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Physics 2203, 2011: Equation sheet for second midterm
1
General properties of Schrödinger’s Equation: Quantum Mechanics
Schrödinger Equation (time dependent)
�
− 2
2m∂ 2Ψ∂x 2
+UΨ = i∂Ψ∂t
Standing wave
�
Ψ(x, t) = Ψ(x)e−iωt
Schrödinger Equation (time independent)
�
−2
2m∂ 2Ψ∂x 2
+UΨ = EΨ
Normalization (one dimensional)
�
Ψ* (x,t)Ψ(x ,t)dx =1−∞
+∞∫
For a constant potential E>U
�
ψ(x) = Ae−ikx + Be+ikx with
�
k = 2m(E −U)2
E<U
�
ψ(x) = Ae−ηx + Be+ηx with
�
η = 2m(U − E)2
Operators:
�
p =−i ∂∂x
,
�
E = −i
∂∂t
,
�
K = −
2
2m∂ 2
∂x2,
�
H = −
2
2m∂ 2
∂x 2+ U
Eigenfunctions and Eigenvalues:
�
p Ψ = pψ then p is the eigenvalue and Ψ is an eigenfunction. 1) One Dimensional Quantum Systems a) Particle in a one-dimensional box, with U=0 for 0<x<L;
Wave functions
�
ψ(x) = 2Lsin(nπx /L) with n= 1, 2, 3, ----
Energies
�
En = n2π22
2mL2
b) Simple Harmonic Oscillator: U=kx2/2
Allowed Energies
�
En = n +1/2( )ωc with
�
ωc = km
First few wave functions
�
Ψ0(x) = A0e− x
2
2b
Ψ1(x) = A1
xbe− x
2
2b
Ψ2(x) = A21− 2x
2
b2⎛
⎝ ⎜
⎞
⎠ ⎟ e
− x2
2b
with
�
b =
mωc
c) Tunneling: Barrier of height U0 with Kinetic Energy E, where within the barrier E<U0 Within the classically forbidden region the wave function is
�
Ψ(x) = Ae+αx + Be−αx
where
�
α = 2m(U0 − E)2
The Tunneling probability
�
T ≈ e−αL
Physics 2203, 2011: Equation sheet for second midterm
2
d) General properties of wave functions The wave function and its derivative must be continuous. In the classically forbidden region (E<U) the wave function curves away from the axis,
exponentially as wave penetrates the classically forbidden region. In the classically allowed region (E<U) the wave function curves toward axis and oscillates
2) Two and Three dimensional systems
a) Two or three dimension box, sides Lx, Ly, Lz
The wave function is
�
ψ(x,y.z) =ψnx(x)ψny
(y)ψnz(z) with
�
ψn p(p) = 2
Lp
sin(nπpLp
) ,
p= x,y or z
The energy is
�
En,n,n = 2π2
2mnxLx
⎛
⎝ ⎜
⎞
⎠ ⎟ 2
+nyLy
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
2
+ nzLz
⎛
⎝ ⎜
⎞
⎠ ⎟ 2⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
b) Hydrogen Atom
In three dimensions
�
∇ 2Ψ =∂ 2Ψ∂x2
+∂ 2Ψ∂y2
+∂ 2Ψ∂z 2
or
�
∇ 2Ψ =1r 2
∂∂r(r2 ∂Ψ
∂r) +
1r 2 sinθ
(sinθ ∂Ψ∂θ) =
1r2 sin2θ
∂ 2Ψ∂φ 2
Hydrogen atom
�
Ψn,l,ml(r,θ,φ) = Rn.l (r)Θ l,ml
(θ)Φml(φ)
with n = 1, 2, →: l= 0, 1,→ n-1: ml=0,±1, ±2, ±l Probability of finding electron between r and r+dr :
�
P(r)dr = r 2 |Rn, l (r) |2 dr
The most probable r is the maximium in
�
P(r)dr = r 2 |Rn, l (r) |2 dr
The expectation of r is given by
�
< rn ,l >= rP(r )dr0
∞∫
!" 0
Physics 2203, 2011: Equation sheet for second midterm
3
3) Electron Spin and Magnetic moment Orbital Spin Quantum number l=0, 1, 2 s = 1/2 Length of vector
�
|L |= l( l + 1)
�
| S|= s(s+ 1) = 3/4 Z component
�
Lz = ml
�
Sz = ms Magnetic Quantum number ml =0, ±1, ±2,,,m ±l ms=±1/2 Magnetic moment µl = -(e/2m)L µs = -(e/m)S Total Angular momentum J=L+S
�
cosθ =Lz|L |
=ml
l (l + 1)
Physics 2203, 2011: Equation sheet for second midterm
4
Magnetic Moment: µB=
�
e2me
