Hydrogen atom

HYD

MOST IMPORTANT SCIENCE TABLE

ALSO A BRIEF HISTORY OF CHEMISTRY

For Hydrogen-like atoms H= -1/2 2∇ - Z/rH = Enl ; = Rnl(r)Y lm( )),,( ϕθrΨ ),,( ϕθrΨ Ψ ϕθ ,

ATOMIC HYDROGEN Atomic No. 1, 1s1 , Atomic Radius 78pm

History of HYD(Gr. hydro: water, and genes: forming)

Hydrogen was prepared many years before it was recognized as a di i b b C di h i 1776 N d b L i idistinct substance by Cavendish in 1776. Named by Lavoisier, hydrogen is the most abundant of all elements in the universe. Hydrogen is estimated to make up more than 90% of all the atoms --three quarters of the mass of the universe! This element is found in the stars and plays an important part in powering thefound in the stars, and plays an important part in powering the universe through both the proton-proton reaction and carbon-nitrogen cycle. Stellar hydrogen fusion processes release massive amounts of energy by combining hydrogens to form Helium. The heavier elements were originally made from hydrogen atoms or from other elementswere originally made from hydrogen atoms or from other elements that were originally made from hydrogen atoms.

CARRIER OF ENERGY IN THE UNIVERSECARRIER OF ENERGY IN THE UNIVERSE

Hydrogen Atom

Atom is a 3D object, and the electron motion is three-dimensional. We’ll start with the simplest casej , p- a hydrogen atom. An electron and a proton (nucleus) are bound by the central-symmetric Coulombinteraction. Because mp>>me, we neglect the proton motion (the reduced mass is very close to me, thecenter of mass of this system is ~2000 times closer to the proton than to the electron). Thus, we cant t thi bl th ti f l t i th 3D t l t i C l b t ti ltreat this problem as the motion of an electron in the 3D central-symmetric Coulomb potential.

Three dimensions – we expect that the motion will becharacterized with three quantum numbers (in fact, therewill be the fourth one, the electron spin, which

d h “i l d f f d ” f

( )2

- eV V r= =

corresponds to the “internal degree of freedom” for anelectron; the spin will be added later “by hand”, it is notdescribed by the non-relativistic quantum mechanics).

( )04

V V rrπε

E < 0 – bound motion, discrete spectrum, E > 0 – unbound motion, continuous spectrum

The task: solve the t-independent S.Eq. for E<0, find the energy eigenvalues (the spectrum) and eigenfunctions (stationary states).

Orbital Motion in Classical Mechanicsr v We consider the case of a central force: the force is directed along the line that

( )

connects the electron and proton or, in our Figure, the electron and the coordinate origin.

Coulomb interaction is responsible for acceleration:2 2

3 204

e mvF r rr rπε

= − = −2mvF

r⎛ ⎞

=⎜ ⎟⎝ ⎠

( )2 22 2

2 2 2r tp pmv pK

m m+

= = =Kinetic energy:

The angular momentum L r p= ×

pr and pt – the radial and tangential components of the momentum

( )L r p r p p r p= × = × + = ×g L r p ( )r t tL r p r p p r p= × = × + = ×

( )2 2 2 2

22 2 2r t r

p p p LK+

= = + For a circularmotion:

2

22LKmr

=22 2 2m m mr motion: 2mr

In general, a non-zero torque leads to the time dependence of L: dLdt

τ =

0dL dpr r Fdt dt

τ = = × = × =

( )is antiparallel tor F

In the field of a central force, the torque is zero, and the angular momentum about the

center is conserved:

This makes L especially useful for analyzing the central force motion.

