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Transcript
Quantum Theory of Atoms
• The Bohr theory of Hydrogen(1913) cannot be extended to other
atoms with more than one electron
• we have to solve the Schrödinger equation(1925)
• since the Coulomb force only depends on r, we should use spherical
coordinates (r,,) instead of Cartesian coordinates (x,y,z)
Quantum Theory of Atoms
• For a particle in a cubic or rectangular box, Cartesian coordinates
are more appropriate and there are three quantum numbers
(n1,n2,n3) needed to label a quantum state
• similarly in spherical coordinates, there are also three quantum
numbers needed
•
n = 1, 2, 3, …..
•
l= 0, 1, 2, …,n-1
=> n values of l for a given value of n
•
m= -l,(-l+1),…0,1,2,…,l => 2l+1 values of m for a given l
• n is the principal quantum number and is associated with the
distance r of an electron from the nucleus
• l is the orbital quantum number and the angular momentum of the
electron is given by L=[l(l+1)]1/2 ħ
• m is the magnetic quantum number and the component of the
angular momentum along the z-axis is Lz = m ħ
Quantum Theory of Atoms
• The fact that both l and m are restricted to
certain values is due to boundary conditions
• in the figure, l=2 is shown
• hence L=(2(2+1))1/2 ħ = ħ(6)1/2
• and m= -2, -1, 0, 1, 2
• the Schrödinger equation can be solved
exactly for hydrogen
• the energies are the same as in the Bohr
theory En = -Z2 (13.6 eV)/n2
• they do not depend on the value of l and m
• this is a special property of an inverse-square law force
• The lowest energy has n=1
=> l=0 and m=0
• the second lowest energy has
n=2 => l=0, m=0
l=1, m=-1,0,1
• hence 4 states!
• Notation:
•
•
•
•
l=0
l=1
l=2
l=3
l=4
S-state
P-state
D-state
F-state
G-state
• When a photon is emitted or
absorbed we must have
m= 0 or ±1
l = ±1
• conservation of angular
momentum and conservation
of energy
Wave Functions
• We denote the solutions of the SE as nlm (r,,)
• the probability of finding the electron at some position (r,,) is
P (r,,)= (nlm)2 dV where dV is the volume element
• in spherical coordinates dV=r2drsindd
R  2
volume of a sphere   
 dV
0 0 0
R  2

0 0
R

2




2
2
sin  d   d 
0 r dr sin  d d   0 r dr 

 0
 0

R3
4

2
2   R 3
3
3
Wave Functions
• the ground state wave function is Cexp(-Zr/a0) where C is
determined from normalization
2

  2
dV  1
  2
2 Zr / a 2
2 2
2

r
dr
sin

d

d


C
e
r dr sin  d d
 
 
0
0 0 0
0 0 0
 C 2 4 (a0 / 4Z 3 )  1
3
hence  100
1 Z

 
  a0 
3/ 2
e Zr / a0
• Hence for the ground state, the probability density 1002 is
independent of  and  and is maximum at the origin
Wave Functions
• What is the probability of finding the electron between r
and r+dr?
• in other words, in a spherical shell of thickness dr
• volume of shell is 4r2 dr (area x thickness)
• P(r)dr = (4r2 2)dr P(r) is the radial probability density
Wave Functions
• Maximum is at r = a0/Z = a0 for Hydrogen (first Bohr orbit)
• not a well defined orbit but rather a cloud
Probability
density
 200  C200 (2  Zr / a0 )e Zr / 2 a
0
 210  C210
Zr  Zr / 2 a0
e
cos
a0
 211  C211
Zr  Zr / a0
e
sin  e  i
a0