Atoms in interstellar space seldom meet. When they do (most likely on ice surfaces), they produce ra — Physical Chemistry — Electrochemistry Chemistry Question
Spectroscopy of interstellar molecules
Atoms in interstellar space seldom meet. When they do (most likely on ice surfaces), they produce radicals and molecules. These species, some of which presumably played a role in the origin of life, have been identified through the use of different spectroscopic methods. Absorption spectra of interstellar species can be observed by using the background radiation as the energy of excitation. Emission spectra from excited species have also been observed. Simple diatomic fragments such as CH and CN were identified in interstellar space over 60 years ago.
When molecules with non–zero dipole moments rotate, electromagnetic radiation can be absorbed or emitted. The spectroscopy related to molecular rotation is called microwave spectroscopy, because the electromagnetic radiation involved is in the microwave region. The rotational energy level of a diatomic molecule is given by EJ = J(J+1)h^2 / 8π^2 where J is the rotational quantum number, h is the Planck constant, I is the moment of inertia, µ R^2. The quantum number J is an integer increasing from 0 and the reduced mass µ is given by m1m2 / (m1+m2) for diatomic molecules (m1 and m2 are masses of the two atoms of the molecule). R is the distance between the two bonded atoms (bond length).
The background electromagnetic radiation in the interstellar space has a characteristic energy distribution related to the temperature of a blackbody source. According to Wien’s law, the wavelength (λ) corresponding to the maximum light intensity emitted from a blackbody at temperature T is given by Tλ = 2.9×10–3 m K. Let’s consider a region near a star where the temperature is 100 K. What is the energy in joule of a photon corresponding to the peak emission from a blackbody at 100 K?
Model Answer
100 λ = 2.9×10–3 m K
λ = 2.9×10–5 m
E(photon) = hc / λ = (6.63 × 10–34 J s) × (3.0 × 108 m s-1) / 2.9 × 10–5 m = 6.9×10–21 J
Carbon monoxide is the second most abundant interstellar molecule after the hydrogen molecule. What is the rotational transition (change of J quantum number) with the minimum transition energy? What is the minimum transition energy of the 12C16O rotation in joules? The bond length of CO is 113 pm. Compare the transition energy of CO with the radiation energy in problem 3.1. What does the result imply?
Model Answer
J: 0 ↔ 1
µ = (12 × 16 / 28) × 1.66×10–27 kg = 1.14×10–26 kg
I = µR2 = (1.14×10–26 kg) (1.13×10–10 m)2 = 1.45×10–46 kg m2
E(0↔1) = h^2 / 8π^2I = (6.63×10–34 J s)2 / (8π^2(1.45×10–46 kg m2)) = 7.68×10–23 J
E(photon) of problem 3.1 = 6.9×10–21 J > E(0↔1) = 7.68×10–23 J
Rotational excitation by the background radiation is feasible.
The distribution of molecules in different energy levels is related to the background temperature, which affects the absorption and emission spectra.
Figure 3–1. Oscillogram for the lowest rotational transition of 12C16O at 115,270 MHz. The upper curve was taken at the temperature of liquid air, the lower at the temperature of dry ice. (Reference: O. R. Gilliam, C. M. Johnson and W. Gordy. Phys. Rev. vol. 78 (1950) p.140.)
The equation for the rotational energy level is applicable to the rotation of the hydrogen molecule. However, it has no dipole moment so that the transition of ∆J = 1 by radiation is not allowed. Instead a very weak radiative transition of ∆J = 2 is observed. Calculate the temperature of interstellar space where the photon energy at the maximum intensity is the same as the transition energy of the hydrogen molecule (1H2) between J = 0 and 2. The H–H bond length is 74 pm.
Model Answer
E(0↔2) = 6 h^2 / 8π^2I = hc / λ
λ = 8π^2cI / 6h
I = µR^2 = [(1/2) × 1.66×10–27 kg] (0.74×10–10 m)2 = 4.55×10–48 kg m2
λ = 8π^2cI / 6h = 8π^2(4.55×10–48 kg m2)(3×108 m s-1) / 6(6.63×10–34 J s) = 2.71×10–5 m
T = 2.9×10–3 m K / λ = 2.9×10–3 m K / 2.71×10–5 m = 107 K
Observation of hydrogen rotational spectra is feasible at 100 K.