<BrianQ> Speaking of driving crazy, what were the homework results... wavelengths for n= 2 to 3 and n=2 to 4
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1 <BrianQ> I have a good reference page for us... <BrianQ> <doos_> yes that is a nice one BrianQ <_Frank_> ok Brian I see it <@Spauwe> got it as well <BrianQ> Speaking of driving crazy, what were the homework results... wavelengths for n= 2 to 3 and n=2 to 4 <BrianQ> um... n = 2 to 4 and n = 2 to 5... <BrianQ> that's what I meant to say <_Frank_> Sorry Brian I haven't done it...busy week (though that s no excuse) <@Spauwe> I 2 was counting on Frank :) <BrianQ> That's ok, we'll use the webpage I just gave you to figure it out, roughly... <_Frank_> ok...sorry <@Spauwe> the websites give me nm values <@Spauwe> what was that conversion again? <BrianQ> we found out last week that transition from n = 1 state to anything else requires what kind of light? <BrianQ> nm is nanometers... which is a length... as in wavelength <_Frank_> fluorescent light <_Frank_> very short wave <@Spauwe> yes... wasn't there a conversion? never mind me <BrianQ> yep, even fluorescent light doesn't get that short wave... <BrianQ> But short wave what )?_ <doos_> 1240/wavelength = energy in ev tim <BrianQ> "short wave "? <_Frank_> ultra violet <BrianQ> Bingo!
2 <BrianQ> Now transition from n = 1 to something else requires very short wave... <BrianQ> transition from n = 2 to higher states require less energy (remember this is true?) <_Frank_> yes <doos_> yes, for hydrogen <BrianQ> why is it true? <doos_> because of the negative <BrianQ> nope <doos_> the secret thingy <BrianQ> nope <doos_> heh <_Frank_> decreasing spaces in energy levels <BrianQ> yes <BrianQ> as n increases, spacing between energies goes down <BrianQ> so, for higher n, energy decreases, which means that frequency... <doos_> goes down <BrianQ> which means that wavelength... <doos_> gets longer <BrianQ> yes... <BrianQ> and moving from ultraviolet to longer wavelengths means you are moving to what kind of light? <doos_> visible <BrianQ> yes <BrianQ> so... <BrianQ> Transitions from n = 2 (except back to n = 1) are in the visible <BrianQ> now... just a guess, what kind of light will facilitate transitions from n = 3 (except back to n = 1 and n = 2)
3 <_Frank_> red <doos_> ir <BrianQ> yep, IR <doos_> oh wait <doos_> oh lol <BrianQ> heh 72.9nm <_Frank_> lucky mugger <doos_> ;) nah.. trying that conversion on that page conversion <_Frank_> I meant Doos <_Frank_> lol...i've done the same but left the answers in ev <BrianQ> Ok, now recall that for hydrogen to absorb a photon, the photon must have exactly the right energy... <BrianQ> what determines the "right energy"? <doos_> the ground state energy <BrianQ> nope <DragonStek> ionization <BrianQ> no[e <BrianQ> um... that is "nope" <@Spauwe> the step it has jump... <_Frank_> the exact EO values for each state <BrianQ> yes Time <BrianQ> Not quite Frank <BrianQ> yes, the energy difference in order to move from one state to another. <BrianQ> Now, most systems start out in a "ground state"
4 <BrianQ> Most hydrogen in the universe is in its ground state <BrianQ> If you have a tube of hydrogen, it is in its ground state... <BrianQ> And so it is going to be difficult to see any visible light from that tube... <BrianQ> And it will also be difficult to see any absorption of visible light from that tube <BrianQ> Tim made the good observation that is why hydrogen doesn't make much of a color center <BrianQ> Now... how to get the hydrogen to emit visible light... that is, how to get it out of the ground state and up to n = 3,4,5..ect states <BrianQ> you are going to need some heavy duty source of UV radiation... or... <@Spauwe> heat it? <BrianQ> you can use something else, like what? <doos_> more likely some long wavelength source <BrianQ> nope, long wavelength source will do nothing... <doos_> x-rays then? <BrianQ> not matched to the energy difference needed <BrianQ> X-rays aren't matched to the energy needed, either <doos_> gamma? <BrianQ> even worse <_Frank_> shrt wave UV <@Spauwe> a lighter? <BrianQ> heh <DragonStek> vuv <doos_> shorter than gamma? <BrianQ> recall that the light has to be exactly right in order to give up its energy... <BrianQ> but not everything has to have exactly the right energy... <BrianQ> so instead of hitting it with light, hit it with... <_Frank_> heat <doos_> neutrons
