notes : _____ quantum mechanics

Notes : _____Quantum Mechanics

AP Physics B

Quantum?Quantum mechanics is the study of processes

which occur at the atomic scale.

The word "quantum" is derived From Latin to mean BUNDLE.

Therefore, we are studying the motion of objects that come in small bundles called quanta. These tiny bundles that we are referring to are electrons traveling around the nucleus.

“Newton, forgive me..”, Albert EinsteinAt the atomic scale Newtonian Mechanics

cannot seem to describe the motion ofparticles. An electron trajectory betweentwo points for example IS NOT a perfectparabolic trajectory as Newton's Lawspredicts. Where Newton's Laws endQuantum Mechanics takes over.....IN ABIG WAY!

One of the most popular concepts concerning Quantum Mechanics is called , “The Photoelectric Effect”. In 1905, Albert Einstein published this theory for which he won the Nobel Prize in 1921.

What is the Photoelectric Effect?In very basic terms, it is when electrons are

released from a certain type of metal upon receiving enough energy from incident light.

So basically, light comes down and strikes the metal. If the energy of the light wave is sufficient, the electron will then shoot out of the metal with some velocity and kinetic energy.

The Electron-Volt = ENERGY

Before we begin to discuss the photoelectric effect, we must introduce a new type of unit.

Recall:

This is a very useful unit as it shortens our calculations and allows us to stray away from using exponents.

The Photoelectric Effect"When light strikes a material, electrons are

emitted. The radiant energy supplies the work necessary to free the electrons from the surface."

Photoelectric Fact #1The LIGHT ENERGY (E) is in the form of quanta

called PHOTONS. Since light is an electromagnetic wave it has an oscillating electric field. The more intense the light the more the field oscillates. In other words, its frequency is greater.

Light Review

versa... viceand,

&between iprelationsh inverse,

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acuum)constant(vlight of speed8

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ffc

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c

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More on Fact #1

hcEf

c

hf

EhfEfE

h hc

6.63x10-34 Js 1.99x10-25 Jm

4.14x10-15 eVs 1.24x103 eVnm

Make sure you USE the correct constant!

Planck’s Constant is the SLOPE of an Energy vs. Frequency graph!

Photoelectric Fact #2The frequency of radiation must be above a certain value

before the energy is enough. This minimum frequency required by the source of electromagnetic radiation to just liberate electrons from the metal is known as threshold frequency, f0.

The threshold frequency is the X-intercept of the Energy vs. Frequency graph!

Photoelectric Fact #3Work function, , is defined as the least energy

that must be supplied to remove a free electron from the surface of the metal, against the attractive forces of surrounding positive ions.

Shown here is a PHOTOCELL. When incident light of appropriate frequency strikes the metal (cathode), the light supplies energy to the electron. The energy need to remove the electron from the surface is the WORK!

Not ALL of the energy goes into work! As you can see the electron then MOVES across the GAP to the anode with a certain speed and kinetic energy.

Photoelectric Fact #4The MAXIMUM KINETIC ENERGY is the energy difference between

the MINIMUM AMOUNT of energy needed (ie. the work function) and the LIGHT ENERGY of the incident photon.

THE BOTTOM LINE: Energy Conservation must still hold true!

Light Energy, E

WORK done to remove the electron

The energy NOT used to do work goes into KINETIC ENERGY as the electron LEAVES the surface.

Putting it all together

bmxy

hfKWhfK

hfWK

hfE

KINETIC ENERGY can be plotted on the y axis and FREQUENCY on the x-axis. The WORK FUNCTION is the y – intercept as the THRESHOLD FREQUNECY is the x intercept. PLANCK‘S CONSTANT is the slope of the graph.

KE max = hf – hfo

KE max = hf ( energy of the original photon ) – hfo (Work function- property of the metal)

Work function is the minimum energy needed to free an electron.

PROBLEM

Light with frequency of 2 X 10 15 hertz is incident on a piece of copper.

A. what is the energy of light in joules and in electron volt?

If the work function for copper is 4.5 eV , what is the maximum kinetic energy , in electron volts of the emitted electrons ?

