Aurelian Isar - Decoherence And Transition From Quantum To Classical In Open ...
Hydrogen atom
1. MOST IMPORTANT SCIENCE TABLE
HYD ALSO A BRIEF HISTORY OF CHEMISTRY
For Hydrogen-like atoms H= -1/2 ∇ 2 - Z/r
H Ψ (r , θ , ϕ ) = EnlΨ (r , θ , ϕ ) ; Ψ = Rnl(r)Y lm( θ , ϕ )
2. 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
distinct b
di i substance by Cavendish in 1776. Named by Lavoisier,
b C di h i 1776 N db L i i
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 the
stars
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were originally made from hydrogen atoms or from other elements
that were originally made from hydrogen atoms.
CARRIER OF ENERGY IN THE UNIVERSE
3. Hydrogen Atom
Atom is a 3D object, and the electron motion is three-dimensional. We’ll start with the simplest case
j , p
- a hydrogen atom. An electron and a proton (nucleus) are bound by the central-symmetric Coulomb
interaction. Because mp>>me, we neglect the proton motion (the reduced mass is very close to me, the
center of mass of this system is ~2000 times closer to the proton than to the electron). Thus, we can
treat this
t t thi problem as th motion of an electron i th 3D central-symmetric C l b potential.
bl the ti f l t in the t l t i Coulomb t ti l
Three dimensions – we expect that the motion will be
characterized with three quantum numbers (in fact, there
will be the fourth one, the electron spin, which
corresponds to the “i
d h “internal d
l degree of f d ” f an
f freedom” for
electron; the spin will be added later “by hand”, it is not
described by the non-relativistic quantum mechanics).
e2
V =V (r) = -
4πε 0r
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).
4. Orbital Motion in Classical Mechanics
r 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:
mv2 p2 ( pr + pt )
2 2
e2 mv2 ⎛ mv2 ⎞ Kinetic
F =− r =− 2 r ⎜F = ⎟ K= = =
4πε 0r 3
r ⎝ r ⎠ energy: 2 2m 2m
pr and pt – the radial and tangential components of the momentum
The angular momentum
g L=r×p L = r × p = r × ( pr + pt ) = r × pt
K=
(p 2
r + pt2 )
=
pr2
+
L2 For a circular K=
L2
2m 2m 2mr 2 motion: 2mr 2
2mr
dL
In general, a non-zero torque leads to the time dependence of L: τ=
dt
In the field of a central force, the torque is dL dp
τ= = r × = r ×F = 0
zero, and the angular momentum about the dt dt
center is conserved:
( )
r is antiparallel to F
This makes L especially useful for analyzing the central force motion.
5. Quantum-Mechanical Approach
As usual, we start with the time-dependent Schrödinger Equation (but now it is 3D case):
∂Ψ ( x, y, z , t ) ⎡ ∂ 2 Ψ ( x , y , z , t ) ∂ 2 Ψ ( x , y , z , t ) ∂ 2 Ψ ( x, y , z , t ) ⎤
2
i =− ⎢ + + ⎥ + V ( x, y , z ) Ψ ( x, y , z , t )
∂t 2m ⎣ ∂x 2 ∂y 2 ∂z 2 ⎦
e2 e2
V ( x, y, z ) = − =− - potential energy of Coulomb interaction
4πε 0 x + y + z
2 2 2 4πε 0r between electron and proton
L l i ∇2
Laplacian ⎛d d d ⎞
p= ⎜ + + ⎟= ∇
ˆ
i ⎝ dx dy dz ⎠ i
∂Ψ ( x, t ) ˆ ⎛ d2 d2 ⎞
2
ˆ d2
i = H Ψ ( x, t ) H ≡− ⎜ 2 + 2 + 2 ⎟ + V ( x, y , z )
∂t 2m dx dy
2 ⎝d d dz ⎠
d
K operator V operator
The potential is time-independent, thus we can separate time and space variables:
dT ( t ) ⎛ −iE ⎞
iEt
i = ET ( t ) Hψ ( x, y, z ) = Eψ ( x, y, z )
ˆ Ψ ( x, t ) = ψ E ( x ) exp ⎜ ⎟
dt ⎝ ⎠
- solution of the t-independent S Eq gives us the eigenfunctions (the orthogonal basis ψi) and
S. Eq.
eigenvalues (spectrum Ei) of the Hamiltonian. Eigenfunctions correspond to the stationary states with
a well-defined energy.
