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General Physics (PHY 2140)

Lecture 17Lecture 17Modern Physics

Atomic PhysicsElectron CloudsThe Pauli Exclusion PrincipleCharacteristic X-RaysAtomic TransitionsLasers and Holography

Chapter 28

http://www.physics.wayne.edu/~alan/2140Website/Main.htm 227/16/20077/16/2007

Lightning ReviewLightning Review

Last lecture:

1.1. Quantum physicsQuantum physicsAtomic DescriptionsAtomic DescriptionsAtomic SpectraAtomic SpectraBohrBohr’’s Atomic Theorys Atomic TheoryQuantum MechanicsQuantum MechanicsQuantum NumbersQuantum Numbers

i fE E hf− =, 1, 2,3,...em vr n n= =

2 2

1 1 1H

f i

Rn nλ

⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠

Review Problem: Suppose that the electron in the hydrogen atom obeyed classical rather then quantum mechanics. Why should such an atom emit a continuous rather then discrete spectrum?

If hydrogen obeyed classical physics, we would have no quantizedelectron orbits. Therefore the transitions between orbits (energy levels) could be arbitrarily large or small. This leads to a continuous spectrum of emitted light.

2 , 1,2,3,...r n nπ λ= =

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Electron CloudsElectron Clouds

The graph shows the solution The graph shows the solution to the wave equation for to the wave equation for hydrogen in the ground statehydrogen in the ground state

The curve peaks at the The curve peaks at the Bohr radiusBohr radiusThe electron is not The electron is not confined to a particular confined to a particular orbital distance from the orbital distance from the nucleusnucleus

The The probabilityprobability of finding the of finding the electron at the Bohr radius is a electron at the Bohr radius is a maximummaximum

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Electron CloudsElectron Clouds

The wave function for The wave function for hydrogen in the ground state is hydrogen in the ground state is symmetricsymmetric

The electron can be found The electron can be found in a spherical region in a spherical region surrounding the nucleussurrounding the nucleus

The result is interpreted by The result is interpreted by viewing the electron as a cloud viewing the electron as a cloud surrounding the nucleussurrounding the nucleus

The densest regions of the The densest regions of the cloud represent the highest cloud represent the highest probability for finding the probability for finding the electronelectron

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90% Probability contours showing relative size of orbitals

radial probability distribution (r 2 ψ2) = probability of finding

electron at a distance r from the center of the nucleus

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Quantum Number SummaryQuantum Number Summary

The values of n can increase from 1The values of n can increase from 1 in in integerinteger stepsstepsThe values of The values of ℓℓ can range from 0 to ncan range from 0 to n--1 in integer steps1 in integer stepsThe values of The values of mm ℓℓ can range from can range from --ℓℓ to to ℓℓ in integer stepsin integer steps

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28.9 The Pauli Exclusion Principle28.9 The Pauli Exclusion Principle

Recall BohrRecall Bohr’’s model of an atom. Why dons model of an atom. Why don’’t all the t all the electrons stay on the lowest possible orbit?electrons stay on the lowest possible orbit?

PauliPauli’’s exclusion principle:s exclusion principle: no two electrons in an atom no two electrons in an atom can ever be in the same quantum statecan ever be in the same quantum state

In other words, no two electrons in the same atom can have In other words, no two electrons in the same atom can have exactly the same values for n, exactly the same values for n, ℓℓ, , mm ℓℓ, and m, and mss

This explains the electronic structure of complex atoms This explains the electronic structure of complex atoms as a as a successionsuccession of filled energy levels with different of filled energy levels with different quantum numbersquantum numbers

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ExamplesExamples

1.1. Hydrogen (one electron), 1sHydrogen (one electron), 1s11

2.2. Helium (two electrons), 1sHelium (two electrons), 1s22

3.3. Lithium (three electrons), 1sLithium (three electrons), 1s222s2s11

See Table 28.4 for other the configurations of other eleSee Table 28.4 for other the configurations of other elements.ments.

