Ch. 29 – Particles and Waves

Newton developed classical physics (kinematics and dynamics) back in the 1600’s.

Here we go! Modern physics in 3 weeks!

p p y ( y )

It was Faraday’s experiments and Maxwell’s mathematics that shaped the field of electromagnetism in the 17 and 1800’s.

From here we discovered that EM waves move at the speed of light, and therefore light itself is an EM wave.

This led to our study of light and optics:This led to our study of light and optics:

Geometric OpticsReflection (mirrors)

Refraction (lenses)

Wave OpticsInterference

Diffraction

So here we are in the late 19th century. It’s 1879 and a very important experiment is being performed by Josef Stefan.

Hi i t d l ith di ti th t i itt d d b b d bHis experiment deals with radiation that is emitted and absorbed by blackbodies.

So what is a blackbody????

All objects are continuously absorbing and emitting radiation.j y g g

When light (or any EM radiation) falls on an opaque body, part of it is reflected, and part of it is absorbed.

Light-colored bodies reflect most of the radiation incident on them.

Dark-colored bodies absorb most of the radiation incident on them.

Choose wisely the color of your tee shirtChoose wisely the color of your tee-shirt.

If an opaque body is in thermal equilibrium with its surroundings, then it must be absorbing and emitting radiation at the same rate (equally).g g ( q y)

It has to, or otherwise it would either heat up or cool off, and then no longer be in thermal equilibrium.

This radiation is known as thermal (heat) radiation.For objects whose temperatures are <~600o C, this radiation is not in the visible part of the EM spectrum, but in the infrared.

The human body is ~37o C, so we emit radiation in the infrared. That’s why you can’t see us in the dark!

At temperatures between 600 – 700o C, there is enough energy in the visible spectrum that a body begins to glow dull red.

At much higher temperatures it will grow bright red, or even white-hot.g p g g ,

A body that absorbs and emits all of the radiation incident on it is called an ideal blackbody.

A blackbody is a piece of matter, and like all matter, it is composed of atoms.

We can treat the atoms in the solid as being connected by invisible springs:

Each atom will vibrate, or oscillate, in 3-dimensions.

This is called the simple harmonic approximation.

It is strictly classical physics.y p y

The vibrating atoms absorb and emit radiation, and classical physics tells us th t th i t it f th di ti itt d b th ill t i ti l tthat the intensity of the radiation emitted by the oscillators is proportional to the temperature of the solid: TI ∝

The equality relationship is:2

4)(λ

λ kTI =λ

Yet again, here is another example of an inverse-square law in physics.

Thi i ll d th R l i h J L It i l i l ltThis is called the Rayleigh-Jeans Law. It is a classical result.k in the above equation is the Boltzmann constant: J/K 1038.1 23−×=k

So if we make a plot of the radiation intensity emitted by the atomicSo, if we make a plot of the radiation intensity emitted by the atomic oscillators versus wavelength, we would get:

The plots are shown here for twoI(λ) T2

The plots are shown here for two temperatures T1 and T2, where T2 > T1.T1

Classical theory predicts that I→ ∞ as λ→0.

λ

Experimentally, however, we find the following result: in blue.

The classical theory only gets it right at large wavelengths, but fails miserably at low wavelengths. This was known as the Ultraviolet Catastrophe.

Thus, we see that the radiation intensity from a perfect blackbody varies from wavelength to wavelength.

At higher temperatures, the intensity per unit wavelength is greater, and the maximum occurs at smaller wavelengthssmaller wavelengths.

Now it’s December, 1900 and a German physicist Max Planck is trying to interpret the blackbody datathe blackbody data.

He worked to come up with a theoretical expression that agreed with the experimental data.

He discovered that he could get good agreement between theory and data if he assumed that the energy of the atomic oscillators was a discreet variable.In other words, the energy could only have certain discreet values, i.e.

εεεε nE ,...3,2,,0=Furthermore, this discreet value, ε, had to be proportional to frequency: f∝εMake this an equality: hf=εMake this an equality: hf=εThe proportionality constant, h, is Planck’s constant: 346.626 10 J sh −= × ⋅

This was Planck’s assumption:.0,1,2,3,.. , == nnhfE

Energy is allowed to only have certain values: It is quantized.

