AP Physics 1 & 2 Atomic Energy Levels

Hi everyone and welcome back to Educator.com.0000

I am Dan Fullerton and today we are going to continue our study of modern physics as we talk about atomic energy levels.0003

Our goals are going to be to explain the significance of Rutherford's gold foil experiment, identify some shortcomings in his model of the atom and describe the assumption of Bohr's model of the hydrogen atom.0009

We will utilize energy level diagrams to determine characteristics of absorbed and emitted photons, we will explain the production of absorption and emission spectra and understand the nature and production of x-rays.0020

Let us start by talking about Rutherford's gold foil experiment.0034

J.J. Thompson had previously shown that atoms contained small negative particles known as electrons.0038

He started to get an idea of charge to mass ratio here, but much beyond that was unknown until in the early 1900s, New Zealand physicist, Ernest Rutherford, shot some alpha particles, which are Helium nuclei, at a very thin sheet of gold foil and then looked at the deflection of those particles.0045

If we have an alpha particle source shot at a gold foil and then look to see what would happen and he saw that most of the particles went through and deflected, but some alpha particles were deflected very large amounts.0065

He was able to make a couple of key conclusions: atoms have a small, massive positive nucleus; electrons orbit the nucleus, and most of the atom, therefore, must be empty space.0077

There were some problems with this model though.0092

The electrons do not emit continuous radiation.0095

What you would expect if charges are moving in a circle; they are accelerating, therefore classical physics predicts that they should release photons.0098

If they release photons, they should be losing energy and if they are losing energy, their orbits should be degrading and eventually they should become unstable.0105

The electron should crash in the nucleus, but that does not happen.0115

Elements were also found to emit and absorb electromagnetic radiation only at specific frequencies, which did not correlate to Rutherford's theory.0119

So, this needed to be refined and along came Danish physicist Niels Bohr.0128

He traveled to the UK to join Rutherford and his research group and he ended up winning the Nobel Prize in 1922 for his work.0132

He said that electrons do not lose energy as they accelerate, the energy of the electron is quantized; it comes in specific amounts, not a continuous spectrum of amounts.0140

And electrons can only exist at specific, discreet energy levels.0149

Now each atom was allowed only a limited number of specific orbits or electrons at each energy level and in order to change energy levels, the electrons must absorb or emit a photon corresponding to that exact difference in energy levels.0154

If they want to move up to a higher energy level, they have to absorb a photon with exactly that amount of energy.0168

If they want to drop down an energy level, they are going to emit a photon with that exact amount of energy.0175

He came up with an equation to tell you what that energy was.0180

The energy required to go to a specific energy level or of a specific energy level is equal to the square of the atomic number times -13.6 electron-volts over the square of the energy level.0185

Now this only worked for small atoms; it does not work for everything, but it was pretty good for small atoms as a starting point.0201

Let us take a look at an energy level diagram.0210

N = 1 down here is the lowest state and you can think of this almost as the energy well we were talking about previously, where if there is an electron at the ground state it is trapped by hydrogen in here and to get out it must give up, it must find, or absorb a photon of at least 13.6 electron-volts to be free.0213

If it absorbed a photon of 14.6 electron-volts, it would have enough to get free and it would have 1 eV left over; that would be its kinetic energy.0232

Now the energy levels here are negative and that indicates that the electron is bound to the nucleus of the atom; it cannot leave and it is not free until it pays at least -13.6 eV or it pays 3.4 eV if it is at the N = 2 state, or whatever state it is in.0242

When the electron reaches 0 state, ionization, it is no longer bound and it can be emitted as a photoelectron.0260

Now while we are here looking at the hydrogen energy level diagram, one of the famous series is called the Balmer series.0267

That is a set of visible spectral lines created by electrons falling in hydrogen to the N = 2 level.0277

Now there is a little bit more to the story than we have talked about so far, there is a lot more to the story.0286

But an electron only has a probability of being located in certain regions surrounding the nucleus and electron orbitals are really these probability regions.0292

So if we want to get much more specific, we have to start looking at probability and statistics to talk about where electrons are, are not, and what their likelihood is of being in a specific position or not being in a specific region or specific position.0300

Let us talk about atomic spectra.0317

If atoms can only emit certain frequency of photons corresponding to the difference in energy as the electron falls from a high energy level to a low energy level, you are going to get only certain frequencies emitted.0319

That is going to give you specific discreet spectra.0330

They can only absorb photons with those energy levels as well, so you get very, very unique spectra for each of these different types of elements.0334

How it works -- If an object is hid to the point where it glows or in incandescence, it emits a continuous energy spectrum.0344

On the other hand, things like gas discharge lamps, they emit light by exciting electrons to higher energy states when the electrons fall back down to their lower energy states, they emit a photon and that photon is emitted at that specific frequency, corresponding to the difference in energy between the energy levels.0350

Therefore, you are going to get a unique discreet spectra; you are only going to get photons emitted at specific frequencies at specific colors.0367

If all colors of light are incident upon a cold gas, the gas can only absorb the frequencies corresponding to increases in energy level, so all colors of light going into a cold gas...0376

...it is only going to absorb the ones that allow electrons to jump to high energy levels and what you are going to get out are all the colors except the ones that are absorbed.0388