1, 0, 0, 1/ 2sn m m= = = = ±

1, 0, 0, 1 21, 0, 0, 1 2

s

s

n m mn m m= = = = += = = = −

1, 0, 0, 1 21, 0, 0, 1 22, 0, 0, 1 2

s

s

s

n m mn m mn m m

= = = = += = = = −

= = = = ±

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The Periodic TableThe Periodic Table

The outermost electrons are The outermost electrons are primarily responsible for the primarily responsible for the chemical properties of the chemical properties of the atomatomMendeleev arranged the Mendeleev arranged the elements according to their elements according to their atomic masses and chemical atomic masses and chemical similaritiessimilaritiesThe electronic configuration of The electronic configuration of the elements explained by the elements explained by quantum numbers and Pauliquantum numbers and Pauli’’s s Exclusion Principle explains Exclusion Principle explains the configuration:the configuration:

1s,2s,2p,3s,3p,4s,3d,4p,5s,4d1s,2s,2p,3s,3p,4s,3d,4p,5s,4d

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Bit of history: MendeleevBit of history: Mendeleev’’s original tables original table

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Problem: electron configuration of OProblem: electron configuration of O

(a) Write out the electronic configuration of the ground state (a) Write out the electronic configuration of the ground state for oxygen (for oxygen (Z Z = 8). (b) Write out values for the set of = 8). (b) Write out values for the set of quantum numbers quantum numbers nn, , ll, , mmll,, and and mmss for each of the for each of the electrons in oxygen.electrons in oxygen.

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(a) Write out the electronic configuration of the ground state f(a) Write out the electronic configuration of the ground state for oxygen (or oxygen (Z Z = 8). (b) = 8). (b) Write out values for the set of quantum numbers Write out values for the set of quantum numbers nn, , ll, , mmll,, and and mmss for each of the for each of the electrons in oxygen.electrons in oxygen.

Given:

Z = 8

Find:

structure

Recall that the number of electrons is the same as the charge of the nucleus. Thus, we have 8 electrons.

1, 0, 0, 1 22, 0, 0, 1 22, 1, (0,1), 1 2

s

s

s

n m mn m mn m m

= = = = ±= = = = ±

= = = = ±

Thus, the electron configuration is:2 2 41 2 2s s p

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QUICK QUIZ

Krypton (atomic number 36) has how many electrons in its Krypton (atomic number 36) has how many electrons in its nextnextto outer shell (to outer shell (n n = 3)?= 3)?

(a) 2(a) 2 (b) 4(b) 4(c) 8(c) 8 (d) 18(d) 18

(d). Krypton has a closed configuration consisting of filled (d). Krypton has a closed configuration consisting of filled nn=1, =1, nn=2, =2, and and nn=3 shells as well as filled 4=3 shells as well as filled 4s s and 4and 4p p subshellssubshells. The filled . The filled nn=3 =3 shell (the shell (the next to outer shellnext to outer shell in Krypton) has a total of 18 electrons, 2 in Krypton) has a total of 18 electrons, 2 in the 3in the 3s s subshellsubshell, 6 in the 3, 6 in the 3p p subshellsubshell and 10 in the 3and 10 in the 3d d subshellsubshell..

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= 2n2

28+

Recall: Kr needs 36 electrons, the remainder are in the N shell.

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Characteristic XCharacteristic X--RaysRays

When a metal target is When a metal target is bombarded by highbombarded by high--energy energy electrons, xelectrons, x--rays are emittedrays are emittedThe xThe x--ray spectrum typically ray spectrum typically consists of a broad continuous consists of a broad continuous spectrum spectrum and a series of sharp and a series of sharp lineslines

The lines are dependent on The lines are dependent on the metalthe metalThe lines are called The lines are called characteristic xcharacteristic x--raysrays

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Explanation of Characteristic XExplanation of Characteristic X--RaysRays

The details of atomic structure can be used to explain The details of atomic structure can be used to explain characteristic xcharacteristic x--raysrays

A bombarding electron collides with an electron in the target A bombarding electron collides with an electron in the target metal that is in an inner shellmetal that is in an inner shellIf there is sufficient energy, the electron is removed from the If there is sufficient energy, the electron is removed from the target atomtarget atomThe vacancy created by the lost electron is filled by an electroThe vacancy created by the lost electron is filled by an electron n falling to the vacancy from a higher energy levelfalling to the vacancy from a higher energy levelThe transition is accompanied by the emission of a photon The transition is accompanied by the emission of a photon whose energy is equal to the difference between the two levelswhose energy is equal to the difference between the two levels