6. Spherical Polar Coordinates
The Coulomb potential has central symmetry (U(r)
depends on neither θ nor φ ). It’s to our advantage to use
spherical polar coordinates (this will allow us to
separate variables).
- distance from
r = x +y +z
2 2 2
origin to point P
z
θ = cos−1 - zenith angle
x +y +z
2 2 2
y
φ = tan −1 - azimuth angle
x
- a (small) volume element used for
integration (no relation to the torque
τ)
7. t-independent Schrödinger Equation in Polar Coordinates
2
⎡ ∂ 2ψ ( x, y, z ) ∂ 2ψ ( x, y, z, t ) ∂ 2ψ ( x, y, z, t ) ⎤
− ⎢ + + ⎥ + V ( x, y, z )ψ ( x, y, z ) = Eψ ( x, y, z )
2m ⎣ ∂x 2
∂y 2
∂z 2
⎦
∂2ψ ( x, y, z ) ∂2ψ ( x, y, z ) ∂2ψ ( x, y, z )
+ +
∂x 2
∂y 2
∂z 2
1 ∂ ⎡ 2 ∂ψ ( r,θ ,φ ) ⎤ 1 ∂ ⎡ ∂ψ ( r,θ ,φ ) ⎤ 1 ∂2ψ ( r,θ , φ )
⎢r ⎥+ 2 ⎢sin θ ⎥+ 2 2
r ∂r ⎣ ∂r ⎦ r sin θ ∂θ ⎣ ∂θ ⎦ r sin θ ∂φ 2
2
2m
+ 2 ⎡ E − V ( r,θ ,φ ) ⎤ψ ( r,θ , φ ) = 0
⎣ ⎦
Af multiplying b r2sinθ
After l i l i by i
∂ ⎡ 2 ∂ψ ⎤ ∂ ⎡ ∂ψ ⎤ ∂2ψ 2m 2 2 ⎡ e2 ⎤
sin θ ⎢r
2
⎥ + sin θ ∂θ ⎣sin θ ∂θ ⎦ + ∂φ 2 + 2 r sin θ ⎢ E + 4πε r ⎥ψ = 0
⎢ ⎥
∂r ⎣ ∂r ⎦ ⎣ 0 ⎦
8. Separation of Variables
∂ ⎡ 2 ∂ψ ⎤ ∂ ⎡ ∂ψ ⎤ ∂2ψ 2m 2 2 ⎡
2m e2 ⎤
sin θ ⎢r
2
⎥ + sin θ ∂θ ⎢sin θ ∂θ ⎥ + ∂φ 2 + 2 r sin θ ⎢ E + 4πε r ⎥ψ = 0
∂r ⎣ ∂r ⎦ ⎣ ⎦ ⎣ 0 ⎦
For a spherically-symmetric potentials (V=V(r)), one can
ψ ( r,θ ,φ ) = R ( r ) Θ (θ ) Φ (φ )
separate the variables by using a trial function
sin 2 θ d ⎡ 2 dR ⎤ sin θ d ⎡
i i d Θ ⎤ 1 d 2Φ 2m 2 2 ⎡ e2 ⎤
R dr ⎣ ⎢r dr ⎥ + Θ dθ ⎢sin θ dθ ⎥ + Φ dφ 2 + 2 r sin θ ⎢ E + 4πε r ⎥ = 0
⎦ ⎣ ⎦ ⎣ 0 ⎦
To show that the variables are separable, let’s rewrite:
sin 2 θ d ⎡ 2 dR ⎤ sin θ d ⎡ d Θ ⎤ 2m 2 2 ⎡ e2 ⎤ 1 d 2Φ
R dr ⎣ ⎢r dr ⎥ + Θ dθ ⎢sin θ dθ ⎥ + 2 r sin θ ⎢ E + 4πε r ⎥ = − Φ dφ 2
⎦ ⎣ ⎦ ⎣ 0 ⎦
Since the left and right sides depends on different variables, both 1 d 2Φ
parts should be equal to the same constant: − = ml2
Φ dφ 2
Divide by
1 d ⎡ 2 dR ⎤ 2m 2 ⎡ e2 ⎤ ml2 1 d ⎡ dΘ ⎤
sin2θ and
