Objectives. We will also get to know about the wavefunction and its use in developing the concept of the structure of atoms.

Similar documents
Three-dimensional systems with spherical symmetry

Lecture 1. time, say t=0, to find the wavefunction at any subsequent time t. This can be carried out by

Jackson 4.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Mechanics Physics 151

Mechanics Physics 151

Physics 2D Lecture Slides Dec 1. Vivek Sharma UCSD Physics

Jackson 3.3 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Physics 505 Electricity and Magnetism Fall 2003 Prof. G. Raithel. Problem Set 7 Maximal score: 25 Points. 1. Jackson, Problem Points.

and Slater Sum Rule Method * M L = 0, M S = 0 block: L L+ L 2

The Solutions of the Classical Relativistic Two-Body Equation

Advanced Quantum Mechanics

Angular Momentum Theory.

Relating Scattering Amplitudes to Bound States

= e2. = 2e2. = 3e2. V = Ze2. where Z is the atomic numnber. Thus, we take as the Hamiltonian for a hydrogenic. H = p2 r. (19.4)

PHYS 705: Classical Mechanics. Central Force Problems I

11) A thin, uniform rod of mass M is supported by two vertical strings, as shown below.

Mechanics Physics 151

Lecture 7: Angular Momentum, Hydrogen Atom

Quantum Mechanics II

Vector Spherical Harmonics and Spherical Waves

Discretizing the 3-D Schrödinger equation for a Central Potential

Doublet structure of Alkali spectra:

( ) into above PDE. ( ), wherec = 1

Energy Levels Of Hydrogen Atom Using Ladder Operators. Ava Khamseh Supervisor: Dr. Brian Pendleton The University of Edinburgh August 2011

Problem Set 10 Solutions

Foundations of Chemical Kinetics. Lecture 9: Generalizing collision theory

20th Century Atomic Theory - Hydrogen Atom

Unit 6 Lecture Number 36

Nuclear models: Shell model

Physics 221 Lecture 41 Nonlinear Absorption and Refraction

Nuclear and Particle Physics - Lecture 20 The shell model

Theorem on the differentiation of a composite function with a vector argument

3.23 Electrical, Optical, and Magnetic Properties of Materials

On- and off-shell Jost functions and their integral representations

Pressure in the Average-Atom Model

Physics 161 Fall 2011 Extra Credit 2 Investigating Black Holes - Solutions The Following is Worth 50 Points!!!

1 Spherical multipole moments

Scattering in Three Dimensions

Anyone who can contemplate quantum mechanics without getting dizzy hasn t understood it. --Niels Bohr. Lecture 17, p 1

A Relativistic Electron in a Coulomb Potential

WOODS-SAXON POTENTIAL IN THE PRESENCE OF A COSMIC STRING

Conducting fuzzy division by using linear programming

An Application of Bessel Functions: Study of Transient Flow in a Cylindrical Pipe

3. Electromagnetic Waves II

Physics 505 Homework No. 9 Solutions S9-1

3D-Central Force Problems I

The Schrödinger Equation in Three Dimensions

On the integration of the equations of hydrodynamics

The geometric construction of Ewald sphere and Bragg condition:

Chapter 6: Rotational and Rovibrational Spectra. A) General discussion of two- body problem with central potential

5.61 Physical Chemistry Lecture #23 page 1 MANY ELECTRON ATOMS

15 Solving the Laplace equation by Fourier method

3.012 Fund of Mat Sci: Bonding Lecture 5/6. Comic strip removed for copyright reasons.

A New Approach to General Relativity

c n ψ n (r)e ient/ h (2) where E n = 1 mc 2 α 2 Z 2 ψ(r) = c n ψ n (r) = c n = ψn(r)ψ(r)d 3 x e 2r/a0 1 πa e 3r/a0 r 2 dr c 1 2 = 2 9 /3 6 = 0.

Mechanics Physics 151

2 Lecture 2: The Bohr atom (1913) and the Schrödinger equation (1925)

1D2G - Numerical solution of the neutron diffusion equation

Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b.