Note : 1 eV = 1.6 X 10-19 J

A. the energy in joules is given by E= hf

E = hf = (6.63X 10 -34 ) ( 2 X 1015) =

E= 1.326X10-16J

Since 1eV = 1.6 X 10-19 J

E= 8.28 eV

B. To find the maximum kinetic energy , we simply subtract the work function from the photon energy .

KEmax = 8.28 V – 4.5 V = 3.79eV

Can we use this idea in a circuit? We can then use this photoelectric effect idea to

create a circuit using incident light. Of course, we now realize that the frequency of light must be of a minimum frequency for this work.

Notice the + and – on the photocell itself. We recognize this as being a POTENTIAL DIFFERENCE or Voltage. This difference in voltage is represented as a GAP that the electron has to jump so that the circuit works

What is the GAP or POTENTIAL DIFFERENCE is too large?

Photoelectric Fact #5 - Stopping PotentialIf the voltage is TOO LARGE the electrons WILL NOT have

enough energy to jump the gap. We call this VOLTAGE point the STOPPING POTENTIAL.

If the voltage exceeds this value, no photons will be emitted no matter how intense. Therefore it appears that the voltage has all the control over whether the photon will be emitted and thus has kinetic energy.

Wave-Particle DualityThe results of the photoelectric effect allowed

us to look at light completely different.

First we have Thomas Young’s Diffraction experiment proving that light behaved as a WAVE due to constructive and destructive interference.

Then we have Max Planck who allowed Einstein to build his photoelectric effect idea around the concept that light is composed of PARTICLES called quanta.

This led to new questions….If light is a WAVE and is ALSO a particle, does

that mean ALL MATTER behave as waves?

That was the question that Louis de Broglie pondered. He used Einstein's famous equation to answer this question.

YOU are a matter WAVE!

Basically all matter could be said to have a momentum as it moves. The momentum however is inversely proportional to the wavelength. So since your momentum would be large normally, your wavelength would be too small to measure for any practical purposes.

An electron, however, due to it’s mass, would have a very small momentum relative to a person and thus a large enough wavelength to measure thus producing measurable results.

This led us to start using the Electron Microscopes rather than traditional Light microscopes.

Problem

Find the de Broglie wavelength for each of the following :

a. A 10g stone moving with a velocity of 20m/s

b. An electron (9.1 X 10-31) kg moving with a velocity of 1 X 107 m/s

a. since m =10g = 0.01 kg and λ = h/mv

λ = 6.63 X 10 -34

----------------- = 3.315 X 10 -33 m

(0.01)( 20)

b. In this part , we have

λ = h / mv = 6.63 X 10 -34

----------------- = 7.3 X 10-11 m

(9.1 X 10-31)(1X 10-7)

The electron microscopeAfter the specimen is prepped. It

is blasted by a bean of electrons. As the incident electrons strike the surface, electrons are released from the surface of the specimen. The deBroglie wavelength of these released electrons vary in wavelength which can then be converted to a signal by which a 3D picture can then be created based on the signals captured by the detector.

CW: Problems

1. What is the Broglie wavelength for a proton ( m = 1.67X 10-27 kg ) with a velocity of

6 X 10 7 m/s

2. What is the momentum associated with yellow light that has wavelength of 5,500 Å

1Å = Å X 10-10 m

1. 6.6 X 10 -15 m

2. 1.2 X 10 -27 kg m/s

Atomic & Nuclear Physics

AP Physics B

Objectives: After completing this module, you should be

able to:• Define and apply the concepts of Define and apply the concepts of mass numbermass number, , atomic numberatomic number, and , and isotopesisotopes..

• Define and apply concepts of Define and apply concepts of radioactive radioactive decaydecay and and nuclear reactionsnuclear reactions..

• Calculate the Calculate the mass defectmass defect and the and the binding energy per nucleonbinding energy per nucleon for a for a particular isotope.particular isotope.

• State the various State the various conservation lawsconservation laws, and , and discuss their application for nuclear discuss their application for nuclear reactions.reactions.

Life and AtomsEvery time you breathe you are taking

in atoms. Oxygen atoms to be exact. These atoms react with the blood and are carried to every cell in your body for various reactions you need to survive. Likewise, every time you breathe out carbon dioxide atoms are released.

The cycle here is interesting.

TAKING SOMETHING IN. ALLOWING SOMETHING OUT!

The Atom

As you probably already know an atom is the building block of all matter. It has a nucleus with protons and neutrons and an electron cloud outside of the nucleus where electrons are orbiting and MOVING.

Depending on the ELEMENT, the amount of electrons differs as well as the amounts of orbits surrounding the atom.

