AP Physics 1 & 2 Wave-Particle Duality

Hi everyone. I am Dan Fullerton and I would like to welcome you back to Educator.com.0000

Today we are going to start talking about modern physics topics, specifically wave particle duality.0004

Our objectives are going to be to explain what is meant by the dual nature of light, to describe evidence indicating both the wave nature and the particle nature of electromagnetic waves, determine the kinetic energy of an emitted photoelectron giving the metals work function and the frequency of incoming electromagnetic radiation...0011

...to calculate the de Broglie wavelength of a moving particle, and explain the significance of the Compton Effect.0028

So talking about the duality of light -- electromagnetic radiation exhibits characteristics and properties of both waves and particles.0037

Particles of electromagnetic radiation are known as photons, a good vocabulary word.0046

Electromagnetic radiation, therefore, has a dual nature, the nature of wave and particles.0053

There is evidence for each of these.0058

Some evidence that it behaves like a wave -- It diffracts, it interferes, the Doppler Effect, and Young's Double Slit experiment, showing interference again.0060

Our particle evidence is blackbody radiation, photoelectric effect, and the Compton Effect.0070

That is what we are going to focus on here today.0076

Blackbody radiation and what is called the UV catastrophe -- radiation emitted from a very hot object known as blackbody radiation did not align with physicist's expectations or their understanding of light as a wave.0080

Very hot objects emitted radiation in a specific spectrum of frequencies and intensities.0094

Hotter objects had higher intensities at lower wavelengths.0099

Cooler objects emitted more intensity at higher wavelengths.0103

Now physicists expected that at very short wavelengths, the energy radiated would become very large and that did not occur and that was known as the UV catastrophe; they did not understand it.0107

What they expected based on classical theory was a graph of wavelength versus intensity that kind of followed that shape.0118

That was their classical expectation, instead, what they found was that it tailed off here.0125

Max Planck proposed that atoms could only absorb or emit radiation in specific non-continuous amounts known as quanta, which solved this puzzle and earned him the Nobel Prize in Physics in 1918.0135

So he was able to explain this, and without going into a lot of depth, that was one piece of evidence that light, electromagnetic radiation, has a particle nature.0146

Now the photoelectric effect is another big one.0158

Electromagnetic radiation striking a piece of metal may emit electrons, which are known as photoelectrons.0161

Not all electromagnetic radiation created photoelectrons though.0169

Each metal has a certain threshold frequency of incoming light below which no photoelectrons were emitted.0172

You had to have a high enough frequency for a specific metal for photoelectrons to come out.0179

If your frequency was lower than that, it did not work and that frequency depended on what metal you were shooting the light at.0184

Increasing the intensity or the brightness of the light did not increase the kinetic energy of any of the emitted electrons.0192

Albert Einstein did some work to explain this phenomenon in 1905.0199

He said "If energy exists in specific, discreet amounts, then electromagnetic radiations exist in specific, discreet amounts," known as photons.0204

A photon has 0 mass and 0 charge and its energy is quanta, as it comes only in specific amounts.0212

The energy of a photon is directly related to its frequency where the energy of a photon is equal to (H), Planck's constant times the frequency or since we know frequency is going to be C/λ, that is HC/λ for an electromagnetic wave.0220

(H), Planck's constant is 6.63 × 10^-34 Joule/seconds (J/s) and if you were to graph this, the energy of a photon versus its frequency, the slope here will give you exactly Planck's constant.0237

Now, how did all of this work?0257

Imagine the electrons and metals are held in energy wells, they are stuck there.0259

The electrons had to absorb at least enough energy to pull the electron out of the well, so that it could be emitted as a photoelectron.0264

They could only be released when they absorbed a single photon that had at least as much energy as was required to get out of the well.0271

The energy required to get out of the well, is known as the work function or phi of the metal and different metals have different work functions.0279

If you could not absorb a photon that would not give you enough energy to jump out of the well.0289

The frequency corresponding to the photon that had enough energy to get the electron out of the well was known as the cut-off frequency and if you absorbed a photon that had more energy than was required to get it out of the well, well, any of that excess energy became kinetic energy of your emitted photoelectron.0294