1.2 Differential cross section

5.111 Lecture Summary #6 Monday, September 15, 2014

However, because the center-of-mass is at the co-ordinate origin, r1 and r2 are not independent, but are related by

PHYS 110B - HW #7 Spring 2004, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II September 15, 2012 Prof. Alan Guth PROBLEM SET 2

Nuclear size corrections to the energy levels of single-electron atoms

7.2. Coulomb s Law. The Electric Force

Key Questions. ECE 340 Lecture 4 : Bonding Forces and Energy Bands 1/28/14. Class Outline: v Crystal Diffraction Crystal Bonding

A Hartree-Fock Example Using Helium

-Δ u = λ u. u(x,y) = u 1. (x) u 2. (y) u(r,θ) = R(r) Θ(θ) Δu = 2 u + 2 u. r = x 2 + y 2. tan(θ) = y/x. r cos(θ) = cos(θ) r.

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

PROBLEM SET #1 SOLUTIONS by Robert A. DiStasio Jr.

221B Lecture Notes Scattering Theory I

As is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3.

Preliminary Exam: Quantum Physics 1/14/2011, 9:00-3:00

Introduction to Nuclear Forces

Question 1: The dipole

Chapter 3 Optical Systems with Annular Pupils

Homework 1 Solutions CSE 101 Summer 2017

Orbital Angular Momentum Eigenfunctions

= ρ. Since this equation is applied to an arbitrary point in space, we can use it to determine the charge density once we know the field.

EFFECTS OF FRINGING FIELDS ON SINGLE PARTICLE DYNAMICS. M. Bassetti and C. Biscari INFN-LNF, CP 13, Frascati (RM), Italy

Chem 453/544 Fall /08/03. Exam #1 Solutions

Phys 201A. Homework 5 Solutions

ON THE TWO-BODY PROBLEM IN QUANTUM MECHANICS

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

An Exact Solution of Navier Stokes Equation

Classical Mechanics Homework set 7, due Nov 8th: Solutions

Physics 235 Chapter 5. Chapter 5 Gravitation

arxiv: v1 [physics.gen-ph] 18 Aug 2018

The Law of Biot-Savart & RHR P θ

Many Electron Atoms. Electrons can be put into approximate orbitals and the properties of the many electron systems can be catalogued

Electrostatic Potential

Electrostatics. 1. Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increase?

MATH 172: CONVERGENCE OF THE FOURIER SERIES

CHAPTER 25 ELECTRIC POTENTIAL

PHYS 705: Classical Mechanics. Central Force Problems II

CHEM1101 Worksheet 3: The Energy Levels Of Electrons

Calculation of Quark-antiquark Potential Coefficient and Charge Radius of Light Mesons

Physics 862: Atoms, Nuclei, and Elementary Particles

Rigid Body Dynamics 2. CSE169: Computer Animation Instructor: Steve Rotenberg UCSD, Winter 2018

Transcription:

Modue "Atomic physics and atomic stuctue" Lectue 7 Quantum Mechanica teatment of One-eecton atoms Page 1 Objectives In this ectue, we wi appy the Schodinge Equation to the simpe system Hydogen and compae the outcome of the Quantum Theoy with the expeimenta esuts obtained fom diffeent expeiments. We wi aso get to know about the wavefunction and its use in deveoping the concept of the stuctue of atoms.

Page Why hydogen? One eecton atom and the simpest bound system in natue Refeing to the adjacent figue -1, the poton (+ve Chage) is at the cente and the eecton (-ve chage) is moving aound the cente. The attactive foce between them is Cooumbic in natue and the potentia is Z V = V ( x, y, z) = 4πε 0 Ze x + y + z x z y Y.Equation -1 Whee Z is the chage of the nuceus, Z = 1 fo hydogen. X Figue - 1 Since it is a two body pobem, we convet it to one body pobem by intoducing educed mass mm µ = m+ M Whee m and M ae the masses of the eecton and poton, espectivey. So, the tota enegy of the system = kinetic enegy + potentia engy 1 x + p y + p z ) + V ( x, y, z) = ( p µ E. Equation - Hee, p x, p y, and p z ae the inea momenta and E is the tota enegy.

Page 3 Fo the quantum mechanica teatment, we wi convet the cassica dynamica quantities (p x, p y, and p z ) to its coesponding quantum mechanica opeatos px i, py i, pz i x y z E i t Substituting in Equation, we get + + V + ( xyz,, ) = i. Equation -3 µ x y z t Hee, we intoduce the WAVEFUNTION to epesent the eecton. ( xyzt,,, ) Ψ=Ψ. Equation -4 The wavefunction contains the infomation about the position and the time evoution of the eectonic motion. We wi undestand moe about this ate. So opeating equation -, on the wavefuntion, ( xyzt,,, ) Ψ=Ψ, ( ) ( ) ( ) ( ) Ψ xyzt,,, Ψ xyzt,,, Ψ xyzt,,, Ψ xyzt,,, + + + V( xyz,, ) Ψ ( xyzt,,, ) = i µ x y z t Ψ Ψ+ VΨ= i µ t. Equation -5 is the Lapacian opeato. x y z = + + This Equation -5 is known as Time dependent Schodinge Equation.

