0:01

In learning objective number 8, we're going to see and understand the connection

Â between wavelength and frequency of the light emitted as an electron

Â transitions from a high energy level down to a lower energy level in an atom.

Â So, we're going to excite an atom, and when you excite an atom,

Â that is meaning you take an electron from a low energy state to a high energy state,

Â and then that electron is going to what to relax back down to ground state and

Â amid a photon.

Â 0:34

Where we see an n equals 1, what we see here is the energy

Â level, which is your first principal energy level.

Â Now, for a hydrogen atom,

Â and that's what we're going to really work on during this lesson.

Â The orbital in there would be the 1s orbital, that'd be true for any atom.

Â 0:55

And when we go up to n equals 2, we have the 2s orbital and

Â we have the 2p orbitals.

Â And they all have exactly the same energy, so

Â that n equals 2, that line there represents their energy.

Â Now we're going to see that when we get into multi electron systems,

Â meaning we aren't dealing with hydrogen,

Â we're dealing with everything else out there with multiple electrons.

Â That the energy level of the 2s and the 2p are not the same, but

Â that's not true for hydrogen, they're exactly the same.

Â And for hydrogen, when you move out to the third shell, third principle energy shell,

Â we have the 3s, the 3p, and the 3d orbitals.

Â They all have the same energy, and that line represents how much energy they have.

Â 1:38

Now, if an electron is here in that n equals 1 level and in that 1s orbital

Â it has the least amount of energy it can have and that's called its ground state.

Â But if we did something to have it absorb energy and

Â it promoted that electron up to the n equals 3, as we see with that gray line.

Â We would say that is a, excited state, okay.

Â So atoms can absorb energy.

Â When they do, electrons are promoted to higher energy levels.

Â Those are called excited states.

Â Now once that electron is up in that excited state,

Â it's not going to want to stay there.

Â It's going to want to come back down to its lowest energy level, okay.

Â So as it returns to what we call the ground state, and

Â I left of a parentheses there.

Â As it returns to the ground state, on those transitions down,

Â it's going to emit photons of light.

Â And we can look at those photons of light.

Â Now I am showing an electron with that red line going from the n

Â equals 3 state down to the n equals 2 state.

Â When it undergoes that transition, it will get rid of a photon of light.

Â And that is what this little squiggly arrow is representing a sine wave for

Â the wavelength of that photon of light emitted.

Â Now the electron won't stop there.

Â It could then relax further down to the n equals 1 state, and

Â if it did that, it would emit another photon of light.

Â 3:03

So that would give two different photons of light.

Â The electron could have chosen to go all the way down to the n equals 1 level and

Â if it did it in one fell swoop it would give away a photon of light for

Â that transition.

Â So every possible conceivable transition that

Â electron can undergo would give a different photon of light.

Â 3:33

Each one of these arcs is representing a general distance away from the nucleus for

Â an electron in one of these levels.

Â And if an electron started off in an n equals, 3, and transitioned

Â to an n equals 2, that'd be a certain amount of energy change for that electron.

Â 3:57

give off a photon of light with wavelength of 657 nanometers.

Â Now, we don't need to memorize that number, but it's just a for instance.

Â If an electron were in the n equals 4 and transitioned down to the n equals 2,

Â that's a bigger change in energy.

Â If it's a bigger change in energy,

Â then it needs to be emitting a photon of light with higher energy.

Â And that would be this green line here, and it has a higher energy photon.

Â 4:24

If an electron went from the n equals 5 state all the way down to the n equals 2

Â state, then that would be an even greater transition of energy for the electron.

Â So the light emitted would have higher energy.

Â Now, what if an electron went from the n equals 5 all the way down to

Â the n equals 1?

Â That would be an even greater energy yet, and that would have to

Â be somewhere down here further away from the visible spectrum and

Â you wouldn't see it as a visible in the visible spectrum but

Â it still would be electromagnetic radiation with a high energy.

Â 5:02

If an electron started at n equals 5 it only went down to the n equals 4 state,

Â that would be a much smaller energy transition and you would expect that to

Â be somewhere off the visible region at a lower energy level.