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Modifications to BohrModifications to Bohr’’s Theorys Theory

For atoms with a larger nuclear charge, but with a single For atoms with a larger nuclear charge, but with a single electron (Heelectron (He++, Li, Li2+2+, Be, Be3+3+), we must modify the energy to ), we must modify the energy to be:be:

(Note the difference is the inclusion of Z, the number of proton(Note the difference is the inclusion of Z, the number of protons in s in the nucleusthe nucleus))

2 2 4 2

2 2 2

(13.6) (eV) 1, 2, 3,2e e

nm k Z e ZE n

n n= − = − = …

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Modifications to BohrModifications to Bohr’’s Theory, cont.s Theory, cont.

Since the energy of an emitted photon is the difference Since the energy of an emitted photon is the difference between energy levels, we can write the wavelength of between energy levels, we can write the wavelength of such a photon as:such a photon as:

2 2 4

3 2 2

1 1 1 1, 2, 3,4e e

f i

m k Z e nc n nλ π

⎛ ⎞= − =⎜ ⎟⎜ ⎟

⎝ ⎠…

2 4

3 2 2

1 1 14

e e

f i

m k eZc n nλ π

⎛ ⎞= −⎜ ⎟⎜ ⎟

⎝ ⎠Or as:Or as:

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Moseley PlotMoseley Plot

λλ is the wavelength of the Kis the wavelength of the Kααlineline

KKαα is the line that is is the line that is produced by an electron produced by an electron falling from the L shell falling from the L shell (n=2) to the K shell (n=1)(n=2) to the K shell (n=1)

From this plot, Moseley was From this plot, Moseley was able to determine the Z values able to determine the Z values of other elements and produce of other elements and produce a periodic chart in excellent a periodic chart in excellent agreement with the known agreement with the known chemical properties of the chemical properties of the elementselements

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Problem: XProblem: X--Rays from NickelRays from Nickel

The KThe Kαα xx--ray is emitted when an electron undergoes a ray is emitted when an electron undergoes a transition form the L shell (n=2) to the K shell (n=1) in a transition form the L shell (n=2) to the K shell (n=1) in a metal. Calculate the wavelength of the Kmetal. Calculate the wavelength of the Kαα xx--ray from a ray from a nickel target, Z=28.nickel target, Z=28.

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The KThe Kαα xx--ray is emitted when an electron undergoes a transition form the ray is emitted when an electron undergoes a transition form the L shell (n=2) L shell (n=2) to the K shell (n=1) in a metal. Calculate the wavelength of thto the K shell (n=1) in a metal. Calculate the wavelength of the Ke Kαα xx--ray from a nickel ray from a nickel target, Z=28.target, Z=28.

Given:

Z = 28

Find:

λ

The atomic number for nickel is Z = 28. Using eq. 28.18 and 28.20 we have:

Thus, the wavelength is:

34 8

16

10

(6.63 10 J s)(3.00 10 m/s)7.78 keV(1.60 10 J/keV)

1.60 10 0.160 mm n

hcEγ

λ−

−

−

× ×= =

×

= × =

i

2 3(28 1) (13.6 eV) 9.91 10 eVKE = − − = − ×2 3

2

(13.6 eV)(28 3) 2.13 10 eV(2)LE = − − = − ×2

2

(13.6) eVeff

n

ZE

n= −

7.78 keVL KhcE E Eγ λ

= − = =

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Atomic Transitions Atomic Transitions –– Energy LevelsEnergy Levels

An atom may have many An atom may have many possible energy levelspossible energy levelsAt ordinary temperatures, most At ordinary temperatures, most of the atoms in a sample are in of the atoms in a sample are in the ground statethe ground stateOnly photons with energies Only photons with energies corresponding to differences corresponding to differences between energy levels can be between energy levels can be absorbedabsorbed

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Atomic Transitions Atomic Transitions –– Stimulated AbsorptionStimulated Absorption

