19-02-2005 10:10									#31
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quote: Originally posted by Japher why is this an astronomy question? This is more of a physics/p-chem question I never took astronomy

the first module is on the spectra of stars, hopefully the second module won't be...

	19-02-2005 10:17									#32
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A few problems:

The plus should be a minus, you're missing 10^-34, a c, and the units are mixed up. Best to rewrite it carefully including all units.

	19-02-2005 10:18									#33
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red... did you even go to the link I gave? scroll down until you see the Hydrogen Spectrum box... they even have a box thing there.

What is c?

	19-02-2005 10:19									#34
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quote: Analyzing light from stars I guess.

I see

	19-02-2005 10:20									#35
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Speed of light.

	19-02-2005 10:25									#36
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quote: Originally posted by Japher red... did you even go to the link I gave? scroll down until you see the Hydrogen Spectrum box... they even have a box thing there. What is c?

I've felt kind of dumb all the way through this thread, now I feel *REALLY* dumb

ramo, 'c' is an applet that answers my question,

I'll still need to get my head around it for the exam.


Japher/Ramo

	19-02-2005 11:39									#37
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quote: Originally posted by reds4ever 1/6.63(1.51+3.4) 0.15(4.9) =0.735 ARRRGGGGGHHHH! *bangs head on keyboard*

No, no. When the electron jumps from n=2 to n=3, it needs to absorb energy equal to -3.4eV - (-1.51eV). Once you have that, you just figure out the wavelength of the photon.

Piece of cake

	19-02-2005 12:53									#38
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It's used for the spectra for stars.

Sorry I was out, or otherwise, I could answer your question.

What happens, is that each electron has levels applying to specific orbitals when energy is induced.

Stars are very hot, therefore, most of the elements on the stars are going to be in this excited state.

Hydrogen, as the most common element has specific orbitals.

You should get a graph like this:

Attachment: spectra1.jpg
This has been downloaded 25 time(s).

	19-02-2005 12:57									#39
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The top line is n = infinity, which is equivalent to the ionisation energy of the hydrogen atom, and the energy required to strip the electron completely away from the atom.

The bottom line is n= 0, which is the rest state of the hydrogen electron.

The states you are interested are n =1 , n =2 , n =3 , n =4 which are the intermediate jumps.

Note, that for the electron to jump from n =1 to n =2 requires less energy than for the electron to jump from n=0 to n=1. This gap narrows, tending to 0 as n tends to infinity.

	19-02-2005 13:00									#40
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There's a couple of equations you are going to need to know here.

When an electron jumps to a higher state, it absorbs energy. When an electron falls from one state to another, it releases energy which we can detect on the spectrograph of the stars.

This is why we get two class of spectra, absorption lines, and emission lines.

Emission lines are bright lines, while absorption lines are 'holes' or dark gaps in the spectra. These lines correspond to the frequency of the energy released or absorbed.

	19-02-2005 13:02									#41
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The equation for the energy of a wave, is E = hc/(lambda)

Where h = Planck's constant and c = the velocity of light.

	19-02-2005 13:04									#42
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quote: Originally posted by Ben Kenobi The equation for the energy of a wave, is E = hc/(lambda)

Therefore, wavelength (lamda) = hc/E. Once he has the energy, he can find out wavelength.

	19-02-2005 13:10									#43
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quote: The energies (En) of the energy levels of the hydrogen atom are approximated by the formula (En/eV) = - (13.6/n^2)

This is the function that tells you how to find any energy level, so long as you know the number of the energy level you are looking for. What is does is give you the distance between the energy levels.

To find this number is very instructive, but that's not what you need right now.

What this means, is for n = 1, E(gap) = -13.6

For n =2 E(gap) = -13.6/4

For n =3 E(gap) = -13.6/9

For n=4 E(gap)= 13.6/16

Now, to find the total energy of each jump, you need to add each gap together.

Therefore, E(total) for n=1 = -13. 6

For n = 2 E(total) = -13.6 - 13.6/4

For n=3 E(total) = - 13.6 - 13.6/4 - 13.6/9

For n=4 E(total) = -13.6 - 13.6/4 -13.6/9 -13.6/16

	19-02-2005 13:15									#44
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quote: I need to show that radiation that has a wavelenght of 656nm (H alpha) is absorbed when hydrogen atoms make the transition from n=2 to n=3.

Okay, this is really simple now.

Your energy will be the difference between the total energy for E3, or as you look below, -13.6/9 eV

So to solve, the equation for energy, as UR has said, h c / E

However, you have to be careful to use the value for h that corresponds to eV, or convert to SI.

The simplest way I know is to convert Energy from eV to Joules.

