Al's notes are useful for answering many of the following questions:
Radio Astronomy Jargon:
http://www.cv.nrao.edu/~awootten/sumstuds/RadAsJargon.pdf
Molecular Emission:
http://www.cv.nrao.edu/~awootten/sumstuds/MolecularEmission.pdf
Question 1: At what frequency does the 2cm line of
H2CO lie?
The frequency of the 2cm line of
H2CO (which has four hyperfine components) is:
Frequency Unc. Formula Quantum Tr(K) Source Telescope Astr. Lab.
(MHz) /Ta(K) ref. ref.
14488.4589( 2) H2CO 2(1,1)-2(1,2) F=1-1 b Sgr B2(M) NRL 26m Eva70 Kuk75
14488.4712( 2) H2CO 2(1,1)-2(1,2) F=1-2 b Sgr B2(M) NRL 26m Eva70 Kuk75
14488.4801( 2) H2CO 2(1,1)-2(1,2) F=3-3 -1.3b Sgr B2(M) NRL 26m Eva70 Kuk75
14488.4899( 2) H2CO 2(1,1)-2(1,2) F=2-2 b Sgr B2(M) NRL 26m Eva70 Kuk75
Table 4: Recommended rest frequencies for observed interstellar molecular lines by Frank J. Lovas
Question 2: What frequency resolution should one use on the Spectrometer? Hint--one would want several resolution elements across the line; the velocity width is given above.
0.1 km/s should be sufficient. This is 4.88 kHz at 14.488 GHz. (e.g. Frequency of 1km/s would be line frequency in Hz divided by speed of light in km/s; to resolve the line one tenth this would be good.)
Question 3: How long should one integrate to obtain a 5 sigma detection of the line at 2'E in the map above?
Well we want frequency resolution of 0.1 km/s = 4830 Hz. So we use the radiometer equation
(see
http://www.cv.nrao.edu/~awootten/sumstudentlecture.zip)
deltaT = K (Tsys/sqrt(bandwidth*time)
For frequency switching, K=sqrt(2). bandwidth = 4830 Hz. Tsys = 30 K deltaT = 0.02 K for a .1 K line at 5 sigma.
Then solving for time, I get 15.5 minutes. This might get us eight points per night.
Question 4: For your measurments of the test source (
L134N), you should have measured a strong absorption line. In this absorption line you should be able to see the hyperfine components of the
H2CO 2(11)-2(12) transition. Mark the location of these hyperfine components on the
L134N spectrum.
From the answer to
Question 1, we know that the relative separation (using the strongest hyperfine component as the reference) between the four hyperfine components is as follows:
Question 5: What are the relative intensities of the hyperfine components?
This involves a little bit of quantum mechanics. The hyperfine structure is due to what is called a "magnetic spin-spin" interaction involving the nuclei (protons) of the two hydrogen atoms in the molecule. The details of the following calculation are a bit involved, so I will just show the result:
Question 6: Using your answer to the last question, what is the optical depth of the absorption line in
L134N?
The observed ratio of the hyperfine transitions yields a measure of the optical depth of the transition. Remember that
Ta = Tex*(1-exp(-tau))
Taking the ratio of two hyperfine lines, and assuming that the excitation temperature (Tex) of the two hypferfine transitions is the same, one gets the following:
Ta,hf1/Ta,hf2 = {1-exp(-tau,hf)}/{1-exp(-a*tau,hf)}
a = I,hf1/I,hf2
Defining R to be the measured ratio of the hyperfine transitions, one gets an equation which can be solved for tau:
R*[1-exp(-a*tau,hf)]+exp[-tau,hf]-1 = 0
Question 7: In the
L134N measurement, what is this line absorbing? _Hint: Remember the
term "optical depth" from the lecture_?
The Cosmic Microwave Background. Remember the term "optical depth" from the lecture? If the optical depth is very large, the line will absorb away ALL of the continuum. But most likely
it is moderate and the depth of the line compared to the strength of the cosmic microwave background gives you the optical depth directly, which leads to the column density of molecules toward the source. If the ratio of the hyperfine lines is what it should be, then the line is not very optically thick. --an independent measure of optical depth comes from
the observed ratio of the hyperfine lines. I went over this way too
fast in the lecture but as you read the notes, think about these things.
--
JeffMangum - 23 Jul 2004