Composition of Matter

All of matter is composed of at least three fundamental particles (approximations):

All of matter is composed of at least three fundamental particles (approximations):

ParticleParticle Fig.Fig.SymSym MassMass ChargeCharge SizeSize

The mass of the proton and neutron are close, but they are about 1840 times the The mass of the proton and neutron are close, but they are about 1840 times the mass of an electron.mass of an electron.

Electron Electron ee-- 9.11 x 10 9.11 x 10-31-31 kg kg -1.6 x 10-1.6 x 10-19 -19 C C Proton Proton pp 1.673 x 101.673 x 10-27-27 kg +1.6 x 10 kg +1.6 x 10-19 -19 C 3 C 3

fmfmNeutron Neutron nn 1.675 x 101.675 x 10-31-31 kg kg 0 0 3 fm3 fm

The Atomic Nucleus

Beryllium AtomBeryllium Atom

Compacted nucleus:Compacted nucleus:

4 protons4 protons

5 neutrons5 neutrons

Since atom is electri-cally Since atom is electri-cally neutral, there must be 4 neutral, there must be 4 electrons.electrons.

4 electrons4 electrons

Modern Atomic Theory

The Bohr atom, which is The Bohr atom, which is sometimes shown with sometimes shown with electrons as planetary electrons as planetary particles, is no longer a valid particles, is no longer a valid representation of an atom, but representation of an atom, but it is used here to simplify our it is used here to simplify our discussion of energy levels.discussion of energy levels.

The uncertain position of an The uncertain position of an electron is now described as a electron is now described as a probability distribution—loosely probability distribution—loosely referred to as an referred to as an electron cloudelectron cloud..

DefinitionsA A nucleonnucleon is a general term to denote a nuclear particle - that is, either a is a general term to denote a nuclear particle - that is, either a proton or a neutron.proton or a neutron.

The The atomic number atomic number ZZ of an element is equal to the number of protons in the of an element is equal to the number of protons in the nucleus of that element.nucleus of that element.

The The mass number mass number AA of an element is equal to the total number of nucleons of an element is equal to the total number of nucleons (protons + neutrons).(protons + neutrons).

The mass number A of any element is equal to the sum of the atomic number Z and the number of neutrons N :

A = N + Z

Symbol Notation

A convenient way of describing an element is by giving its mass number and A convenient way of describing an element is by giving its mass number and its atomic number, along with the chemical symbol for that element.its atomic number, along with the chemical symbol for that element.

A convenient way of describing an element is by giving its mass number and A convenient way of describing an element is by giving its mass number and its atomic number, along with the chemical symbol for that element.its atomic number, along with the chemical symbol for that element.

A Mass numberZ Atomic numberX Symbol

A Mass numberZ Atomic numberX Symbol

For example, consider beryllium (Be): 94Be

Example 1: Describe the nucleus of a lithium atom which has a mass number of 7 and an atomic number of 3.

Lithium AtomLithium Atom

N = A – Z = N = A – Z = 7 - 3 7 - 3

A = A = 7; Z = 3; 7; Z = 3; NN = ? = ?

Protons: Z = 3Protons: Z = 3

neutrons: neutrons: NN = 4 = 4

Electrons: Electrons: Same as ZSame as Z

73 Li73 Li

When the atom gets excited or NOT

To help visualize the atom think of it like a ladder. The bottom of the ladder is called GROUND STATE where all electrons would like to exist. If energy is ABSORBED it moves to a new rung on the ladder or ENERGY LEVEL called an EXCITED STATE. This state is AWAY from the nucleus.

As energy is RELEASED the electron can relax by moving to a new energy level or rung down the ladder.

Energy LevelsYet something interesting happens as

the electron travels from energy level to energy level.

If an electron is EXCITED, that means energy is ABSORBED and therefore a PHOTON is absorbed.

If an electron is DE-EXCITED, that means energy is RELEASED and therefore a photon is released.

We call these leaps from energy level to energy level QUANTUM LEAPS.

Since a PHOTON is emitted that means that it MUST have a certain wavelength.

Energy of the PhotonWe can calculate the ENERGY

of the released or absorbed photon provided we know the initial and final state of the electron that jumps energy levels.

