General Physics (PHY 2140)

Lightning Review Last lecture:

Lecture 17

1. Quantum physics 9 Atomic Descriptions 9 Atomic Spectra 9 Bohr’ Bohr’s Atomic Theory 9 Quantum Mechanics 9 Quantum Numbers

¾ Modern Physics 9Atomic Physics 9Electron Clouds 9The Pauli Exclusion Principle 9Characteristic X-Rays 9Atomic Transitions 9Lasers and Holography

http://www.physics.wayne.edu/~alan/2140Website/Main.htm

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λ

⎛ 1 1 ⎞ = RH ⎜ 2 − 2 ⎟ ⎜ n f ni ⎟ ⎝ ⎠

If hydrogen obeyed classical physics, we would have no quantized electron orbits. Therefore the transitions between orbits (energy levels) could be arbitrarily large or small. This leads to a continuous spectrum of emitted light. 1

Electron Clouds

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Electron Clouds

The graph shows the solution to the wave equation for hydrogen in the ground state „ The curve peaks at the Bohr radius „ The electron is not confined to a particular orbital distance from the nucleus The probability of finding the electron at the Bohr radius is a maximum

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me vr = n , n = 1, 2,3,...

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?

Chapter 28 7/16/2007

2π r = nλ , n = 1, 2,3,... Ei − E f = hf

The wave function for hydrogen in the ground state is symmetric „ The electron can be found in a spherical region surrounding the nucleus The result is interpreted by viewing the electron as a cloud surrounding the nucleus „ The densest regions of the cloud represent the highest probability for finding the electron

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radial probability distribution (r 2 ψ2) = probability of finding electron at a distance r from the center of the nucleus 90% Probability contours showing relative size of orbitals 7/16/2007

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Quantum Number Summary

The values of n can increase from 1 in integer steps The values of ℓ can range from 0 to nn-1 in integer steps The values of m ℓ can range from -ℓ to ℓ in integer steps 7/16/2007

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28.9 The Pauli Exclusion Principle

Examples

Recall Bohr’ Bohr’s model of an atom. Why don’ don’t all the electrons stay on the lowest possible orbit? Pauli’ Pauli’s exclusion principle: no two electrons in an atom can ever be in the same quantum state „

In other words, no two electrons in the same atom can have exactly the same values for n, ℓ, m ℓ, and ms

1. Hydrogen (one electron), 1s1

n = 1, = 0, m = 0, ms = ±1/ 2

2. Helium (two electrons), 1s2

n = 1, = 0, m = 0, ms = + 1 2 n = 1, = 0, m = 0, ms = −1 2

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

3. Lithium (three electrons), 1s22s1 n = 1, = 0, m = 0, ms = + 1 2 n = 1, = 0, m = 0, ms = −1 2 n = 2, = 0, m = 0, ms = ± 1 2 See Table 28.4 for other the configurations of other elements. elements.

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The Periodic Table

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Bit of history: Mendeleev’s original table

The outermost electrons are primarily responsible for the chemical properties of the atom Mendeleev arranged the elements according to their atomic masses and chemical similarities The electronic configuration of the elements explained by quantum numbers and Pauli’ Pauli’s Exclusion Principle explains the configuration: 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d 7/16/2007

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(a) Write out the electronic configuration of the ground state for for oxygen (Z (Z = 8). (b) Write out values for the set of quantum numbers n, l, ml, and ms for each of the electrons in oxygen.

Problem: electron configuration of O

Given: Z=8

(a) Write out the electronic configuration of the ground state for oxygen (Z (Z = 8). (b) Write out values for the set of quantum numbers n, l, ml, and ms for each of the electrons in oxygen.

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

n = 1, = 0, m = 0, ms = ± 1 2 n = 2, = 0, m = 0, ms = ± 1 2 n = 2, = 1, m = (0, 1), ms = ± 1 2

Find:

Thus, the electron configuration is:

1s 2 2s 2 2 p 4

structure

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QUICK QUIZ Krypton (atomic number 36) has how many electrons in its next to outer shell (n (n = 3)? (a) 2 (c) 8

= 2n2

(b) 4 (d) 18

+

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

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

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Characteristic X-Rays

Explanation of Characteristic X-Rays

When a metal target is bombarded by highhigh-energy electrons, xx-rays are emitted The xx-ray spectrum typically consists of a broad continuous spectrum and a series of sharp lines „ The lines are dependent on the metal „ The lines are called characteristic xx-rays

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The details of atomic structure can be used to explain characteristic xx-rays A bombarding electron collides with an electron in the target metal that is in an inner shell If there is sufficient energy, the electron is removed from the target atom The vacancy created by the lost electron is filled by an electron electron falling to the vacancy from a higher energy level The transition is accompanied by the emission of a photon whose energy is equal to the difference between the two levels

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Modifications to Bohr’s Theory

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Modifications to Bohr’s Theory, cont.