The electron in the metal absorbs a photon with energy -- E = hf -- greater than the metals work function phi.0314

So then the electron is emitted as a photoelectron.0324

Any absorbed energy beyond that of the work function, beyond what was required to free it from the well, became the kinetic energy of the photon.0327

So the kinetic energy of the emitted electron is equal to the energy of the absorbed photon minus the work function -- anything you had left over.0335

Kind of like -- imagine you are in debtor's prison.0346

If you are going to absorb a check from somebody to get out of prison, you can only take that check if it is enough to get you completely out of debt and if you have any excess left over after you pay off your debt, that is your kinetic energy, that is your extra energy to move around, your extra freedom.0349

Now a typical graph that shows this, shows frequency of the incoming photon on one axis and the kinetic energy of the photoelectrons on the other axis.0365

The work function of the metal is down here and until you have a photon with enough frequency, enough energy to free the electron from the well, nothing happens, but once you get to that cut-off frequency, any excess energy caused by having a higher frequency becomes the kinetic energy of the photoelectron.0374

Let us take a look at how the photoelectric effect might work in a circuit.0397

Let us assume that we are going to take an inner vacuum -- there is our vacuum chamber and we are going to put some electrodes with a metal on them.0401

We will put one electrode here and we will put a metal on it over here -- make that a little thicker to show that.0412

We will hook this up running it outside into an ammeter and then bring that over to a source of potential difference and hook up the other side.0418

We have this (V), positive and negative -- and what we are going to do is we are going to shine a light on our metal.0429

There is our flashlight -- shining photons on our metal.0436

Now if those photons have enough energy to overcome the work function to free the electrons from the well, if they are absorbed, then we are going to have the electron get emitted.0443

Any excess energy of the electron, anything that was not spent in freeing the electron from the well, becomes the electron's kinetic energy and we have a positive voltage over here, to help collect the electron, so that is sometimes called the 'collector' here.0454

Now, if you reverse this and you make the voltage here too large, then the electron cannot jump across anymore and that is going to happen when you have the maximum kinetic energy of the emitted photoelectron is going to equal the energy that you have here, (Q) times that voltage.0469

Therefore they call the voltage where this happens -- kinetic energy max divided by the charge -- they are going to call that the stopping potential.0485

At that point, you cannot collect any of those electrons; the voltage is off-setting whatever kinetic energy you happen to have there.0496

But that is what the photoelectric effect typically looks like in a circuit.0504

Now in 1922, US physicist, Arthur Compton, shot an x-ray photon at a graphite target to observe the collision between the photon and the electrons of the graphite atom.0509

He observed some very interesting things.0519

The x-ray was scattered and it was emitted with a longer wavelength.0523

If it had a longer wavelength it must have lost some energy and it lost some momentum, but the energy and momentum lost by that x-ray, a piece of electromagnetic radiation was exactly equal to the energy and momentum gained by that photoelectron.0528

We are talking about conservation of momentum, while we are dealing with photons.0542

He proposed this wavelength, the Compton wavelength, equal to the mass of the electron times the speed of light, which is 2.4 × 10^-12 m.0549

The maximum wavelength shift that they saw on the x-ray was twice this constant, it could be anywhere between 0 and twice this value, so the Compton wavelength is half the maximum wavelength shift of the x-ray following this event.0557

What does all of this means?0575

Well, the major conclusions were that photons have momentum and photons obey conservation laws of energy in momentum.0577

Now Einstein, previously, had predicted that E = mc² or if we rearrange this, E/C = mc.0584

But we also know that E = HC/λ or (hf), therefore we could rewrite this as HC/λC = mc, but look at this (mc) over here, that is a mass times a velocity; that is a momentum -- P = mc.0595

So if I divide out -- the (C)'s make a ratio of 1 -- what you found was that H/λ = momentum or wavelength is equal to H/P (the momentum).0619

Compton's work verified Einstein's predictions that a photon would have momentum.0637

Now let us talk about the de Broglie wavelength.0645

Electromagnetic waves can behave as moving particles -- we have just shown that.0648