=5.78x10-5eV/T (Tesla)
Zeeman Effect: Δml=0, ±1 Magnetic Energy Um=mlµBB Anomalous Zeeman Effect: Δ(ml+ms)=0, ±1: Total magnetic moment µ = µ L+ µ S and the energy
is given by
�
Um =e2m
B(ml + 2ms )
Spin Magnetic Moment: µ spin=-(e/me)S, µ total=µ orb+µ spin=-(e/2me)[L+2S] 4) Multielectron Atoms: Pauli Principle, Hund’s Rule Pauli: Only one electron in each quantum state (n,l,ml,ms) Hunds Rule: Electrons fill different orbitals with unpaired spins as long as possible. This
maximizes S. If S is maximium the lowest state has the largest L. Fermions have antisymmetric wave functions with respect to exchange of indistinguishable
particles. Bosons have symmetric wave functions with respect to exchange. Notation: The ground state configuration of Ne with Z=10 is 1s22s22p6 ml =1 ml =0 ml =-1 1s 2s 2p Filled shells: Every filled shell is spherical with zero angular momentum. Ionization Potential: The ionization potential can be approximated by I=(Zeff)2(13.6eV). Zeff
accounts for the effective screening of the Z protons by the Z-1 electrons. Zeff can be determined from the ionization potential, the excitation energies or the effective radius.
4) Old Stuff Waves as particles: Photon has energy hf E = hf,
�
E = ω , with
�
ω = 2πf E = hc/λ = 1240/λ eV nm E=pc Particles as waves P = h/λ de Broglie wave length λ = h/p
�
ke2 =1.44nm(eV ) Uncertainty principle
�
ΔkΔx ≥1/2 and
�
ΔωΔt ≥1/2 which can also be written as
�
ΔpΔx ≥ /2 and
�
ΔEΔt ≥ /2
Physics 2203, 2011: Equation sheet for second midterm
5
Bohr Atom The angular momentum is quantized
�
L = mvr = n or
�
2πrn = nλ yielding the following equations:
�
rn =n22
ke2m where the Bohr orbit radius is defined
�
a0 =2
ke2m= 0.05292nm
�
En = −m2n2
(ke2
)2 giving
�
En = −13.6eVn2
�
vn =ke2
n
�
ke2 =e2
4πε0=1.440ev.nm and
�
hc =1240eV .nm and
�
c =197.3eV .nm
Excitation series
�
1λ
= R[ 1n2 f
−1n2i] i is the initial state and f the final state
�
R =mek
2e4
4πc3=E1hc
=1.097x10−7m−1 = 0.001097nm−1 This is for a mass me
Bohr Atom with m=center of mass: For a real system m should be the center of mass m’
where
�
m'= mMm + M
=M
1+ M /m
Then
�
r'n =n22
ke 2m'= n2
me
m'a0 and
�
E 'n = −m'2n2
( ke2
)2 = −
m'me
E1n2
Bohr Atom with nuclear charge Z:
�
En (Z) =−mZ 2
2n 2(ke 2
)2 and
�
rn (Z) =n 2 2
Zke2m: Replace a0 with a0/Z is every equation
5) Constants c = 2.998 x 10+8 m/s h = 6.626 x 10-34 J.sec =4.136 x 10-15 eV.sec
�
=1.055x10−34 J.s = 6.582x10−16eV .s
�
k =14πε0
= 8.988x109N .m 2 /C2
me = 9.109 x x 10-31 kg mec2 = 0.511 MeV mp = 1.673 x 10-27 kg mpc2 = 938.28 MeV mn = 1.675 x 10-27 kg mnc2 = 939.57 MeV mp= 1836me mn=1839me Å = 10-10 m nm = 10-9 m e = 1.6 x 10-19 coul E = hc/λ = 1240/λ eV nm
�
ke2 =e2
4πε0=1.440ev.nm µB=
�
e2me
=5.78x10-5eV/T (Tesla)
R=
�
mek2e4
4πc 3=1.097x10-2nm (Rydberg const.)
(6) Math
binomial expansion: (1 + x)n = 1 + nx + n(n-1)x2/2! + n(n-1)(n-2)x3/3! µm=10-6m ,nm=10-9m, pm=10-12m, fm=10-15m
Physics 2203, 2011: Equation sheet for second midterm
6
integrals
�
z n0
∞∫ e− zdz = (n)! for n= integer>0
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