Main content

## MCAT

### Course: MCAT > Unit 8

Lesson 23: Electronic structure- Electronic structure questions
- Photoelectric effect
- Bohr model radii (derivation using physics)
- Bohr model radii
- Bohr model energy levels (derivation using physics)
- Bohr model energy levels
- Absorption and emission
- Emission spectrum of hydrogen
- Heisenberg uncertainty principle
- Quantum numbers
- Quantum numbers for the first four shells
- Electron configurations for the first period
- Electron configurations for the second period
- Electron configurations for the third and fourth periods
- Electron configurations of the 3d transition metals
- Paramagnetism and diamagnetism
- Electron configurations article

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# Absorption and emission

Using shell model diagram to relate absorption to emission. Derives relationship between emitted photon and energy levels, the Balmer-Rydberg equation. Created by Jay.

## Want to join the conversation?

- Can we replace 'mu' with f since it is actually the frequency?(20 votes)
- Yes, it is the same thing. The formal letter for frequency is 'nu' but f is used just as often in scientific notation(19 votes)

- A photon is emitted during emission. What exactly is a photon?(7 votes)
- Short answer: a photon is a particle of light.

Longer answer: light is energy. Sometimes we think of light as being a wave in the form of an electro-magnetic wave but other times it can be described as a particle. A photon in this case, is 1 unit of light with a variable amount of energy which depends on its frequency.(24 votes)

- At about1:15, the speaker says that the electron at a higher energy level will eventually fall back down to its ground state. What causes the electron to fall back down and release energy in the form of a photon? Or is it some sort of random phenomenon?(6 votes)
- It's not random. Everything tries to attain stability and as long as the lower orbits are not filled up, the electron has a tendency to return to the ground state and in the process, it has to release the excess energy it possesses(12 votes)

- Wait I thought the Probability model stated that you have different chances of finding the electron at different places? How do we attempt to find the radius or the wavelength unless we assume the Probability model is false?(5 votes)
- You are correct that the position of the electron is not precise, and is given as a probability distribution. What is precise however are the quantized energy levels and angular momentum of the electron. This comes out of the solution of Schrodinger's equation.(15 votes)

- What exactly do we mean by 'energy levels' ?

A hydrogen atom is in possession of a single 1s2 orbital, wherein there exists a single electron. If i were to provide this electron with a certain amount of energy, as i understand it, it would move to a higher energy level, but doesn't a hydrogen atom contain only a single orbital? where does this electron actually go? does it still remain within the orbital?(7 votes)- Every atom has every possible orbital, the electron configuration is simply which orbitals have electrons in the lowest energy state. If the atom gains some energy the electron can be promoted to higher energy orbitals.(5 votes)

- When an electron goes through the process of emission, can we see the light with our own eyes?(6 votes)
- Even dispersion of white light is an example. White light contains all seven colours of the spectrum. So it provides electrons at different energy levels with the photon that they require to jump to the next level.(2 votes)

- Do the concepts of absorption and emission only apply to hydrogen?(4 votes)
- Luckily these concepts apply to all gaseous elements and molecules. Helium, for example, will also have absorption and emission spectra(5 votes)

- at9:08why are you able to just "pull out a negative one" Where did negative one come from?(6 votes)
- The emission of an electron results in the emission of photon, producing light. (1:30) But what happens to the atom when it absorbs the extra enegry in absorption? At what point will emission occur?(3 votes)
- It is not the emission of an electron that produces a photon. A photon is produced when an electron releases energy by moving to a lower energy level within the atom.

When the atom absorbs energy, it re-emits it in a random direction, almost instantaneously. The exact time delay is random, but always so fast that it seems pretty much instantaneous.(5 votes)

- Every step is very clear. From higher level to lower level, I do understand the election releases energy. But why is the energy in the form of light? And since lights are particles, how can I imagine where the particles are from?(5 votes)
- Electrons can only emit energy in the form of photons.(0 votes)