( )is antiparallel to r F

Quantum-Mechanical ApproachAs usual, we start with the time-dependent Schrödinger Equation (but now it is 3D case):

( ) ( ) ( ) ( ) ( ) ( )2 2 22

2 2 2

, , , , , , , , , , , ,, , , , ,

2x y z t x y z t x y z t x y z t

i V x y z x y z tt m x y z

⎡ ⎤∂Ψ ∂ Ψ ∂ Ψ ∂ Ψ= − + + + Ψ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦

2 2

( )2 2

2 2 200

, ,44

e eV x y zrx y z πεπε

= − = −+ +

- potential energy of Coulomb interaction between electron and proton

L l i ∇2 d d d⎛ ⎞

2 2 2 2

2 2 2ˆ ( , , )

2d d dH V x y zd d d

⎛ ⎞≡ − + + +⎜ ⎟

⎝ ⎠

( ) ( ), ˆ ,x t

i H x tt

∂Ψ= Ψ

∂

Laplacian ∇2ˆ d d dp

i dx dy dz i⎛ ⎞

= + + = ∇⎜ ⎟⎝ ⎠

2 2 22m dx dy dz⎜ ⎟⎝ ⎠

K operator V operator

( )t∂

The potential is time-independent, thus we can separate time and space variables:

( )dT t iE⎛ ⎞( ) ( )ˆ , , , ,H x y z E x y zψ ψ=( ) ( )dT t

i ET tdt

=

- solution of the t-independent S Eq gives us the eigenfunctions (the orthogonal basis ψ ) and

( ) ( ), expEiEtx t xψ −⎛ ⎞Ψ = ⎜ ⎟

⎝ ⎠

- solution of the t-independent S. Eq. gives us the eigenfunctions (the orthogonal basis ψi) and eigenvalues (spectrum Ei) of the Hamiltonian. Eigenfunctions correspond to the stationary states with

a well-defined energy.

Spherical Polar CoordinatesThe Coulomb potential has central symmetry (U(r)depends on neither θ nor φ ). It’s to our advantage to usespherical polar coordinates (this will allow us toseparate variables).

2 2 2r x y z= + +

z

- distance from origin to point P

1

2 2 2cos z

x y zθ −=

+ +

y

- zenith angle

1tan yx

φ −= - azimuth angle

- a (small) volume element used for integration (no relation to the torque

τ)

t-independent Schrödinger Equation in Polar Coordinates

( ) ( ) ( )2 2 22 x y z x y z t x y z tψ ψ ψ⎡ ⎤∂ ∂ ∂

( ) ( ) ( )2 2 2x y z x y z x y zψ ψ ψ∂ ∂ ∂

( ) ( ) ( ) ( ) ( ) ( )2 2 2

, , , , , , , ,, , , , , ,

2x y z x y z t x y z t

V x y z x y z E x y zm x y z

ψ ψ ψψ ψ

⎡ ⎤∂ ∂ ∂− + + + =⎢ ⎥∂ ∂ ∂⎣ ⎦

( ) ( ) ( )2 2 2

, , , , , ,x y z x y z x y zx y z

ψ ψ ψ∂ ∂ ∂+ +

∂ ∂ ∂

( ) ( ) ( )22

2 2 2 2 2

, , , , , ,1 1 1sinsin sin

r r rr

r r r r rψ θ φ ψ θ φ ψ θ φ

θθ θ θ θ φ

∂ ∂ ∂⎡ ⎤ ⎡ ⎤∂ ∂+ +⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦

( ) ( )2

2 , , , , 0m E V r rθ φ ψ θ φ+ − =⎡ ⎤⎣ ⎦

Af l i l i b 2 i θAfter multiplying by r2sinθ

2 22 2 2 22sin sin sin sin 0m er r Eψ ψ ψθ θ θ θ ψ

⎡ ⎤∂ ∂ ∂ ∂ ∂⎡ ⎤ ⎡ ⎤+ + + +⎢ ⎥⎢ ⎥ ⎢ ⎥ 2 20

sin sin sin sin 04

r r Er r r

θ θ θ θ ψθ θ φ πε

+ + + + =⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Separation of Variables2 2

2 2 2 22m eψ ψ ψ ⎡ ⎤∂ ∂ ∂ ∂ ∂⎡ ⎤ ⎡ ⎤2 2 2 22 2

0

2sin sin sin sin 04

m er r Er r r

ψ ψ ψθ θ θ θ ψθ θ φ πε

⎡ ⎤∂ ∂ ∂ ∂ ∂⎡ ⎤ ⎡ ⎤+ + + + =⎢ ⎥⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦ ⎣ ⎦