Energy Level DiagramsTo represent these transitions we can construct an ENERGY LEVEL DIAGRAM

Note: It is very important to understanding that these transitions DO NOT have to occur as a single jump! It might make TWO JUMPS to get back to ground state. If that is the case, TWO photons will be emitted, each with a different wavelength and energy.

ExampleAn electron releases energy

as it moves back to its ground state position. As a result, photons are emitted. Calculate the POSSIBLE wavelengths of the emitted photons.

Notice that they give us the energy of each energy level. This will allow us to calculate the CHANGE in ENERGY that goes to the emitted photon.

This particular sample will release threedifferent wavelengths, with TWO beingthe visible range ( RED, VIOLET) andONE being OUTSIDE the visible range(INFRARED)

Energy levels Application: SpectroscopySpectroscopy is an optical technique by which we can

IDENTIFY a material based on its emission spectrum. It is heavily used in Astronomy and Remote Sensing. There are too many subcategories to mention here but the one you are probably the most familiar with are flame tests.

When an electron gets excited inside a SPECIFIC ELEMENT, the electron releases a photon. This photon’s wavelength corresponds to the energy level jump and can be used to indentify the element.

Different Elements = Different Emission Lines

Emission Line SpectraSo basically you could look at light

from any element of which the electrons emit photons. If you look at the light with a diffraction grating the lines will appear as sharp spectral lines occurring at specific energies and specific wavelengths. This phenomenon allows us to analyze the atmosphere of planets or galaxies simply by looking at the light being emitted from them.

Nuclear Physics - RadioactivityBefore we begin to discuss the specifics of radioactive

decay we need to be certain you understand the proper NOTATION that is used.

To the left is your typical radioactive isotope.Top number = mass number = #protons + neutrons. It is represented by the letter "A“

Bottom number = atomic number = # of protons in the nucleus. It is represented by the letter "Z"

Nuclear Physics – Notation & IsotopesAn isotope is when you have

the SAME ELEMENT, yet it has a different MASS. This is a result of have extra neutrons. Since Carbon is always going to be element #6, we can write Carbon in terms of its mass instead.

Carbon - 12Carbon - 14

Isotopes of Elements

IsotopesIsotopes are atoms that have the same number of protons ( are atoms that have the same number of protons (ZZ11= Z= Z22), but a ), but a

different number of neutrons (N). (different number of neutrons (N). (AA11 A A22))

Helium - 4Helium - 4

42He

Helium - 3Helium - 3

32He Isotopes of Isotopes of

heliumhelium

NuclidesBecause of the existence of so many isotopes, the term Because of the existence of so many isotopes, the term elementelement is sometimes confusing. The term is sometimes confusing. The term nuclidenuclide is better. is better.

A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes.

The following are best described as nuclides:The following are best described as nuclides:

32He

42He

126C

136C

Atomic Mass Unit, uOne One atomic mass unitatomic mass unit (1 u)(1 u) is equal to one-twelfth of the mass of the is equal to one-twelfth of the mass of the most abundant form of the carbon atom--most abundant form of the carbon atom--carbon-12carbon-12..

Atomic mass unit: 1 u = 1.6606 x 10-27 kg

Common atomic masses:

Proton: 1.007276 u Neutron: 1.008665 u

Electron: 0.00055 u Hydrogen: 1.007825 u

Einstein – Energy/Mass EquivalenceIn 1905, Albert Einstein publishes a 2nd major

theory called the Energy-Mass Equivalence in a paper called, “Does the inertia of a body depend on its energy content?”

Einstein – Energy/Mass EquivalenceLooking closely at Einstein’s equation we see that he

postulated that mass held an enormous amount of energy within itself. We call this energy BINDING ENERGY or Rest mass energy as it is the energy that holds the atom together when it is at rest. The large amount of energy comes from the fact that the speed of light is squared.

Energy Unit Check

2

2

2

2

2

22

s

mkgm

s

mkgWE

s

mkgNmaF

NmJouleFxWs

mkgJoulemcE

net

B

Mass Defect

The nucleus of the atom is held together by a STRONG NUCLEAR FORCE.

The more stable the nucleus, the more energy needed to break it apart.Energy need to break to break the nucleus into protons and neutrons is called the Binding EnergyEinstein discovered that the mass of the separated particles is greater than the mass of the intact stable nucleus to begin with.This difference in mass (m) is called the mass defect.

