Lecture 30: The Hydrogen Atom - Michigan State University · •We know that a hydrogen atom...
Transcript of Lecture 30: The Hydrogen Atom - Michigan State University · •We know that a hydrogen atom...
Lecture 30:The Hydrogen Atom
Phy851 Fall 2009
Example 2: Hydrogen Atom
• The Hamiltonian for a system consisting of anelectron and a proton is:
• In COM and relative coordinates, theHamiltonian is separable:
• The energy eigenvalue equations are:
pep
p
e
e
RR
e
m
P
m
PH rr
−−+=
0
222
422 πε
rCM HHH +=
R
ePHr
0
22
42 πεµ−=
M
PH CMCM 2
2
=
( ) )()()()( R
r
C
CMrCM
R
r
C
CM EEEEEEH ⊗+=⊗
)()()( C
CMCM
C
CMCCM EEEH =
)()()( R
rr
R
rRr EEEH =
Central Potential Ansatz
• For a spherically symmetric potential, we canchoose to form simultaneous eigenstates ofHr, L2, and Lz:
• Separate the radial and angular degrees offreedom:
• The Radial wave equation is then:
)(1
42
)1(
2)( ,
0
2
2
2
2
22
, rRr
e
rrrRE nnn ll
llhh
−
++
∂
∂−=
πεµµ
)()()(,,,,
Ω⊗= mnmn
rRlll
)(, ,1)(
rRrnr n
r
ll −=
),(,,)(
φθφθ mYm ll =Ω
)()(,,
RR
r mnE l=
)()(,,,,
R
n
R
r mnEmnH ll =
( ) )(2)(2 ,,1,,RR
mnmnL lllhl +=
)()(,,,,
RR
z mnmmnL lhl =
Natural Units
• Introduce dimensionless units:
• The radial equation becomes:
)(1
42
)1(
2)( ,
0
2
2
2
2
22
, rRr
e
rrrRE nnn ll
llhh
−
++
∂
∂−=
πεµµ
λρ=r
nnE εµλ22
2
h−=
)()(, ρλρ uRn =l
0)1(1
4
222
0
2
=
−
+−+′′ u
eu nερρπε
µλ ll
h
Consult the Sacred Text
• From p.781, Handbook of MathematicalFunctions:– IF
– THEN
– We just need:
0)1(1
4
222
0
2
=
−
+−+′′ u
eu nερρπε
µλ ll
h
04
1
4
1
2
122
2
=
−
−+
+++′′ u
nu r
ρα
ρα
20
2
4
2
2
12
hπεµλα enr =
++
Generalized Laguerre Polynomial
)1(4
1 2
+−=−
llα
nε=41
∞= ,,3,2,1,0 Krn # nodes in radialwavefunction
)()( )(21
2 ρρρ ααρ
rnLeCu
+−
=
Sort Out the Details
• Solve for εn:
• Solve for α:
• Solve for λ:
20
2
4
2
2
12
hπεµλα enr =
++)1(
4
1 2
+−=−
llα
nε=41
4
1=nε
144 22 ++= llα
12 += lα
( )12
42
20 ++= lh
rneµπε
λ
2
20
0
4
ea
µπε h
= `Bohr Radius’
( )120 ++= lrna
λ
Original Units
• The energy levels are given by:
• The principle quantum number is therefore:
• For a given n, the allowed l values are:
nnE εµλ22
2
h−=
( )220
2
1
1
2 ++−=
l
h
r
nna
Eµ
( )120 ++= lrna
λ4
1=nε
1++= lrnn
220
2 1
2 naEn µ
h−=
∞= ,,2,1,0 Krn
∞= ,,2,1,0 Kl
∞= K,3,2,1n
0min, =rn 11min,max −=−−= nnn rl
1,,0 −= nKl
J1018.2eV6.132
1820
2
1−×=−=−=
aE
µh
Radial Wavefunction:
€
ψn,l,m r,θ,φ( ) =8 n − l −1( )!
2n a0n( )3(n + l)!e−r / a0n 2r
a0n
l
Ln−l−1(2l+1) 2r
a0n
Yl
m θ,φ( )
)()( )(21
2 ρρρ ααρ
rnLeCu
+−
=
=
0,
2)(
an
rurRn l
,20 na
=λ
= +
−−
+−
0
121
1
0,
22)( 0
an
rL
an
reCrR n
anr
nll
l
l
r=λρ
)()(, ρλρ uRn =l
12 += lα
0
2
na
r=ρ
12
1+=
+l
α
Degeneracy of the nth level:
• We have:
lKll
Kl
K
,,1,
1,,1,0
,3,2,1
+−−=
−=
=
m
n
n
( )∑−
=
+=1
1
12n
ndl
l
2ndn =
)1(21
1
−+= ∑−
=
nn
l
l
( ) )1(12
12 −+−= nnn
2n=
2
eV6.13
nEn −=
Energy Level Diagram
• By taking spin into account (two spin statesper orbital), and using the Pauli principle, thedegeneracies of these levels explain much ofthe structure of the periodic table:– 2, 8, 18, 32,….
E
0
n=1
n=2
n=3n=4
ll=0 l=1 l=2 l=3
1s
2s 2p
3s4s
3p4p
3d4d 4f
(7)(5)(3)(1)
(1)
(4)
(9)(16)
(d)
Hydrogen Spectrum
• We know that a hydrogen atom absorbs andemits light only at specific frequencies
• This is due to the presence of quantumresonances
• There is a resonance associated with eachtransition between two energy levels– Resonance frequencies:
n=2
n=1
n=3n=4
Lymanseries(UV)
ω41ω31
ω21
Balmerseries(visible)
ω42 ω32
ω43
hnn
nn
EE ′′
−=ω