5 <_Frank_> photons a hammer... gehhe <BrianQ> no.. heat isn't a particle and neutrons... well the neutron source would cost more than the UV laser <BrianQ> and not photons, because they have to have the right energy... <@Spauwe> what else is there around? <@Spauwe> hmmmm <BrianQ> and Tim, has anyone ever told you that a hydrogen atom is very very small? <_Frank_> electrons then <BrianQ> yes! <BrianQ> electrons! <BrianQ> and this is what is done in spectral tubes such as neon signs (or fluorescent tubes such as you see in offices) <_Frank_> we spoke about this several months ago <BrianQ> electrons are boiled off a filament at one end of the tube and pulled by a strong electric field <BrianQ> yes. <BrianQ> and so they get a lot of kinetic energy... they collide with hydrogen, and lose some energy.. <BrianQ> which goes into exciting the hydrogen to something other than the ground state. <BrianQ> The first direct observation of transferring electron energy to create excited states of atoms <BrianQ> I think it was called the <BrianQ> Franck-Hertz experiment [22:02] <BrianQ> But anyways, we don't need to directly observe this, we can just witness the results. <@Spauwe> so it's electrons used as hammers <@Spauwe> kinetic energy after all <BrianQ> yes, the electrons are our hammers
6 <BrianQ> indeed <BrianQ> So now we imagine we've got a tube full of hydrogen being bombarded by electrons <BrianQ> so we have all kinds of atoms in all varieties of energy states... <BrianQ> What happens is that we start seeing emission of photons... the atoms all want to go back to ground state. <BrianQ> or at least to lower energy states (spontaneous emission, as it is known) <BrianQ> Now... UV light is decay back to n = 1 state... <BrianQ> visible light is decay back to n = 2 state <BrianQ> and IR light is decay back to higher n states (from even higher n states) <BrianQ> Got that? <doos_> yes <_Frank_> question <BrianQ> yes Frank? <_Frank_> the energy released from the higher states n3, n4 etc actually release energy of a lower value than the return from say n2 to n1 <BrianQ> yes, that is correct <_Frank_> ok thanks <@Spauwe> the steps are smaller... <BrianQ> Yes, n =99 to n = 2 requires smaller energy than n = 2 to n = 1 <@Spauwe> less in - less out <BrianQ> Now visible light transitions in hydrogen are referred to as the Balmer series... <BrianQ> With the n = 3 to n =2 being the first Balmer transition... and we worked out last time that occurs at red light <BrianQ> The next is n = 4 to n = 2, and that is cyan (blue-green) light <BrianQ> Roughly speaking, because I can't remember the wavelengths off the top of my head... <_Frank_> would these be quoted as nm or as ev <BrianQ> The n=3 to n=2 is around 650 nm, the n=4 to n=2 is around 450 nm <BrianQ> whichever... <_Frank_> 0k
7 <BrianQ> Then the next is n=5 to n=2, and that is violet... <BrianQ> and in fact, all higher ones are violet... so that the n=99 to n = 2 is just a little below 400 nm <BrianQ> Well, guys, I apologize, but I've got to end this a bit early :( <doos_> oh a shame brian <doos_> that is fun though <DragonStek> oh ok thanks brian <Crystal> thanks y'all :) <@Spauwe> righto <BrianQ> we'll pick this up again next time... but notice in the link I gave you... <_Frank_> thanks again Brian...we'll continue next week <BrianQ> you'll see a red, cyan, and violet line or two <doos_> yes thanks Brian.. have it bookmarked <BrianQ> ciao ciao <_Frank_> me2 signed off") 02[22:17] * BrianQ (n=fn-javac@ ) Quit ("Java user
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