,3) levels.

(c) If only metastable (J = K) levels are populated, use the definition of Z as a sum over all populated states, and

$$\displaystyle \begin{aligned}n(J)/n(\mathrm{total})=(2J+1)\mathrm{e}^{(B J(J+1)+ K^2(C-B))/kT}/Z\end{aligned}$$

and the B and C values for NH3 to obtain the ratio between the population of the (3,3) levels and all metastable levels.

12. The selection rules for dipole transitions of the doubly deuterated isotopomer D2CO differ from that of H2CO since D2CO has two Bosons, so the symmetry of the total wavefunction must be symmetric. Determine these rules following the procedure in Sect. 15.6.2.