So should it not make sense that moving particles can exhibit wave properties and they do.0652

de Broglie shot electrons through double slits and observed the diffraction patterns, similar to the Davisson-Germer experiment.0657

The smaller the particle, the more apparent the wave properties and the wavelength of a moving particle then is known as its de Broglie wavelength.0663

As we just said, the momentum is H/λ then λ wavelength is H/P or H/MV, so moving particles have a wavelength, known as their de Broglie wavelength.0670

The Davisson-Germer experiment, that we just said, further verified the wave nature of moving particles.0689

Shoot electrons at a double slit and you can see the interference pattern here.0693

We saw that previously with monochromatic light.0697

Now we can see it by shooting small things like electrons; you can actually pick up that interference pattern.0700

A couple of examples -- A photon of light traveling through space with a wavelength of 600 nanometers (nm) -- λ = 600 nm or 600 × 10^-9 m has what energy?0707

Well, energy is HC/λ which is going to be 6.63 × 10^-34 (Planck's constant) × 3 × 10⁸ m/s (C) all over the wavelength, 600 × 10^-9 m or 3.31 × 10^-19 J.0722

As we are talking about these very small amounts of energy it is probably useful to come back to electron volts, so let us go convert that into electron volts while we are here.0746

We have 3.31 × 10^-19 J and we want to convert that to electron volts, so I will put joules in the denominator, so joules will cancel out.0754

I want electron volts and I know that 1 eV = 1.6 × 10^-19 J.0766

Joules will make a ratio of 1 and as I multiply that through, I come up with about 2.07 eV, so those are equivalent answers.0775

Example 2 -- Find the frequency of a photon whose energy is 3 × 10^-19 J.0787

Well, E = HF, therefore Frequency is E/H or 3 × 10^-19 J/H (6.63 × 10^-34 J-s), Planck's constant...0794

...which is going to be 4.53 × 10¹⁴ Hz.0814

Another example -- The alpha line in the Balmer series of the hydrogen spectrum consists of light having a wavelength of 656 nm.0828

Find the frequency of this light and determine its energy in joules and electron volts.0836

Let us find its frequency first.0842

If V = F(λ), then frequency = V/λ...0844

...which is going to be its speed, 3 × 10⁸ m/s over its wavelength 656 × 10^-9 m or 4.57 × 10¹⁴ Hz.0851

Now let us determine its energy in joules.0871

Energy is Planck's constant times frequency, so that will be 6.63 × 10^-34 × 4.57 × 10¹⁴ Hz (frequency), which will give us 3.03 × 10^-19 J.0875

Let us find that energy in electron volts.0902

To convert 3.03 × 10^-19 J, we are going to multiply by JeV...0905

...and we know 1 eV is 1.6 × 10^-19 J, and joules make a ratio of 1 and we will come up with about 1.89 eV.0916

Let us take a look at a slightly more involved example.0933

Given the work function of gold is 5.1 eV, answer the following questions.0936

What minimum frequency of electromagnetic radiation is required to produce photoelectrons from gold?0942

Let us start there.0947

If we have 5.1 eV -- let us convert that work function into joules first.0949

5.1 eV × -- well 1 eV = 1.6 × 10^-19 J, so that will give us 8.16 × 10^-19 J as our work function.0955

What frequency is required?0972

Well, if E = Hf, then frequency is equal to E/H, where our energy is going to be 8.16 × 10^-19 J/6.63 × 10^-34 (Planck's constant)...0974

...therefore, the frequency that we need is going to be 1.23 × 10¹⁵ Hz.0992

What is the speed of a photoelectron emitted from gold after it absorbs a photon with frequency 1.5 × 10¹⁵ Hz?1008

Well, let us start by figuring out what the energy is of that photon in joules.1016

That energy is going to be Hf, which is 6.63 × 10^-34 × 1.5 × 10¹⁵ (frequency), which will give us an energy of about 9.94 × 10^-19 J.1020

All right, so the work function was 8.16 × 10^-19 J and the energy of the photon was 9.94 × 10^-19 J -- the kinetic energy must be the difference between those.1040

