- [Voiceover] I'm sure that most of you
know the famous story of Isaac Newton
where he took a narrow beam of light
and he put that narrow beam of light through a prism
and the prism separated the white light
into all the different colors of the rainbow.
And so if you did this experiment,
you might see something like this rectangle up here
so all of these different colors of the rainbow
and I'm gonna call this a continuous spectrum.
It's continuous because you see all
these colors right next to each other.
So they kind of blend together.
So that's a continuous spectrum
If you did this similar thing with hydrogen,
you don't see a continuous spectrum.
So, if you passed a current through
a tube containing hydrogen gas,
the electrons in the hydrogen atoms
are going to absorb energy
and jump up to a higher energy level.
When those electrons fall down to a lower energy level
they emit light and so we talked
about this in the last video.
This is the concept of emission.
If you use something like a prism or a defraction
grading to separate out the light,
for hydrogen, you don't get a continuous spectrum.
You'd see these four lines of color.
So, since you see lines, we call this a line spectrum.
So this is the line spectrum for hydrogen.
So you see one red line and it turns out that
that red line has a wave length.
That red light has a wave length of 656 nanometers.
You'll also see a blue green line
and so this has a wave length of 486 nanometers.
A blue line, 434 nanometers,
and a violet line at 410 nanometers.
And so this emission spectrum is unique to hydrogen
and so this is one way to identify elements.
And so this is a pretty important thing.
And since line spectrum are unique,
this is pretty important to explain
where those wavelengths come from.
And we can do that by using the equation
we derived in the previous video.
So I call this equation the Balmer Rydberg equation.
And you can see that one over lamda,
lamda is the wavelength of light that's emitted,
is equal to R, which is the Rydberg constant,
times one over I squared, where I is talking
about the lower energy level,
minus one over J squared, where J is
referring to the higher energy level.
For example, let's say we were considering
an excited electron that's falling
from a higher energy level n is equal to three.
So let me write this here.
So we have an electron that's falling
from n is equal to three down to
a lower energy level, n is equal to two.
All right, so it's going to emit light
when it undergoes that transition.
So let's look at a visual representation of this.
Now let's see if we can calculate
the wavelength of light that's emitted.
All right, so if an electron is falling
from n is equal to three to n is equal to two,
I'm gonna go ahead and draw an electron here.
So an electron is falling from n is equal
to three energy level down to n is equal to two,
and the difference in those two energy levels
are that difference in energy is equal
to the energy of the photon.
And so that's how we calculated the
Balmer Rydberg equation in the previous video.
All right, let's go ahead and calculate
the wavelength of light that's emitted
when the electron falls from the third
energy level to the second.
So, we have one over lamda is equal
to the Rydberg constant, as we saw in the
previous video, is one point zero nine seven
times ten to the seventh.
The units would be one over meter, all right?
One over I squared.
So, I refers to the lower energy level, all right?
So the lower energy level is when n is equal to two.
So we plug in one over two squared.
And then, from that, we're going to
subtract one over the higher energy level.
That's n is equal to three, right?
So this would be one over three squared.
So one over two squared minus one over three squared.
Let's go ahead and get out
the calculator and let's do that math.
So one over two squared, that's one fourth,
so that's point two five, minus one over
three squared, so that's one over nine.
So, one fourth minus one ninth gives us
point one three eight repeating.
And if we multiply that number by the
Rydberg constant, right, that's one point
zero nine seven times ten to the seventh,
we get one five two three six one one.
So let me go ahead and write that down.
So now we have one over lamda is equal to
one five two three six one one.
So to solve for lamda, all we need to
do is take one over that number.
So one over that number gives us
six point five six times ten to the negative seven
and that would now be in meters.
So we have lamda is equal to six point five
six times ten to the negative seventh meters.
So let's convert that into, let's go like this,
let's go 656, that's the same thing
as 656 times ten to the negative ninth meters.
