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

between wavelength and frequency of the light emitted as an electron

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

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

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

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

amid a photon.

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

level, which is your first principal energy level.

Now, for a hydrogen atom,

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

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

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

we have the 2p orbitals.

And they all have exactly the same energy, so

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

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

meaning we aren't dealing with hydrogen,

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

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

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

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

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

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

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

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

But if we did something to have it absorb energy and

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

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

So atoms can absorb energy.

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

Those are called excited states.

Now once that electron is up in that excited state,

it's not going to want to stay there.

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

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

I left of a parentheses there.

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

it's going to emit photons of light.

And we can look at those photons of light.

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

equals 3 state down to the n equals 2 state.

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

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

the wavelength of that photon of light emitted.

Now the electron won't stop there.

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

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

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So that would give two different photons of light.

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

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

that transition.

So every possible conceivable transition that

electron can undergo would give a different photon of light.

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

an electron in one of these levels.

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

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

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

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

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

that's a bigger change in energy.

If it's a bigger change in energy,

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

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

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

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

So the light emitted would have higher energy.

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

the n equals 1?

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

be somewhere down here further away from the visible spectrum and

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

it still would be electromagnetic radiation with a high energy.

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

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

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

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

the transition of the electron.

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

Look at,

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

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

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

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

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

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

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

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

And they are actually in the visible range.

And they're called the Balmer series there, but

they happen to fall on the visible range.

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

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

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

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

the photons emitted.

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

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

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

And that's with the delta E.

That's a change in energy of that electron.

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Whatever the size of that energy change,

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

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

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

Now the electron is going down.

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

Photons can't have negative energies.

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

It's always going to be a positive.

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

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

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

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

further away from the nucleus.

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

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

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

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

closer, and closer together until we completely remove it,

and that's what the n equals infinity means.

We have removed it from the atom altogether, and

now it's no longer associated with that atom.

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

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

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

electron so that you can keep that in mind.

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This is the Rydberg Constant.

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

we've seen the Rydberg constant before.

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

similar to this and see what it was for.

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

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

So we're taking the equation and subtracting it.

So n sub f would be the final energy level and

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

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

Now you must, must,

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

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

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

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

get some information down about that.

First thing is, like I said,

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

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

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

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

It is losing energy as it transitions down.

So this delta is a negative sign.

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

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

So there's two equations that we learned earlier.

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

And we've got c equals lambda nu.

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

the light emitted in this little photon.

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

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So now, you get to try.

Take your equation, this is for a hydrogen atom.

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

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

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

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

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

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Well, hopefully you selected your answer.

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

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

If you chose this one,

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

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

So it's losing energy, and

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

So the right answer is this.

Now if you got 4, or

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

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

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

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

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

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

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

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

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

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

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

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Okay, that's what we just calculated.

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

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

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

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

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

Positive value because photons have positive energy.

So now you take that energy of the photon and

you calculate the wavelength of that light that was emitted.

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

Select your answer and then resume.

[SOUND] Well, did you select 1280?

Now this is 1280 nanometers.

That's what I asked it for in.

Now if you calculated it and you got 1.28 times 10

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

And that would be the correct answer.

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

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

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

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

I can combine them and do it in one equation.

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

I'm going to solve for nu.

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

E equals hc over lambda.

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

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

times 10 to the minus 34 Joules times seconds.

3.00 times 10 to the 8 meters per second.

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

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

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

Remember 1 nanometer equals 10 to the minus 9 meters.

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So this is the end of our learning objective 8.

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

the energy of the photons emitted.

And the energy change of the electron as

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

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

the information that delta E is, absolute value of delta

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

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

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