So our kinetic energy is going to be the energy of the photon, 9.94 × 10^-19 J - 8.16 × 10^-19 J (work function) or about 1.178 × 10^-19 J.1054

That is the left over energy after we have gotten the electron out of the energy well.1074

Now we need to figure out the speed of the photoelectron.1079

To do that, I am going to go back to my formula for kinetic energy.1083

If kinetic energy is 1/2 mv², then the velocity is going to be the square root of 2 times the kinetic energy over the mass...1087

...or 2 × 1.78 × 10^-19 J/9.11 × 10^-31 kg (mass of electron) square root or about 625,000 m/s.1096

Let us take this a little bit further.1123

What potential difference would be required to stop this photoelectron, or what is its stopping potential?1129

Well, remember stopping potential is its maximum kinetic energy divided by the charge.1135

We just said its maximum kinetic energy was 1.78 × 10^-19 J and the charge is going to be 1.6 × 10^-19 coulombs, so our stopping potential then must be 1.11 volts.1141

Finally, part D -- What would happen to the maximum kinetic energy of the photoelectron if the intensity of the EM radiation were doubled?1160

Well, if you increase the intensity, the brightness, you are not going to do anything to the maximum kinetic energy, it is the frequency that tells you how much energy is involved.1168

More photons of the same energy, more intensity is not going to do anything to the maximum kinetic energy.1179

It might give you more of them, but it is not going to touch the maximum kinetic energy, so for part D -- nothing.1184

Which phenomenon provides evidence that light has a wave nature?1196

The emission of light from an energy level transition in a hydrogen atom -- I do not know if that tells me anything about the wave nature.1200

The diffraction of light, waves diffract -- That has to be it.1208

Absorption of light by a black sheet of paper -- It does not say that it must have a wave nature.1213

Reflection of light from a mirror -- Well, yes waves reflect, but also, so do tennis balls for example.1218

Throw a tennis ball at an angle of incidence of 30 degrees against a mirror and assuming you do not break the mirror it is going to bounce off at an angle of reflection of 30 degrees.1224

So reflection is not purely a wave property.1233

Our best answer must be (B), the diffraction of light, waves diffract.1236

Moving electrons are found to exhibit properties of...?1243

...Particles only -- No.1247

...Waves only -- No.1249

...Both particles and waves -- Yes, that is our whole topic in this video -- Wave particle duality.1250

The wave particle duality of light -- Electrons behave like particles and waves as do photons.1257

What is the de Broglie wavelength of an electron with a kinetic energy of 6 eV?1272

Well, the de Broglie wavelength, λ = H/P, which is going to be (H) over momentum as (mv), so that is 6.63 × 10^-34/9.11 × 10^-31 kg (mass of electron) × its velocity, which I do not have yet.1277

Let us figure out its velocity.1299

If its kinetic energy is 6 eV, let us convert that to joules first.1301

So, 6 eV -- we want to convert that to joules.1305

I know 1 eV is 1.6 × 10^-19 J, so our electron volts will cancel out and I will get 9.6 × 10^-19 J.1310

Now, kinetic energy is equal to 1/2 mv², therefore velocity is 2 × kinetic energy divided by mass square root or 2 × 9.6 × 10^-19 J/9.11 × 10^-31 kg square root (mass of particle)...1322

...or about 1.45 × 10⁶ m/s.1343

I can plug that in up here for my velocity -- 1.45 × 10⁶ m/s and get an answer of about 5.02 × 10^-10 m.1348

Let us take a look at one last problem in this unit.1367

Find the de Broglie wavelength of a 700 kg wrecking ball traveling with velocity 12 m/s.1371

λ = H/P, which is H/mv, where (H) is 6.63 × 10^-34, our mass of our wrecking ball is 700 kg at its velocity of 12 m/s is going to give us a wavelength of 7.89 × 10^-38 m.1379

A tiny, tiny, tiny wavelength, which is why we do not observe the wavelength of large objects, larger particles.1404

They have to be really, really, really, really, really tiny to observe those effects.1412

Hopefully, that gets you good understanding of the wave particle duality.1419

Thanks so much for watching here at Educator.com. We will talk to you soon. Make it a great day everyone!1422