Particle Charge Mass (g) Mass (amu)

Proton +1 1.6727 x 10-24 g 1.007316

Neutron 0 1.6750 x 10-24 g 1.008701

Electron -1 9.110 x 10-28 g 0.000549

Mass Defect - Explained

The extra mass turns into energy holding the atom together.

Binding Energy is proportional to the mass defect .

The energy equivalent of 1 atomic mass unit υ

Is equal to 931.5 million electron volts ( MeV)

Mass Defect = Predicted – Isotope Mass or experimental value

Binding Energy = Mass Defect X 931.5 MeV

Avg BE = BE / A A = mass number

Exampe 2: The average atomic mass of Boron-11 is 11.009305 u. What is the mass of the nucleus of one boron atom in kg?

Electron: 0.00055 uElectron: 0.00055 u115B

= 11.009305= 11.009305

The mass of the nucleus is the atomic mass less the mass of Z = 5 The mass of the nucleus is the atomic mass less the mass of Z = 5 electrons:electrons:

Mass = 11.009305 u – 5(0.00055 u)Mass = 11.009305 u – 5(0.00055 u)

1 boron nucleus = 11.00656 u1 boron nucleus = 11.00656 u

-271.6606 x 10 kg11.00656 u

1 um

m = 1.83 x 10-26 kgm = 1.83 x 10-26 kg

2 8; 3 x 10 m/sE mc c 2 8; 3 x 10 m/sE mc c

Mass and EnergyRecall EinsteinRecall Einstein’’s equivalency formula for m and E:s equivalency formula for m and E:

The energy of a mass of 1 u can be found:The energy of a mass of 1 u can be found:

EE = (1 u) = (1 u)cc22 = = (1.66 x 10(1.66 x 10-27 -27 kg)(3 x 10kg)(3 x 1088 m/s) m/s)22

E = 1.49 x 10-10 J OrOr E = 931.5 MeV

When converting amu to When converting amu to energy:energy: 2 MeV

u931.5 c 2 MeVu931.5 c

Example 3: What is the rest mass energy of a proton (1.007276 u)?

EE = = mcmc22 = = (1.00726(1.00726 u)(931.5 MeV/u)u)(931.5 MeV/u)

Proton: E = 938.3 MeVProton: E = 938.3 MeV

Similar conversions show other rest mass energies:Similar conversions show other rest mass energies:

Electron: Electron: EE = 0.511 MeV = 0.511 MeVElectron: Electron: EE = 0.511 MeV = 0.511 MeV

Neutron: E = 939.6 MeVNeutron: E = 939.6 MeV

The Mass Defect

The The mass defectmass defect is the difference between the rest mass of a nucleus is the difference between the rest mass of a nucleus and the sum of the rest masses of its constituent nucleons.and the sum of the rest masses of its constituent nucleons.

The The mass defectmass defect is the difference between the rest mass of a nucleus is the difference between the rest mass of a nucleus and the sum of the rest masses of its constituent nucleons.and the sum of the rest masses of its constituent nucleons.

The whole is less than the sum of the parts!The whole is less than the sum of the parts! Consider the carbon-12 Consider the carbon-12 atom (12.00000 u):atom (12.00000 u):

Nuclear mass = Mass of atom – Electron Nuclear mass = Mass of atom – Electron masses = 12.00000 u – masses = 12.00000 u – 6(0.00055 u) = 11.996706 6(0.00055 u) = 11.996706 uuThe The nucleusnucleus of the carbon-12 atom has this mass. of the carbon-12 atom has this mass.

(Continued . . .)(Continued . . .)

Mass Defect (Continued)Mass of carbon-12 nucleus: Mass of carbon-12 nucleus: 11.99670611.996706

Proton: 1.007276 uProton: 1.007276 u Neutron: 1.008665 uNeutron: 1.008665 u

The nucleus contains 6 protons and 6 neutrons:The nucleus contains 6 protons and 6 neutrons:

6 p = 6(1.007276 u) = 6.043656 u6 p = 6(1.007276 u) = 6.043656 u6 n = 6(1.008665 u) = 6.051990 u6 n = 6(1.008665 u) = 6.051990 u

Total mass of parts: = Total mass of parts: = 12.095646 u12.095646 u

Mass defect Mass defect mmDD = 12.095646 u – 11.996706 u = 12.095646 u – 11.996706 u

mD = 0.098940 umD = 0.098940 u

The Binding Energy

The The binding energy binding energy EEBB of a nucleus is the energy required to of a nucleus is the energy required to

separate a nucleus into its constituent parts.separate a nucleus into its constituent parts.