For a spherically-symmetric potentials (V=V(r)), one canseparate the variables by using a trial function ( ) ( ) ( ) ( ), ,r R rψ θ φ θ φ= Θ Φ

2 2 2i i 1 2d dR d d dθ θ ⎡ ⎤Θ Φ⎡ ⎤ ⎡ ⎤2 2 22 2 2

2 20

sin sin 1 2sin sin 04

d dR d d d m er r ER dr dr d d d rθ θ θ θ

θ θ φ πε⎡ ⎤Θ Φ⎡ ⎤ ⎡ ⎤+ + + + =⎢ ⎥⎢ ⎥ ⎢ ⎥Θ Φ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

To show that the variables are separable, let’s rewrite:

2 2 22 2 2sin sin 2 1sin sind dR d d m e dr r Eθ θ θ θ

⎡ ⎤Θ Φ⎡ ⎤ ⎡ ⎤+ + + = −⎢ ⎥⎢ ⎥ ⎢ ⎥ 2 20

sin sin4

r r ER dr dr d d r d

θ θθ θ πε φ

+ + + =⎢ ⎥⎢ ⎥ ⎢ ⎥Θ Φ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Since the left and right sides depends on different variables, both 221 d mΦ

− =parts should be equal to the same constant: 2 lmdφ

− =Φ

222 21 2 1 ilmd dR m e d dθ

⎡ ⎤ Θ⎡ ⎤ ⎡ ⎤⎢ ⎥

Divide by i 2θ d 2 2

2 20

1 2 1 sin4 sin sin

lmd dR m e d dr r ER dr dr r d d

θπε θ θ θ θ

⎡ ⎤ Θ⎡ ⎤ ⎡ ⎤+ + = −⎢ ⎥⎢ ⎥ ⎢ ⎥Θ⎣ ⎦ ⎣ ⎦⎣ ⎦sin2θ and regroup:

Separation of Variables (cont’d)

( )2

2 21 2d dR m e⎡ ⎤⎡ ⎤Thus( )2 2

20

1 2 14

d dR m er r E l lR dr dr rπε

⎡ ⎤⎡ ⎤ + + = +⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦2 1m d ∂Θ⎡ ⎤

Equation for R(r)

21 d Φ

( )2

1 sin 1sin sin

lm d l ld

θθ θ θ θ

∂Θ⎡ ⎤− = +⎢ ⎥Θ ∂⎣ ⎦Equation for Θ(θ)

22

1l

d mφΦ

− =Φ ∂ Equation for Φ(φ)

Or, in a more conventional

form: ( )22

2 2 2

11 2 04

l ld dR m er E Rr dr dr r rπε

⎡ ⎤+⎛ ⎞⎡ ⎤ + + − =⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦04r dr dr r rπε⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

( )2

2

1 sin 1 0lmd l lθ⎡ ⎤∂Θ⎡ ⎤ + + − Θ=⎢ ⎥⎢ ⎥⎣ ⎦( ) 2sin 1 0

sin sinl l

dθ

θ θ θ θ+ + Θ⎢ ⎥⎢ ⎥∂⎣ ⎦ ⎣ ⎦

22 0d mΦ

+ Φ=2 0lmφ

+ Φ =∂

The initial equation in 3 polar coordinates has been separated into threeone dimensional equations.

22

2

d mdϕΦ

= − Φdϕ

21 sin ( 1) 0d d mθ⎛ ⎞Θ Θ

− + + Θ =⎜ ⎟ 2sin ( 1) 0sin sind d

θθ θ θ θ

+ + Θ =⎜ ⎟⎝ ⎠

( 1)1 2d d R +⎛ ⎞ { }22 2 2

( 1)1 2 ( ) 0d d Rr R E V r Rr d r d r r

μ+⎛ ⎞− + − =⎜ ⎟

⎝ ⎠

Solve Φ equation. Find it is good for only certain values of m. (0,+-1,..+- ℓ )

Solve Θ equation. Find it is good for only certain values of β=ℓ(ℓ+1); ℓ =0,1,2…]q g y β ( ); , , ]

