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### Course: AP®︎/College Chemistry > Unit 3

Lesson 12: Electronic transitions in spectroscopy# Electronic transitions and energy

Electronic transitions occur in atoms and molecules due to the absorption or emission of electromagnetic radiation (typically UV or visible). The energy change associated with a transition is related to the frequency of the electromagnetic wave by Planck's equation,

*E*=*h*𝜈. In turn, the frequency of the wave is related to its wavelength and the speed of light by the equation*c*= 𝜆𝜈. Created by Sal Khan.## Want to join the conversation?

- One question on the video, when Sal Khan stated that there is no such thing as a 3 and a half shell, aren't there sp3 "hybrid" shells that form half-shells between one energy orbital and the next?(6 votes)
- So when referring to electron shells, or energy levels, we are referring to the principal quantum number, n. Quantum mechanics states that energy, and by extension electrons, are quantized around the nucleus in distinct bands or levels. Meaning that electrons do no exist in between these energy levels. Principal quantum numbers are integer values, so n= 1,2,3, etc., but not fractional values like 1.5 or 3.25.

sp3, sp2, and sp refer to hybrid orbitals which compose an electron shell. Orbitals are the subunits of an electron shell. So there are 2sp3 and 3sp3 orbitals (again the number in front represents the principle quantum number), but not 2.5sp3 orbitals.

Hope that helps.(16 votes)

- Why did he change 486 nm to 486 * 10 ^-19? Doesn't that kind of complicate things?(2 votes)
- The reason Sal has to convert from nm to meters is because of the constants he is using in the bottom right of the screen. The values he's using for Planck's constant and the speed of light have units of J•s and m/s respectively. These units are using the base SI units which includes using meters (and not nanometers) for length, seconds for time, and kilograms for mass. And when we're doing math with scientific values and constants we have to have unit agreement, meaning they have to same units for all these properties or else we get wrong answers.

So even though 486 nm is simpler compared to 486•10^(-19) m, we have to have the wavelength in meters for the math to work out correctly.

If we really wanted to use 486 nm, we would just need new values for the constants which use nanometers instead of meters. However this usually makes things more complicated compared to simply converting the 486 nanometers to meters.

Hope that helps.(6 votes)

- What will a hydrogen atom with an excited electron do to us or our environment?(2 votes)
- An excited hydrogen atom is simply one with excess energy. We would observe it emitting light so the excited electron can relax to a more stable position in the hydrogen atom.

Hope that helps.(3 votes)

- Does amplitude of light affect electron excitement?It doesn't seem so when just observing the equation.

But it just feel strange that two waves with same frequency but different amplitude will havr the same effect in electron excitement.

Thank you.(2 votes)- The amplitude of light corresponds to the amount of photons in a beam of light. Higher amplitude means more photons. The energy of those photons however depends on the frequency of the light. The amplitude and the frequency of light are independent of each other.

If we considering electron excitements, these can only happen when light of sufficient energy is absorbed by the atom. And since frequency and energy are linked, this means only light above a certain frequency threshold is able to cause these excitations. Since the amplitude of light is linked to the number of photons, if light with high amplitude by insufficient frequency is absorbed by an atom then we simply have a large amount of photons with insufficient energy making contact, but not being able to cause excitations.

Hope that helps.(2 votes)

- I have a super interesting question, what if we bombard ourselves with high-energy light that is so small that it doesn't damage our DNA but it is able to excite the electrons!

Then we could walk around quite literally excited, and we should be able to radiate back light in a lower wavelength out right? Which should be able to be detected I hope(2 votes) - A little error: The final answer was in joules when it should have been in joule-seconds.(1 vote)
- A concept check here is helpful. We’re trying to determine the energy associated with an electron transition. Does it make sense for energy to be in joule-seconds?

The final equation Sal uses to determine the energy is: E = (h∙c)/λ

If we examine just the units, Planck’s constant is in J∙s, the speed of light is in m/s, and wavelength is in m. If we substitute these units into the equation, it’ll tell us the units of E.

E = ((J∙s)(m/s))/m = (J∙m)/m = J

So the energy is correctly reported as joules.(1 vote)