## Video transcript

- We've been talking about the Bohr model for the hydrogen atom, and we know the hydrogen atom has one positive charge in the nucleus, so here's our positively charged nucleus of the hydrogen atom and a
negatively charged electron. If you're going by the Bohr model, the negatively charged
electron is orbiting the nucleus at a certain distance. So, here I put the
negatively charged electron a distance of r1, and
so this electron is in the lowest energy level, the ground state. This is the first energy level, e1. We saw in the previous video
that if you apply the right amount of energy, you can
promote that electron. The electron can jump up
to a higher energy level. If we add the right amount of energy, this electron can jump up
to a higher energy level. So now this electron is a distance of r3, so we're talking about the
third energy level here. This is the process of absorption. The electron absorbs energy and jumps up to a higher energy level. This is only temporary though, the electron is not going
to stay there forever. It's eventually going to fall
back down to the ground state. Let's go ahead and put that
on the diagram on the right. Here's our electron, it's
at the third energy level. It's eventually going to fall
back down to the ground state, the first energy level. Here's the electron going back to the first energy level here. When it does that, it's
going to emit a photon. It's going to emit light. When the electron drops
from a higher energy level to a lower energy level, it emits light. This is the process of emission. I could represent that photon here. This is how you usually
see it in textbooks. We emit a photon, which is going to have a certain wavelength. Lambda is the symbol for wavelength. We need to figure out how to relate lambda to those different energy levels. The energy of the photon is, the energy of the emitted photon is equal to the difference in energy between those two energy levels. We have energy with the third energy level and the first energy level. The difference between those... So, the energy of the third energy level minus the energy of
the first energy level. That's equal to the energy of the photon. This is equal to the
energy of that photon here. We know the energy of a
photon is equal to h nu. Let me go ahead and write that over here. Energy of a photon is equal to h nu. H is Planck's constant,
this is Planck's constant. Nu is the frequency. We want to think about wavelength. We need to relate the
frequency to the wavelength. The equation that does that is of course, C is equal to lambda nu. So, C is the speed of light,
lambda is the wavelength, and nu is the frequency. So, if we solve for the frequency, the
frequency would be equal to the speed of light divided by lambda. Then, we're going to take all of that and plug this in to here. We get the energy of a photon is equal to Planck's constant, h,
I'll write that in here, times the frequency, and
the frequency is equal to c over lambda. Now we have the energy of the photon is equal to hc over lambda. Instead of using E3 and
E1, let's think about a generic high energy level. Let's call this Ej now, so this is just a higher energy level, Ej. The electron falls back down
to a lower energy level, which we'll call Ei. Instead of using E3 and E1,
let's make it more generic, let's do Ej and Ei. Let's go ahead and plug that in now. The energy of the photon would be equal to the higher energy level, Ej minus the lower energy which is Ei. So now we have this equation, let me go and highlight it here. We have hc over lambda
is equal to Ej minus Ei. Let's get some more room, and let's see if we can solve that a
little bit better here. So let me write this down here. We have hc over lambda
is equal to the energy of the higher energy
level, minus the energy of the lower energy level, like that. Alright, so in an earlier
video, I showed you how you can calculate the
energy at any energy level. We derived this equation. The energy at any energy
level, n, is equal to E1 divided by n squared. So, if we wanted to know the energy when n is equal to j, that
would be just E1 over j squared. We could take that and we
could plug it in to here. Alright, if I wanted to know the energy for the lower energy level, that was Ei, and that's equal to E1
divided by i squared. I could take all of this, I could take this and I
could plug it into here. Let's, once again, get some more room. Let's write what we have so far. We have hc over lambda is equal to Ej was E1 over j squared and
Ei was E1 over i squared. Okay, we could pull
out an E1 on the right. So we have hc over lambda is equal to E1, and so that would give
us one over j squared minus one over i squared, like that. We could divide both sides
by hc, so let's do that. So on the left, we could
have one over the wavelength is equal to E1 divided by hc, one over j squared minus
one over i squared. Again, from an earlier
video, we calculated what that E1 is equal to. So, one over lambda is equal
to E1 was negative 2.17 times 10 to the negative 18 joules. So once again, you can
see that calculation in an earlier video. It took us a while to get there. We are going to divide by hc,
and this is one over j squared minus one over i squared. Well let's look a little
bit more closely at what we have right here. So, if I, for right now, not
worry about the negative sign, and just think about what we have. This is all equal to a constant. H is Planck's constant, and
c is the speed of light, so we have all these constants here. We could rewrite all of these as just R. I'm going to rewrite this as R, so this would be one
over lambda is equal to negative R times one over j squared minus one over i squared. So, R is called the Rydberg constant so let's see if we can solve for that. Over here R would be equal to 2.17, times 10 to the negative 18, over h is Planck's constant, that's 6.626 times 10 to the negative 34, and then c is the speed of light. So we could use 2.9979
times 10 to the eighth meters per second as the speed of light. If you do all that math, I won't do it here just to save time, but if you do all that, you'll get 1.09 times 10 to the seventh and
this is one over meters. I think I might have, it's possible I had a rounding issue in here, because if you use 2.18,
you get a better number. It's 1.097 times 10 to the seventh, which is the Rydberg constant. This is just, again I'm just
trying to show you the idea behind the Rydberg constant
here so we get this value for the Rydberg constant. So you could plug that in
for R if you needed to, and we're going to be doing
that in the next video. You could stop right there, and you have related the wavelength, right to your different energy levels. You could go a little bit further, let's just go a little bit further here. So, one over lambda is
equal to negative R. If we pull out a negative one, that would give us one over i squared minus one over j squared. Now we have these two
negative signs, right? These two negative signs out here which gives up a positive. This is now equal to one over lambda, one over the wavelength
is equal to positive R, the Rydberg constant,
times one over i squared minus one over j squared. Remember what i and j represented. I represented the lower energy level, and j represented the higher energy level. This is an extremely useful equation, so usually you see this called
the Balmer-Rydberg equation. We've derived this equation
using the assumptions of the Bohr model, and this equation is extremely useful because it explains the entire emission spectrum of hydrogen. This is again, this is
why we were exploring the Bohr model in the first place. We got this equation,
and in the next video we're going to see how this explains the emissions spectrum of hydrogen. Think about lambda or the wavelength, right as the light is
emitted when the electron falls back down to a lower energy state.