maths/astronomy question

[reds4ever]

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...

[Ramo]

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.

[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?

[Japher]

Analyzing light from stars I guess.

I see

[Ramo]

Speed of light.

[reds4ever]

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

[Urban Ranger]

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

[Ben Kenobi]

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:

[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.

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.

[Ben Kenobi]

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.

[Ben Kenobi]

The equation for the energy of a wave, is E = hc/(lambda)

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

[Urban Ranger]

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.

[Ben Kenobi]

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

[Ben Kenobi]

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.

[Ben Kenobi]

I don't know, someone help me on this.

Are the CGS units of nm and eV equivalent?