i d r + 2 r ⎢E + ⎥ = 2 − sin θ
i
regroup: R dr ⎢ dr ⎥
⎣ ⎦ ⎣ 4πε 0r ⎦ sin θ Θ sin θ dθ ⎢
⎣ dθ ⎥
⎦
9. Separation of Variables (cont’d)
Thus 1 d ⎡ 2 dR ⎤ 2m 2 ⎡ e2 ⎤
⎢r dr ⎥ + 2 r ⎢ E + 4πε r ⎥ = l ( l + 1)
Equation for R(r)
R dr ⎣ ⎦ ⎣ 0 ⎦
ml2 1 d ⎡ ∂Θ ⎤
− sin θ ⎥ = l ( l + 1)
Equation for Θ(θ)
2 ⎢
sin θ Θ sin θ dθ ⎣ ∂θ ⎦
1 d 2Φ
− = ml2
Φ ∂φ 2 Equation for Φ(φ)
Or, in a more
e2 ⎞ l ( l + 1) ⎤
conventional
1 d ⎡ 2 dR ⎤ ⎡ 2m ⎛
⎢r dr ⎥ + ⎢ 2 ⎜ E + 4πε r ⎟ − r 2 ⎥ R = 0
form:
2
r dr ⎣ ⎦ ⎣ ⎝ 0 ⎠ ⎦
1 d ⎡ ∂Θ ⎤ ⎡ ml2 ⎤
sin θ + ⎢l ( l + 1) − 2 ⎥ Θ = 0
sin θ dθ ⎢
⎣ ∂θ ⎥ ⎣
⎦ sin θ ⎦
d 2Φ
+ ml2Φ = 0
∂φ 2
10. The initial equation in 3 polar coordinates has been separated into three
one dimensional equations.
d 2Φ
= −m 2Φ
dϕ 2
1 d ⎛ d Θ ⎞ m2Θ
⎜ sin θ ⎟ − 2 + ( + 1) Θ = 0
sin θ d θ ⎝ d θ ⎠ sin θ
1 d ⎛ 2 d R ⎞ ( + 1) 2μ
2 ⎜r ⎟− 2
R + 2 { E − V (r )} R = 0
r dr⎝ dr ⎠ r
Solve Φ equation. Find it is good for only certain values of m. (0,+-1,..+- ℓ )
good for only certain values of β=ℓ(ℓ+1); ℓ =0,1,2…]
Solve Θ equation. Find it is g
q y ( ); , , ]
Solve R equation. Find it is good for only certain values of E. Quantization with
Principal q
p quantum number “n” ; n=1,2,3,….; ℓ=n-1 ]
, , , ;
Copyright – Michael D. Fayer, 2007
11. Since β = ( + 1) , we have
1 d ⎛ 2 dR ⎞ ⎡ ( + 1) 2 μ ⎤
r + ⎢− + 2 ( E − V ( r )) ⎥ R = 0
r 2 dr ⎜ dr ⎟ ⎣
⎝ ⎠ r2 ⎦
Ze 2
V (r ) = − The potential only enters into the R(r) equation. Z is the
4πε 0 r
charge on the nucleus. One for H atom. Two for He+, etc.
Make the substitutions
2μ E
α2 = − 2
μ Ze 2
λ=
4πε 0 2α
Introduce the new independent variable
p
ρ = 2α r ρ is the the distance variable in units of 2α.
Copyright – Michael D. Fayer, 2007
12. Making the substitutions and
with
S ( ρ ) = R( r )
yields
1 d ⎛ 2 dS ⎞ ⎛ 1 ( + 1) λ ⎞ 0≤ρ ≤∞
ρ
⎜ dρ ⎟ +⎜− − + ⎟S = 0
ρ dρ ⎝
2
⎠ ⎝ 4 ρ 2
ρ⎠
To solve - look at solution for large ρ, r → ∞ (like H. O.).
