Modern Physics

Thanks to: Dr. P. Bertrand Oak Ridge HiS 1

Quantum Physics

    Physics on a very small (atomic) scale is “ quantized”.

Quantized phenomena are discontinuous discrete , and generally very small.

and Quantized energy can be throught of an existing in packets of energy of specific size.

Atoms can absorb and emit atom.

quanta of energy, but the energy intervals are very tiny, and not all energy levels are “allowed” for a given 2

Electromagnetic Spectrum

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Light is a ray

  We know from geometric optics that light behaves as a ray; it travels in a straight line.

When we study ray optics, we ignore the nature light, and focus on how it behaves when it hits a boundary and reflects or refracts at that boundary.

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But light is also a wave!

  We studied this earlier in the year and we will use this equation again here.

C = f l    C: 3 x 10 8 m/s (speed of light in a vacuum) f: frequency (Hz or s -1 ) l : wavelength (m) (distance from crest to crest) 5

In quantum physics, we focus on how light behaves as a particle!

   Light has a dual nature. In addition to behaving as a wave , it also behaves as a particle .

It has energy and momentum, just like particles do. Particle behavior is pronounced on a very small level, or at very high light energies.

A particle of light is called a “photon”.

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Calculating photon energy

   The energy of a photon is calculated from the frequency of the light.

E = hf = hc / l because c = f l E = nhf (for multiple n # of photons)    E: energy (J or eV) h: Planck’s constant  6.625 x 10 -34 J s (SI system)  4.14 x 10 -15 eV s (convenient) f: frequency of light (s -1 , Hz) 7

Checkpoint

  Which has more energy in its photons, a very bright, powerful red laser or a small key-ring red laser?

 Neither! They both have the same energy per photon; the big one has more power.

Which has more energy in its photons, a red laser of a green laser?

 The green one has shorter wavelength and higher frequency. It has more energy per photon.

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The “electron volt” (eV)

   The electron volt is the most useful unit on the atomic level.

If a moving electron is stopped by 1 V of electric potential, we say it has 1 electron volt (or 1 eV) of kinetic energy.

1 eV = 1.602 x 10 -19 J 9

Sample Problem

What is the frequency and wavelength of a photon whose energy is 4.0 x 10 -19 J?

E = hf f = E/h = c = f l l = c/f = 10

Sample Problem

      The bonding energy of H 2 is 104.2 kcal/mol. Determine the frequency and wavelength of a photon that could split one atom of H 2 into two separate atoms. (1 kcal = 4l86 J).

E = (104.2 kcal)(4186 J)( 1 mol mol kcal 6.02x10

23 ) mol’cls E = 7.24 x 10 -19 f = 7.24 x 10 -19 J = hf J/ 6.625 x 10 -34 Js f = 1.09 x 10 15 Hz *** 11



Atomic Transitions

How many photons are emitted per second by a He-Ne laser that emits 3.0 mW of power at a wavelength of 632.8 nm?

P = E tot /t E = hf  P = n(hf)/t c = f l P = nh(c/ l ) f = c/ l n = (P)(t)( t (c)(h) l ) 12

Solution

Find total energy in one second from power P = W/t = E tot / t E tot = Pt = 3.0 x 10 Now see how many photons, n, will produce this energy E = hf (one photon) E tot = nhf (for n photons) E = nhc/ l -3 J (since wavelength is given and not frequency) 3.0x10

-3 = n (6.625x10

-34 Js)(3.0x10

8 m/s)/632.8x10

-9 m n = 9.55 x 10 16 13

General Info re the Atom

   Atoms are composed of  Nuclei (protons and neutrons) and electrons When an atom encounters a photon  It usually ignores the photon, but sometimes absorbs the photon If an atom absorbs a photon  The photon disappears and gives all its energy to the atom’s electrons 14

Quantized atomic energy levels

*This graph shows allowed quantized energy levels in a hypothetical atom.

*The more stable states are those in which the atom has lower energy.

*The more negative the state, the more stable the atom.