Solve R equation. Find it is good for only certain values of E. Quantization with

Principal quantum number “n” ; n=1,2,3,….; ℓ=n-1 ]p q ; , , , ; ]

Copyright – Michael D. Fayer, 2007

( 1)β = +

1 ( 1) 2d dR μ+⎛ ⎞ ⎡ ⎤

Since , we have

( )22 2 2

1 ( 1) 2 ( ) 0d dRr E V r Rr dr dr r

μ+⎛ ⎞ ⎡ ⎤+ − + − =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

2

0

( )4ZeV r

rπε= − The potential only enters into the R(r) equation. Z is the

charge on the nucleus. One for H atom. Two for He+, etc.

2 2 Eμα

Make the substitutions

22

μα = −

2Zeμλ = 204

λπε α

=

Introduce the new independent variable

2 rρ α=

p

ρ is the the distance variable in units of 2α.


Making the substitutions andwith

( ) ( )S R rρ =yields

⎛ ⎞ ⎛ ⎞22 2

1 1 ( 1) 04

d dS Sd d

λρρ ρ ρ ρ ρ

⎛ ⎞ ⎛ ⎞++ − − + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠0 ρ≤ ≤ ∞

To solve - look at solution for large ρ, (like H. O.).r →∞

Consider the first term in the equation aboveConsider the first term in the equation above.2

2 22 2 2

1 1 2d dS d S dSd d d d

ρ ρ ρρ ρ ρ ρ ρ ρ

⎛ ⎞ ⎛ ⎞⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ρ ρ ρ ρ ρ ρ⎝ ⎠ ⎝ ⎠⎝ ⎠

2

2

2dd

dSd

S= +2d dρ ρ ρ

This term goes to zero asr →∞r →∞

The terms in the full equation divided by ρ and ρ2 also go to zero as . r →∞Copyright – Michael D. Fayer, 2007

Then, as r →∞

2 1d S2

2

1 .4

d S Sdρ

=

/ 2S e ρ+=/ 2S e ρ−=

The solutions are

This blows up asNot acceptable wavefunction.

r →∞

The full solution is

/ 2( ) ( )S e Fρρ ρ−=( ) ( )S e Fρ ρ

⎛ ⎞ ⎛ ⎞

Substituting in the original equation, dividing by and rearranging gives/ 2e ρ−

2

2 ( 1) 11 0F F Fλρ ρ ρ ρ

⎛ ⎞ ⎛ ⎞+′′ ′+ − + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0 ρ≤ ≤ ∞

The underlined terms blow up at ρ = 0. Regular point.Copyright – Michael D. Fayer, 2007

Singularity at ρ = 0 - regular point,to remove, substitute

( ) ( )F Lρ ρ ρ=

′′ ′

Gives

(2( 1) ) ( 1) 0.L L Lρ ρ λ′′ ′+ + − + − − =

Equation for L. Find L, get F. Know F, have . ( ) ( )S R rρ =q , g , ( ) ( )ρ

Solve using polynomial method.


20 1 2( )L a a a aν

νν

ρ ρ ρ ρ= = + + +∑ Polynomial expansion for L.Get L' and L'' by term by termν Get L and L by term by termdifferentiation.

Following substitution, the sum of all the terms in all powers of ρ equal 0.

00 1{ } ( 1) 2( 1) 0a aρ λ − − + + =

o ow g subs u o , e su o e e s powe s o ρ equ 0.The coefficient of each power must equal 0.

0 1{ } ( 1) 2( 1) 0a aρ λ + + =

[ ]11 2{ } ( 1 1) 4( 1) 2 0a aρ λ − − − + + + =

Note not separate oddand even series.

[ ]22 3{ } ( 1 2) 6( 1) 6 0a aρ λ − − − + + + =

Recursion formula

1( 1 )

[2( 1)( 1) ( 1)]aa ν

νλ ν

ν ν ν+− − − −

=+ + + +

Recursion formulaGiven a0, all other terms coefficientsdetermined. a0 determined bynormalization condition.[2( 1)( 1) ( 1)]ν ν ν+ + + + normalization condition.