Quantizing the energy has some radical effects. Take conservation of energy for example:for example:EM waves carry energy, so a vibrating atom emitting radiation must be losing energy. Conservation of energy tells us that the energy carried away by the EM waveConservation of energy tells us that the energy carried away by the EM wave must be equal to the energy lost by the vibrating atom. Let’s say a vibrating atom has energy, E = 3hf.

Since it’s emitting radiation it must be losing energy and according to PlanckSince it’s emitting radiation, it must be losing energy, and according to Planck, the next lowest energy it could have would be E = 2hf.

That means the EM wave must have an energy E = hfThat means the EM wave must have an energy, E = hf.

So energy comes in discreet packets of (hf) called quanta, or quantum of energy.

Planck’s work paved the way for the development of the New Physics, or Quantum Mechanics.

Planck wasn’t satisfied with his results and conclusion. He tried for several years to reconcile his results with classical physics, but never could.The significance of his work was never realized until 1905.The significance of his work was never realized until 1905.

1905 – The Miracle Year in Physics

A young former patent clerk named Albert Einstein published 4 papers between March 17 and Sept. 27, 1905.

1 On a Heuristic Viewpoint of the Generation and Conversion of Light1. On a Heuristic Viewpoint of the Generation and Conversion of Light

2. A New Measurement of Molecular Dimensions and on the Motion of Small Particles Suspended in a Stationary Liquid

The arguments and calculations are among the most difficult in hydrodynamics and could only be approached by someone who possesses understanding and talent for the treatment of mathematical and physical problems, and it seems to me that Herr Einstein has provided evidence that he is capable of occupying himself successfully with scientific problems. I have examined the most important part of the calculations and have found them to be correct in every respect, and the manner of the treatment testifies to mastery of the mathematical methods concerned.

-Professor KleinerUniversity of Zurich

3. On the Electrodynamics of Moving Bodies

4. Does an Object’s Inertia Depend on its Energy Content?

It is the topic of Einstein’s first paper that we will focus on next.He worked with Planck’s assumption of quantized energy and assumed that it was a universal characteristic of light.was a universal characteristic of light.

He postulated that the energy of light, instead of being equally distributed throughout space, consisted of discrete quanta of energy called photons.

The energy of light (or a photon) is given by:

λhchfE ==

Thus, if you know the frequency or wavelength of light, you can calculate its energy.

Th t th i t it f th li ht th h t it t i b t hThe greater the intensity of the light, the more photons it contains, but each photon has an energy, E = hf.

Einstein will use these ideas to explain something called the Photoelectric Effect.

Which light bulb is emitting radiation?

Question 29 - 1

c g t bu b s e tt g ad at o

1 A1. A2. B3 B th3. Both4. Neither

A Quantum Mechanic!!!

29.3 The Photoelectric Effect

In 1887 Heinrich Hertz produced and detected electromagnetic waves, thus p g ,proving Maxwell’s theory.He also discovered something called the Photoelectric Effect.

When light shines on a metal plate, some of the electrons in the metal get ejected from its surface and then are accelerated by a potential differenceby a potential difference.

This results in a current flow in the circuit as shownas shown.

Important Characteristics of the photoelectric effect

1. Only light with a frequency above some minimum value, fo, will result in electrons u a ue, fo, esu t e ect o sbeing ejected – regardless of the light’s intensity.

Let’s look at a plot of the KE of the ejected electrons vs. the frequency of the light shining on the metal:

Notice: No electrons are ejected from the metal for frequencies below some fo.

f < fo, no ejected electrons.

f ≥ fo, electrons are ejected from the metal’s surface.

fo is called the Threshold Frequency.

Now choose some constant value for the frequency f ≥ fo, so that electrons are being ejected from the metalbeing ejected from the metal.

2. The maximum KE of the ejected electrons remains constant, even if the intensity of the light is increased.

Classically, we would expect higher intensity light to eject electrons with greater KE. It doesn’t happen.