So find a table, that lets you convert eV to Joules. Then you can use SI units to calculate the answer, and get your wavelength in m, which can be easily converted to nm.

	19-02-2005 13:20									#45
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I don't know, someone help me on this.

Are the CGS units of nm and eV equivalent?

	19-02-2005 13:52									#46
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quote: Originally posted by Ramo It's the same thing. En isn't unitless - it's expressed in Joules, eV's, whatever. So you divide 5 J (for example) by 1 eV, and you get a unitless quantity.

A rather large unitless quantity, however...

	19-02-2005 14:00									#47
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quote: Originally posted by Ben Kenobi I don't know, someone help me on this. Are the CGS units of nm and eV equivalent?

Why the **** would they be? One's a length and one's an energy. There are no unitless fundamental constants in CGS to get you from one to the other.

Even the "God-given" system makes energy = 1/distance, and that's as close as you'll get in any reasonable units system.

For everything non electrodynamical in nature the only difference between CGS and SI is unitless constants.

	19-02-2005 14:03									#48
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And ffs you don't need a table to go from eV to J

Energy (in Joules) = Energy (in eV) * 1.6 * 10^-19

That 1.6*10^-19 is the fundamental unit of charge expressed in coulombs ( known as e, about number 5 on the most important physical constants to remember)

1) c
2) hbar
3) G
4) k
5) e
6) permeability of free space
7) fine structure constant
8) Avogadro's number
9) stefan-boltzmann constant
10) R

Last edited by KrazyHorse on 19-02-2005 at 14:12

	19-02-2005 14:20									#49
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quote: Originally posted by Ben Kenobi The top line is n = infinity, which is equivalent to the ionisation energy of the hydrogen atom, and the energy required to strip the electron completely away from the atom. The bottom line is n= 0, which is the rest state of the hydrogen electron

No it isn't. The spectrum of n, the primary quantum number, runs from n = 1 to n = infinity. The ground state of hydrogen is the n=1, l=m=0 state.

	19-02-2005 14:22									#50
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quote: That 1.6*10^-19 is the fundamental unit of charge expressed in coulombs ( known as e, about number 5 on the most important physical constants to remember)

You might remember that number, but I don't. That's why we have tables in the first place, so we can look these things up.

quote: Why the **** would they be? One's a length and one's an energy. There are no unitless fundamental constants in CGS to get you from one to the other.

That's not what I was asking. I just needed to know whether Joules is to metres as nm are to eV. I don't use cgs, but many astronomers do because it makes for simpler calculations.

	19-02-2005 14:24									#51
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No, many astronomers use it because they're dense mother****ers.

	19-02-2005 14:26									#52
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It doesn't actually make for simpler calculations, Ben. There's no reason it would unless you were using the Gaussian brand of CGS and spending a lot of time doing classical electrodynamics. Astronomers are just dumb and refuse to conform.

	19-02-2005 14:26									#53
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Ah, I was thinking of ergs - Joules.

I don't disagree with you, KH.

	19-02-2005 14:28									#54
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Well, then why have a unit like eV for energy? Why not use Joules?

Is it because the partical physicists are dense, or is it because it makes no sense to use Joule for the energies they deal with?

	19-02-2005 14:29									#55
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Hell, I'll agree to use parsecs if they would just bite the bullet and go to SI like they should.

	19-02-2005 14:31									#56
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Parsecs are hella easier than metres.

All the astronomers I know love to make you frusterated with the units, just because.

	19-02-2005 14:32									#57
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quote: Originally posted by Ben Kenobi Well, then why have a unit like eV for energy? Why not use Joules? Is it because the partical physicists are dense, or is it because it makes no sense to use Joule for the energies they deal with?

a) I hate using eV too, but at least that number is obvious from e

b) Ben, do you mind explaining to me what happens on the scale of ergs that astronomers are interested in? Depending on what they're doing they need stuff on the scale of eV (12 orders of magnitude off), GeV (4 orders of magnitude off) or stuff on the scale of billions of joules (many many many orders of magnitude off)

	19-02-2005 14:33									#58
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quote: Originally posted by Ben Kenobi Parsecs are hella easier than metres.

I meant for astronomical scales. At least that unit makes sense.

	19-02-2005 14:51									#59
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quote: b) Ben, do you mind explaining to me what happens on the scale of ergs that astronomers are interested in? Depending on what they're doing they need stuff on the scale of eV (12 orders of magnitude off), GeV (4 orders of magnitude off) or stuff on the scale of billions of joules (many many many orders of magnitude off)

It's all about units of luminosity, as far as I can tell.

	19-02-2005 15:51									#60
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quote: Originally posted by Ben Kenobi Well, then why have a unit like eV for energy?

Because it's easier to use in some instances - like this one.