maths/astronomy question

[Ramo]

Thats it! E sub n, isn't E sub 2 it's E sub -3.4

Cheers, stick around I've got 9 more months of this

No, Japher's notation isn't right... E_n are the energy eigenvalues of the hydrogen atom. n is a positive integer. The energy loss for an electron is E_nfinal - E_ninitial. That's equated to hc/lambda, the energy of the photon.

Inflate your Upload Space

Does then the Expression (En/eV) literally mean En expressed as eV? I've never come across that before.

i was looking at dividing En (a value) by eV (a unit)

It's the same thing. E_n 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.

[Japher]

http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

that explains it pretty well IMO, at least the first box

[Ramo]

isn't that what Einstein got his nobel prize for?

No, Einstein didn't have anything to do with the Hydrogen atom. He got his Nobel Prize for the photoelectric effect.

[Japher]

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

Yes. I think how you would be better off viewing it is

En = -(13.6/n^2) eV, which you get by multipying each side by eV, it is only a unit value meant to apply to the value of the calculation since n, the only factor, is off itself a unitless value.

[reds4ever]

Right, hang on a minute.

E sub 2 = -13.6/(2*2)
E sub 2 = -13.6/4
E sub 2 = - 3.4 eV

E sub 3 = -13.6/(3^2)
E sub 3 = -13.6/9
E sub 3 = 1.51 eV

I take it Plancks constant figures in the next stage, or have I got it horribly wrong again?

[Japher]

No, Einstein didn't have anything to do with the Hydrogen atom. He got his Nobel Prize for the photoelectric effect.

E=hv(nu)

right?

[reds4ever]

[QUOTE] Originally posted by Ramo


Inflate your Upload Space

/QUOTE]

hey, gimme a break, it's late and I'm old

[Japher]

I take it Plancks constant figures in the next stage, or have I got it horribly wrong again?

Nope, your on the right track. You have the energies of the electron at their current energy levels (2 and 3). For an electron to move fron n=3 to n=2 it would release an energy with the a value equal to their differences of?

[reds4ever]

Originally posted by Ramo
So you divide 5 J (for example) by 1 eV, and you get a unitless quantity.

Thats what was f$%king me up! thanks

Like japher said En = -(13.6/n^2)eV would v'e made more sense to me.

[Japher]

Like japher said En = -(13.6/n^2)eV would v'e made more sense to me.

Yes it would I am so with you

When all else fails list the units and make sure that the units on one side of the equation can cancel out with the other side...

[Ramo]

E=hv(nu)

right?

Yeah, but not the math (which Planck figured out). The idea of a photon, and why high frequency light shined on metals eject electrons.

[Japher]

that's what I always relate it too

[Japher]

why is this an astronomy question?

This is more of a physics/p-chem question

I never took astronomy

[reds4ever]

1/6.63(1.51+3.4)
0.15(4.9)
=0.735

ARRRGGGGGHHHH!

*bangs head on keyboard*

[Ramo]

Analyzing light from stars I guess.