Bohr Model of the Hydrogen Atom.!
Rohlf, p85-87"
(This was covered in Honors Physics II.)"
This is a semi-classical model which assumes the electron has well defined orbits "
(particle properties) and interference phenomena (wave properties) "
F(r)r̂ = mar̂
e2
ke2
F(r) =
= 2
2
4!" 0 r
r
v2
a=
r
ke2
r2
mv 2
=
r
where k =
1
4!" 0
k is not
2!
!!!
#
!"
"
F = ma
ke2
mv 2
=
2
r
r
p2
ke = rmv = r
m
2
2
Resonance condition for a stable orbit - constructive interference:
n! = 2" r
n!
r=
2"
r=
n!
2"
h
p=
!
Also, p =
h
!
p2
ke = r
m
2
!=
h
p
Momentum Quantization"
Begin with
n!
r=
2"
h
!=
p
p2
ke = r
m
2
# n! & p
Substitute for r : ke = %
.
$ 2" (' m
2
# nh & p 2 nhp
Now substitute for ! : ke = %
=
$ 2" p (' m 2" m
2
2" mke2
) pn =
nh
2
Energy Quantization"
2! mke2
Momentum: pn =
nh
2! 2 mk 2 e4
Show K =
n2h2
pn2
Kinetic energy: K =
2m
Potential energy: U =
4! 2 mk 2 e4
U="
n2h2
"ke
r
2
# 1 & # 2! mke2 &
Show U = " % ( %
$ m ' $ nh ('
2! 2 mk 2 e4
) "2K !!!!!!!!!!!!!!!K )
n2h2
2
Bohr Model: Energy levels of Hydrogen"
E = K +U
2! mke2
Momentum: pn =
nh
pn2
1 " 2! mke2 %
2
!!!!ke = r
=!r $
m
m # nh '&
2
2
pn2
1 " 2! mke2 %
1 4! 2 m 2 k 2 e 4 2! 2 mk 2 e 4
Kinetic energy: K =
=
=
=
2m 2m $# nh '&
2m
n2 h2
n2 h2
2
2
(ke2
" 1 % " 2! mke %
Potential energy: U =
= ($ ' $
= (2K
# m & # nh '&
r
Bohr Model (cont. 2)"
2! 2 mk 2 e 4
K=
n2 h2
4! 2 mk 2 e 4
U="
n2 h2
2! 2 mk 2 e 4
C
E = K + U = K " 2K = "K = "
=
"
n2 h2
n2
Exercise: Find the numerical value of En in units of
Jouls and convert to eV."
Exercise: Calculate E in mks units (J) and in eV units."
2! me k
2
C=
4
h
2
=
2
2! me
2
1
16! " 0
2
2
h
4
=
2
9.109 # 10 ) (1.602 # 10 )
(
=
8 ( 8.854 # 10 ) ( 6.626 # 10 )
$31
$12
En = $
2.179 # 10
n
2
$19
2
me
8" 0 h
2
$34
2
= 2.179 # 10
En (eV)= $
En = $
13.60
"2
n
2
4
$18
J
4
2.179 # 10
eV
n
2
$18
$18
J
1
1.602 # 10
$19
J/eV
=$
13.60
n
2
eV
Radius and speed of electron in lowest orbit"
ke2
p2
ke2 m
!!!!!
=! !!!!!!!!!r = 2
r
m
p
2! mke2
p=
nh
=
n2 h2#0
"!!!!r =
! me2
(
) ( 8.854 $ 10 ) = 0.529 $ 10
! ( 9.109 $ 10 ) (1.602 $ 10 )
n 2 6.626 $ 10 %34
%31
2
%12
%19 2
%10
o
$ n m & 0.53n A = 0.053n 2 nm
2
2
o
r1 = 0.53 A
Exercise: find the speed of ab electron in the lowest orbit, ie find !=v/c."
Speed of electron in lowest orbit"
Speed of electron in lowest orbit"
v
p 2" mke2 1
1
e2
!= =
=
!!!!!k =
!!!! ! =!!!!
c mc
nh m
4"# 0
2# 0 nhc
=
(
(
1.602 $ 10 %19
)(
)
2
)(
)
2 8.854 $ 10 %12 6.626 $ 10 %34 3 $ 10 8 n
For n = 1
!&
&
1 1
137 n
1
137
e2
e2
1
Fine structure constant: ' =
=
&
2# 0 hc
4"# 0 !c 137
Homework 8!
Due Oct.9!
Bohr model: Ch. 3, problems 36, 42, 55 !
Schroedinger’s Equation -1925!
Rohlf, Chapter 7, p. 191-218
"
Classical, non-relativistic physics relates kinetic, potential and total energy: "
K +V = E
! 2
P (r )
2m
!
+ V (r ) = E
In quantum physics the particles are treated as waves and obey a wave equation."
This is the Schroedinger equation. "
"
$#( r ,t)
!
"
"
"
2
!
" #( r ,t) + V ( r )#( r ,t) = i!
2m
$t
2
Symbolically "
P2
! +V! = H!
