My peak wavelength is 4020.5435. Using Wien's Displacement Law : T = 2.897E7 divided by the peak wavelength ( lambda max), I got a temperature of 7205 K. My estimated value for the temperature of 41 cygnus star is 7200K.
Following the Spectroscopy Guidelines, the peak wavelength is found using a best fit curve (without specific absorption data) and I find it to be near 4300 angstrom for Cygnus 41. This correlates to a temperature of 6700K (using the applet program).
My peak wavelength of the spectrum is 4018.1062. Applying Wien's law, the computed temperature is 7192 K. Using 2 significant figure, my temperature is 7200K.
Using the exmine tool of GA, I found out the highest peak wavelength to be 4018 angstroms. using Wien's law, my temperature for this star is 7190. Since the spectrum is consist of different spectral lines, I used 2 significant figure to express my temperature. For 41_cyg, its temperature is 7200K
I got the same peak using Graphical Analysis. Using wien's law, my calculated temperature is 7192K. I compared the spectrum with the graph in the simulator and I got a lesser temperature. I analyzed the absorption lines and determine the prominent lines. There is a dominance of Ca II absorption and the metallic lines become noticeable. These are characteristics of a class F star. Then I looked up the SIMBAD then confirmed my temperature. Using 2 sigfig, I got 7200K for 41_cyg
Great work! and this week you should receive a check for the first 3 modules, assuming I have you SSN (a few of you still haven't replied to my emails)
When I found the temperature, I looked for the highest peak, at first. I decided this is wrong. I think I should look at the peak of the curve fit (which we don't have, but can approximate one just by looking), which gives a slightly different peak than others.
So, the peak of an approximated curve fit (which doesn't necessarily have to correspond to an actual peak)has a wavelength of 4300 angstroms. This means theres a temperature of 6720 K.
I thought the peak was around 4200 Angstroms, yielding a temp of about 7142 degrees. According to simbad, the temp is 6526, which puts the peak wavelength around 4597. At that wavelength, the spectrum is clearly past its peak. What gives?
I thought the peak wavelength was roughly 4,050 A. This would correlate with 7150 K.
At first I didn't understand why this link: http://www.noao.edu/education/astrobits/files/Spectroscopy-GA3-Abits.pdf said that the peak was at 4200 until I read Triple D's post about using the line of best fit. So that means a better estimate for the temperature is 6900 K.
Module 5 Assignment 7 After looking at the plot on GA and comparing it to the curve on pg 10 of Stellar Spectroscope, I estimated the λ at about 4100 Å. Using Wien’s Law I estimate the temperature at about 7066° K.
I estimated the peak wavelength at 4019 Angstroms. Using Wien's Law to two significant figures, I computer the temperature of 41 Cygnus to be 7200K.
Using the SIMBAD website I pulled up the information for 41 Cygnus. I found where it said that the Spectral Type is F5 Iab. There is sure a lot of data about a lot of things that I have no idea about. I could not find where it lists the temperature.
I analyzed the star and didn't come up with anything too different than anyone else. I took the peak of the wavelength as being 4020 A and so the temperature is 7,200 K.
One thing that I can perhaps add to the discussion is that I had my students go off and analyze the spectra of Cygnus 41. They all pretty much universally concluded that the elements in it are:
balmer series (so H), He, N III, Mg, Si III, Ca, Sc, Ti, Mn, Fe, Sr, Hg, CH, TiO, Telluric absorption bands
I tried to connect this with looking at emission spectra and having them calculate that different emission lines corresponded to different wavelengths. The whole lesson was, I think, successful in instilling some sense of what emission lines and absorption lines are.
Cygnus 41 has a spectrum peak at 4000A with this I estimated the temperature was 7300K
ReplyDeleteMy peak wavelength is 4020.5435. Using Wien's Displacement Law : T = 2.897E7 divided by the peak wavelength ( lambda max), I got a temperature of 7205 K. My estimated value for the temperature of 41 cygnus star is 7200K.