Page 4 Since the potentia V ( x, y, z) does not depend on time t, and we ae inteested to evauate the enegy of the stationay (time independent) states, we can take this wavefunction as the expicit dependence of time such as iet/ ( xyzt,,, ) ψ ( xyze,, ) Ψ = Substituting this to Equation -5, we get µ + = ( xyz,, ) V ( xyz,, ) E ( xyz,, ) ψ ψ ψ This is the Time Independent Schodinge Equation.. Equation -6 Now we have to evauate the stationay state enegies ( E ) of the eecton in this system by soving this equation to expain the obseved specta of hydogen. Since the potentia is spheicay symmetic Ze Ze V ( x, y, z) = = = V ( ) 4πε x + y + z 4πε 0 0 Z We can convet Equation -6 to its spheica poa coodinate (Figue -) fom, µ + = (,, ) V ( ) (,, ) E (,, ) ψ θφ ψ θφ ψ θφ. Equation -7 Radia distance, Poa ange θ and X x θ z φ y Y Azimutha ange φ Figue - The fom of the Lapacian opeato in Spheica poa coodinates is 1 1 1 = + + sinθ θ θ sin θ φ sinθ

Page 5 Using the sepaation of the vaiabes, we can wite the wavefunction as the poduct fom ( ), Θ( ), Φ ( ) of the independent vaiabes R θ φ ψ(, θφ, ) = R( ) Θ( θ) Φ( φ) Substituting in equation 7, 1 R ΘΦ 1 1 sin R ΘΦ R ΘΦ θ V ( ) R ER µ + + + ΘΦ = ΘΦ sinθ θ θ sin θ φ Caying out the patia diffeentiation, ΘΦ R R Φ Θ sin R Θ Φ θ V ( ) R ER µ + + + ΘΦ = ΘΦ sinθ θ θ sin θ φ Mutipying the this equation by µ RΘΦ and eaanging we get, 1 R 1 Θ 1 Φ µ + sinθ + E V + ( ) = 0 R Θ sinθ θ θ Φ sin θ φ Sepaating the adia ( R( ) ) and angua ( ( θ), ( φ) Θ Φ ) pat 1 d dr µ 1 d dθ 1 d Φ + E V ( ) sinθ R d d = Θ sinθ dθ dθ Φ sin θ dφ Note hee that the patia deivative foms ae conveted to tota deivative fom. Now each equation shoud be equa to a constant, et s take as λ.

Page-6 So, the adia pat is 1 d dr µ + E V ( ) R d d = λ 1 d dr µ λ + ( E V ( ) ) R= 0 d d. Equation -8 And the angua pat is 1 d dθ 1 d Φ sinθ + λ = Θ sinθ dθ dθ Φ sin θ dφ sinθ 1 + = Θ dθ dθ Φ dφ dθ d Φ sinθ λ sin θ. Equation -9 Now, sepaating the poa and the azimutha pat sinθ d 1 sinθ dθ sin d Φ + λ θ = m =. Equation -10 Θ dθ dθ Φ dφ Hee, we have taken that both sides shoud be equa to a constant m. Note that we have thee independent equations fom equation 9 and 10.

Page 7 Now et us fist conside the azimutha pat (Equation -10) 1 d Φ = m Φ dφ 1 d Φ + m Φ= 0 Φ dφ The genea soution of this equation may be witten in the fom 1 im Φ m ( φ ) = e φ... Equation -11 π Which satisfies the othonomaity condition π 0 Φ ( φ) Φ ( φ) dφ = δ * m m' mm ' As the eigen functions must be singe vaued, i.e., Φ(0) = Φ(π) which means e im 0 im π = e And using Eue s fomua 1 = Cos( m π) + isin( m π) This is satisfied ony if m = 0, ±1, ±,.. Theefoe, acceptabe soutions ony exist when m can ony have cetain intege vaues, i.e. it is a quantum numbe. Thus, m is caed the magnetic quantum numbe in spectoscopy because it pays oe when atom inteacts with magnetic fieds.