And so that's 656 nanometers.
656 nanometers, and that should sound familiar to you.
All right, so let's go back up here
and see where we've seen 656 nanometers before.
656 nanometers is the wavelength
of this red line right here.
So, that red line represents the light
that's emitted when an electron falls
from the third energy level down to the second energy level.
So let's go back down to here
and let's go ahead and show that.
So we can say that a photon, right,
a photon of red light is given off
as the electron falls from the third
energy level to the second energy level.
So that explains the red line in the
line spectrum of hydrogen.
So how can we explain these other lines that we see, right?
So we have these other lines over here, right?
We have this blue green one,
this blue one, and this violet one.
So if you do the math, you can use the
Balmer Rydberg equation or you can do this
and you can plug in some more numbers
and you can calculate those values.
So those are electrons falling from
higher energy levels down to the second energy level.
So let's go ahead and draw them on our diagram, here.
So, let's say an electron fell from the
fourth energy level down to the second.
All right, so that energy difference,
if you do the calculation, that turns
out to be the blue green line in your line spectrum.
So, I'll represent the light emitted like that.
And if an electron fell from the fifth energy
level down to the second energy level,
that corresponds to the blue line
that you see on the line spectrum.
And then, finally, the violet line must be
the transition from the sixth energy level
down to the second, so let's go ahead and draw that in.
And so now we have a way of explaining
this line spectrum of hydrogen that we can observe.
And since we calculated this Balmer Rydberg
equation using the Bohr equation,
using the Bohr model, I should say,
the Bohr model is what allowed us to do this.
So the Bohr model explains these different
energy levels that we see.
So when you look at the line spectrum of hydrogen,
it's kind of like you're seeing energy levels.
At least that's how I like to think about it
'cause you're, it's the only real way
you can see the difference of energy.
All right, so energy is quantized.
We call this the Balmer series.
So this is called the Balmer series for hydrogen.
But there are different transitions that you could do.
For example, let's think about an electron
going from the second energy level to the first.
All right, so let's get some more room here
If I drew a line here, again, not drawn to scale.
Think about an electron going from the
second energy level down to the first.
So from n is equal to two to n is equal to one.
Let's use our equation and let's
calculate that wavelength next.
So this would be one over lamda
is equal to the Rydberg constant, one
point zero nine seven times ten to the seventh,
that's one over meters, and then we're
going from the second energy level to the first,
so this would be one over the lower energy level squared
so n is equal to one squared minus
one over two squared.
All right, so let's get some more room,
get out the calculator here.
So, one over one squared is just one,
minus one fourth, so that's point seven five
and so if we take point seven five of the Rydberg constant,
let's go ahead and do that.
So one point zero nine seven times ten
to the seventh is our Rydberg constant.
Then multiply that by point seven five, right?
So three fourths, then we should get that number there.
So that's eight two two seven five zero zero.
So let's write that down.
One over the wavelength is equal to
eight two two seven five zero.
So to solve for that wavelength,
just take one divided by that number
and that gives you one point two one
times ten to the negative seven and that'd be in meters.
So the wavelength here is equal to one point,
let me see what that was again.
One point two one five.
One point two one five times ten
to the negative seventh meters.
And so if you move this over two, right,
that's 122 nanometers.
So this is 122 nanometers, but this
is not a wavelength that we can see.
So 122 nanometers, right, that falls into
the UV region, the ultraviolet region,
so we can't see that.
We can see the ones in the visible spectrum only.
And so this will represent a line in a different series
and you can use the Balmer Rydberg equation
to calculate all the other possible
transitions for hydrogen and
that's beyond the scope of this video.
So, here, I just wanted to show you that
the emission spectrum of hydrogen
can be explained using the Balmer Rydberg equation
which we derived using the Bohr model of the hydrogen atom.
So even thought the Bohr model of the hydrogen
atom is not reality, it does allow us to figure
some things out and to realize that energy is quantized.