The The binding energy binding energy EEBB of a nucleus is the energy required to of a nucleus is the energy required to

separate a nucleus into its constituent parts.separate a nucleus into its constituent parts.

EB = mDc2 where c2 = 931.5 MeV/u

The binding energy for the carbon-12 example is:The binding energy for the carbon-12 example is:

EEBB = (= (0.098940 u)(931.5 MeV/u)

EB = 92.2 MeVBinding EB for C-12:

Binding Energy per Nucleon

An important way of comparing the nuclei of atoms is finding their binding An important way of comparing the nuclei of atoms is finding their binding energy per nucleon:energy per nucleon:

An important way of comparing the nuclei of atoms is finding their binding An important way of comparing the nuclei of atoms is finding their binding energy per nucleon:energy per nucleon:

Binding energy per nucleon

MeV =

nucleonBE

A

For our C-12 example A = 12 and:For our C-12 example A = 12 and:

MeVnucleon

92.2 MeV7.68

12BE

A MeV

nucleon

92.2 MeV7.68

12BE

A

Formula for Mass DefectThe following formula is useful for mass defect:The following formula is useful for mass defect:

D H nm Zm Nm M D H nm Zm Nm M Mass defectMass defect mmDD

mmHH = 1.007825 u; m = 1.007825 u; mnn = 1.008665 u = 1.008665 u

Z is atomic number; N is neutron number; M is mass of atom Z is atomic number; N is neutron number; M is mass of atom (including electrons).(including electrons).

By using the mass of the hydrogen atom, you avoid the necessity of subtracting electron masses.

By using the mass of the hydrogen atom, you avoid the necessity of subtracting electron masses.

Example 4: Find the mass defect for the nucleus of helium-4. (M = 4.002603 u)

D H nm Zm Nm M D H nm Zm Nm M Mass defectMass defect mmDD

ZmZmHH = (2)(1.007825 u) = 2.015650 u = (2)(1.007825 u) = 2.015650 u

NmNmnn = (2)(1.008665 u) = 2.017330 u = (2)(1.008665 u) = 2.017330 u

MM = 4.002603 u (From nuclide tables) = 4.002603 u (From nuclide tables)

mmDD = (2.015650 u + 2.017330 u) - 4.002603 u = (2.015650 u + 2.017330 u) - 4.002603 u

mD = 0.030377 umD = 0.030377 u

42He

Example 4 (Cont.) Find the binding energy per nucleon for helium-4. (mD = 0.030377 u)

EB = mDc2 where c2 = 931.5 MeV/uEB = mDc2 where c2 = 931.5 MeV/u

EEBB = = (0.030377 u)(931.5 MeV/u) = (0.030377 u)(931.5 MeV/u) = 28.3 MeV28.3 MeV

MeVnucleon

28.3 MeV7.07

4BE

A MeV

nucleon

28.3 MeV7.07

4BE

A

A total of 28.3 MeV is required To tear apart the nucleons from the He-4 atom.

A total of 28.3 MeV is required To tear apart the nucleons from the He-4 atom.

Since there are four nucleons, we find thatSince there are four nucleons, we find that

Mass Defect – Example

Problem :

Helium -4 nucleus contains 2 protons( each with a mass of 1.0087u ) and 2 neutrons ( each has a mass of 1.0087 u)

The total predicted mass would be 4.0330 u .

Experimentally , the Helium -4 is found to be 4.0026u .

What is the Mass defect ?

What is the Binding energy in MeV?

RadioactivityWhen an unstable nucleus releases energy and/or

particles.

Radioactive DecayThere are 4 basic types of

radioactive decay Alpha – Ejected Helium Beta – Ejected Electron Positron – Ejected Anti-

Beta particle Gamma – Ejected Energy

You may encounter protons and neutrons being emitted as well

n

p

e

e

He

10

11

00

01

01

42

Alpha Decay

HeUPu 42

23692

24094

Alpha Decay Applications

?42

24195

AZHeAm

Americium-241, an alpha-emitter, is used in smoke detectors. The alpha particles ionize air between a small gap. A small current is passed through that ionized air. Smoke particles from fire that enter the air gap reduce the current flow, sounding the alarm.