Page 8 Now we wi discuss the poa pat in Equation 10 Reaanging, we get sinθ d dθ sinθ + λ sin θ = m Θ dθ dθ 1 1 sin d d ( θ ) d Θ m sin θ + λ Θ ( θ) = 0 d sin θ θ θ θ... Equation - Let us intoduce new vaiabe ω = Cosθ, the Equation-1 becomes, ( ω) d d Θ m (1 ) ω + λ Θ ( ω) = 0 dω dω 1 ω... Equation -13 The Equation-13 has singuaities at ω = ± 1, which may be eiminated by having a soution Θ in the fom of Θ= Then Equation-13 becomes / (1 ω ) s u s ω m dω dω 1 ω d u du (1 ω ) ω( s+ 1) + λ+ u = 0 Which can be witten as (1 ω d u du s m ) ( s 1) s s u 0 d ω ω dω λ + + + =... Equation -14 1 ω

Page-9 The ast singua tem in Equation-14 can be emoved by taking s = ± m 0, and then we have (1 ω d u ) ω du ( m 1) [ λ m( m 1) ] u 0 dω + dω + + =... Equation -15 Which is a egua equation and hence its seies soution may be witten as u = aω... Equation -16 = 0 Substituting in Equation-15, yieds the ecusion eation a + ( + m)( + m + 1) λ = a ( + 1)( + ) Requiing that the seies be (Equation-16) be imited by a cetain powe = q, i.e. equiing that it be a poynomia of ode q, we have to intoduce the condition a+ = 0, a 0 which equies λ = ( q+ m)( q+ m + 1) Whee q = 0, 1,, 3,. Now we intoduce = q+ m = 0,1,,3,... (obita quantum numbe) such that m λ = ( + 1)

Page-10 Then Equation-1 becomes 1 sin 17 ( θ ) d sin d Θ ( 1) m θ + + Θ ( θ) = 0 d d sin θ θ θ θ... Equation - This is we known associated Legende diffeentia equation with its soution as associated Legende poynomias m m Θ, = C P ( ω)... Equation -18 m Whee m m C is the nomaization facto and P ( ω) is the associated Legende poynomia defined by + m m / ( 1) m d ω P = (1 ω ) + m dω! m m / d = (1 ω ) P ( ω) m dω Whee P ( ω) is the odinay Legende Poynomia. This expession hods fo negative vaues of m aso P ( + m )! ( ω) = ( 1) P ( ω) ( m )! m m m Fom this we can estabish the ange of vaiation of the azimutha (magnetic) quantum numbe m = 0, ±1, ±, ±3,, ± The associated Legende Poynomias satisfy the othonomiization popety 1 1 ( + m )! P ( ω) P ( ω) dω = δ m m, + 1 ( m )! So, C m = + 1( m )! ( + m )!

Page-11 Then Equation-18 becomes Θ = m, + 1( m )! P ( + m )! m ( ω)... Equation -19 So, we have now two quantum numbes, namey, Obita () and magnetic (m ) = 0, 1,, 3,... m =, + 1,..., 0,..., 1, Depending of the vaue of and m, some of the Associated Legende poynomias ae P P 0 0 = 1 ( θ) ( ) = cosθ P = 1 cos 0 ± 1 1 1 P P = 1 3cos θ P = 1 cos θ cosθ 0 ± 1 = 1 cos θ ± 1/ 1/

Page-1 These functions ae 1.5 1.0 0.5 0.5 45 90 135 180 in Deg

Now, we can have the tota angua pat, fom Equation-11 and -19 Y ( θφ, ) =Θ ( θ) Φ ( φ) m m, m = + 1( m )! P 4 π ( + m )!.Equation -0 m (cos θ ) e im φ

Recap In this ectue, we have eant the quantum mechanica teatment of hydogen atom pobem. To do this, we have used the sepaation of vaiabes method and in this method we have witten the tota wavefunction as a poduct function of the adia pat and the angua pat. Since the couomb inteaction between the nuceus and the eecton is adiay symmetic, we have soved the angua pat sepaatey. Whie soving the angua pat of the wavefunction, we have intoduced two quantum numbes namey, azimutha o obita quantum numbe and magnetic quantum numbe m. (****Check the symbo in the ectue*********) In the next ectue we wi sove the adia pat of the wavefunction and wi constuct the tota wavefunction of the hydogen atom. We wi aso cacuate the enegy eves of the hydogen atom and wi compae it with the obseved spectum.