Consider the first term in the equation above.
above
1 ⎛ d ⎛ 2 dS ⎞ ⎞ 1 ⎛ 2 d 2 S dS ⎞
⎜ ⎜ρ ⎟⎟ = ⎜ρ
ρ2 ⎝ d ρ ⎝ d ρ ⎠⎠ ρ2 ⎝ d ρ2
+ 2ρ ⎟
dρ ⎠
d 2 S 2 dS
= +
dρ 2
ρ dρ
This term goes to zero as
r →∞
The terms in the full equation divided by ρ and ρ2 also go to zero as r → ∞ .
Copyright – Michael D. Fayer, 2007
13. Then, as r → ∞
d 2S 1
= S.
dρ 2
4
The solutions are
S = e− ρ / 2 S = e+ ρ / 2
This blows up as r → ∞
Not acceptable wavefunction.
The full solution is
S( ρ ) = e−ρ / 2 F ( ρ )
Substituting in the original equation, dividing by e − ρ / 2 and rearranging gives
⎛2 ⎞ ⎛λ ( + 1) 1 ⎞
F ′′ + ⎜ − 1 ⎟ F ′ + ⎜ − − ⎟F = 0 0≤ρ≤∞
⎝ρ ⎝ρ ρ ρ⎠
2
⎠
The underlined terms blow up at ρ = 0. Regular point.
Copyright – Michael D. Fayer, 2007
14. Singularity at ρ = 0 - regular point,
to remove, substitute
F ( ρ ) = ρ L( ρ )
Gives
ρ L′′ + (2( + 1) − ρ ) L′ + (λ − − 1) L = 0.
Equation for L.
q Find L, g F.
, get Know F, have S ( ρ ) = R( r ) .
,
Solve using polynomial method.
Copyright – Michael D. Fayer, 2007
15. L( ρ ) = ∑ aν ρ ν = a0 + a1 ρ + a2 ρ 2 + Polynomial expansion for L.
ν Get L' and L'' by term by term
L L
differentiation.
Following subs u o , the su o all the terms in all powe s o ρ equal 0.
o ow g substitution, e sum of e e s powers of equ
The coefficient of each power must equal 0.
{ρ 0 } (λ − − 1)a0 + 2( + 1)a1 = 0
Note not separate odd
{ρ1} (λ − − 1 − 1)a1 + [ 4( + 1) + 2] a2 = 0 and even series.
{ρ 2 } (λ − − 1 − 2)a2 + [ 6( + 1) + 6] a3 = 0
Recursion formula
Given a0, all other terms coefficients
−(λ − − 1 − ν )aν determined. a0 determined by
aν +1 =
[2(ν + 1)( + 1) + ν (ν + 1)] normalization condition.
Copyright – Michael D. Fayer, 2007
16. −(λ − − 1 − ν )aν Provides solution to differential equation,
aν +1 =
[2(ν + 1)( + 1) + ν (ν + 1)] but not good wavefunction if infinite
g
number of terms.
Need to break off after the n' term by taking
y g
λ − − 1 − n′ = 0
or integers
ege s
λ=n with n = n '+ + 1 n is an integer.
n' radial quantum number
n total quantum number
n=1 s orbital n ' = 0, l = 0
n=2 s, p orbitals n ' = 1, l = 0 or n ' = 0, l = 1
n=3 s, p, d orbitals n ' = 2, l = 0 or n ' = 1, l = 1 or n ' = 0, l = 2
Copyright – Michael D. Fayer, 2007
17. Thus,
R( r ) = e − ρ / 2 ρ L( ρ )
with
ith
L(ρ)
defined by the recursion relation,
relation
and
λ=n
n = n '+ + 1
integers
Copyright – Michael D. Fayer, 2007
18. n=λ n = 1, 2, 3,
μ Ze 2
λ=
4πε 0 2α
2μ E
α =−
2
2
μ Z 2e 4
n =λ =−
2 2
32π 2ε 0 2 E
2
μ Z 2e 4
En = − 2 2 2 Energy levels of the hydrogen atom.