1( 1 )

[2( 1)( 1) ( 1)]aa ν

νλ ν

ν ν ν+− − − −

=+ + + +

Provides solution to differential equation,but not good wavefunction if infinite[2( 1)( 1) ( 1)]ν ν ν+ + + + gnumber of terms.

Need to break off after the n' term by takingy g

1 0nλ ′− − − =

or integersor

with ' 1n n nλ = = + +

ege s

n is an integer.

n' radial quantum numbern total quantum number

1 s orbital ' 0, 02 s, p orbitals ' 1, 0 or ' 0, 1

n n ln n l n l= = == = = = =2 s, p orbitals 1, 0 or 0, 1

3 s, p, d orbitals ' 2, 0 or ' 1, 1 or ' 0, 2n n l n ln n l n l n l= = = = = = =


Thus,

/ 2( ) ( )R r e Lρ ρ ρ−=

ithwithL(ρ)

defined by the recursion relationdefined by the recursion relation,

and

' 1n

n nλ == + +

integers


1, 2, 3,n nλ= =

2

24Zeμλ

πε α=

22

2 Eμα = −

04πε α

2 42 2 Z en μλ= =

2

2 2 2032

nE

λπ ε

= = −

2 42 4

2 2 208nZ eEh n

με

= − Energy levels of the hydrogen atom.0

Z is the nuclear charge 1 for H; 2 for He+ etcZ is the nuclear charge. 1 for H; 2 for He+, etc.


20

0 2

hae

επμ

= 110 5.29 10a m−= ×

Bohr radius - characteristic length in H atom problem.

I t f B h di2 2

28nZ eE = −

In terms of Bohr radius

20 08 a nπε

Lowest energy, 1s, ground state energy, -13.6 eV.

-1109 677cmR

Rydberg constant

109,677cmHR =

2

2n HZE R hc= − 2n Hn

4em eR = Rydberg constant if proton had infinite mass.2 308

Rh cε∞ =

y g pReplace μ with me. R∞ = 109,737 cm-1.


( )ϕΦ ( )R r

Have solved three one-dimensional equations to get

( )θΘ( )m ϕΦ ( )nR r

The total wavefunction is

( )m θΘ

( , , ) ( ) ( ) ( )n m m m nr R rϕ θ ϕ θΨ = Φ Θ

1 2 31, 2, 3n =

1, 2, 0n n= − −

, 1m = − ⋅ ⋅ ⋅ −


• First few radial wave functions Rnℓ

Hydrogen Atom Radial Wave Functionsnℓ

• Subscripts on R specify the values of n and ℓ.

21

Solution of the Angular and Azimuthal S gEquations

• The solutions for Eq (7.8) are .• Solutions to the angular and azimuthal equations are linked because both

have mℓhave mℓ.• Group these solutions together into functions.

spherical harmonics---- spherical harmonics

22

Normalized Spherical HarmonicsNormalized Spherical Harmonics

23

Solution of the Angular and Azimuthal gEquations

• The radial wave function R and the spherical harmonics Ydetermine the probability density for the various quantum states. The total wave function depends on n ℓ and mℓThe total wave function depends on n, ℓ, and mℓ. The wave function becomes

24

Principal Quantum Number nPrincipal Quantum Number n• It results from the solution of R(r) . Because R(r) includes the potential

energy V(r).The result for this quantized energy is

• The negative means the energy E indicates that the electron and proton are bound together.

25

O bit l A l M t Q tOrbital Angular Momentum Quantum Number ℓ

• It is associated with the R(r) and f(θ) parts of the wave function• It is associated with the R(r) and f(θ) parts of the wave function.

• Classically, the orbital angular momentum with L =Classically, the orbital angular momentum with L mvorbitalr.

• ℓ is related to L by .

I ℓ 0• In an ℓ = 0 state, .

It disagrees with Bohr’s semiclassical “planetary” model of electrons orbiting a nucleus L = nħ.

26

O bit l A l M t Q tOrbital Angular Momentum Quantum Number ℓ

• A certain energy level is degenerate with respect to ℓ when the energy is independent of ℓ.

• Use letter names for the various ℓ values.– ℓ = 0 1 2 3 4 5 . . .– Letter = s p d f g h . . .