## Video transcript

- [Lecturer] In this video, we're going to be talking
about exciting electrons. We can interpret that both ways that electrons can be exciting and that we're going to excite them into higher energy levels. Or we're going to think about what happens when they get unexcited, when they go back into
lower energy levels. And to help us understand this, I'll start with a simple atom. Hydrogen is the simplest I know, and we're gonna think about
the version of a hydrogen that we typically see the isotope that only has one proton in its nucleus. And it typically has one
or will have one electron if it's a neutral hydrogen atom and it would normally
be in its ground state, if it isn't excited yet. So it's going to be in that first shell but it can be excited to other shells. It could be excited to the second shell or the third shell or the fourth shell. And this is obviously hand drawn and not hand drawn that well, and this is really just to help us for visualization purposes. We know that electrons don't orbit nuclei the way planets orbit stars. They have both particle
and wave like properties, and they're more of a
probability density function of where you might find them. And these energy levels are associated with different probability
densities of various energies. But this is what an electron
will typically look like. If we're thinking about just
a neutral hydrogen atom, where the electrons in its ground state. Now let's say we're dealing
with a hydrogen atom where the electron has already
been excited a little bit. So instead of it being in the first shell, it's already in the second
shell right over here. And what we're going to
do is we're gonna hit it with a photon that excites it even more. And the photon. And we know that light has both particle and wave like properties. When we think of it as a particle, we think of it as a photon, but I will depict it like this. So this light has a
wavelength of 486 nanometers. And we know that that photon that hits it with a wavelength of 486 nanometers has sufficient energy to excite
this electron in this case, actually from N equals
two from the second shell to the fourth shell. So it'll go all the way over there. So it will absorb that photon. And then after some time
it can come back down and when it comes back down,
I could do it over here. So after some time it can,
that electron right over here, that excited electron, it can
go back from the fourth shell to the second shell. And when it does it, it will emit a photon
of that same wavelength. So why is that does that it will emit a photon of 486 nanometers. So just like that, we already are starting to understand that photons of the right energy can excite an electron by a
shell or more than one shell. When we talk about quantum mechanics, is this notion that photons
need a certain amount of energy in order to be able to excite the electron to the next energy level or
the energy level after that. Things in between don't work. And the same thing is true when
you're emitting the energy, the electrons is not gonna go
from the fourth energy level to someplace in between
the fourth and the third. It can't do that. It has these quantum states. It's going to be there in
the fourth or the third or the second or the first, there's no such thing as
a three and a half shell. And we can actually answer based on this, we can think about what
is the energy difference between these shells. And the energy difference
between the shells is essentially the energy
of the photon that we emit when we go from the fourth energy shell from the fourth shell to the second shell. To figure out the energy of that photon, we just have to think
about some useful formulas in quantum mechanics. The first and I'm gonna just
look at it right over here is that the energy is
equal to Planck's constant times the frequency. So this thing that looks like a V this is actually the Greek letter, the lowercase Greek letter Nu, and this is what we
typically use for frequency, especially when we're
talking about frequencies of things like light. And we also know how to go
between frequency and wavelength, because we see that the speed of light is equal to whatever the
wavelength of that light is times the frequency of that light. So how would we figure out
the energy of one photon, of 486 nanometer light? Well, we could think about it this way. We can first figure out its frequency using C is equal to Lambda times new. Let me write this down. So we know that see the speed of light is equal to the wavelength of the light times the frequency of that light. And so if we know the wavelength, we can figure out the frequency by dividing both sides by Lambda. So let's do that. So if we divide both sides by Lambda, we get that the frequency of the light is going to be equal to the speed of light divided by the wavelength of the light. Remember, they've given us the
wavelength of the light here, 486 nanometers, or at least
I have given it to you. And then you could take
this and plug it back into Planck's equation up here, that energy is equal to Planck's constant times the frequency to
figure out the energy. So let me write that down. So the energy is going to be
equal to Planck's constant times the frequency, well we know the
frequency right over here. So it's going to be equal
to Planck's constant times, the speed of light divided by
the wavelength of our light, which we know is 486 nanometers, 486 nanometers. So we could say, just
scroll down a little bit, that the energy is going to
be equal to Planck's constant times the speed of light divided by instead of writing the wave
length is 486 nanometers, I can write it as 486 times 10
to the negative ninth meters a nanometer is just one
billionth of a meter, and then we can just
get our calculator out and we know what Planck's constant is. They give it right over here. We know what the speed of
light is right over here. And we know that we have a maximum, over here they're giving
us four significant figures in each of these. And then we have three
significant figures here. So our answer's going to be in terms of three significant figures. I'm going to get Planck's
constant, which is 6.626 times 10 to the negative
34th joule seconds. So let me write that down. So times 10 to the negative 34th, and I'm gonna multiply that
times the speed of light. So times 2.998 times 10 to the eighth meters per second gets me this business. And then I'm gonna divide that by 486 times 10 to the negative ninth gives me, I think we deserve a
little bit of a drum roll, gives me this. And if we were to look at
three significant figures, this would be 4.09 times
10 to the negative 19th. And Planck's constant here
has given in terms of jewels, 4.09 times 10 to the negative 19 joules. So what that tells us
is that the difference in these energy levels is this many joules or the energy of that photon that has a wavelength of 486 nanometers. That energy is 4.09 times 10
to the negative 19 joules.