Beta Decay

AceRa 22889

01

22888

There aren’t really any applications of beta decay other than Betavoltaics which makes batteries from beta emitters. Beta decay, did however, lead us to discover the neutrino.

Beta Plus Decay - Positron

ThePa 23090

01

23091

Isotopes which undergo this decay and thereby emit positrons include carbon-11, potassium-40, nitrogen-13, oxygen-15, fluorine-18, and iodine-121.

Beta Plus Decay Application - Positron emission tomography (PET) Positron emission tomography

(PET) is a nuclear medicine imaging technique which produces a three-dimensional image or picture of functional processes in the body. The system detects pairs of gamma rays emitted indirectly by a positron-emitting radionuclide (tracer), which is introduced into the body on a biologically active molecule. Images of tracer concentration in 3-dimensional space within the body are then reconstructed by computer analysis.

Gamma Decay

00

24094

24094 PuPu

Gamma Decay ApplicationsGamma rays are the most dangerous type of radiation

as they are very penetrating. They can be used to kill living organisms and sterilize medical equipment before use. They can be used in CT Scans and radiation therapy.

Gamma Rays are used to view stowaways inside of a truck. This technology is used by the Department of Homeland Security at many ports of entry to the US.

Significant Nuclear Reactions - Fusion

nHeHH 10

42

31

21

nuclear fusion is the process by which multiple like-charged atomic nuclei join together to form a heavier nucleus. It is accompanied by the release or absorption of energy.

Fusion Applications - IFEIn an IFE (Inertial Fusion Energy) power plant, many (typically

5-10) pulses of fusion energy per second would heat a low-activation coolant, such as lithium-bearing liquid metals or molten salts, surrounding the fusion targets. The coolant in turn would transfer the fusion heat to a power conversion system to produce electricity.

CW: PROBLEMS

COMPLETE THE EQUATIONS:1. 238 4

92U 2He + ?

What decay is this ?

2.234 234

90 Th 91 Pa + ?

What kind of decay is this ?

3. 64 o ?

29Cu + -1e

What process is the above equation ?

4.64 o

29 Cu 1 e + ?

What decay is this ?

1. 234

91 Th

Alpha particle decay

2. o

-1 e

Beta decay

3. 64

28 Ni

electron capture

3. Transmutation4.64

28 Ni

positron emission

Significant Nuclear Reactions - Fission

Nuclear fission differs from other forms of radioactive decay in that it can be harnessed and controlled via a chain reaction: free neutrons released by each fission event can trigger yet more events, which in turn release more neutrons and cause more fissions. The most common nuclear fuels are 235U (the isotope of uranium with an atomic mass of 235 and of use in nuclear reactors) and 239Pu (the isotope of plutonium with an atomic mass of 239). These fuels break apart into a bimodal range of chemical elements with atomic masses centering near 95 and 135 u (fission products).

energynKrBaUn 10

9236

14156

23592

10 3

Fission BombOne class of nuclear weapon, a fission

bomb (not to be confused with the fusion bomb), otherwise known as an atomic bomb or atom bomb, is a fission reactor designed to liberate as much energy as possible as rapidly as possible, before the released energy causes the reactor to explode (and the chain reaction to stop).

A nuclear reactor is a device in which nuclear chain fission reactions are initiated, controlled, and sustained at a steady rate, as opposed to a nuclear bomb, in which the chain reaction occurs in a fraction of a second and is uncontrolled causing an explosion.

Atomic Physics & Quantum Effects

AP Physics B

Blackbody radiation A blackbody absorbs all incident light rays.

All bodies, no matter how hot or cold, emit electromagnetic waves. We can see the waves emitted by very hot objects because they are within the visible spectrum (light bulb filament; red-hot metal). At lower temps we can’t see the waves but they are still there. For example, the human body emits waves in the infrared range. This is why we can use infrared detecting devices to “see” in the dark.

The distribution of energy in blackbody radiation is independent of the material from which the blackbody is constructed – it depends only on the temperature.

The diagrams below show the intensity of various frequencies of EM radiation emitted by blackbodies of various temperatures. Note that as the temperature increases, the energy emitted (area under curve) increases and the peak in the radiation shifts to higher frequencies.

This is important because classical physics predicts a completely different curve that increases to infinite intensity in the ultraviolet region. (thus called the Ultraviolet Catastrophe). The only way to make sense of this finding is by saying energy is quantized (Planck’s quantum hypothesis)