Transcript Prezentacja programu PowerPoint

Lecture V
Hydrogen Atom
dr hab. Ewa Popko
Niels Bohr
1885 - 1962
Bohr Model of the Atom
• Bohr made three assumptions (postulates)
• 1. The electrons move only in certain circular orbits, called
STATIONARY STATES. This motion can be described
classically
• 2. Radiation only occurs when an electron goes from one
allowed state to another of lower energy.
• The radiated frequency is given by
hf = Em - En
where Em and En are the energies of the two states
• 3. The angular momentum of the electron is restricted to
integer multiples of h/ (2p) = 
mevr = n  (1)
Erwin Schrödinger
1887 - 1961
28
The hydrogen atom
The Schrödinger equation
Hˆ  x, y, z  = E x, y, z 


2
 2 
 2   2









V
x
,
y
,
z

  x, y, z  = E  x, y, z 
2
2
2 

z 


 2m  x y

Partial differential equation with three independent variables
The potential energy in
spherical coordinates
(The potential energy
function is spherically
symmetric.)
1 e2
V (r ) = 
4p 0 r
The spherical
coordinates
(alternative to
rectangular coordinates)
The hydrogen atom
For all spherically symmetric potential-energy functions:
( the solutions are obtained by a method called separation of variables)
 ( x, y, z )   r , ,  = Rnl r Ylm  ,  = Rnl r  l ,m ( ) m ( )
l
Radial function
Angular function of  and 
Thus the partial differential equation with three independent variables
three separate ordinary differential equations
The functions  and  are the same for every spherically
symmetric potential-energy function.
l
The solution
The solution is determined by boundary conditions:
- R(r) must approach zero at large r (bound state electron localized near the nucleus);
  and  must be periodic:
(r,, and (r,,2p describe the same point, so
=2p;
  and  must be finite.
Quantum numbers:
n - principal
l – orbital
ml - magnetic
Principal quantum number: n
The energy En is determined by n = 1,2,3,4,5,…;
Ionized atom
 e4
1
En = 
 2
2 2 2
32p  0  n
n=3
- 3.4 eV
1
En = 13.6eV  2
n
  reduced mass
me mN
=
me  mN
E = - 13.6 eV
n=2
n=1
Energy Levels of the Hydrogen Atom
Series limits
E (eV)
0.00
-0.85
-1.51
n=
n =4
Paschen
-3.40
n =3
n =2
Balmer
-13.6
Lyman
n =1

n =5
Balmer series
4000
5000
6000
7000 Å
(Å)
H
4000
5000
6000
7000 Å
k(cm-1)
Red
15234
6565
n1 = 3
n1 = 4
H
greenish-blue 4862
20565
H
blue
23033
4342
n1 = 5
H
violet
24374
4103
n1 = 6
Electron-Volt
• There are many different units used to describe energy
• One of the most useful in quantum physics is the electronvolt
• The electron-volt is defined as the energy needed to move
an electron through a potential difference of one volt
1(eV)=1.6*10-19J
The Rydberg constant
1
En = 13.6eV  2
n
E = En i  En f
•The Rydberg is a measure of energy
normally expressed in m-1 but we can
convert to other forms
R (J) = R (m-1)h c = R (eV) e
1R = 13.6 eV
1

1
= 1R 2  2 
 ni n f 
wavelength
 1

1
= R 2  2 
n


n
f
i


1
T. W. Hansch
1941 -
Rydberg measured (1998) to be
550 (84) m-1
R = 10 973 731.568
Quantization of the orbital angular momentum.
The possible values of the magnitude L of the orbital angular
momentum L are determined by the requirement, that the  function
must be finite at =0 and =p.
L =  l (l  1) l = 0, 1, 2, ... Orbital quantum number

There are n different possible values of L for the n th energy level!
Quantization of the component of the orbital angular
momentum
Lz = ml 
ml    l (l  1)

Lz  L =  l (l  1)
ml  l
ml = 0,  1,  2, ...  l
Quantum numbers: n, l, m
n – principal quantum number
n – determines permitted values of the energy
n = 1,2,3,4...
l – orbital quantum number
l - determines permitted values of the orbital
angular momentum
l = 0,1,2,…n-1;
ml - magnetic quantum number
ml – determines permitted values of the z-component
of the orbital angular momentum
ml = 0,  1,  2,...  l
Wave functions
n=1
l=0
n,l,m
n=2
 r , ,  = Rnl r  l ,m ( ) m ( )
l
l
l = 0,1
R(r ) ~ e
r
  polynomial
 ~
e

 i
l = 1 m = ±1
n=3
l = 0,1,2
Quantum number notation
Degeneracy : one energy level En has different
quantum numbers l and ml
l = 0 : s states
n=1 K shell
l = 1 : p states
n=2 L shell
l = 2 : d states
n=3 M shell
l = 3 : f states
n=4 N shell
l = 4 : g states
n=5 O shell
.
.
.
.
Electron states
n=3
n =1
K
1s
2s
L
l = 0 & ml = 0
n=2
l = 0 & ml = 0
l = 1 & ml =  1
2p
l = 1 & ml = 0
l = 1 & ml = 1
l = 0 & ml = 0
3s
l = 1 & ml =  1
l = 1 & ml = 0
l = 1 & ml = 1
3p
l = 2 & ml =  2
M
l = 2 & ml =  1
l = 2 & ml = 0
l = 2 & ml = 1
l = 2 & ml = 2
3d