• Atomic states are referred to by their n and ℓ.• A state with n = 2 and ℓ = 1 is called a 2p state.

Th b d diti i ℓ• The boundary conditions require n > ℓ.

27

Magnetic Quantum Number mℓMagnetic Quantum Number mℓ

• The angle is a measure of the rotation about the z axis• The angle is a measure of the rotation about the z axis.• The solution for specifies that mℓ is an integer and related to the z

component of L.The relationship of L, Lz, ℓ, and mℓfor ℓ = 2.

i fi d

p

is fixed because Lz is quantized.Only certain orientations of areOnly certain orientations of are possible and this is called space quantization.

28

Magnetic Quantum Number mℓMagnetic Quantum Number mℓ

• Quantum mechanics allows to be quantized along only one direction in space. Because of the relation L2 = Lx

2 + Ly2 + Lz

2 the knowledge of a second component would imply a knowledge of the third component because we know .

• We expect the average of the angular momentum components squared to be .

29

7.4: Magnetic Effects on Atomic Spectra—The

• The Dutch physicist Pieter Zeeman showed the spectral lines

7.4: Magnetic Effects on Atomic Spectra The Normal Zeeman Effect

The Dutch physicist Pieter Zeeman showed the spectral lines emitted by atoms in a magnetic field split into multiple energy levels. It is called the Zeeman effect.

Anomalous Zeeman effect:• A spectral line is split into three lines.• Consider the atom to behave like a small magnet.

Thi k f l t biti i l t l f I d /• Think of an electron as an orbiting circular current loop of I = dq / dt around the nucleus.

• The current loop has a magnetic moment μ = IA and the period T =The current loop has a magnetic moment μ IA and the period T 2πr / v.

• where L = mvr is the magnitude of the orbital

30angular momentum.

Since there is no magnetic field to The Normal Zeeman Effect

align them, point in random directions. The dipole has a potential energygy

• The angular momentum is aligned with the magnetic moment, and the torque between and causes a precession of .

Where μB = eħ / 2m is called a Bohr magneton.• cannot align exactly in the z direction and

h l t i ll d ti d i t ti31

has only certain allowed quantized orientations.

The Normal Zeeman Effect• The potential energy is quantized due to the magnetic quantum number• The potential energy is quantized due to the magnetic quantum number

mℓ.

• When a magnetic field is applied, the 2p level of atomic hydrogen is split into three different energy states with energy difference of ΔE = μ B Δminto three different energy states with energy difference of ΔE = μBB Δmℓ.

mℓ Energy1 E0 + μBB

0 E0

−1 E0 − μBB1 E0 μBB

32

The Normal Zeeman Effect• A transition from 2p to 1s.

33

Probability Distribution Functions• The differential volume element in spherical polar coordinates is

h fTherefore,

• We are only interested in the radial dependence.

• The radial probability density is P(r) = r2|R(r)|2 and it depends only on n and l.

34

Ground State (n = 1, l = 0, ml = 0)

( )2 2 2 0d R dR m E U r R+ + ⎡ ⎤⎣ ⎦

For all s-states (l=0), the angular momentum is zero, and ( )2 2 0E U r R

dr r dr+ + − =⎡ ⎤⎣ ⎦the radial equation is reduces to

The ground state (n = 1, l = 0, ml = 0):

( ) ( ) ( ) ( ) 11,0,0 1, 0 0, 0 0, , ,

1l l

iE tn l l m mr t R r eθ φ θ φ −= = = = =Ψ = Θ Φ

( )R0 1/

3/20

1 r a iE te eaπ

− −= 20

0 2

4ameπε

=- the Bohr radius

0 0.53A=0.053nma ≈ o

( )1, 0n lR r= =

- recall that we got this estimate from the uncertainty principle

( ) 2P r d R dτ τ= ΘΦ 2 sind r drd dτ θ θ φ=The probability of finding an electron in a spherical shell

between r and r+dr from the nucleus (§6.7):

( )P r d R dτ τ= ΘΦ

( )P r ∝

sind r drd dτ θ θ φ=

(§ )2 2R r ( )

22 2 2 22 2

0 0

sinP r dr R r dr d d R r drπ π

θ θ φ= Θ Φ =∫ ∫r2dr captures the volume of the shell

0 0

Other l = 0 States (s-states)For all s-states, the angular momentum is zero: 1, 2, 3,...n = 0l = 0lm =, , , l

n=2

n=3n 3

“Boundary” sphere: r1s = 1.4 Å, r2s = 3.3 Å, r3s = 10 Å

Probability Distribution Functions

R(r) and P(r) for the lowest-lying states oflowest-lying states of the hydrogen atom.

