Unformatted text preview: chem 260/261 F11 Friday Septemeber 16, 2011 Chem 260/261 Today in Chemistry 260/261 • Continue the PIB • What has the particle in a line taught us? • Particle in a Box in 2 and 3 dimensions • Harmonic oscillator • Describing the hydrogen atom with wave functions • Origins of quantum numbers • Origins of orbitals Lecture 5 September 16, 2011 Next in Chemistry 260/261 • Reading: 5.1
5.4 • More on properties of the hydrogen atom orbitals • Many
electron atoms, periodic properties • HW2 due Today (OWL and MS) Example: 1D Particle
in
a
box Example: 1D Particle
in
a
box A particle is con.ined in a region of space in one
dimension, x. V(x) ∞ ∞ 0 e– ∞ 0, if 0 < x < L ∞, otherwise V= x L V(x) The potential function is: d2
! ( x ) = " k 2! ( x )
dx 2 ∞ The wave equation problem all over again… 0 e– x L ! ( x ) = A sin kx + B cos kx We can write the Schrödinger equation: ! !2 d 2 " ( x ) + V" ( x ) = E" ( x )
2 m dx 2
!2 d 2 " ( x ) = E" ( x )
2 m dx 2 d2
2 mE
! ( x ) = " 2 ! ( x ) = " k 2! ( x )
dx 2
! Need a function that after you take two derivatives gives you itself back times a constant… Since m, E and h are all positive, we know that the constant must be negative (from the minus sign). k2 = ∞ 0 e– L x First case, x=0: Lecture 5 Example: 1D Particle
in
a
box ! ( x = L ) = A sin kL = 0 0 kL = n! , n = 1, 2, 3, ... Discretezation! (Voila!!) No assumptions, besides boundary conditions! ! ( x ) = A sin V(x) n" x
, n = 1, 2, 3, ...
L Already, we can calculate the energy: ∞ ∞ 0 e– L k=
2 x 2 mE " n! %
=$ '
!
2 # L & 9h2
8 mL2 h2
8 mL2 "h%
n2 $ ' ! 2
# 2! &
n2h2
=
, n = 1, 2, 3, ...
2 mL2
8 mL2 Look at energy level trends: E ! n2 4h2
8 mL2 2 2 n 22! 2
!
E=
=
2 mL2 The same boundary condition we had for a string. ! ( x = 0 ) = A sin k 0 + B cos k 0 = 0 + B " 1 = B n!
k=
, n = 1, 2, 3, ...
L Hey, that worked! We get ! Ak 2 sin kx ! Bk 2 cos kx = ! k 2 ( A sin kx + B cos kx ) our function back times a • Now we have a function that solves the Schrödinger equation, • We know k, but not A and B. (in terms of energy and system parametrs) • What else do we know about the wave function? Since it is related to probability, and we know the electron cannot be outside the ! ( x = 0, x = L ) = 0
box, we have boundary conditions Second case, x=L: d2
d
! ( x ) = ( Ak cos kx " Bk sin kx ) = " Ak 2 sin kx " Bk 2 cos kx
dx 2
dx
negative constant. ! ( x ) = A sin kx + B cos kx ∞ Take second derivative: 2 mE
!
2 Example: 1D Particle
in
a
box: Boundary Conditions V(x) d
d
! ( x ) = ( A sin kx + B cos kx ) = Ak cos kx " Bk sin kx
dx
dx energy ! Take .irst derivative: E∝ 1
L2 Look at differences between levels: !Enm = En " Em = n2h2 m2h2
h2
"
=
n2 " m2
8 mL2 8 mL2 8 mL2 ( ) 1 chem 260/261 F11 Friday Septemeber 16, 2011 ! ( x ) = A sin V(x) n" x
, n = 1, 2, 3, ...
L Example: 1D Particle
in
a
box V(x) x L L L " ! ( x ) dx = A " sin
2 9h2
8 mL2 2 0 2 0 n# x
dx = 1
L We call this NORMALIZATION. L energy ) sin 2 0 n! x
# 1 sin 2 n! &
dx = L % "
(
$2
L
4 n! ' 0 L 4h2
8 mL2 nπx
⎛ L⎞
A ∫ sin
dx = A2 ⎜ ⎟ = 1
L
⎝2⎠
0 h2
8 mL2 A2 = 2 Example: 1D Particle
in
a
box 2 2
L ! ( x) = A= 2
n" x
sin
L
L ∞ e– 0 x n=2 x e– 0 probability density 2
! n ( x) L x Prob(x1 < x < x2) n=2 n2h2
, n = 1, 2, 3, ...