Also, we would expect that if we used very low intensity light, that it would take a long time for the electrons to build up enough energy to be ejected from the metal’s surface.

That doesn’t happen either! Even if the light intensity is very low, electrons are still ejected from the metal’s surface, almost instantaneously, as long as f ≥ fo.

As we mentioned previously, Einstein assumed that light was composed of discrete packets (particles) of energy called photons.

And the photon energy is given by: hcp gy g y

λhchfE ==

The more intense the light is, the more photons it carries, but each photon still has an energy: E = hf.

Now let’s examine the photoelectric effect in a little more detail.

Surface

Free electrons occupy the entire volume of the metal.

H l t l t th

Light, hf Ejected e-, KE

MetalSurface electrons However, electrons close to the

metal’s surface (surface electrons) are more weakly bound to the metal than the deep electrons.

Deep electrons

metal than the deep electrons.

Therefore, it is the surface electrons that are ejected during the photoelectric effect.

But even though the surface electrons are more weakly bound, there is still a minimum “binding energy” I must overcome to get them out of the metal.

This is called the Work Function (Wo) of the metal.This is called the Work Function (Wo) of the metal.

It is an energy, and it is typically on the order of a few eV.

During the effect, a photon of light (f > f ) with energy hf strikes the metal andDuring the effect, a photon of light (f > fo) with energy hf strikes the metal and electrons are ejected with energy KE.

By conservation of energy, the following relationship must be true: The photon energ The binding energ KE of the ejected electronsThe photon energy – The binding energy = KE of the ejected electrons

−=− eo KEWhf This is the Einstein equation for the Photoelectric Effect. or

It was for this work on the photoelectric effect that Einstein received the Nobel Prize in Physics in 1921.The wave description of EM radiation (light) fails to describe theThe wave description of EM radiation (light) fails to describe the photoelectric effect. We need to use the particle (photon) picture.

…But don’t abandon the wave description yet!

The photoelectric effect was one of the earliest indications of the Particle/Wave Duality of Light.

So light is composed of particles called photonsSo light is composed of particles called photons.Einstein showed that the total energy of an object is: 2mcE γ=

1Where:

2

211

cv−

=γ Notice, that if v = 0, then γ = 1, and 2mcE =

This is just his famous rest-mass equation.

Rewrite the general energy equation: ( ) 222

21 mcEmcEcv =−⇒= ( )cγ

What if we apply this equation to photons? Well, for photons, v = c.

( ) 22 01 2

2 mcmcEcc =⇒=− Therefore, for the equation to be correct,

m must equal 0 for the photon!

A photon is a massless particle that moves at the speed of light!

In the photoelectric effect, electrons are ejected from the surface of a metal when light shines on it Which of the

Question 29 - 2

surface of a metal when light shines on it. Which of the following would lead to an increase in the maximum KE of

the ejected electrons?

1. Increase the frequency of the lightthe light.

2. Increase the intensity of the light.g

3. Use a metal with a larger work function.

=− KEWhf −= eo KEWhf

29.5 Particle Waves – de Broglie

Working as a graduate student in 1923, de Broglie hypothesized that if light waves can have particle-like properties, then maybe particles (i.e.

l t ) h lik ti

Louis de Broglie

electrons) can have wave-like properties. This was a very radical assumption, since at this time, there was no evidence to support this hypothesishypothesis.

From Einstein’s relativity equations, it is found that the momentum of a photon is:

h

λh

chfp == de Broglie rewrote this as:

ph

=λp is the relativistic momentum.

This is known as the de Broglie wavelength of a particle.

Thi ll t l l t th l th f bj t ithThis allows us to calculate the wavelength for any object with nonzero momentum, p.

Now let’s repeat Young’s double-slit experiment, but this time lets shoot electrons (particles) at the slits instead of light.

What would we expect to see???

Well, we might expect the screen to appear as it does to the left two bright fringes one directlydoes to the left – two bright fringes, one directly behind each slit.