37

The 2s Hydrogen orbital

/2 10/2

2 0 30

1(2 / )4 2

−= − =r as B r a e B

aψ

πProbabilityamplitude

When r = 2a0, this term goes to zero.There is a “node” in the wave function.

a0 = 0.529, the Bohr radius

022 /2

2 0(2 / ) −⎡ ⎤= −⎣ ⎦r a

s B r a eψ Absolute value of the wavefunctionsquared – probability distribution.

0.5

0 2

0.25

0 2

0.3

0.4

0 1

0.15

0.2

ψψ2

node node

2 4 6 8

0.1

0.2

0.05

0.1

2 4 6 8 1 2 3 4

r (Å) r (Å)Copyright – Michael D. Fayer, 2007

[ ]2 2( ) 4 ( )D R d

Radial distribution function

[ ] 2( ) 4 ( )nl nD r R r r drπ=

Probability of finding electron distance r from the nucleus in a thin spherical shell.

0 41s

0.2

0.4r2 d

r

0.01 2 3 4 5 6 7

π[R

n(r

)]

0.22s

4π

0

0.1

r/a00 2 4 6 8 10 12 14 16


Example 20.2Example 20.2

a. Consider an excited state of the H atom with the electron in the 2s orbital Is the wave functionthe electron in the 2s orbital.Is the wave function that describes this state,an eigenfunction of the kinetic energy? Of the potential energy?kinetic energy? Of the potential energy?

( ) 02/2/3

211 arerr −⎟⎟⎞

⎜⎜⎛

⎟⎟⎞

⎜⎜⎛

=ψ

b C l l h l f h ki i d

( ) 0

00200 2

32e

aar ⎟⎟

⎠⎜⎜⎝

−⎟⎟⎠

⎜⎜⎝

=π

ψ

b. Calculate the average values of the kinetic and potential energies for an atom described by this

f tiwave function.

SolutionSolution

a. We know that this function is an eigenfunction of ththe total energy operator because it is a solution of the Schrödinger equation. You can convince yourself that the total energy operator does not commute with either the kinetic energy operator or the potential energy operator by extending the discussion of Example Problem 20.1. Therefore, this wave function cannot,be an eigenfunction of either of these operators.

SolutionSolution

b. The average value of the kinetic energy is given by

( ) ( )⎞⎛ ⎤⎡ ⎫⎧ ⎞⎛⎞⎛

= ∫∞2

ˆ* kinetickinetic

ddh

dEE ττψτψ

ππ

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−= ∫∫∫

∞−−

0

22/

0

22

2/

00

2

030

00 212sin32

12

arar drrear

drdr

drd

re

ardd

ah θθφ

πμ

ππ

( )+−⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−= ∫

∞ −−

0

220

203

0

2/2/

030

00 1016

42

81

2

arar drrrraa

raee

ar

ah

πμ

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−−= ∫ ∫∫∫

∞−−−−

0

/440

/330

/

0

/220

30

0000

41389

81

2arararar er

aer

are

aer

aah

πμ

SolutionSolution

We use the standard integral,1

0

:/! andxex naxn = +∞

−∫

Using the relationship

20

2 8/ ahEkinetic μ=

g p

2

2for ,32 00

2

=−== nEa

eEkinetic πε

SolutionSolution

The average potential energy is given by( )ˆ∫ ( )

