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Absorption Spectra (1 Viewer)

Zarathustra

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It's the sodium which produces the dark or 'd' lines. Put a Na spectrograph on top of the Sun spectrograph and you get a continuous spectra.

The trig parallax is caused by the relative movememnts of the stars i the sky - it's covered in Jacaranda.
 

deadnature

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~JiNnIE~ said:
excuse me, sorry to bug in ppl, big apologies
but i need help with Trigonometric parallax? what da heck is it?

Combine the words to get the definition - Trigonometry - the form of math that deals with determining either the length of a side of a triangle or an anlge inside the triangle

Parallax - Parallax referring to the apparent shift of an object when viewed at different angles. Normally bi-anually (6 monthly for Earth).

Trigonometric Parallax - The method in which to determine the distance of an object using trigonmetry.
 

peeasoup

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Trig parallax is where you measure the parallax shift of a star by observing it 6 months apart against a background of stars, then using the parallax shift (the angle P), and the distance between the sun and earth as 1 AU, you can use trigonometry to find the distance. You go Sin p = earths orbital radius / d.
you end up with d = 1/p. (d = distance and p = parallax shift angle).
Do you get the absorbtion spectrum from stars because basically some of the wavelengths are absorbed and scattered in the atmoshpere because of the relative coolness ? ? ? ? ? ?
 

~JiNnIE~

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I get it now.Thank you all. btw u noe for d 2nd dot point in d syllabus, it says to discuss why some wavebands can be easily detected from space, do ya write about the wavelengths being transparent to certian wavelengths and optical windows and stuff?
u noe what? I still habn't finished my summary for Astrophysics, man im gonna fail physics big time.
 

mcstump

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the reason you can detect some wavelengths from space and not from earth is because the earth's atmosphere absorbs some of the wavelengths. it is suposed to stop radiation from our own sun affecting us, but also means that we cannot detect measurements from other stars from within our atmosphere
 

mcstump

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It's the sodium which produces the dark or 'd' lines. Put a Na spectrograph on top of the Sun spectrograph and you get a continuous spectra.

wtf was Zarathustra on about? dark lines are caused from nebula between the star and our observation point. The energy which is contained in the light exites electrons in the nebula, causing them to jump to a higher energy band. as this energy is lost from the light that we recieve, dark lines apear on the spectrum depending on what atom it hits and from what energy level the elactron jumps from and to.
the inverse is also true. when the electrons loose the energy, they drop down an energy band(s) this releases the energy in a random direction, and if it is towards us, it appears on the spectrum as emission lines.
ps Zarathustra your signature is about 7 lines too long
 

~JiNnIE~

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Thanks ^^. uh oh... i hab 2 more qns; Syllabus 5 investigation qn, it says 'perform an investigation to model the light curves of eclipsing binaries using computer simulation' what if u never did that at school? the other qn-> Do we need to know the Wien's Law and Stefan's law? its not in the syllabus i dun fink and the constants are not given to us on the physics Formula sheet either!
i dun fink i will get my summaries done in time. Im starting to panic, i fink im askin these qns abit too late.
 

mcstump

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if its not in tha sylubus they not allowed to ask it
i have never heard of either of their laws even tho i might no what they are
you have still got a few weeks to sumarise im sure ull do fine
just get a few lines for each of the dot points n u should do fine :)
where u from?
 

~JiNnIE~

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thanks mcstump, thanks for making me feel better.^^ ; but i dun quite get ur qn, im a lil slow...eheh
U mean what skool i go to? or u askin me my nat?
...where are you from?
 

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Tecchnically your right neither Law is required, however like the pendulum formula in the gravity prac, one or the other is required to analyse information to predict the surface temperature of a star from its intensity/wavelength graph.
A description is required but not the name of the Law. It also could come up as a process style question
 

helper

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According to the board the whole syllabus will be examined over a period of time. So somehow at sometime
 

mcstump

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are caused from nebula between the star and our observation point. The energy which is contained in the light exites electrons in the nebula, causing them to jump to a higher energy band. as this energy is lost from the light that we recieve, dark lines apear on the spectrum depending on what atom it hits and from what energy level the elactron jumps from and to.
the inverse is also true. when the electrons loose the energy, they drop down an energy band(s) this releases the energy in a random direction, and if it is towards us, it appears on the spectrum as emission lines.
 

peeasoup

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I guess that makes sense. I dont get this bit, "The energy which is contained in the light exites electrons in the nebula". Im assuming that a nebula emmits light. So at what point between the observer and the nebula is the energy lost in the photons ? ?
 

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The original light is from other stars. A nebulae itself doesn't produce energy by fusion etc. Rather it absorbs the light of other objects.

The light that is absorbed by nebulae results in electrons jumping up. The nebulae then re-emits photons as the electrons drop back down. The lights is radiated in all directions resulting in emmision spectra in any direction except along the path of the original light beam

Light Source ---> Nebulae --> Observer sees absorption spectra as direct line.
/ | \​
Observers see emmision spectra as only see re-radiated energy​
 
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peeasoup

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aahh right, i get ya. So light energy from stars hits the nebula and excites electrons, making them jump up energy band, then back down which re-radiates light, which we view as being emission spectrum, where you only see thin lines of wavelengths - most the spectrum is not represented. Like, the emission lines shown correspond to the radiated photons of discrete energy levels. Does that sound right ?
 
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mcstump

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that sounds right to me sory about the confusion i know what it is but its a bit hard to explain

"in any direction except along the path of the original light beam" why cant it be re-emited in the same direction?
 

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