Chapter 4 One-Electron Atoms
One-electron atoms: H, He+, Li2+, etc.
4-1 Quantum Theory of the One-electron Atom
 2▽2Ψ/2m+(E-V)Ψ=0, where V=-e2/4πε0r
1   2  
1
 
 
1
 2  2m
e2
r

sin



(
E

)  0




  r 2 sin 2   2  2
40 r
r 2 r  r  r 2 sin   
Let Ψ(r,θ,φ)=R(r)Θ(θ)Φ(φ)


1 d  2 dR  2m
e2
l (l  1)
r

[
(
E

)
]R  0 …(1),


n
2
2
40 r
r dr  dr  
r2
ml
1 d 
d 
d 2
2
 ml   0 …(3)
)]  0 …(2), and
 sin 
  [l (l  1) 
2
2
sin  d 
d 
sin 
d
2
En is the eigenvalue of (1), and we can prove that En=
 me 4
32n 2 2  0  2
2
, where n=1, 2, …
(Principal quantum number).
l(l+1) is the eigenvalue of (2), and we can prove that l=0, 1, 2, …, n-1 (Orbital
quantum number).
ml is the eigenvalue of (3), and we can prove that ml=0, ±1, ±2, …, ±l (Magnetic
quantum number).
1 d  2 dR  2m
 2 l (l  1)
r

[(
E

V
)

]R  0 , and set E=Kradial+Korbital+V.


n
r 2 dr  dr   2
2mr 2
1 d  2 dR  2m
 2 l (l  1)
K radial R  0 is dependent on
If Korbital =
, then (1)  2
r

2
r dr  dr   2
2mr
only r and the energy of radial motion. It is in agreement with the physical depiction.
By (1):
∴ Korbital
 2 l (l  1) (mvr) 2
L2
 Angular momentum of an electron is
=
=
=
2mr 2
2mr 2
2mr 2
L= l (l  1) .

Eg. The wave function of a 1s electron is  
e
r
a0
3
 a0 2
1/r for an electron in the hydrogen atom. [台大電研]
. Please find the average of
Selection rules of transitions: If
∫Ψn*xΨmdr≠0, the electron can transfer
from quantum state Ψm (energy level
Em) to state Ψn (energy level En).
(Proof)
Consider
a
composite
wavefunction Ψ=aΨn+bΨm, where
Ψn=Ψne-i(2πEn/h)t, Ψm=Ψme-i(2πEm/h)t, and
|a|2+|b|2=1.
<x>=∫Ψ*xΨdr=a2 ． ∫x|Ψn|2dr+b2 ．
∫x|Ψm|2dr+2a*bcos(2πνt) ． ∫Ψn*xΨmdr,
where ν=(En-Em)/h
If ∫Ψn*xΨmdr=0, the average electron
position <x> is independent of t.
Therefore, it is a forbidden transition.
∴ Transition from quantum state Ψm to
state Ψn is allowed if ∫Ψn*xΨmdr≠0.
For hydrogen, selection rules are △l=l’-l=±1 and △ml=ml’-ml=0, ±1 if an electron is
transferred from Ψn,l,ml to Ψn’,l’,ml’.
4-2 Bohr’s Model of One-electron Atoms


n  2rn

h

For a one-electron atom, the stability conditions of orbit are 
,

mv
 2
2
 mv  1  Ze
 rn
4 0 rn 2

where n=1, 2, 3, …, and Z is atomic number. We can solve the above equations an
obtain v=
n 2 h 2 0
h 40 rn
, λ= 
, rn =
, n=1, 2, 3, …
e
mZ
Zme 2
40 mrn
Ze
Total energy: En=K+V=mv2/2-Ze2/4πε0rn=Ze2/8πε0rn-Ze2/4πε0rn=-Ze2/8πε0rn
=-mZ2e4/8n2h2ε2=-E1Z2/n2, where E1=13.6eV
Eg. (a) An electron of H atom drops from the second excited state to the ground
state, how much energy does it lose? (b) For He+ atom, repeat the calculation.
(Sol.) (a) n: 3→1, E1-E3=-13.6(1/12-1/32)=-12.08eV
(b) -12.08×22=-48.355eV
4-3 Zeeman Effect
The torque τ on a magnetic dipole μ=IA=-eL/2m of an orbiting electron in B:
τ=μ×B=μBsinθ. Hence the potential energy is Um=∫μdθ=-μBcosθ=eLBcosθ/2m. And
utilizing cosθ=
ml
l (l  1)
and L= l (l  1) , we have Um=mlB(e  /2m), where
μB=e  /2m is called Bohr’s magneton.
Normal Zeeman effect: ν1=ν0-μBB/h=ν0-eB/4πm, ν2=ν0, ν3=ν0+μBB/h=ν0+eB/4πm.