8ε 0 h n
Z is the nuclear charge. 1 for H; 2 for He+, etc
charge etc.
Copyright – Michael D. Fayer, 2007
19. ε 0 h2
a0 = a0 = 5.29 × 10−11 m
πμ e 2
Bohr radius - characteristic length in H atom problem.
In terms of B h radius
I t f Bohr di
Z 2e 2
En = −
8πε 0 a0 n 2
Lowest energy, 1s, ground state energy, -13.6 eV.
Rydberg constant
RH = 109 677cm -1
109,677cm
Z2
En = − 2 RH hc
n
me e 4 Rydberg constant if p
y g proton had infinite mass.
R∞ = 2 3
8ε 0 h c Replace μ with me. R∞ = 109,737 cm-1.
Copyright – Michael D. Fayer, 2007
20. Have solved three one-dimensional equations to get
Φ m (ϕ ) Θ m (θ ) Rn ( r )
The total wavefunction is
Ψ n m (ϕ ,θ , r ) = Φ m (ϕ ) Θ m (θ ) Rn ( r )
n=1 2 3
1, 2,
= n − 1, n − 2, 0
m = , − 1 ⋅ ⋅⋅ −
Copyright – Michael D. Fayer, 2007
21. • First few radial wave functions Rnℓ
Hydrogen Atom Radial Wave Functions
• Subscripts on R specify the values of n and ℓ.
21
22. S
Solution of the Angular and Azimuthal
g
Equations
• The solutions for Eq (7.8) are .
• Solutions to the angular and azimuthal equations are linked because both
have mℓ.
• Group these solutions together into functions.
---- spherical harmonics
22
24. Solution of the Angular and Azimuthal
g
Equations
• The radial wave function R and the spherical harmonics Y
determine the probability density for the various quantum states.
The total wave function depends on n ℓ and mℓ.
n, ℓ,
The wave function becomes
24
25. Principal 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
26. Orbital A
O bit l Angular Momentum Quantum
l M t Q t
Number ℓ
• It is associated with the R(r) and f(θ) parts of the wave function.
function
• Classically, the orbital angular momentum with L =
mvorbitalr.
• ℓ is related to L by .
• I an ℓ = 0 state,
In .
It disagrees with Bohr’s semiclassical “planetary” model of electrons
orbiting a nucleus L = nħ.
26
27. Orbital A
O bit l Angular Momentum Quantum
l M t Q t
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
The boundary conditions require n > ℓ
d diti i ℓ.
27
28. Magnetic Quantum Number mℓ
• The angle is a measure of the rotation about the z axis.axis
• The solution for specifies that mℓ is an integer and related to the z
component of L.
p
The relationship of L, Lz, ℓ, and mℓ
for ℓ = 2.
is fi d
i fixed
because Lz is quantized.
Only certain orientations of are
possible and this is called space
quantization.
28
29. Magnetic Quantum Number mℓ
• Quantum mechanics allows to be quantized along only one direction in
space. Because of the relation L2 = Lx2 + Ly2 + Lz2 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
30. 7.4: Magnetic Effects on Atomic Spectra The
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 of an electron as an orbiting circular current loop of I = d /
Think f l t biti i l tl f dq
dt around the nucleus.
• The current loop has a magnetic moment μ = IA and the period T =
2πr / v.
• where L = mvr is the magnitude of the orbital
angular momentum.
30
31. The Normal Zeeman Effect
Since there is no magnetic field to
align them, point in random
directions. The dipole has a potential
energy
gy
• 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
has l
h only certain allowed quantized orientations.
t i ll d ti d i t ti
31
32. The Normal Zeeman Effect
• 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 = μBB Δmℓ.