What we actually see is shown in the figure at the at e actua y see s s o t e gu e at t elower left – alternating dark and bright fringes.

In other words, the electrons have acted like waves and interfered with each other to produce the classic interference pattern!

Our notion of the electron as being a tiny discrete particle of matter does not account for the fact that the electron can behave as a wave in some circumstances.

It hibit d l t b h i tiIt exhibits a dual nature – behaving sometimes like a particle, and sometimes like a wave.

Things are even weirder than this!!!.....

The previous scenario is also true for light. Sometimes it’s wave-like and sometimes it’s particle-like.

This is referred to as the Particle-Wave Duality of Light.

The first experiment to show wave-like properties of particles, and thus prove de Broglie’s hypothesis, was a diffraction experiment.

Two American physicists (Davisson and Germer) in 1927 demonstrated electron diffraction with a crystal of nickel.

Wave properties are displayed by all particles, such as protons and neutrons too.

NaCl – Neutron diffraction NaCl – X-ray diffraction

So all objects have a de Broglie wavelength – baseballs, cars, even you and me!

But remember, in order for wave effects to be seen, such as interference and diffraction, the wavelength must be comparable to the size of the opening or obstacle.

For fun, let’s calculate my de Broglie wavelength:

ph

=Dr.Youngλmvh

= ( )( )346.626 10 J s

60 kg 9.05 m/s

−× ⋅= m 101 36−×=

p mv ( )( )g

So what does this number mean???

W ll th i f t i hl 1 × 10 10 S d B li l thWell, the size of an atom is roughly 1 × 10-10 m. So my deBroglie wavelength is some 26 orders of magnitude smaller than the size of an atom!!!

Which means….we don’t observe wave-like properties with everyday objects, b b ll h tbaseballs, humans, etc.Thus, we need really small masses and high speeds to observe the wave-like properties…..sub-atomic particles!

But, just for fun, consider if my de Broglie wavelength was say 1 meter.

What might happen if I ran into a forest of pine trees???

You could interfere with yourself! And diffract!

So what are these particle-waves really???They are waves of probability!

In the double-slit experiment done with electrons, the bright fringes on the screen are regions of high probability.In other words, it’s more probable that the electrons will hit the screen there.

In the double-slit experiment done with light, the intensity of the bright fringes is proportional to either E2 or B2.

For particle waves, the intensity of the maxima (bright fringes) is proportional to:

2Ψ

Ψ represents the wave function of the particle.

It was Erwin Schrödinger and Werner Heisenberg who independently developed the theory of how to determine a particle’s wave function.

Once you have , you can predict how the particle will evolve over time.Ψ

They established a new branch of physics called Quantum Mechanics!

y , y p pΨ

The characteristic interference pattern becomes evident after a sufficient number of electrons haveevident after a sufficient number of electrons have struck the screen.

It (the double slit experiment) encapsulates the centralIt (the double-slit experiment) encapsulates the central mystery of quantum mechanics. It is a phenomenon which is impossible, absolutely impossible, to explain in any classical way and which has in it the heart of quantum mechanics. In reality, it contains the only mystery…the basic peculiarities of all quantum mechanics.

-Richard P. Feynman

29.6 The Heisenberg Uncertainty PrincipleSo last time we saw that the bright fringes on the screen from the double-slit experiment with electrons were regions of high probabilityexperiment with electrons were regions of high probability.An individual electron could go to any bright fringe; we don’t know which one.

To tear into this a little more, let’s just consider single-slit diffraction with l t d t t th l t th t f th t l b i htelectrons, and concentrate on those electrons that form the central bright

fringe.The electrons enter the slit with momentum px.Once they pass thru the slit, some of the electrons must gain momentum in the y-direction.The maximum any electron could gain would be Δpy.Thus, Δpy represents the uncertaintyin the momentum in the directionin the momentum in the y-direction.

From single-slit diffraction with light, we know that, specifies the location of the first dark fringe. W

λθ =sinlocation of the first dark fringe.