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Quantized atomic energy levels

 The highest allowed energy is 0.0 eV. Above this level, the atom loses its electron, This level is called the ionization or dissociation level.

 The lowest allowed energy is called the ground state.

This is where the atom is most stable.

 States between the highest and lowest state are called excited states.

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 Transitions of the electron within the atom must occur from one allowed energy level to another.

 The atom CANNOT EXIST between energy levels.

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Absorption of photon by atom

   

When a photon of light is absorbed by energy of the atom.

The photon disappears.

The energy of the atom increases by exactly the amount of energy contained in the photon.

The photon can be absorbed ONLY if it

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Absorption of photon by atom

 When a photon is absorbed, it excites the atom to higher quantum energy state.

 The increase in energy of the atom is given by D E = hf.

19 Ground state

Absorption Spectrum

• When an atom absorbs photons, it removes the photons from the white light striking the atom, resulting in dark bands in the spectrum.

• Therefore, a spectrum with dark bands in it is called an absorption spectrum.

Absorption spectrum seen through hand held spectroscope 20

Absorption Spectrum

Absorption spectra always involve atoms going up in energy level. 21

Emission of photon by atom

   

When a photon of light is emitted by an atom, it causes a decrease in the energy of the atom.

A photon of light is created.

The energy of the atom decreases by exactly the amount of energy contained in the photon that is emitted.

The photon can be emitted ONLY if it can produce an “allowed” energy decrease in an excited atom.

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Emission of photon by atom

• When a photon Is emitted from An atom, the Atom drops to Lower Quantum Energy state.

• The drop in energy can be computed by D E = -hf.

D

E = -hf

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Emission Spectrum

• When an atom emits photons, it glows! The photons cause bright lines of light in a spectrum.

• Therefore, a spectrum with bright bands in it is called an emission spectrum.

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Emission of photon by atom

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Sample Problem

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Solution

*** 27

Photoelectric Effect #1 • Sample Problem 28

Solution

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Sample Problem

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Solution

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Atoms absorbing photons increase in energy 32

Question

  Now, suppose a photon with TOO MUCH ENERGY encounters an atom?

If the atom is “photo-active”, a very interesting and useful phenomenon can occur…  This phenomenon is called the

Photoelectric Effect.

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Photoelectric Effect

E = work function + K 34

Photoelectric Effect

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Sample Problem

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Solution

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Photoelectric Effect #2

• Sample Problem 38

Sample Problem

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Solution

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Review of Photoelectric Effect

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Question

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The Photoelectric Effect Experiment 43

Photoelectric Effect

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Strange results in the Photoelectric Effect experiment 45

Voltage versus current for different intensities of light 46

Voltage versus current for different frequencies of light 47

Experimental determination of the Kinetic Energy of a photoelectron 48

Graph of Photoelectric Equation

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Photoelectric simulations

• http://lectureonline.cl.msu.edu/~mmp/kap28/Pho toEffect/photo.htm

• This is a link for a simulated photoelectric effect experiment • • Another link: http://zebu.uoregon.edu/%7Esoper/Light/atomspec tra.html

*** 50

Mass of a Photon

• Photon do not have “rest mass”. They must travel at speed of light and nothing can travel at the speed of light unless it has mass = zero.

• A photon has a fixed amount of energy (E = hf) • We can calulate how much mass would have to be destroyed to create a photon (E=mc 2 ) 51

Sample Problem

• Calculate the mass that must be destroyed to create a photon of 340 nm light.

E mass mc 2 = E photon = hf = hc / l m = h = h c l c l m = (6.625 x 10 -34 kgm 2 /s 2 x s) = (3x 10 8 m/s)(340 x 10 -4 m) 52

Momentum of Photon p = mv = mc(c/c)=mc 2 /c = E/c = hf/c = h/ l • Photon do not have “rest mass”, yet they have momentum! This momentum is evident in that, given a large number of photons, they create a pressure?