The blue dots represent The blue dots represent electronselectronsWhen a photon with energy When a photon with energy ΔΔE is absorbed, one electron E is absorbed, one electron jumps to a higher energy jumps to a higher energy levellevel

These higher levels are These higher levels are called called excited statesexcited statesΔΔE = hE = hƒƒ = E= E22 –– EE11In general, In general, ΔΔE can be the E can be the difference between any two difference between any two energy levelsenergy levels

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Atomic Transitions Atomic Transitions –– Spontaneous EmissionSpontaneous Emission

Once an atom is in an excited Once an atom is in an excited state, there is a constant state, there is a constant probability that it will jump back probability that it will jump back to a lower state by emitting a to a lower state by emitting a photonphotonThis process is called This process is called spontaneous emissionspontaneous emission

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Atomic Transitions Atomic Transitions –– Stimulated EmissionStimulated Emission

An atom is in an excited stated An atom is in an excited stated and a photon is incident on itand a photon is incident on itThe incoming photon The incoming photon increases the probability that increases the probability that the excited atom will return to the excited atom will return to the ground statethe ground stateThere are two emitted There are two emitted photons, the incident one and photons, the incident one and the emitted onethe emitted one

The emitted photon is in The emitted photon is in exactly in phase with the exactly in phase with the incident photonincident photon

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Population InversionPopulation Inversion

When light is incident on a system of atoms, both stimulated When light is incident on a system of atoms, both stimulated absorption and stimulated emission are equally probableabsorption and stimulated emission are equally probableGenerally, a net absorption occurs since most atoms are in the Generally, a net absorption occurs since most atoms are in the ground stateground stateIf you can cause more atoms to be in excited states, a net emissIf you can cause more atoms to be in excited states, a net emission ion of photons can resultof photons can result

This situation is called a This situation is called a population inversionpopulation inversion

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LasersLasers

To achieve laser action, three conditions must be metTo achieve laser action, three conditions must be met

The system must be in a state of population inversionThe system must be in a state of population inversionThe excited state of the system must be a The excited state of the system must be a metastable statemetastable state

Its lifetime must be long compared to the normal lifetime of Its lifetime must be long compared to the normal lifetime of an excited statean excited state

The emitted photons must be confined in the system long The emitted photons must be confined in the system long enough to allow them to stimulate further emission from other enough to allow them to stimulate further emission from other excited atomsexcited atoms

This is achieved by using reflecting mirrorsThis is achieved by using reflecting mirrors

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Production of a Laser BeamProduction of a Laser Beam

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Laser Beam Laser Beam –– He Ne ExampleHe Ne Example

The energy level diagram for NeThe energy level diagram for NeThe mixture of helium and neon is The mixture of helium and neon is confined to a glass tube sealed at confined to a glass tube sealed at the ends by mirrorsthe ends by mirrorsA high voltage applied causes A high voltage applied causes electrons to sweep through the electrons to sweep through the tube, producing excited statestube, producing excited statesWhen the electron falls to EWhen the electron falls to E22 in in Ne, a 632.8 nm photon is emittedNe, a 632.8 nm photon is emitted

(3s(3s22 →→ 2p2p44))

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A More Detailed Diagram of a HeA More Detailed Diagram of a He--Ne Ne Laser OperationLaser Operation

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HolographyHolography

Holography is the production of Holography is the production of threethree--dimensional images of an dimensional images of an objectobjectLight from a laser is split at BLight from a laser is split at BOne beam reflects off the object One beam reflects off the object and onto a photographic plateand onto a photographic plateThe other beam is diverged by The other beam is diverged by Lens 2 and reflected by the Lens 2 and reflected by the mirrors before striking the filmmirrors before striking the film

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Holography, contHolography, cont

The two beams form a complex interference pattern on the The two beams form a complex interference pattern on the photographic filmphotographic film

It can be produced only if the phase relationship of the two wavIt can be produced only if the phase relationship of the two waves es remains constantremains constantThis is accomplished by using a laserThis is accomplished by using a laser

The hologram records the intensity of the light and the phase The hologram records the intensity of the light and the phase difference between the reference beam and the scattered beamdifference between the reference beam and the scattered beamThe image formed has a threeThe image formed has a three--dimensional perspectivedimensional perspective