Â So that's the connections between the light emitted, and

Â the transition of the electron.

Â Now all these possibilities can happen and we're going to take this diagram.

Â Look at,

Â this is flattening out those curves, showing us the energies of each of those.

Â 5:31

So if an electron is down here it has a certain amount of energy.

Â And we're focusing in on the hydrogen atom only, so it's sitting somewhere in

Â the 1s orbital when it's down here, and that would be its ground state.

Â 5:44

If an electron transition from the n equals 5 all the way down to

Â the n equals 1 it would emit a photon of light.

Â And this line is not written in a color because it's outside of the visible range.

Â These are the three transitions that we saw on the previous slide.

Â And they are actually in the visible range.

Â And they're called the Balmer series there, but

Â they happen to fall on the visible range.

Â There's not other transition that's going to,

Â for the hydrogen atom, that's going to fall in this visible range.

Â 6:13

Okay, so what we really have to grasp here, this is a very key statement as

Â we do spectroscopy in the studies of the transitions of the electrons and

Â the photons emitted.

Â That the energy of that transition, and let's examine this one right here for now.

Â That transition from the n equals 2, to the n equals 1,

Â as the electron transitions down, it has a change of energy.

Â And that's with the delta E.

Â That's a change in energy of that electron.

Â 6:45

Whatever the size of that energy change,

Â that has to equal the energy of the photon that's emitted.

Â So when that happens and that transition occurs and a photon of light is emitted,

Â the energy of that photon is equal to the energy change of the electron.

Â Now the electron is going down.

Â It's losing energy so it's changing energy is actually negative.

Â Photons can't have negative energies.

Â They either don't have energy or they have energy.

Â It's always going to be a positive.

Â A change can be positive or negative, and as an electron transitions down,

Â its change energy is negative, but a photon must have a positive energy.

Â 7:27

The other thing I want you to notice about this diagram is somewhat drawn to scale,

Â and that the energy levels get closer and closer together as you move further and

Â further away from the nucleus.

Â So the transition from the n equals 2 to n equals 1,

Â which I represented here, okay, is a much bigger gap than from the 3 to the 2.

Â Which is bigger than the, 4 to the 3, and so forth.

Â 7:54

And so they'll just keep getting closer, and closer, and closer, and

Â closer, and closer together until we completely remove it,

Â and that's what the n equals infinity means.

Â We have removed it from the atom altogether, and

Â now it's no longer associated with that atom.

Â [SOUND] So this is the equation that gives you the change in energy of the electron.

Â This is the change in energy of electron, that's what delta E stands for.

Â So I'm going to put a subscript, elec for

Â electron so that you can keep that in mind.

Â 8:29

This is the Rydberg Constant.

Â Now, we've seen this equation, a similar equation, and

Â we've seen the Rydberg constant before.

Â 8:37

And you might want to pause and look back in your notes where you saw an equation

Â similar to this and see what it was for.

Â It was to determine the energy of an electron in a hydrogen atom.

Â So now we're going to determine the energy change.

Â So we're taking the equation and subtracting it.

Â So n sub f would be the final energy level and

Â n sub i would be the initial energy level of that electron as it

Â transitions from a high energy state down to the low, final minus initial.

Â Now you must, must,

Â must keep in mind that this equation only applies to hydrogen atom.

Â Now for any atom whatsoever, whatever the transition of the electron,

Â if you can know that, has to be the same as the energy of the photon as it

Â 9:43

Okay, so let's look at this connection a little more closely and

Â get some information down about that.

Â First thing is, like I said,

Â the change in energy of the electron is equal to the energy of the photon.

Â But lets look at those absolute value signs for just a moment.

Â As the electron is going from the n equals 3, in this diagram,

Â down to the n equals 2, its change in energy is going to be negative.

Â It is losing energy as it transitions down.

Â So this delta is a negative sign.

Â 10:40

The other thing that we need to keep in mind, is that we've learned how to

Â take the energy of the photon and know the wavelength and the frequency.

Â So there's two equations that we learned earlier.

Â We've got E, and, equals h times nu.