If the screen distance is large, then , or θθ tansin ≈ .tanx

y

pp

WΔ

==λθ

Thus,Wpp x

yλ

=Δ But, from de Broglie, we know that:λhpx =

Thus, .Whpy =Δ So what does this equation tell us???

Th ll W i th t l k th l f th l t itThe smaller W is, the more accurately we know the y-value of the electron as it passes thru the slit.

But, the smaller W is, the greater Δpy becomes. In other words, the more accurately we know the particle’s position in the slit, the larger the uncertainty in its momentum. Thus W represents the uncertainty in the position of thee-

ySlit

Thus, W represents the uncertainty in the y-position of the electron: .yW Δ≡Plugging into above, we find:

2≥ΔΔ ypy h

=W

Slit 2 π2This is a statement of the Heisenberg Uncertainty Principle (HUP) for Momentum and Position:

It is impossible to specify precisely the position and momentum of a particle at the same time.

Another way to write the HUP is the following:

2≥ΔΔ tE

This is a statement of the Heisenberg Uncertainty Principle for Energy and Time:

We cannot know exactly the energy of a particle when it’s in a particular state of a physical system and also know precisely the time interval that it remains in that state

Principle for Energy and Time:

that state.

Or, the shorter the lifetime of a particle in a particular state, the greater the uncertainty in its energy.

Example:

Suppose an electron is confined to a box that is roughly the size of an atom, say 0.5 A on each side. Take this distance to be equal to the uncertainty in the electron’s position and calculate the minimum uncertainty in its momentum. Then, assume that this uncertainty in momentum is equal to the magnitude of the electron’s momentum at some arbitrary time and calculate the electron’s kinetic energysome arbitrary time and calculate the electron s kinetic energy.

Solution:2

≥ΔΔ ypyy

py Δ=Δ⇒

2 )105.0)(4(10626.6

10

34

−

−

××

=π

m/skg 1006.1 24 ⋅×= −

Now assume p = Δpy: mpmvKE 2

22

21 == J 1017.6

)1011.9(2)1006.1( 19

31

224−

−

−

×=××

=)(

eV 4=

This is on the order of atomic energies! So the HUP tells us that an electron confined to a box the size of an atom must have an energy ~ 5 eV Thus it isconfined to a box the size of an atom must have an energy 5 eV. Thus, it is bouncing around quite rapidly, ~1.2 × 106 m/s.

What if we repeat this calculation for a tennis ball (m = 0.06 kg) confined to a cardboard box that is 0.5 m on each side?box that is 0.5 m on each side?

2≥ΔΔ ypy

ypy Δ

=Δ⇒2 )5.0)(4(

10626.6 34

π

−×= m/skg 1006.1 34 ⋅×= −

2 234Now assume p = Δpy: m

pmvKE 22

221 == J 1028.9

)06.0(2)1006.1( 68

234−

−

×=×

=

eV 107.5 49−×=This is equivalent to a speed of ~1 × 10-33 m/s!

In other words, the tennis ball just sits there!

In a Young's double-slit experiment that uses electrons, the angle that locates the first-order bright fringes is θA = 2.7 × 10-4 degrees when the magnitude of the electron momentum is p = 1 2 × 10-22 kg · m/s With the same double slit whatelectron momentum is pA = 1.2 × 10 kg m/s. With the same double slit, what momentum magnitude pB is necessary so that an angle of θB = 7.0 ×10-4 degrees locates the first-order bright fringes?

sin hd m mp

θ λ= = m = 1 for 1st- order bright fringe

STM: peek in atoms & electron waves

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Fe on Cu(111)Fe on Cu(111)

electron density of stateselectron density of statesBe(0001)Be(0001)

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Ch. 30 The Nature of the Atom30.1 Rutherford Scattering

We know there are two types of charge, + and -, and that matter is composed of both.English physicist J.J. Thompson discovered the electron and measured its

i 1897mass in 1897.He then suggested the Plum-Pudding Model of the atom:

Negative electrons were scattered like plums in a pudding of positive charge egat e e ect o s e e scatte ed e p u s a pudd g o pos t e c a gethat was equally distributed throughout the atom

electrons

Smeared-out positive charge

To this point, no one had detected the positively-To this point, no one had detected the positivelycharged particles in an atom, but since they knew what the mass of the electron was, they knew most of the atom’s mass was due to the positive

hcharges.