8 mL2 V V(x) = 0 V(x) = ∞ L E α n2 16 !
2 d 2" ( x )
= E" ( x )
2 m dx 2 ! ( x ) = A sin( kx )
wavefunctions 2
# n" x &
sin %
$L(
'
L x  ψ (x)2 n=1 x ! n (x) = V(x) = ∞ ! 2 d 2" ( x ) ˆ !
+ V ( x ) " ( x ) = E" ( x )
2m dx 2
! x x n=1 0 x1 ψ (x)2 x n=2 Summary: What we know from solving the particle in a box problem x2 = ∫ ψ 2 ( x)dx
This is the probability distribution ψx) ( x) x Look at the distributions. We can calculate the probability of .inding the electron between two points as: ∞ n2h2
, n = 1, 2, 3, ...
8 mL2 x V(x) ∞ 2
n n=1 2
L En = ! n=3 ! n ( x) En = n=3 x L probability density 0 ∞ Wave function Born: the square of the wave function is the probability, so if we add up the probability over the whole box, we had better .ind the electron in there! ∞ e– 2
n" x
sin
L
L Now we know everything! We can look at the wave functions and probability distributions: What about A? ∞ ! ( x) = E= n2 h2
8mL2 Energy Example: 1D Particle
in
a
box 9 4 quantized energies 1 BOUNDARY CONDITIONS Relevance to chemistry So What! What does this have to do with anything real? 1 2 A(! ) 3 Chlorophyll a λ
ΔEnm = En − Em = Lecture 5 ( h2
n2 − m2
8mL2 ) Chlorophyll b Carotenoid 4 Structures from wikipedia 2 chem 260/261 F11 Friday Septemeber 16, 2011 What we know from solving the particle in a box problem What we know from solving the particle in a box problem V L V(x) = 0 V(x) = ∞ 0 0 0 ψ n, m (x, y ) = ψ n (x )ψ m (y ) = 2
⎛ nπ x ⎞ 2
⎛ mπ y ⎞
sin ⎜
sin
⎟
⎟
Lx ⎝ Lx ⎠ Ly ⎝ Ly ⎠ En, m = En + Em =
0 L x 2
⎛ nπ x ⎞ 2
⎛ mπ y ⎞
sin ⎜
sin ⎜
⎝L⎟ L
⎠
⎝L⎟
⎠
L [ 8h 2
=
8mL2 h
n2 + m2
8mL2 E c a b E1,1 = 2
2
2⎞
⎛
ny
nz ⎟
⎜
⎟
+
+
2
2⎟
8m ⎜ 2
b
c⎠
⎝a Lx 5h 2
8mL2 2h 2
8mL2 Transmission electron microscopic image 2
h ⎜nx http://web.mit.edu/chemistry/nanocluster/home.html Paul Alivisatos http://www.cchem.berkeley.edu/~pagrp/ Paul O Brien 1+4=5 1+1=2 2
⎛ nπ x ⎞ 2
⎛ mπ y ⎞
sin ⎜
sin ⎜
⎝L⎟ L
⎠
⎝L⎟
⎠
L Lz Ly Enx ,n y ,nz = 2
2
h 2 ⎡ nx n y nz2 ⎤
⎢++⎥
8m ⎢ L2 L2 L2 ⎥
y
z⎦
⎣x nx = 1, 2, 3, … n y = 1, 2, 3, …
nz = 1, 2, 3, … ⎛ n yπ y ⎞
⎛n π x⎞
⎛
⎞
2
2
⎟ × 2 sin⎜ nzπ z ⎟
ψ nx ,n y ,nz (x, y, z ) = ψ nx (x )ψ n y ( y )ψ nz (z ) =
sin⎜ x ⎟ ×
sin⎜
⎜L ⎟
⎜L ⎟
⎜L ⎟ L
Lx
Ly
z
⎝ z⎠
⎝ x⎠
⎝ y⎠
~1
100 nm 1 dimension, 1 quantum number 2 dimensions, 2 quantum numbers 3 dimensions, 3 quantum numbers We can take these ideas and use them to understand the qualitative features of electrons in nanoparticles – particles with length scales in the nanometer range. Cartoon of a cadmium selenide (CdSe) nanoparticle Another 1
D system – the Harmonic Oscillator Recall the potential function we drew for the hydrogen molecule (lecture 2) Nanometer and subnanometer clusters Moungi Bawendi 4+4=8 ⎧0 for 0 < x < Lx , 0 < y < Ly , 0 < z < Lz
V ( x, y, z ) = ⎨
⎩∞ otherwise (2,1) (1,1) 1+9=10 n,m=2,2 n,m=1,1 2 E1, 2 = E2,1 = 3