That is called an absorption spectrum.0397

Let us take a look at these.0400

If we have something like incandescence white light, we are going to put it through a prism, then we can see that we have all frequencies and all colors of lights.0401

On the other hand, if we shine white light through a cold gas and then separate it out, we are going to see that the only colors missing are going to be the ones that correspond to jumps in energy levels for the electrons because that is what is absorbed in the cold gas.0410

Or if we create an emissions spectrum, we take a hot gas, we excite the gas, the electrons in the gas, the higher energy levels, and when they fall down they emit photons at those specific frequencies.0427

So when we separate that spectrum, what we are going to see is lines of light at specific frequencies, or specific colors.0440

These are going to correspond to exactly the holes in the absorption spectrum assuming it is the same element in your gas.0447

Now x-ray production is basically the photoelectric effect in reverse.0457

X-rays can pass through light materials and we know that; we have done that at the doctor's office when we have broken a bone and they go show you how it is passed through your skin and a part gets absorbed by the bone and they can see what is inside of you.0462

Now electrons, the way they are formed, you accelerate electrons through a large potential difference and you have them collide with a plate made of Molybdenum or Platinum.0472

When you do that you knock electrons out of the lowest and equals one energy state, then electrons in the atoms of the metal must fall down to feel that in equal's one state.0482

As they fall down, they emit a photon and that photon has enough energy that it is an x-ray.0492

It has a large amount of energy, so that is the typical way we are going to form x-ray photons.0497

Some examples here -- An electron in a hydrogen atom drops from the N = 3 to the N = 2 state.0505

Determine the frequency of the emitted radiation.0514

Well the energy of the photon that is emitted is going to be its initial energy minus its final energy.0517

Its initial was -1.51 eV and its final is going to be -3.4 eV, so the energy of our photon is 1.89 eV.0526

Now, we need to convert that to joules, so 1.89 eV times -- well we want electron-volts to go away and we want joules and 1 eV = 1.6 × 10^-19 J.0545

Electron-volts make a ratio of 1 and I come up with an energy in joules of about 3.02 × 10^-19.0560

Now that we have that, let us determine the frequency.0569

If energy equals Hf, then frequency is energy over (h) or 3.02 × 10^-19 J over Planck's constant, 6.63 × 10^-34, which implies that the frequency is going to be about 4.56 × 10¹⁴ Hz.0573

Let us take a look at another one.0603

Determine the energy in electron-volts and joules of a photon emitted by an electron as it moves from the N = 6 to the N = 2 state.0605

Is this the only energy that an electron in the N = 6 energy level could emit?0616

Well, absolutely we know the answer to this part is no. It could go from 6 to 5, 6 to 4, 6 to 3.0621

It could go in many other ways to emit other energies of photons.0626

Now let us figure out what the energy is of the photon as it goes from 6 to 2.0630

Well, our energy of our photon is going to be the initial minus the final energy.0635

It starts at -.38 eV minus the final, which is -3.4 eV, therefore, the energy of the photon and electron-volts is just going to be about 3.02 eV.0642

To convert that to joules, we will start off with 3.02 eV and we want electron-volts to go away, so we will put that in the denominator.0660

We want joules and 1 eV = 1.6 × 10^-19, therefore the energy in joules is going to be about 4.83 × 10^-19 J.0669

Another one -- According to the Bohr model of the hydrogen atom, an electron dropping from the 5th energy level and eventually landing in the 2nd energy level could produce photons of how many different frequencies?0691

Let us draw a poor man's energy diagram -- 5th level, 4th level, 3rd level, 2nd level and let us see how many paths we have from 5 down to 2.0702

We could go from 5 to 4; there is 1, 2, 3; we could go from 5 to 3 and there is 4.0715

We could go from 5 all the way to 2; there is 5 and we could go all the way from 4 to 2, so 6 different photon frequencies could be produced by this transition depending on the path the electron took on its way down to that final state.0724

The bright line emission spectrum of an element can best be explained by electrons transitioning between discreet energy levels in the atoms of that element -- well that sounds promising.0745

Protons acting as both particles and waves -- not sure what that has to do with our problem.0756

Photons scattering electrons demonstrating the Compton Effect -- it had nothing to do with our bright line emission spectrum.0763

Protons being dispersed uniformly throughout the atoms of that element -- not really sure even what that means.0769

So our best answer must be A, the bright line emission spectrum is caused by electrons transitioning between discreet energy levels in the atoms of that element.0774

Let us try one last sample problem.0785

The diagram below represents the bright line spectra of 4 elements, A, B, C, and D and the spectrum of an unknown gas sample.0789

Based on the spectra, which two elements are found in the unknown sample?0797

Well, what I am going to do here is if this is my unknown sample, I need to find where we have lines that match up from these different elements.0801

On the unknown sample, I have a line right there and that looks like it could correspond to element (B) and that also would correspond to (B), so I am pretty sure that we have some (B) in our sample.0809

Let us see what else we have.0824

We have two lines here in (C) and we have those two lines and that also corresponds here and here and that covers all of the lines, so I would say that we must have (B) and (C) in our unknown sample.0827

Our unknown sample would be made up of elements of (B) and (C) and this is one way you can do gas analysis to figure what an unknown sample is made out of.0843

Hopefully that gets you a good start on atomic energy levels.0853

Thanks so much for watching us here at Educator.com. Make it a great day everyone!0856