212sin1

*

22/2/22

00 ⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

⎟⎞

⎜⎛⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

=

∫∫∫

∫∞

−− drererdde

dEE

arar

potentialkinetic

φθθφ

ττψψππ

141

22sin324

/3/2/2

0 0000300

00

⎟⎞

⎜⎛

⎥⎦

⎢⎣

⎟⎟⎠

⎜⎜⎝

−⎟⎠

⎜⎝⎥⎦

⎢⎣

⎟⎟⎠

⎜⎜⎝

−−=

∫ ∫ ∫

∫∫∫∞ ∞ ∞e

drear

ea

dda

ararar

φθθφππε

( )1

1448

14

22

0 0 0

/320

/2

0

/300

000

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−= ∫ ∫ ∫ −−−

e

drera

drera

drea

e ararar

ππε

( )281

42

203

00

−=

e

aa

eπε

2for 216 00

==−= nEa

enπε

SolutionSolution

We see that

kineticpotentialtotalpotential EEEE 2 and 2 −==

The relationship of the kinetic and potential energies is a specific example of the virial theorem and holds for any system in which the potential is Coulombic.

2<T> = n <V> V( r ) = (r) **(n)( ) ( ) ( )

SHAPES OF HYDROGEN ATOM WAVE FUNCTIONS

( )ϕΦ ( )R r

Have solved three one-dimensional equations to get

( )θΘ( )m ϕΦ ( )nR r

The total wavefunction is

( )m θΘ

( , , ) ( ) ( ) ( )n m m m nr R rϕ θ ϕ θΨ = Φ Θ

1 2 31, 2, 3n =

1, 2, 0n n= − −

, 1m = − ⋅ ⋅ ⋅ −


• First few radial wave functions Rnℓ

Hydrogen Atom Radial Wave Functionsnℓ

• Subscripts on R specify the values of n and ℓ.

48

Solution of the Angular and Azimuthal S gEquations

• The solutions for Eq (7.8) are .• Solutions to the angular and azimuthal equations are linked because both

have mℓhave mℓ.• Group these solutions together into functions.

spherical harmonics---- spherical harmonics

49

Normalized Spherical HarmonicsNormalized Spherical Harmonics

50

Nodal Characteristics of

“Hydrogen Atom Wave Functions”Hydrogen Atom Wave Functions

One Electron Wave Functions=ORBITALS

INTERLACING OF RADIALINTERLACING OF RADIAL NODESNODES

Nodes of Hydrogenic wave functionsψnℓ (r,ө,φ)= Rnℓ(r) Yℓm(ө,φ)ψnℓ (r,ө,φ) Rnℓ(r) Yℓm(ө,φ)

Total # of RADIAL NODES [ Spherical Surfaces ]= n-ℓ-1

Total number of ANGULAR NODES [ Planes ] = ℓ

Total number of Nodes = n - 1Total number of Nodes = n - 1

Orbitals of Hydrogenic Atom

Close to the nucleus,

Wave Functions of Hydrogen

Close to the nucleus,p orbitals are proportional to r,d orbitals are proportional to r2,

3and f orbitals are proportional to r3.Electrons are progressively excludedfrom the neighbourhood of the nucleusfrom the neighbourhood of the nucleusas l increases. An s orbital has afinite, nonzero value at the nucleus.

Orbital near nucleus

00r as r →

The only radial node conditionThe only radial node condition2s, 3p,4d,5f….orbital of Hydrogen atom, p, , y g

The Only Radial NodeThe Only Radial NodeIn the first excited state

of H atom−,

of H atomfor any given is given exactly

( 1)(,

. .2)only node

f y g g ya ur − = + +y

Radial Nodes → Spherical Surface(s)→ n-ℓ+1

ℓ=0; 1s,2s,3sψψ

4πr**2 R(r)**2

Radial Nodes Contd.

Angular Nodes→ Planes → ℓ

-1/32 a.u.

-1/8 a u

-1/18 a.u.

1/8 a.u. Degeneracy=n*n

-0.5a.u.

The most probable value of moments of “ r “

; 0,1, 2,3... 1, 2, 3,...nr n< > = − − −; , , , , , ,

SPHERICALLY CONFINED HYDROGEN ATOM

RADIAL EXPECTATION VALUES <rn>

3d 3p 3s

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Hydrogen atom