ReplyDeleteFollowing the Spectroscopy Guidelines, the peak wavelength is found using a best fit curve (without specific absorption data) and I find it to be near 4300 angstrom for Cygnus 41. This correlates to a temperature of 6700K (using the applet program).
ReplyDeleteMy peak wavelength of the spectrum is 4018.1062. Applying Wien's law, the computed temperature is 7192 K. Using 2 significant figure, my temperature is 7200K.
ReplyDeleteUsing the exmine tool of GA, I found out the highest peak wavelength to be 4018 angstroms. using Wien's law, my temperature for this star is 7190. Since the spectrum is consist of different spectral lines, I used 2 significant figure to express my temperature. For 41_cyg, its temperature is 7200K
ReplyDeleteI got the same peak using Graphical Analysis. Using wien's law, my calculated temperature is 7192K. I compared the spectrum with the graph in the simulator and I got a lesser temperature. I analyzed the absorption lines and determine the prominent lines. There is a dominance of Ca II absorption and the metallic lines become noticeable. These are characteristics of a class F star. Then I looked up the SIMBAD then confirmed my temperature. Using 2 sigfig, I got 7200K for 41_cyg
ReplyDeleteGreat work! and this week you should receive a check for the first 3 modules, assuming I have you SSN (a few of you still haven't replied to my emails)
ReplyDeleteWhen I found the temperature, I looked for the highest peak, at first. I decided this is wrong. I think I should look at the peak of the curve fit (which we don't have, but can approximate one just by looking), which gives a slightly different peak than others.
ReplyDeleteSo, the peak of an approximated curve fit (which doesn't necessarily have to correspond to an actual peak)has a wavelength of 4300 angstroms. This means theres a temperature of 6720 K.
I thought the peak was around 4200 Angstroms, yielding a temp of about 7142 degrees. According to simbad, the temp is 6526, which puts the peak wavelength around 4597. At that wavelength, the spectrum is clearly past its peak. What gives?
ReplyDeleteI found the peak at 4020 A, which corresponds to 7200 K.
ReplyDeleteI thought the peak wavelength was roughly 4,050 A. This would correlate with 7150 K.
ReplyDeleteAt first I didn't understand why this link: http://www.noao.edu/education/astrobits/files/Spectroscopy-GA3-Abits.pdf said that the peak was at 4200 until I read Triple D's post about using the line of best fit. So that means a better estimate for the temperature is 6900 K.
From John Webster
ReplyDeleteModule 5 Assignment 7
After looking at the plot on GA and comparing it to the curve on pg 10 of Stellar Spectroscope, I estimated the λ at about 4100 Å. Using Wien’s Law I estimate the temperature at about 7066° K.
Module 5 Assign 7
ReplyDeleteUsing the Graphic Analysis, the highest peak I recorded is 4021, the temperature is approximate 7200K. Please email for screen shot.
I estimated the peak wavelength at 4019 Angstroms. Using Wien's Law to two significant figures, I computer the temperature of 41 Cygnus to be 7200K.
ReplyDeleteUsing the SIMBAD website I pulled up the information for 41 Cygnus. I found where it said that the Spectral Type is F5 Iab. There is sure a lot of data about a lot of things that I have no idea about. I could not find where it lists the temperature.
I analyzed the star and didn't come up with anything too different than anyone else. I took the peak of the wavelength as being 4020 A and so the temperature is 7,200 K.
ReplyDeleteOne thing that I can perhaps add to the discussion is that I had my students go off and analyze the spectra of Cygnus 41. They all pretty much universally concluded that the elements in it are:
balmer series (so H), He, N III, Mg, Si III, Ca, Sc, Ti, Mn, Fe, Sr, Hg, CH, TiO, Telluric absorption bands
I tried to connect this with looking at emission spectra and having them calculate that different emission lines corresponded to different wavelengths. The whole lesson was, I think, successful in instilling some sense of what emission lines and absorption lines are.