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in this video I'm gonna focus on Bohr's.model of the atom now he proposed that.the electrons move around the nucleus in.circular orbits.so let's say this is the nucleus and so.this is the first energy level this is.the second this is the third and so.forth.now the electrons can only be in these.energy levels it could be in the fourth.or fifth but it can't be like anywhere.in between and so the energy levels of.electrons are quantized because they can.only have discrete values the electron.can't be in the one point for energy.level or 1.86 energy level it has to be.in the first second or third or fourth.energy level so these energy levels.there are integer based values so the.electrons they can only occupy certain.orbits and in that sense they're.quantized now I need to know that when.electron falls from a high energy level.to a low energy level it's going to emit.energy in the form of a photon now when.an electron jumps let's say from a low.energy level to a high energy level it.can only do so if it absorbs a photon.with the right energy now in this.problem we're going to talk about how to.calculate how much energy is released or.absorbed when an electron moves from one.energy level to another so let's read.the first question how much energy is.released when an electron falls from the.N equal four to the N equal to energy.level inside a hydrogen atom now the.equation that you need to calculate it.is this equation e is equal to negative.two point one seven eight.times 10 to the minus 18 joules.multiplied by 1 over and final squared.minus 1 over an initial squared an.initial is the original energy level.which is.the fourth energy level and final is.where the electron is going to that is.the second energy level so all we needs.to do is plug those numbers into the.equation and we can find out how much.energy is released from an electron.within a hydrogen atom so let's go ahead.and do that.so and finally in this problem is to.don't forget to square it and an initial.is 4 so 1 over 2 squared 2 squared is 4.and 4 squared is 16 so we got 1/4 minus.1 over 16 and that's equal to 3 over 16.which is as a decimal point 1 8 7 5 so.let's go ahead and multiply these two.numbers so you should get negative 4.point 0 8 4 times 10 to the negative 19.joules so that's how much energy is.released when an electron falls from the.fourth to the second energy level now.has a question for you why is this.answer negative recall that when H falls.from let's say a high energy level to a.low energy level a photon is emitted so.the electron is losing energy and that's.why it's negative if the electron jumped.from let's say the second to the fourth.it would be absorbing energy so it would.be positive so that's what we have a.negative sign to show that energy is.being released as the electron falls to.a lower energy level now let's move on.to Part B how can we calculate the.frequency and now that we have the.energy of the photon now if you recall.from earlier videos the energy of a.photon is equal to the product of.Planck's constant and the frequency of.that photon so to calculate the.frequency it's simply the energy divided.by Planck's constant so it's gonna be.four point zero eight four times ten to.the negative 19 joules you don't need.the negative sign because frequency is.always positive the purpose of the.negative sign is to tell us if a photon.is being absorbed or released by an.electron.so let's go ahead and end the FIDE those.two numbers so you should get 6 point 1.6 4 times 10 to the 14 Hertz or 1 over.seconds so that's the frequency now that.we have the frequency let's move on to.Part C let's calculate the wavelength of.the photon and nanometers now the speed.of light is the product of the.wavelength and the frequency so the.wavelength is the speed of light divided.by the frequency the speed of light is 3.times 10 to the 8 meters per second and.the frequency is 6 point 1 6 4 times 10.to the 14 with the unit's run over.seconds.now let's go ahead and put that in the.calculator so I got four point eight.seven times ten to the negative seven.meters now we need to convert this to.nanometers so how can we do that now you.need to know that one nanometer is.equivalent to ten to the minus nine.meters so you can take the answer that.we have for point eight seven times ten.to the minus seven and divide it by one.times 10 to the negative nine and so.this is equal to 487 natives so that's.how you can calculate the wavelength of.this photon number two an electron in.the third energy level or an equal three.State absorbed as a photon with a.wavelength of twelve eighty three point.four five nanometers and so what energy.level will the electron jump to so we.know what equation we need to use but.we're given the wavelength the first.thing we need to do is calculate the.energy of the photon using the.wavelength now the energy of a photon is.Planck's constant times the frequency.and we know the frequency is the speed.of light divided by the wavelength so.that equation comes from this equation.if you divide both sides by lambda so.what I'm gonna do is replace the.frequency with this expression so in.terms of the wavelength the energy of.the photon is Planck's constant times.speed of light divided by the wavelength.so let's use that equation.now we need to convert nanometers to.meters so just replace nm with 10 to the.minus 9 meters and this will give you.the energy of the photon so the energy.of the photon is 1.5 for 9 times 10 to.negative 19 joules so now that we have.the energy of the photon we can now use.this equation.our goal is to find the value of an.final and it should be an integer value.like two three four five six seven it.shouldn't be like six point four eight.point seven we should get a whole number.so we have an initial it's dream and we.have the energy now well we need to.determine is is the energy positive or.negative because if you get this wrong.it can mess up the entire problem so our.key expression is absorbs the electron.absorbs a photon so it's absorbing.energy which means the energy of the.electron is increasing so E is positive.so we're gonna replace e with this.number with a positive sign so one point.five four nine times 10 to the negative.19 and that's a Joule symbol is equal to.this number multiplied by 1 over n final.squared minus 1 over an initial squared.which an initial street so all we got to.do at this point is we need to solve for.the missing variable so I'm gonna walk.you through this step by step.so the first thing we want to do is take.this number and divide it by that number.so one point five four nine times 10 to.the negative 19 divided by a negative.two point one seven eight times 10 to.the negative 18.you should get negative point zero seven.one one two and that's equal to one over.and final squared minus three squared is.nine so that's 1 over nine so next let's.add one over nine to both sides so if.you have a calculator