• A photon’s momentum is calculated by p = E = hf = h c c l 53

Experimental proof of the momentum of photons • Compton scattering – High-energy photons collided with electrons exhibit conservation of momentum – Work Compton problems just like other conservation of momentum problems - except the momentum of a photon uses a different equation.

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Sample Problem

• What is the momentum of photons that have a wavelength of 620 nm?

p = h = 6.625 x 10 -34 l kgm 620 x 10 -9 m 2 /s 2 x s = __________________kgm/s kgm/s  mass x velocity 55

Sample Problem

• What is the frequency of a photon that has the same momentum as an electron with speed of 1200 m/s?

p e m e v e = p p = h/ l = h/c/f = hf/c f = m e v e c/h f = (9.11x10 -31 kg)(1200m/s)(3x10 8 m/s) 6.625 x 10 -34 kgm 2 /s 2 x s f = ________________s -1 56

Wave-Particle Duality

• Waves act like particles sometimes and particles act like waves sometimes.

• This is most easily observed for very energetic photons (gamma or x-Ray) or very tiny particles (electrons or nucleons) 57

Particles and Photons both have Energy • A moving particle has kinetic energy.

– E = K = ½ mv 2 • A particle has most of its energy locked up in its mass.

– E = mc 2 • A photon’s energy is calculated using its frequency.

– E = hf 58

Particles and Photons both have Momentum • For a particle that is moving – p = mv • For a photon – p = h/ l – Check out the units! They are those for momentum.

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Particles and Photons both have Wavelength • For a photon – l = c/f • For a particle, which has an actual mass, this equation still works – l = h/p where p = mv – This is referred to as the

deBroglie wavelength

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Experimental proof that particles have wavelength • Davisson-Germer Experiment – Verified that electrons have wave properties by proving that they diffract – Electrons were “shone” on a nickel surface and acted like light by diffraction and interference 61

Problem

• What is the momentum of photons that have a wavelength of 620 nm?

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Sample problem

What is the wavelength of a 2,200 kg elephant running at 1.2 m/s?

p = h/ l l  l = h/p = h/mv = 6.625 x 10 -34 Js (2200kg)(1.2m/s) l elephant = 2.51 x 10 -37 m 63

Nuclear Decay

• http://library.thinkquest.org/27954/welcom e.htm

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Naming a Nucleus

• Physicists Chemists • • Mass # Electronic Chg on atom or molecule • Charge # # atoms in molec.

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Most common isotope of carbon

12 C Mass Number: protons plus neutrons Element symbol 6 Atomic number: protons 66

Isotopes

• Isotopes have the same atomic number but different atomic mass.

• Isotopes have similar or identical chemistry.

• Isotopes have different nuclear behavior.

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Uranium Isotopes

238 235

U

92 92

U

Low Radioactive Fission 68

Nuclear Particles

Proton Mass: 1 amu Charge: +e Neutron Mass: 1 amu Charge: 0 (1 amu = 1/12 mass of Carbon 12 atom) 69

Negative -1 0

e

Electrons

Positive 0 +1

e

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Nuclear Reactions

• Nuclear Decay - a spontaneous process in which an unstable nucleus ejects a particle and changes to another nucleus.

– Alpha decay – Beta decay • Beta Minus • Positron • Fission - a nucleus splits into two fragments of roughly equal size • Fusion - Two nuclei combine to form a heavier nucleus.

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Decay Reactions

• Alpha decay – A nucleus ejects an alpha particle, which is just a helium nucleus • Beta decay – A nucleus ejects a negative electron • Positron decay – A nucleus ejects a positive election • Simulations – http://library.thinkquest.org/17940/texts/radioa ctivity/radioactivity.html

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Alpha Decay

• This occurs when a helium nucleus is released.

• This occurs only with very heavy elements.

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Beta (

b -

) Decay

• A beta particle (negative electron) is released when a nucleus has too many neutrons for the protons present. A neutron converts to a proton and electron leaving a greater number of protons. An antineutrino is also released.

+

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Positron (

b +

) Decay

• Positron decay occurs when a nucleus has too many protons for the neutrons present. A proton converts to a neutron. A neutrino is also released.