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

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

me ke2 Z 2 e 4 Z 2 (13.6) En = − =− (eV) n = 1, 2, 3, … 2 2 n2 2 n

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λ

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

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Or as: 7/16/2007

=

me ke2 Z 2 e 4 ⎛ 1 1 ⎞ ⎜⎜ 2 − 2 ⎟⎟ n = 1, 2, 3, … 3 4π c ⎝ n f ni ⎠

1

λ

=Z

me ke2 e 4 ⎛ 1 1 ⎞ ⎜ − ⎟ 4π c 3 ⎜⎝ n 2f ni2 ⎟⎠ 20

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Moseley Plot

Problem: X-Rays from Nickel

λ is the wavelength of the Kα line „ Kα is the line that is produced by an electron falling from the L shell (n=2) to the K shell (n=1) From this plot, Moseley was able to determine the Z values of other elements and produce a periodic chart in excellent agreement with the known chemical properties of the elements

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

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The Kα x-ray is emitted when an electron undergoes a transition form the L shell (n=2) to the K shell (n=1) in a metal. Calculate the wavelength of the the Kα x-ray from a nickel target, Z=28. The atomic number for nickel is Z = 28. Using eq. 28.18 and 28.20 we have: Given: Z = 28 En = −

Z eff2 (13.6) n2

eV

EK = −(28 − 1) 2 (13.6 eV) = −9.91×103 eV (13.6 eV) EL = −(28 − 3) 2 = −2.13 ×103 eV 2 (2) hc Eγ = EL − EK = 7.78 keV =

λ

Find: λ

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Atomic Transitions – Energy Levels An atom may have many possible energy levels At ordinary temperatures, most of the atoms in a sample are in the ground state Only photons with energies corresponding to differences between energy levels can be absorbed

Thus, the wavelength is:

λ=

hc (6.63 ×10−34 Jis)(3.00 ×108 m/s) = Eγ 7.78 keV(1.60 ×10−16 J/keV)

= 1.60 ×10−10 m = 0.160 nm 7/16/2007

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Atomic Transitions – Stimulated Absorption

Atomic Transitions – Spontaneous Emission Once an atom is in an excited state, there is a constant probability that it will jump back to a lower state by emitting a photon This process is called spontaneous emission

The blue dots represent electrons When a photon with energy ΔE is absorbed, one electron jumps to a higher energy level „

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These higher levels are called excited states ΔE = hƒ hƒ = E2 – E1 In general, ΔE can be the difference between any two energy levels

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Atomic Transitions – Stimulated Emission

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Population Inversion

An atom is in an excited stated and a photon is incident on it The incoming photon increases the probability that the excited atom will return to the ground state There are two emitted photons, the incident one and the emitted one „ The emitted photon is in exactly in phase with the incident photon

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When light is incident on a system of atoms, both stimulated absorption and stimulated emission are equally probable Generally, a net absorption occurs since most atoms are in the ground state If you can cause more atoms to be in excited states, a net emission emission of photons can result „

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This situation is called a population inversion

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Lasers

Production of a Laser Beam

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

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The system must be in a state of population inversion The excited state of the system must be a metastable state Its lifetime must be long compared to the normal lifetime of an excited state The emitted photons must be confined in the system long enough to allow them to stimulate further emission from other excited atoms ¾ This is achieved by using reflecting mirrors

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A More Detailed Diagram of a HeHe-Ne Laser Operation

Laser Beam – He Ne Example The energy level diagram for Ne The mixture of helium and neon is confined to a glass tube sealed at the ends by mirrors A high voltage applied causes electrons to sweep through the tube, producing excited states When the electron falls to E2 in Ne, a 632.8 nm photon is emitted (3s2 → 2p4)

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Holography

Holography, cont

Holography is the production of threethree-dimensional images of an object Light from a laser is split at B One beam reflects off the object and onto a photographic plate The other beam is diverged by Lens 2 and reflected by the mirrors before striking the film

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

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It can be produced only if the phase relationship of the two waves waves remains constant This is accomplished by using a laser

The hologram records the intensity of the light and the phase difference between the reference beam and the scattered beam The image formed has a threethree-dimensional perspective

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