d PIAB: n,m=1,3;3,1 ) What we know from solving the particle in a box problem E2 , 2 En,m = ( n,m=1,2;2,1 (2,2) (1,2) n2h2 m2h2
h2
+
=
n2 + m2
8 mL2 8 mL2 8 mL2
x y Symmetry (i.e. a square box) Energy levels may become equal (degenerate) ψ n, m (x, y ) = ψ n (x )ψ m (y ) = Wave functions for the 2
D Particle in a Square ψ n, m (x, y ) = ψ n (x )ψ m (y ) = 0 (see for yourself ‒
plug back the solution to the SE and see if it works..) L x Energy is equal to the sum of the 1D energies V(x,y) = 0 y Potential is zero inside the 2D box: X position has no effect on the Y potential (vice versa) Solution is the product of the 1D x and y solutions: V(x,y) = 0 Multiply the wavefunctions to Zind the 2D wavefunction x L L y ! n (x) = V(x) = ∞ 2
# n" x &
sin %
(
$L'
L HA RAB HB Zoom in on the potential at RAB = Re http://www.chemistry.manchester.ac.uk/groups/pob/index.html ~ 5 nm
Cadmium Selenide (CdSe )Quantum Dots: Luminescence (light emission is related to the difference in energy levels) Excited State Which CdSe Quantum Dots are the largest? Ground State A B C D Increasing size red
shifts lowest transition. Quantum ConZinement blue
shifts energy levels. Re http://www.sitemason.vanderbilt.edu/physics/matphys/research Lecture 5 3 chem 260/261 F11 Friday Septemeber 16, 2011 Another 1
D system – the Harmonic Oscillator Recall the potential function we drew for the hydrogen molecule (lecture 2) Another 1
D system – the Harmonic Oscillator RAB HA Recall the potential function we drew for the hydrogen molecule (lecture 2) HB Zoom in on the potential at RAB = Re Re The parabolic potential: Hooke s Law Another 1
D system – the Harmonic Oscillator Recall the potential function we drew for the hydrogen molecule (lecture 2) RAB HA This simple potential model is widely used (after all we do not have many systems that can be solved exactly). unfortunately) 1
Reduced mass HB V ( RAB ) = Re µ V ( RAB ) = m1 1
2
k ( RAB ! Re )
2 The green line is a parabola: right near Re, it overlaps almost perfectly, there are some differences at the edges. 2
! n" x $
sin #
&
"L%
L n=5 25 1
2" Ψn(x) is more a more complex formula, but the behavior is similar. k
µ n=4 7.5 n=3 5.5 Energy n=3 9 Energy n=4 16 n=2 3.5 1.6 1.5 n=1 n=2 4 n=1 1 L n=0 0.5 0 F = ! k ( RAB ! Re )
Stiffness of the spring: k (spring constant). 1635
1703 x = R ! Re
" !2 d 2
ˆ%
$ ! 2 µ dx 2 + V ( x )'! ( x ) = E! ( x )
#
& The energy levels will be quantized because the wave function is bounded… ∫ ∞ ψ ( x) 2 dx = 1 −∞ Phase oscillation – like the oscillation of the string, but, as for a standing wave, the nodes are stationary for the energy eigenstates. x n=1,2,3… n=0,1,2,… In both cases the lowest energy state has kinetic energy >0. This is a quantum phenomenon. No kinetic energy would mean Δx=0 and Δp=0 This cannot be! Lecture 5 2 http://phet.colorado.edu/ 1$
!