Â And we've got c equals lambda nu.

Â So, you could use the first equation to get the frequency of

Â the light emitted in this little photon.

Â And you could use sequels lambda nu to get the wavelength of that light.

Â 11:09

So now, you get to try.

Â Take your equation, this is for a hydrogen atom.

Â An electron is transitioning from the n equals 5 to the n equals 3 state.

Â Watch your sines very closely because you see two of these answers have

Â are the same except they're only different by their sine, so

Â be thinking about what that transition is and what a sine should be.

Â And stop and try to calculate it and then choose your answer.

Â 11:36

Well, hopefully you selected your answer.

Â If you selected this answer, number 2, then you forgot to square your ns.

Â Don't forget that it is n squared so your phi has to be 25, and your 3 becomes 9.

Â If you chose this one,

Â then you either put them in the wrong order or you just dropped the minus sign.

Â The electron is going from the n equals 5 down to the n equals 3.

Â So it's losing energy, and

Â when you decrease your amount of energy, it has to be negative.

Â So the right answer is this.

Â Now if you got 4, or

Â you understand where your mistake was, you can just kind of skip ahead.

Â But if you'd like to see me work through it, just keep watching.

Â Since it's a hydrogen atom, we know this is the equation that we would use.

Â [SOUND] And then we can plug in our values,

Â Rydberg constant is 2.18 times 10, there's a 1 there,

Â 2.18 times 10 to the minus 18, and that is joules.

Â Final, where we ended up, was 3, we square that.

Â Initial, where we started, was 5, we square that.

Â Now you'll do the subtraction first and then multiply and

Â then don't forget to change the sine and that will give you that answer.

Â So now we know the change in energy of the electron.

Â 12:59

Okay, that's what we just calculated.

Â Well we know that whatever that change of energy of the electron is,

Â the energy of the photon is the absolute value of that.

Â So as this electron transitions down to the n equals 3,

Â a photon of light will be emitted and the energy of that photon of light is going

Â 13:21

to be equal to 1.55 times 10 to the minus 19 joules.

Â Positive value because photons have positive energy.

Â So now you take that energy of the photon and

Â you calculate the wavelength of that light that was emitted.

Â Pause, select your answer and make sure you look for your units carefully.

Â Select your answer and then resume.

Â [SOUND] Well, did you select 1280?

Â Now this is 1280 nanometers.

Â That's what I asked it for in.

Â Now if you calculated it and you got 1.28 times 10

Â to the minus 6 meters, okay, you didn't convert it on over to nanometers.

Â And that would be the correct answer.

Â Now if you got that and you want to move on, go ahead.

Â Otherwise you can watch what I do to come up with that wavelength.

Â I would use E equals h nu, and I could calculate my frequency.

Â And then c equals lambda nu to calculate my wavelength, or

Â I can combine them and do it in one equation.

Â To combine them I'm going to take the equation on the right and

Â I'm going to solve for nu.

Â So c over lambda is equal to nu, and I'm going to plug that into the nu I see here.

Â E equals hc over lambda.

Â Then I'm going to solve for lambda, lambda equals hc over E.

Â And then I'll plug my values in, lambda is equal to 6.626

Â times 10 to the minus 34 Joules times seconds.

Â 3.00 times 10 to the 8 meters per second.

Â And the E was 1.55 times 10 to the minus 19 Joules.

Â So when you calculate all of that, you're going to get this number here,

Â 1.28 times 10 to the minus 6 meters, which is this many nanometers.

Â Remember 1 nanometer equals 10 to the minus 9 meters.

Â 15:34

So this is the end of our learning objective 8.

Â We are seeing the connection between the wave length and frequency of the light or

Â the energy of the photons emitted.

Â And the energy change of the electron as

Â it transitions from a high energy level down to a low energy level.

Â We were given an equation for a hydrogen atom, but the equation or

Â the information that delta E is, absolute value of delta

Â E of the electron is equal to the E of the photon.

Â And these two equations here would apply no matter what the atom is, but

Â the equation with the Rydberg constant only applies to the hydrogen atom.

Â