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General Physics (PHY 2140) Lecture 17 Modern Physics Atomic Physics Electron Clouds The Pauli Exclusion Principle Characteristic X-Rays Atomic Transitions Lasers and Holography Chapter 28 5/24/2017 http://www.physics.wayne.edu/~alan/2140Website/Main.htm 1 Lightning Review Last lecture: 1. Quantum physics Atomic Descriptions Atomic Spectra Bohr’s Atomic Theory Quantum Mechanics Quantum Numbers 2 r n , n 1, 2,3,... Ei E f hf mevr n , n 1, 2,3,... 1 1 RH 2 2 n f ni 1 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 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. 5/24/2017 2 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 5/24/2017 3 Electron Clouds 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 5/24/2017 4 radial probability distribution (r 2 2) = probability of finding electron at a distance r from the center of the nucleus 90% Probability contours showing relative size of orbitals 5/24/2017 5 5/24/2017 6 5/24/2017 7 Quantum Number Summary The values of n can increase from 1 in integer steps The values of ℓ can range from 0 to n-1 in integer steps The values of m ℓ can range from -ℓ to ℓ in integer steps 5/24/2017 8 28.9 The Pauli Exclusion Principle Recall Bohr’s model of an atom. Why don’t all the electrons stay on the lowest possible orbit? 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 This explains the electronic structure of complex atoms as a succession of filled energy levels with different quantum numbers 5/24/2017 9 Examples 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 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. 5/24/2017 10 The Periodic 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’s Exclusion Principle explains the configuration: 1s,2s,2p,3s,3p,4s,3d,4p,5s,4d 5/24/2017 11 Bit of history: Mendeleev’s original table 5/24/2017 12 Problem: electron configuration of O (a) Write out the electronic configuration of the ground state for oxygen (Z = 8). (b) Write out values for the set of quantum numbers n, l, ml, and ms for each of the electrons in oxygen. 5/24/2017 13 (a) Write out the electronic configuration of the ground state for oxygen (Z = 8). (b) Write out values for the set of quantum numbers n, l, ml, and ms for each of the electrons in oxygen. Given: Z=8 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 5/24/2017 14 QUICK QUIZ Krypton (atomic number 36) has how many electrons in its next to outer shell (n = 3)? (a) 2 (c) 8 (b) 4 (d) 18 (d). Krypton has a closed configuration consisting of filled n=1, n=2, and n=3 shells as well as filled 4s and 4p subshells. The filled n=3 shell (the next to outer shell in Krypton) has a total of 18 electrons, 2 in the 3s subshell, 6 in the 3p subshell and 10 in the 3d subshell. 5/24/2017 15 = 2n2 + 5/24/2017 Recall: Kr needs 36 electrons, the remainder are in the N shell. 28 16 Characteristic X-Rays When a metal target is bombarded by high-energy electrons, x-rays are emitted The x-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 x-rays 5/24/2017 17 Explanation of Characteristic X-Rays The details of atomic structure can be used to explain characteristic x-rays 5/24/2017 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 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 18 Modifications to Bohr’s Theory For atoms with a larger nuclear charge, but with a single electron (He+, Li2+, Be3+), we must modify the energy to be: me ke2 Z 2e4 Z 2 (13.6) En (eV) n 1, 2, 3, 2 2 2 2 n n (Note the difference is the inclusion of Z, the number of protons in the nucleus) 5/24/2017 19 Modifications to Bohr’s Theory, cont. 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 e4 1 1 2 2 n 1, 2, 3, 3 4 c n f ni 1 Or as: 5/24/2017 me ke2e4 1 1 Z 2 3 2 4 c n f ni 1 20 Moseley Plot λ is the wavelength of the K line K 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 5/24/2017 21 Problem: X-Rays from Nickel The K 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 K x-ray from a nickel target, Z=28. 5/24/2017 22 The K 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 K 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) n 2 eV EK (28 1) 2 (13.6 eV) 9.91103 eV 2 (13.6 eV) 3 EL (28 3) 2.13 10 eV 2 (2) hc E EL EK 7.78 keV Find: Thus, the wavelength is: hc (6.63 1034 J s)(3.00 108 m/s) E 7.78 keV(1.60 1016 J/keV) 1.60 1010 m 0.160 nm 5/24/2017 23 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 5/24/2017 24 Atomic Transitions – Stimulated Absorption The blue dots represent electrons When a photon with energy ΔE is absorbed, one electron jumps to a higher energy level 5/24/2017 These higher levels are called excited states ΔE = hƒ = E2 – E1 In general, ΔE can be the difference between any two energy levels 25 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 5/24/2017 26 Atomic Transitions – Stimulated Emission 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 5/24/2017 27 Population Inversion 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 of photons can result 5/24/2017 This situation is called a population inversion 28 Lasers To achieve laser action, three conditions must be met 5/24/2017 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 This is achieved by using reflecting mirrors 29 Production of a Laser Beam 5/24/2017 30 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) 5/24/2017 31 A More Detailed Diagram of a He-Ne Laser Operation 5/24/2017 32 Holography Holography is the production of three-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 5/24/2017 33 Holography, cont The two beams form a complex interference pattern on the photographic film It can be produced only if the phase relationship of the two 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 three-dimensional perspective 5/24/2017 34