mℓ Energy
1 E0 + μBB
0 E0
−1
1 E0 − μBB
32
34. Probability Distribution Functions
• The differential volume element in spherical polar coordinates is
Therefore,
h f
• 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
35. Ground State (n = 1, l = 0, ml = 0)
For all s-states (l=0), the angular momentum is zero, and d 2 R 2 dR 2m
the radial equation is reduces to 2
+ + 2 ⎡E −U ( r )⎤ R = 0
⎣ ⎦
dr r dr
The ground state (n = 1, l = 0, ml = 0):
Ψ1,0,0 ( r,θ ,φ , t ) = Rn=1,l =0 ( r ) Θl =0,ml =0 (θ ) Φml =0 (φ ) e−iE1t
Rn=1,l =0 ( r ) 1
= e−r / a0 e−iE1t 4πε 0 2 - the Bohr radius
π a0
3/2 a0 = 2
o
a0 ≈ 0.53A=0.053nm
me
- recall that we got this estimate from the uncertainty principle
P ( r ) dτ = RΘΦ dτ
2
dτ = r 2 sin θ drdθ dφ
P(r) ∝ The probability of finding an electron in a spherical shell
between r and r+dr from the nucleus (§6.7):
(§ )
2 π 2π
R r2
P ( r ) dr = R r 2dr ∫ Θ sin θ dθ ∫ Φ dφ = R r 2dr
2 2 2 2
0 0
r2dr captures the volume of the shell
36. Other l = 0 States (s-states)
For all s-states, the angular momentum is zero: n = 1, 2, 3,... l = 0
, , , ml = 0
n=2
n 3
n=3
“Boundary” sphere: r1s = 1.4 Å, r2s = 3.3 Å, r3s = 10 Å
38. The 2s Hydrogen orbital
1
ψ 2 s = B(2 − r /a0 )e − r/2 a0
B= Probability
4 2π a0
3 amplitude
When r = 2a0, this term goes to zero. a0 = 0.529, the Bohr radius
There is a “node” in the wave function.
2 2 Absolute value of the wavefunction
ψ 2 s = ⎡ B(2 − r /a0 )e
⎣
− r /2 a0
⎤
⎦ squared – probability distribution.
0.25
0.5
0.2
02
0.4
ψ2
ψ 0.3 0.15
node 0.1
01 node
0.2
02
0.1 0.05
2 4 6 8 1 2 3 4
r (Å) r (Å)
Copyright – Michael D. Fayer, 2007
39. Radial distribution function
Dnl ( r ) = 4π [ Rn ( r )] r 2 d
2
dr
Probability of finding electron distance r from the nucleus in a thin spherical shell.
1s
4π[Rn (r)]r2dr 0.4
04
0.2
0.0
1 2 3 4 5 6 7
2s
0.2
0.1
0
0 2 4 6 8 10 12 14 16
r/a
0
Copyright – Michael D. Fayer, 2007
40. Example 20.2
a. Consider an excited state of the H atom with
the electron in the 2s orbital Is the wave function
orbital.Is
that describes this state,an eigenfunction of the
kinetic energy? Of the potential energy?
3/ 2
1 ⎛1⎞ ⎛ r ⎞
ψ 200 (r ) = ⎜ ⎟
⎜a ⎟ ⎜ 2 − ⎟e − r / 2 a 0
⎜
32π ⎝ 0⎠ ⎝ a0 ⎟
⎠
b. C l l
b Calculate the average values of the kinetic and
h l f h ki i d
potential energies for an atom described by this
wave f ti
function.
41. Solution
a. We know that this function is an eigenfunction of
the
th
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.