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Neutrino and Anti Neutrino …

• Proposed to make beta and positron decay obey conservation of energy.

• Possess energy and spin, but do not possess mass or charge.

• Do not react easily with matter and are extremely hard to detect.

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Gamma Radiation, h

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1.

2.

3.

4.

5.

Calculating energy released in nuclear reactions

Add up the mass (in atomic mass units, u_ of the reactants. Use your book.

Add up the mass ( in amu’s) of the products.

Find the difference between reactant and product mass. The missing mass has been converted into energy.

Convert mass to kg (1 u = 1.66x10

-27 kg) Use E = mc 2 to calculate energy released.

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Complete the reaction and identify the type of decay: 79

Complete the reaction for the alpha decay of Thorium-232.

+ 80

Nuclear Bombardment

• http://library.thinkquest.org/27954/nuclear.

htm 81

Fission and Fusion

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Fission

      Occurs when an unstable, heavy nucleus split apart into two lighter nuclei (new elements).

Can be induced by free neutrons.

Mass is destroyed and energy produced according to E = mc 2 .

http://library.thinkquest.org/17940/texts/fission/fissio n.html

http://www.atomicarchive.com/Movies/index.shtml

www.atomicarchive.com/Movies/Movie4.shtml

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Neutron-induced fission

   Produces a chain reaction Nuclear power plants operate by harnessing the energy released in fission by controlling the chain reaction Nuclear weapons depend upon the initiation of an uncontrolled fission reaction 84

 Leads to an exponential growth of chain reactions.

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Critical Mass

  The neutrons released from an atom that has undergone fission cannot immediately be absorbed by other nearby fissionable nuclei until they slow down to “thermal” levels.

A critical mass is the smallest amount of

fissile

material needed for a sustained

nuclear chain reaction

. 86

Nuclear Reactors

Nuclear reactors produce electrical energy through fission.

Advantages: a large amount of energy is produced without burning fossil fuels or creating greenhouse gases.

Disadvantage: produces highly radioactive waste.

Simulation: http://video.google.com/videosearch ?q=nuclear+power+plant+operation& hl=en&emb=1&aq=f# http://www.howstuffworks.com/nucle ar-power.htm

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Fission

   Occurs only with very heavy elements since fissionable nuclei are too large to be stable.

A charge/mass calculation is performed to balance the nuclear equation.

Mass is destroyed and energy is produced according to E = mc 2 .

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Problem: complete the following reaction and determine the energy released.

?

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    

Fusion

Occurs when 2 light nuclei come together to form a new nucleus of a new element.

The most energetic of all nuclear reactions.

Produced on the sun.

Fusion of light elements can result in non-radioactive waste.

Proton-Proton Reaction www.chemistrydaily.com/chemistry/Nuclear_reaction 91

Fusion

   The reaction that powers the sun. It has not been reliably sustained on earth in a controlled reaction.

Advantages: tremendous energy produced and lack of radioactive waste products.

Disadvantages: too much energy to control.

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Problem

 When a free proton is fused with a free neutron to form a deuterium nucleus, how much energy is released?

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Mass defect…

   How much mass is destroyed when a nucleus is created from its component parts.

Generally much less than the mass of a proton or neutron, but it is still significant.

This loss of energy results in the creation of energy according to E = mc 2 .

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What is the mass defect of energy in eV and J?

12 C in atomic mass units? How does this relate to mass in kg and 6 1 n + 6 l H  12 C 1u=1.66x10

-27 kg 0 0 E =eV, e =1.6x10

-19 C Mass of reactants 6(1.008665)+6(1.0078225) =12.09894

Mass of products: 1(12.000) =12.00000

D m 0.09894

D

m=(0.09894u)(1.66x10

-27 kg/u)

D

m=1.4x10

-28 kg

E=1.476x10

-11 J=92MeV

E =

D

mc 2 = (1.4x10

-28 kg)(3x10 8 m/s)2

1.6x10

-19 J/eV

E = 1.476 x 10 -11 J

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