take negative.point zero seven one one two plus one.over nine and you should get point zero.three nine nine nine which should.probably be point zero four now let's.put this over one and let's cross.multiply one times one is one and then.we have point zero three nine nine nine.times and final squared next divide both.sides by point zero three and nine nine.nine.so and final squared is 1/2 point zero.three nine nine nine so that's basically.about twenty-five it's like twenty five.point zero zero six and if we take the.square root of both sides and final is.five if you get a whole number as an.answer then chances are you have the.right answer if you don't get a whole.number somewhere along the lines you.might have made a mistake but this.answer is reasonable because if it.absorbs a photon we should get an energy.value that's greater than three and five.is greater than three so this makes.sense and that's the answer it's gonna.jump into the fifth energy level so.here's a MOS which waste problem for you.which electron transition involves the.greatest release of energy is it a b c d.or e so let's focus on the term release.of energy that means that the electron.has to fall from a high energy level to.a low energy level so n has to be.decreasing not increasing if energy is.being released so we can eliminate.answer choice a because it's going from.low energy level to a high energy level.and we don't want that now we could.eliminate answer choice C as well so now.is it going to be B D or e a quick way.to tell is to look at the final energy.level here the final energy level is.four here it's one and here it's two.typically the answer is going to be the.one with the lowest energy level so it's.going to be D because N equals 1 is less.than 2 or 4 and it really doesn't depend.on the difference between the energy.levels if you're looking at e if you'd.pick a as an answer is not correct but.if you're thinking because 7 to 2 that's.the difference of 5 that must mean more.that must be greater than three to one.which is a difference of two it's not.gonna work out that way now instead of.drawing a circular orbit I'm gonna draw.a horizontal line to represent the.energy levels so let's say if this is.the first energy level the second would.probably somewhere over here and then.this would be like the third and then.this would be like the fourth and then.this would be like the fifth do you see.what's happening as you move further.away from the nucleus the distance.between the energy levels they get.smaller and smaller and smaller so a.transition from two to one involves a.lot more energy than three to two it.even involves a lot more energy than.four to two now granted this is not.drawn to scale but here's how you can.find out for sure which one is going to.be the highest you can use the energy.equation that we've talked about but you.could ignore the two point one seven.eight times ten to the negative eighteen.value focus on this portion of the.equation 1 over n final squared minus 1.over an initial squared so the choice.that has the greatest value of this.expression will be the one that's gonna.release the greatest amount of energy so.let's calculate it for B so that's 1.over 6 squared actually and finals 4 so.2 1 / 4 squared minus 1 over 6 squared.go ahead and type that in your.calculator and get the decimal value for.it so you should get point 0 3 4 7 2 now.let's do the same for answer choice D so.it's going to be 1 over 1 squared minus.1 over 3 squared.so this is going to give you point eight.eight nine is basically point eight.repeating and then for answer choice E.is going to be one over two squared.minus one over seven squared and so this.is going to be point two two nine six so.as you can see the one with the largest.energy value is the one with the lowest.final state that's the answer is d the.second highest was the one with the next.lowest final energy level which was N.equals two the lowest was the one that.had the highest final energy level which.was N equals four so a quick way to get.the answer is to look at the final state.the one with the lowest final state is.the one that's gonna release the most.energy and it usually works out that way.but if you feel dubious about that just.do what I did here use this equation and.see which one gives you the largest.value and so in this problem our answer.is answer choice D it's going to release.the greatest amount of energy which are.the following electron transitions will.limit a photon in the visible light.spectrum so how can we find the answer.to this question the only way to do it.is just to know your stuff so let's go.over some stuff that you need to know so.just like before I'm gonna draw the.different energy levels so this is gonna.be the first energy level let's call.this one the second and this is gonna be.the third and then the fourth and let's.put the fish here.now any time an electron falls to the.first energy level it can go from two to.one it can go from three to one four to.one and so or five to one the final.energy level has to be one so anytime it.falls to the first energy level this is.known as the Lyman series now anytime it.falls to the second energy level you can.go from three to two they can go from.four to two or five to two or six to two.this is known as the Balmer series.and if the final state is the third.energy level let's say from 5 to 3 or 4.to 3 this is no one has the passion.series or passion or I'm not sure.exactly how you say its name thing is.passion you can look it up though and.then if it falls let's say from 6 to 5 I.mean 6 to 4 or 5 to 4 that's known as.the bracket series and then if it goes.down to the fifth I'm gonna put this in.yellow that's known as the Pfund series.now the most common ones that you're.going to deal with are the first three.your teacher may expect you to know all.five but typically you probably tested.on the first three if you have to choose.which one to memorize now his we need to.know the Lyman series is associated with.the ultraviolet spectrum the Balmer.series is associated with the visible.light spectrum most of it is visible.live I think one of them might be using.but the majority of it is visible light.and everything else like the passion.bracketed the P fun it's all in the.infrared or the IR region so to answer.this question all we got to do is look.for a photon that's in the Balmer series.so the final state has to be n equal to.so it's not a it's not be it's C because.the final state is N equals 2 so C is.the answer so answer choice a is.associated with the passion series so.this would emit a photon in the infrared.region B is associated with the Lyman.series that ends at N equals 1 so it's.going to emit a photon in the UV.spectrum C is going to be visible light.D is associated with the bracket series.so that's going to be an infrared region.and E is associated with the Pfund.series which is also in the infrared.region.

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