42. Solution
b. The average value of the kinetic energy is given by
Ekinetic = ∫ψ * (τ )Ekineticψ (τ )dτ
ˆ
r ⎞ − r / 2 a0 ⎛ 1 d ⎡ 2 d ⎧⎛ r ⎞ − r / 2 a0 ⎫⎤ ⎞ 2
2π π ∞
h 1 ⎛ ⎜
=− ∫ 0
dφ ∫ sin θdθ ∫ ⎜ 2 − ⎟e
⎜ ⎢r ⎨⎜ 2 − ⎟e ⎬⎥ ⎟r dr
2 μ 32πa0
3
0⎝
a0 ⎟
⎠ ⎜ r 2 dr ⎢ dr ⎩⎜
⎣ ⎝ a0 ⎟
⎠ ⎭⎥ ⎟
⎦⎠
0 ⎝
∞
⎛ r ⎞ − r / 2 a 0 ⎛ e − r / 2 a0 ⎞
=−
h 1
2 μ 8πa0
3 ∫ ⎜ a0 ⎟
⎜ 2 − ⎟e ⎜−
⎜ 4a 3 r ⎟ ( )
⎟ 16a0 − 10a0 r + r 2 r 2 dr
2
0⎝ ⎠ ⎝ 0 ⎠
h 1 ⎛ 9 ⎞
∞
8 3 3 − r / a0 1
3 ⎜ 2 ∫
=− ⎜ 2 − r / a0
r e − ∫ re − r / a0
− 3 ∫r e + 4 ∫r e 4 − r / a0
⎟
⎟
2 μ 8πa0 ⎝ a0 0 a0 a0 4 a0 ⎠
43. Solution
We use the standard integral,
∞
∫
0
x n e − ax dx = n! / a n +1 :
Ekinetic = h 2 / 8μa0
2
Using the relationship
g p
e2
Ekinetic = = − E , for n = 2
32πε 0 a0
44. Solution
The average potential energy is given by
Ekinetic = ∫ψ *E potentialψ (τ )dτ
ˆ
e2 1
2π π
⎡⎛
∞
r ⎞ − r / 2 a0 ⎤⎛ 1 ⎞ ⎡⎛ r ⎞ − r / 2 a0 ⎤ 2
=−
4πε 0 32πa0
3 ∫ dφ ∫ sin θdθ ∫ ⎢⎜ 2 − a0 ⎟e ⎥⎜ r ⎟⎢⎜ 2 − a0 ⎟e ⎥ r dφ
⎜
0 ⎣⎝
⎟
⎠ ⎝ ⎠ ⎣⎜⎝
⎟
⎠
0 0 ⎦ ⎦
1 ⎛ ⎞
∞ ∞ ∞
e2 4 2 − r / a0 1
=− ⎜ 4∫ e − r / a0 dr − ∫ r e
3 ⎜
dr + 2 ∫ r e
3 − r / a0
dr ⎟
4πε 0 8πa0 ⎝ 0 a0 0 a0 0 ⎟
⎠
=−
e2 1
4πε 0 8a03
( )
2
2 a0
e2
=− = 2 En for n = 2
16πε 0 a0
45. Solution
We see that
E potential = 2 Etotal and E potential = −2 Ekinetic
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 ) = ( ) **(n)
( (r) ( )
47. Have solved three one-dimensional equations to get
Φ m (ϕ ) Θ m (θ ) Rn ( r )
The total wavefunction is
Ψ n m (ϕ ,θ , r ) = Φ m (ϕ ) Θ m (θ ) Rn ( r )
n = 1 2, 3
1, 2
= n − 1, n − 2, 0
m = , − 1 ⋅ ⋅⋅ −
Copyright – Michael D. Fayer, 2007
48. • First few radial wave functions Rnℓ
Hydrogen Atom Radial Wave Functions
• Subscripts on R specify the values of n and ℓ.
48
49. S
Solution of the Angular and Azimuthal
g
Equations
• The solutions for Eq (7.8) are .
• Solutions to the angular and azimuthal equations are linked because both
have mℓ.
• Group these solutions together into functions.
---- spherical harmonics
49
53. Nodes of Hydrogenic wave functions
ψnℓ (r,ө,φ)= Rnℓ(r) Yℓm(ө,φ)
(r,ө,φ)
Total # of RADIAL NODES [ Spherical Surfaces ]= n-ℓ-1
Total number of ANGULAR NODES [ Planes ] = ℓ
Total number of Nodes = n - 1
54. Orbitals of Hydrogenic Atom Wave Functions of Hydrogen
Close to the nucleus,
p orbitals are proportional to r,
d orbitals are proportional to r2,
and f orbitals are proportional to r3.
Electrons are progressively excluded
from the neighbourhood of the nucleus
as l increases. An s orbital has a
finite, nonzero value at the nucleus.
Orbital near nucleus
r as r → 0
55. The only radial node condition
2s, 3p,4d,5f….orbital of Hydrogen atom
, p, , y g
The Only Radial Node
In the first excited state
of H − atom
f
for any given , is given exactly
yg g y
ronly − node = ( + 1)( + 2) a.u.
y