Then Rutherford comes along (a New Zealander), working in London in 1907.

H h d b t d i ith M i d Pi C i i M t l h di dHe had been studying with Marie and Pierre Curie in Montreal who discovered radioactivity.

They observed two types of radiation. Rutherford called them α and β.

Rutherford soon realized that β-radiation or β-rays were really just electrons.

But α-particles were much heavier and positively chargedBut α-particles were much heavier and positively charged.

(α-particles are really the nucleus of a helium atom. They consist of 2 protons and 2 neutrons. No one knew this at the time.)

R th f d d id d t th ti l t i ti t th i t l t t fRutherford decided to use the α-particles to investigate the internal structure of atoms.

He shot the α-particles at a thin film of gold:

If Thompson’s model of the atom was correct, very little should happen to the α-particles. They should basically travel p y ystraight thru the film.

Most of them did just that, but some of the a-particles were deflected at large angles Inparticles were deflected at large angles. In some cases almost 180o!

This was completely shocking! Rutherford would later say that, “It was as if

So how could these large deflections occur?

p y g ywe were firing shells at tissue paper and having them bounce straight back at you!”

Really only one solution presented itself, and that was to have all the positive charges in the atom concentrated in a very small region – called the atomic nucleus.

The nucleus was much smaller than the entire atom, but it accounted for almost all of the atom’s mass.Rutherford named the positively charged particles in the nucleus ProtonsRutherford named the positively-charged particles in the nucleus Protons.

This led to the model of the atom that is still widely accepted today – the nuclear model (planetary model).

(The model is qualitatively accurate, although the electrons do not orbit the nucleus in pretty, constant, circular orbits. Th t ll i t i l d f b bilit )They actually exist in a cloud of probability.)

An atom is mostly empty space!!!

Nuclear radii ~1 × 10-15 m

Electron orbit ~1 × 10-10 mElectron orbit 1 × 10 m

But, there are problems with the Planetary Model of the Atom:

Consider an electron orbiting some nucleus at a radius r:

The electron has both KE and EPE.

And…..it is accelerating: vma e2

=r Nucleusr

But we know from previous chapters that accelerating charges produce electromagnetic waves, and those waves carry energy.

v

rElectron

waves carry energy.This carries some of the electron’s energy away, it slows down, its orbit decreases, since opposite charges attract, and the electron spirals into the nucleus……..The atom collapses!

The planetary model is not stable based on classical electrodynamics!

We’ll come back to this soon, but first let’s talk about line spectra.p

30.2 Line Spectra

Remember, all objects emit EM radiation. Hot filaments in incandescent lightRemember, all objects emit EM radiation. Hot filaments in incandescent light bulbs emit a continuous range of wavelengths, some of which are in the visible spectrum.

This range of different wavelengths results from the electrons being “excited” in all the different atoms that make up the solid.In contrast to this, single atoms only emit radiation at certain wavelengths.

Low-pressure gases behave this way, for example, and only emit certain wavelengths of light when exited.

And, we can use a diffraction grating to separate these wavelengths and produce a characteristic line spectra for those atoms.

Take mercury (Hg) for example:Take mercury (Hg) for example:

Hg sourceThis is mercury’s line spectra.

Diffraction grating

Hydrogen gas, being the simplest element, also has the simplest line spectraspectra.

It forms 3 groups of lines, called the Lyman, Balmer, and Paschen Series.

Only the Balmer series lies in the visible spectrum.

Balmer series for hydrogen gas.

Lyman Balmer and PaschenLyman, Balmer, and Paschen discovered relationships between the different wavelengths of light in the line spectra.