E = h! # n + &
"
2%
!= x k ( RAB ! Re ) Harmonic Oscillator Not Zinite unless E = hv (n+ ½) The boundary condition determines E Particle in a Box 0 Reduced mass ⎡ 2 d 2 1 2 ⎤
+ kx ⎥ψ ( x) = Eψ ( x)
⎢−
2
2
⎣ 2µ dx
⎦ Harmonic Oscillator Wave functions ! n ( x) = m2 m1 m2
m1 + m2 2 F = ma =
k(R
Re) Re ! h2 $
E = n2 #
&
" 8mL % HB Zoom in on the potential at RAB = Re Re If we zoom in enough, the potential takes on a familiar shape. RAB HA For the Harmonic Oscillator Simulator: http://phet.colorado.edu/en/simulation/bound
states 4 chem 260/261 F11 Friday Septemeber 16, 2011 Another Example of Quantum Weirdness … Tunneling ! ( x )2 1$
!
E = h! # n + &
"
2% n=4 7.5 n=0 n=3 5.5 Energy Scanning Tunneling Microsope P=0.843 n=2 3.5 Gerd Binning P=0.157 1.5 n=1 0.5 n=0 16% of the time the oscillator would be measured in the forbidden zone x 0 The probability distribution goes beyond the
classical limit. In this region potential energy > total energy! (if the oscillator is classical) Tunneling is responsible for a number of strange phenomena – but it is also useful… Nobel Prize 1986 Scanning Tunneling Microsope Images Molecules on surfaces Build your own – see www.geocities.com/spm_stm/Project.html Heinrich Rohrer Applying Schrödinger's equation to the hydrogen atom Electron Waves Coulomb s Law e
Perylene on graphite r p+ z y x Ze2
ˆ
V (r ) = !
4"# 0 r Schrödinger's equation for an electron in a 3D box with a proton: ! IBM Research: Fe on Cu surface http://eels.kuicr.kyoto
u.ac.jp/stm.en.html 48 iron atoms in a circular ring "corral" some surface state electrons and force them into "quantum" states of the circular structure. The ripples in the ring of atoms are the density distribution of a particular set of quantum states of the corral. The artists were delighted to discover that they could predict what goes on in the corral by solving the classic eigenvalue problem in quantum mechanics
a particle in a hard
wall box. http://www.almaden.ibm.com/vis/stm/
corral.html#stm7 Cartesian vs. spherical coordinates But the H atom has a spherically symmetric potential (assuming the proton remains stationary)… it would be very inconvenient to describe the electron as being in a box Therefore, it is natural to describe the atom using SPHERICAL (POLAR) COORDINATES math vs. physics convention for
spherical coordinates: θ and ϕ
are de.ined oppositely under
these conventions
Particle in a Sphere !
2 # d 2" d 2" d 2" & ˆ
+ 2 + 2 ( + V ( x, y, z ) " = E"
2me % dx 2
dy
dz '
$ Schrödinger's equation in polar coordinates ! !2 # 1 d # r 2 d" & + 1 d # sin ) d" & + 1 # d 2" & & ! e2 " = E"
%
(
%
(
%
('
2me % r 2 dr $ dr ' r 2 sin ) d) $
d) ' r 2 sin 2 ) $ d 2* ' ( 4+, o r
$
Must solve the Schrödinger Equation to Zind wave functions and energies of states Requires more advanced mathematical techniques than we have time to discuss in Chemistry 260 TRANSFORMATIONS x = r sinθ cosφ
y = r sinθ sinφ polar angle θ: 0 to π x + y +z =r Lecture 5 2 2 As a result, we will concentrate mainly on the results, and only sketchily consider the methods ! (r," ,# ) = R(r )$(" )%(# )
Radial Component z = r cosθ
2 r: 0 to ∞ 2 azimuthal angle ϕ: 0 to 2π Angular Component (Y) (Spherical Harmonic) The Schrödinger Equation is exactly solvable for the H atom
5 chem 260/261 F11 Friday Septemeber 16, 2011 eimφ = cos mφ + i sin mφ The angular components of the atomic wave function Spherical Harmonics When separating a differential equation to more manageable portions ★ must set each of those sections equal to a constant. The boundary conditions then result in quantization. The complete angular component of the H atom wave function isn t that scary: ! (r," , # ) = R(r )$(" )%(# ) 2ℓ + 1 (ℓ $ m)! m
!
!
Pl (cos! )eim"
4# (ℓ + m)! ℓ ! Ylm (!," ) =
ℓ Let s consider the angular parameters Zirst Separating out the Φ and Θ terms from the H atom Schrödinger equation gives rise to two new equations and two new constants: d 2!
= # m2!
d" 2 magnetic quantum number $ 1 m2
1
d$
d# '
! 2!
