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2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger Equation for Hydrogen Atom b) Angular Momentum in Quantum Mechanics c) Quantum Number and Wave Functions Roadmap for Exploring Hydrogen Atom Today Lec. (Quantum) Monday Lec. (semi-classical) Review of Bohr’s Model of the Atom ܧ ܭ గఌ ସ గఌ గఌ “quantization condition”. ܮ ݎݒൌ ሺ ൌ 1,2. . ሻ ܨ ସ గఌ (Z=1 for hydrogen) ݎ ସ గఌ and ܧ ଷଶ ሺగఌ Monday In Class Problem 22. Prove it! Bohr radius . Ground State of H - 13.60 eV Bohr Model based on “ semiclassical theory”. Orbit quantized: Quantum Orbit deterministic: Classic Limits of Bohr’s Model Model based on Quantum Mechanics. Orbit replaced by probability (wavefunction)

“quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

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Page 1: “quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

2/28/2016

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PHYS 342 Modern Physics Atom II: Hydrogen Atom

Today Contents:

a) Schrodinger Equation for Hydrogen Atomb) Angular Momentum in Quantum Mechanicsc) Quantum Number and Wave Functions

Roadmap for Exploring Hydrogen Atom

Today Lec.

(Quantum)

Monday Lec.

(semi-classical)

Review of Bohr’s Model of the Atom

“quantization condition”. 1,2. .

(Z=1 for hydrogen)

and

Monday In Class Problem 22.

Prove it!

Bohr radius . Ground State of H

- 13.60 eV

Bohr Model based on “ semi‐classical theory”.Orbit quantized: QuantumOrbit deterministic: Classic

Limits of Bohr’s Model

Model based on Quantum Mechanics.Orbit replaced by probability(wavefunction)

Page 2: “quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

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Electron Wave Function(determined by

3D Schrodinger Equations)

Schrodinger Equation for Hydrogen Atom

3D wave function in spherical polar coordinates

Review of Solving Schrodinger Equation

The same procedure as you solve second‐order differentialequations: A y’’(x)+ B y(x)=01) Write down the equation.2) Find the general solution (GE) which contains somearbitrary constants.3) Find the particular solution (PE): Use the boundarycondition and the normalization condition to determine theenergy E and the constants in GE.

How to solve  one‐dimensional time independent Schrodinger equation for arbitrary U(x).

2 ) 0

Schrodinger Eq. in Spherical Polar Coordinates 3D wave function in Cartesian coordinates

0)()],,([2),,(),,(),,(22

2

2

2

2

2

xzyxUEz

zyxy

zyxx

zyx

3D wave function in spherical polar coordinates

rek

rerU

2

0

2

4)(

Separating the Hydrogen Equation

Radial Equation Angular Equation

Page 3: “quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

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Separating the Hydrogen Equation

.equantions aldifferenti partialfor

constants arbitrary are C and )1( ....0,1,2,3... φ llCl r

Radial Equation Angular Equation

Radial Equation of Hydrogen

radius.Bohr the is 4

number) quantum (orbital 1...3,2,1,0

number) quantum (principal ...3,2,1 function. Laguerre associated the is is

2

20

0 ea

nlnLn,l

Plots of Radial Wave Function

, )

Radial Probability Density for Hydrogen

,

https://www.youtube.com/watch?v=Q9Sl1PYSyOw(1:20-4:20)

Page 4: “quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

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Radial Probability Density for Hydrogen

1) For the larger n, the average distance between electron and nuclei is ????.

3) For the same n, if l is larger, the average distance between electron and nuclei is ???, electron has ???? negative potential energy.

2) For the larger n, the electron has ??? negative potential energy, meaning ???? the total energy.

In-class Problem 23. Discuss and find answer (larger, small, more, less…)

4) The different l states for the same n has the same total energy because E only depends on n, so the larger l means ???? kinetic energy.

Today Contents:

a) Schrodinger Equation for Hydrogen Atomb) Angular Momentum in Quantum Mechanicsc) Quantum Number and Wave Functions

Angular Equation of Hydrogen

)1( and .0,1,2..... llCl r

functions Legendre associated normalized are )(, mlP ...2,1,0 ,

21)( lmeF im

m

mechanics. quantum in momentumangular to related are )( and )(, mml FP

Angular shape of the wave function

Classical Picture of Angular Momentum

mechanics. quantum in momentumangular to related are )( and )( FPl

Orbits have the same energy, but have different angular momentum (the angular shape of the obits are different).

Page 5: “quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

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Angular Momentum in Quantum Mechanics

Orbit Angular Momentum in Hydrogen atom is determined by

two quantum number: l and ml=0,1,2,...n‐1

m=‐l,‐l+1,…0,1,2,...l

functions Legendre associated normalized are )(lP ....2,1,0 ,

21)( meF im

Different l and m = different angular momentum = different shapes of the wave function.

Plots of Angular Wave Function

p orbitall=1

d orbitall=2

s orbitall=0

p orbital, l=1

Different m determines the orientation of the double-lobe orbitals.

m=0 is

m=1 is

m=-1 is

Angular Probability Density

, ,

https://www.youtube.com/watch?v=LkPK-DuDWx0 (1417)

Page 6: “quantization condition”. . L I R N L L1,2..webphysics.iupui.edu/342/phy342sp16/Lec12.pdf2/28/2016 1 PHYS 342 Modern Physics Atom II: Hydrogen Atom Today Contents: a) Schrodinger

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Quantum Numbers

Quantum numbers that emerge from the wave functions of Schrodinger

1st n = energy level2nd  l= shape of orbital (s, p, d or f)3rd ml = orientation  of orbital 

Quantum Numbers Hydrogen Wave Function

Hydrogen Wave Function

ICP 24: For a hydrogen atom in the ground state (n=1,l=0,m=0), what is the probability of the electron between 1.00a0 and 1.01 a0.

(Hint: the wave function is smooth near a0, so it is not necessary to evaluate any integrals to solve this problem.The wave function is on Page 205)

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Space Density of Hydrogen Wave Function

, , , ,

sin

Space Density of Hydrogen Wave Function n=3

Hydrogen Atom Animation

http://www.youtube.com/user/terningphysics?feature=watch

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ICP 25: Find the directions in space in which the angular probability density of the l=2, m=1 electron in hydrogen has its maxima and minima.

( Hint: the angular wave function l=2 are described by the following five d wave functions. Which one is m=1?

, , to calculate P )