2m
Schroedinger’s Equation"
"
$#( r ,t)
!
"
"
"
2
!
" #( r ,t) + V ( r )#( r ,t) = i!
2m
$t
2
P2
! +V! = H!
2m
Momentum operator: P " #i!$
"
Potential energy operator: V " V (r )
%
Energy operator: H " i!
%t
"
$#( r ,t)
!
"
"
"
2
!
" #( r ,t) + V ( r )#( r ,t) = i!
2m
$t
2
Solve the Schroedinger equation by separation of variables."
!
!
!(
r
,t)
=
"
(
r
)T (t)
Assume:"
$(# T )
!2 2
"
Obtain:" !
" (# T ) + V ( r )(# T ) = i!
2m
$t
!2
$T
"
2
=!
T " # + V ( r )# T = i!#
2m
$t
Divide the entire equation by
!T,
and set each side to a constant
!2 1 2
1 $T
"
!
# " + V ( r ) = i!
=%
2m "
T $t
Time dependent equation: !
!T
i!
= "T
!t
!T i"
+ T =0
!t
!
Space dependent equation: !
!2 1 2
"
!
# " + V (r ) ! $ = 0
2m "
#"+
2
2m
!
2
(
)
"
$ ! V (r ) " = 0
Time equation:
!T i"
+ T =0
!t !
Space equation:
2m
"
! " + 2 # $ V (r ) " = 0
!
2
(
)
Solution to the t dependent equation:
!T i"
+ T =0
!t !
Assume an exponential solution:
Insert into the diff. eq. above:"
i! bt
bAebt +
Ae = 0
!
i!
b+
=0
!
The solution is then
T = Aebt
A!ebt i" bt
+ Ae = 0
!t
!
i!
b="
!
T = Ae
! i" t
T = Ae
" =
#
!
"
i!
t
!
Quantum mechanics relates the energy of a particle and the "
frequency of the associated wave according to: "
h
E = hf =
! = !!
2"
!T iE
+ T =0
!t !
T=
E
"i t
Ae !
The space dependent part of Schroedinger’s equation."
! 2" +
!"+
2
2m
"
#
$
V
(
r
) " =0
2
!
2m
!
2
(
)
(
"
E # V (r ) " = 0
)
Schroedinger’s equation in one dimension:"
" 2 ! ( x ) 2m
+ 2 (E # V ( x) )! ( x) = 0
2
"x
!
Simple example:"
Assume a simple one dimensional “toy” atom "
in which the electron is free to move between
and outside
x=0 and x=L,
"=0. This is called the “particle in the box”.
V ( x) = 0,
" ! ( x ) 2m
+ 2 E! ( x) = 0
2
"x
!
2
Particle in the Box"
Rohlf Sec. 7.2, p193"
Inside the box the particle is free; V(x)=0.
!
" 2# (x) 2m
+ 2 E# (x) = 0
2
"x
!
The equation is a simple linear equation. "
Assume a solution:"
! ( x) = Be
cx
Insert into Schroedinger’s equation.
!2 Becx 2m
cx
+
EBe
!x 2
!2
c2 +
2m
E=0
2
!
c=± "
2m
E
2
!
= ±i
2m
E
2
!
Assumed solution:"
with"
! ( x) = Becx
2m
c=± ! 2 E
!
2m
= ±i 2 E
!
Write it in terms of momentum p .
E=
Then
1 2
mv
2
!p$
2
v =# &
"m%
2
2
'
1 !p$
p2
E = m# & =
2 "m%
2m
2m p 2
p
c .= ±i
E
=
±i
=
±i
!
!2
! 2 2m
The solution is:
2m
!=
p
i x
Fe !
p
"i x
+ Ge !
p
E
i
i
$ i !p x
#i x ' #i t
( px# Et )
# ( px+ Et )
! = " T = & Fe + Ge ! ) e ! = Fe !
+ Ge !
%
(
Superposition of wave moving to right and left: standing wave."
Exercise: Show these are oscillating functions which can be written:"
( )
( )
! = Acos kx + B sin kx
Note:
with"
p
k!
!
h
h 2! h
p = !k =
k=
= : the deBroglie relation
2!
2! "
"
! (x) = Feikx + Ge"ix
{ ( )
( )}
{ ( )
( )}
! = F cos kx + i sin kx + G cos kx " i sin kx
(
) ( ) (
) ( )
= F + G cos kx + i F " G sin kx
( )
( )
= Acos kx + B sin kx
( ))
( ( )
= ( Acos ( kx ) + B sin ( kx )) ( cos ($ t ) + i sin ($ t ))
#(x,t) = Acos kx + B sin kx e"i$ t
( ) ( )
( ) ( )
#(x,t) = Acos kx cos $ t + B sin kx cos $ t i i i i
Standing waves"
Solution"
Inside the box " !
= Asin kx + B cos kx
Outside the box "!
=0
Apply boundary conditions."
! (0) = 0 = Asin(0) + B cos 0
! ( L) = 0 = Asin kL
$ n#
! (x) = Asin &
% L
'
x)
(
"B=0
" kL = n#
k=
n#
L