!!
& sin "
) = ℓ (ℓ + 1)
sin " # sin " d" %
d" ( angular momentum quantum number !(" ) = ( 2# ) 2 (cos m" + i sin m" )
!( " ) = 2 #ℓ !# 1
2 Φ(φ ) = Φ(φ + 2π ) ⇒ m = 0, ± 1, ± 2,
ℓ ! 1 m
!% !
!
!
ℓ !$( 2ℓ + 1)(ℓ # m )!(ℓ + m )!& 2 sin "
' ( (#1) !#
ℓ m (1 # cos ")ℓ m !(ℓ # m )!
!
!# m The angular components require Zinite single
valued (quantized) solutions The radial component of the H atom wave function Just as we did for the angular components, we collect r terms in the Schrödinger
equation and set them equal to a constant: 1 d ⎛ 2 dR ⎞ 2me r 2
(E − V (r )) = ( + 1)
⎜r
⎟+
R dr ⎝ dr ⎠
2
energy is related only to the radial component Like the angular component, the solutions to this equation are in the form of polynomials (Laguerre polynomials). The Laguerre polynomials contain a parameter, n, which can only take on non
zero integer values. 1 3 ⎛ 2 ⎞ 2 ⎛ (n − − 1)! ⎞ 2 ⎛ 2r ⎞ na0 2 +1 ⎛ 2r ⎞
⎟⎜
⎜
⎟⎜
⎟
⎜
⎟
R(r ) = ⎜
3⎟ ⎜
⎟⎜
⎟ e Ln+ ⎜ na ⎟
⎝ na0 ⎠ ⎝ 2n [(n + )!] ⎠ ⎝ na0 ⎠
⎝ 0⎠
principle quantum number associated Legendre polynomials (polynomials in cos(θ)) m=
ℓ,(
ℓ+1),…,0,…,(ℓ
1),ℓ (ℓ
m)! ★ m canNOT be greater than ℓ, or the Legendre Polynomial is unde.ined ℓ Examples 0 1
Y00 (!," ) =
4# !
ℓ = 0, m = 0 1 !
ℓ = 1, m = ±1 Y11 (!," ) = 3
sin ! ei"
8# !
ℓ = 1, m = 0 Y10 (!," ) = 2 3
cos!
8# 3 m: 1 1 2 3 ! (r," , # ) = R(r )$(" )%(# ) H atom wave function summary 3 0 ⎛ 2 ⎞ 2 ⎛ (n − − 1) ! ⎞ 2 ⎛ 2r ⎞ na0 2+1 ⎛ 2r ⎞
⎟⎜
⎜
⎟⎜
⎟
⎜
⎟
R( r ) = ⎜
3⎟ ⎜
⎟⎜
⎟ e Ln+ ⎜ na ⎟
⎝ na0 ⎠ ⎝ 2n [(n + ) !] ⎠ ⎝ na0 ⎠
⎝ 0⎠
!( " ) = 2 #ℓ !# 1
2 1 −r ℓ ! m
!% !
!
!
ℓ !$( 2ℓ + 1)(ℓ # m )!(ℓ + m )!& 2 sin "
' $ ( (#1) !#
ℓ m (1 # cos ")ℓ m !(ℓ # m )!
!
!# m 1 !(" ) = ( 2# ) 2 (cos m" + i sin m" )
Solutions depend on three constants: n, ℓ , m QUANTUM NUMBERS −r n = 1, 2, 3, 4... (n
ℓ
1)! n
ℓ
1 ≥ 0 n
1 ≥ ℓ spherical harmonic
describes the angular part of the wave function in any central potential
ℓ = 0, 1, 2, 3, …, (n
1) Quantum numbers can take on only certain allowed, integer values based on the above wave function: n = 0, 1, 2, 3, … ℓ = 0, 1, 2, 3, …, (n
1) m=
ℓ,(
ℓ+1),…,0,…,(ℓ
1),ℓ Today in Chemistry 260/261 • What has the particle in a line taught us? • Particle in a Box in 2 and 3 dimensions • Harmonic oscillator • Describing the hydrogen atom with wave functions • Origins of quantum numbers • Origins of orbitals Next in Chemistry 260/261 • Reading: 5.1
5.4 • More on properties of the hydrogen atom orbitals • Many
electron atoms, periodic properties • HW2 due today (OWL and MS) Lecture 5 6 ...
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