Lyman Series: 2,3,4,... ,11122 =⎟⎠⎞

⎜⎝⎛ −= nRy , , ,,1 22 ⎟

⎠⎜⎝ nλ

⎞⎛Balmer Series: 3,4,5,... ,1

211

22 =⎟⎠⎞

⎜⎝⎛ −= n

nR

λ

Paschen Series: 4,5,6,... ,1311

22 =⎟⎠⎞

⎜⎝⎛ −= n

nR

λ 3 ⎠⎝ nλ

-17 m 101.097Constant Rydberg ×==R

Since only the Balmer series is in the visible spectrum, what would be the longest and shortest wavelengths we can see from hydrogen emission?longest and shortest wavelengths we can see from hydrogen emission?

The longest would occur for n = 3, and the shortest will occur for n = ∞.

For n = 3, the Balmer series is: ⎟⎠⎞

⎜⎝⎛ −= 22 3

1211 R

λ⎟⎠⎞

⎜⎝⎛ −=⇒

91

411 R

λ

nm 6562.7==⇒

Rλ Red part of visible spectrum

For n = ∞, the Balmer series is: ⎟⎠⎞

⎜⎝⎛

∞−= 22

1211 R

λ⎟⎠⎞

⎜⎝⎛=⇒

411 R

λ⎠⎝ ∞2λ ⎠⎝ 4λ

nm 3654==⇒ λ UViolet part of visible spectrum

Rp p

D

30.3 Bohr Model of the Atom

Ni l B h di d i 1962 H ti f D k h t t E l dNiels Bohr died in 1962. He was a native of Denmark who went to England on scholarship to work with J.J. Thompson. He and Thompson didn’t get along, so Bohr went to work for Rutherford.

H k d th l t d l d t i d t d t d h ttHe worked on the planetary model and tried to understand how matter could be stable if electrons are accelerating around the nucleus.

Bohr started with hydrogen, since it is the simplest of all elements. We have just 1 electron orbiting 1 protonhave just 1 electron orbiting 1 proton.

Hydrogen atom

v

r Proton

Electron

Like Einstein did with the photoelectric effect, Bohr adopted Planck’s idea

v

of quantized energy levels, and assumed that the electron in the hydrogen atom could only have certain values of energy, i.e. they are quantized.

Each of these energy levels corresponds to a unique orbit that the electron moves in around the proton. The larger the orbit, the larger the energy.

So here are Bohr’s assumptions:

1. Electrons travel in fixed orbits around the proton, each orbit being defined by a unique radius and energy. These orbits are called stationary orbits or states, and while in these orbits, the electrons do not emit radiation.

How can we have radiationless orbits??? This goes against classicalHow can we have radiationless orbits??? This goes against classical physics!!!!!

Furthermore, we know that all matter absorbs and emits radiation, so how do Bohr’s atoms do this??

To explain this Bohr incorporated Einstein’s photon model:To explain this, Bohr incorporated Einstein s photon model:

2. An electron in an atom emits radiation (emits photons) when and only when it moves from a higher energy level to a lower one i e it moves fromwhen it moves from a higher energy level to a lower one, i.e. it moves from a larger radius orbit to a smaller one.

As the electron makes a transition from a higher energy level to a lower one, it emits a photon.

By conservation of energy, the energy of the photon must be equal to the difference in the energies between the two stationary states Ei and Ef:

⇒−= fi EE Energy Photon hcλhcEEhf fi =−=

R b hi i ll h i l d B h ld hi iRemember, this is all theoretical, and Bohr could use this equation to calculate the wavelengths (colors) of light emitted by a hydrogen atom.

But first, he needed a way to calculate the energies, Ei and Ef.

To do this, he ends up using both classical and quantum physics.

An electron sitting in the n = 2 energy level of a hydrogen atom has an energy of -3.4 eV. This electron makes a

Question 30 - 1

atom has an energy of 3.4 eV. This electron makes a transition to the n=1 state by emitting a photon. The n=1

state has an energy of -13.6 eV. What is the energy of the emitted photon?emitted photon?

1. 10.2 eV2. 17.0 eV3. -10.2 eV4. -17.0 eV

fi EE −= Energy Photon

eV 2.10)eV 6.13(eV 4.3 =−−−=