19-02-2005 08:24									#1
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i need a hand (ie. the answer, with workings) with this question.


The energies (En) of the energy levels of the hydrogen atom are approximated by the formula

(En/eV) = - (13.6/n^2)

Where eV is electron volts and n is the energy subscript.

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.

I've got as far as E2= -3.4eV and E3=-1.51eV and presumably I need to run the above equation for both energy levels and come up with a difference of 656nm (H alpha)

it's the first set of brackets I don't understand, one (En) is a value and the other (eV) is a unit? How do I arrive at a sensible answer?

I'm 36 and signed on a for an Astronomy course, not this!

	19-02-2005 08:39									#2
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eV is a unit of energy ("electron-volt"). It's the energy an electron gains after being accelerated through a 1 V potential. The formula just means that E_n is -13.6/n² eV.

	19-02-2005 08:48									#3
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In this problem, the idea is if an electron loses energy by switching "orbits," it emits a photon which has an energy of h(nu) (Planck's constant multiplied by the frequency), which is hc/lambda (since the wave length times the frequency of light is the speed of light).

	19-02-2005 09:01									#4
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quote: Originally posted by Ramo In this problem, the idea is if an electron loses energy by switching "orbits," it emits a photon which has an energy of h(nu) (Planck's constant multiplied by the frequency), which is hc/lambda (since the wave length times the frequency of light is the speed of light).

Ramo, I know i need Plancks constant, and I understand the emitted energy is at 656 nM and so blocking H alpha, I just don't know where to plug the numbers in.

Whats the deal with eV? Does (En/eV) mean En expressed as electron volts? And n^2 is that 2 and 3 or the values of E2 and E3?

	19-02-2005 09:05									#5
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quote: Originally posted by Ramo eV is a unit of energy ("electron-volt"). It's the energy an electron gains after being accelerated through a 1 V potential. The formula just means that En is -13.6/n2 eV.

Ah, sorry i didn't read the above properly (playing RTW and ALT tabbing), thanks a lot.

The n^2 is 2 or 3 in this case, right?

	19-02-2005 09:06									#6
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eV is the energy expressed in electron volts

n is the energy level

calculate the energy values at each level (each n) E=13.6/n^2

take the difference to give you the energy emitted

then plug it into E=hv, and solve for v (the wavelenght)

or something

	19-02-2005 09:08									#7
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v= whatever Ramo said

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

	19-02-2005 09:09									#8
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quote: Originally posted by Japher eV is the energy expressed in electron volts n is the energy level calculate the energy values at each level (each n) E=13.6/n^2 take the difference to give you the energy emitted then plug it into E=hv, and solve for v (the wavelenght) or something

Cheers Japher

it's more the notation i'mtrying to understand, I know whats happening and what i need to prove, but......

	19-02-2005 09:12									#9
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am I wrong? I have no idea.

	19-02-2005 09:13									#10
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quote: Originally posted by Japher am I wrong? I have no idea.

Neither have I, thats my problem

	19-02-2005 09:14									#11
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What? I understand the equations, what don't you understand?

	19-02-2005 09:22									#12
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the problem I think you are seeing is that the value n has no units. Thus, by them putting that eV there it supplies the units. What they should of said is

E sub n = (1/n2^2)-(1/n1^2)... with the resulting value having the units eV

	19-02-2005 09:30									#13
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quote: Originally posted by Japher the problem I think you are seeing is that the value n has no units. Thus, by them putting that eV there it supplies the units. What they should of said is E sub n = (1/n2^2)-(1/n1^2)... with the resulting value having the units eV

Thats it! n = -3.4 or whatever.

The way you put it is much clearer,

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

	19-02-2005 09:33									#14
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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)

	19-02-2005 09:38									#15
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quote: Thats it! n = -3.4 or whatever. The way you put it is much clearer, Cheers, stick around I've got 9 more months of this

No! It's not

(E sub n) is a value, the sub n indicates that it is E (energy) at a certain level (n)... At level 1 it is Esub1 at level 2 it is Esub2.... the sub part is only an indicator to denote that it E being "represented" at that state.

Thus, at n=2 you have Esub2 at n=3 you have Esub3

the equation says that Esub2=-13.6/(2*2) and it has the units of eV

Esub3 = -13.6/(3*3) and has the units eV

rarely would you see a n=2.3, or whatever. since there are not intervals of energy levels.

	19-02-2005 09:42									#16
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quote: 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.

quote: 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.

	19-02-2005 09:42									#17
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http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html

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

	19-02-2005 09:45									#18
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quote: 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.

	19-02-2005 09:46									#19
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quote: 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.

	19-02-2005 09:46									#20
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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?

	19-02-2005 09:46									#21
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quote: 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?

	19-02-2005 09:51									#22
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[QUOTE] Originally posted by Ramo


Inflate your Upload Space

/QUOTE]

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

	19-02-2005 09:53									#23
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quote: 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?

	19-02-2005 09:55									#24
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quote: 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.

	19-02-2005 09:57									#25
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quote: 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...

	19-02-2005 10:04									#26
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quote: 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.

	19-02-2005 10:05									#27
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that's what I always relate it too

	19-02-2005 10:06									#28
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why is this an astronomy question?

This is more of a physics/p-chem question

I never took astronomy

	19-02-2005 10:09									#29
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1/6.63(1.51+3.4)
0.15(4.9)
=0.735

ARRRGGGGGHHHH!

*bangs head on keyboard*

	19-02-2005 10:10									#30
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Analyzing light from stars I guess.