bohr model of the hydrogen atom. · bohr model of the hydrogen atom.! rohlf, p85-87" this is a...
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F(r)r̂ = mar̂
F(r) =e
2
4!"0r
2=ke
2
r2
where k =1
4!"0
k is not 2!
#!!!
a =v
2
r
ke2
r2=mv
2
r
Bohr Model of the Hydrogen Atom.!Rohlf, p85-87"
This is a semi-classical model which assumes the electron has well defined orbits "
(particle properties) and interference phenomena (wave properties) "
(This was covered in Honors Physics II.)"
F!"= ma"
ke2
r2=mv
2
rke
2= rmv
2= r
p2
m
Resonance condition for a stable orbit - constructive interference:
n! = 2"r r =n!
2"Also, p =
h
!! =
h
p
r =n!
2"p =
h
!ke
2= r
p2
m
Begin with r =n!2"
! =h
pke
2= r
p2
m
Substitute for r : ke2=
n!2"
#$%
&'(p
2
m.
Now substitute for ! : ke2=
nh
2" p#$%
&'(p
2
m=nhp
2"m
) pn =2"mke2
nh
Momentum Quantization"
Energy Quantization"
Momentum: pn =2!mke2
nh
Kinetic energy: K =pn
2
2mShow K =
2! 2mk2e4
n2h2
Potential energy: U ="ke2
rShow U = " 1
m
#$%
&'(
2!mke2
nh
#
$%&
'(
2
U = " 4! 2mk2e4
n2h2 ) "2K !!!!!!!!!!!!!!!K ) 2! 2mk2e4
n2h2
Bohr Model: Energy levels of Hydrogen"
E = K +U
Momentum: pn =2!mke2
nh!!!!ke
2= r
pn2
m=!r
1
m
2!mke2
nh
"#$
%&'
2
Kinetic energy: K =pn
2
2m=
1
2m
2!mke2
nh
"#$
%&'
2
=1
2m
4! 2m
2k
2e
4
n2h
2=
2! 2mk
2e
4
n2h
2
Potential energy: U =(ke2
r= (
1
m
"#$
%&'
2!mke2
nh
"#$
%&'
2
= (2K
Bohr Model (cont. 2)"
K =2!
2mk
2e4
n2h2
U = "4!
2mk
2e4
n2h2
E = K +U = K " 2K = "K = "2!
2mk
2e4
n2h2
= "C
n2
Exercise: Find the numerical value of En in units of
Jouls and convert to eV."
"
C =2!
2me
4k
2
h2
=1
16!2"
0
2
2!2me
4
h2
=me
4
8"0
2h
2
=9.109 #10
$31( ) 1.602 #10$19( )
4
8 8.854 #10$12( )
2
6.626 #10$34( )
2= 2.179 #10
$18
En= $
2.179 #10$18
n2
J En(eV)= $
2.179 #10$18
J
n2
1
1.602 #10$19
J/eV= $
13.60
n2
eV
En= $
13.60
n2
eV
Exercise: Calculate E in mks units (J) and in eV units."
Radius and speed of electron in lowest orbit"
!!!!!ke
2
r=!p
2
m!!!!!!!!!r =
ke2m
p2
p =2!mke
2
nh"!!!!r =
n2h
2#
0
!me2
=n
26.626 $10
%34( )2
8.854 $10%12( )
! 9.109 $10%31( ) 1.602 $10
%19( )2= 0.529 $10
%10$ n
2 m & 0.53n
2A
o
= 0.053n2
nm
r1= 0.53 A
o
Speed of electron in lowest orbit"
Exercise: find the speed of ab electron in the lowest orbit, ie find !=v/c."
! =v
c=
p
mc=
2"mke2
nh
1
m!!!!!k =
1
4"#0
!!!!! =!!!!e
2
2#0nhc
=1.602 $10
%19( )2
2 8.854 $10%12( ) 6.626 $10
%34( ) 3$108( )n
&1
137
1
n
For n = 1 ! &1
137
Fine structure constant: ' = e
2
2#0hc
= e
2
4"#0!c
&1
137
Speed of electron in lowest orbit"
Homework 8!
Due Oct.9!Bohr model: Ch. 3, problems 36, 42, 55 !
Schroedinger’s Equation -1925!Rohlf, Chapter 7, p. 191-218"
( )2
( )2
P rK V E V r E
m+ = + =
!!
Classical, non-relativistic physics relates kinetic, potential and total energy: "
In quantum physics the particles are treated as waves and obey a wave equation."
This is the Schroedinger equation. "
Symbolically "
P2
2m! +V! = H!
!!
2
2m"
2#("r , t) +V (
"r )#(
"r , t) = i!
$#("r , t)
$t
P2
2m! +V! = H!
Momentum operator: P " #i!$
Potential energy operator: V " V ("r )
Energy operator: H " i!%
%t
Schroedinger’s Equation"
!!
2
2m"
2#("r , t) +V (
"r )#(
"r , t) = i!
$#("r , t)
$t
Solve the Schroedinger equation by separation of variables."
!(!r , t) =" (
!r )T (t)Assume:"
!!
2
2m"2 (#T ) +V (
"r )(#T ) = i!
$(#T )
$t
= !!
2
2mT"2# +V (
"r )#T = i!#
$T
$t
Obtain:"
!!
2
2m"
2#("r , t) +V (
"r )#(
"r , t) = i!
$#("r , t)
$t
!!
2
2m
1
"#2" +V (
"r ) = i!
1
T
$T
$t= %
Divide the entire equation by !T, and set each side to a constant
i!!T
!t= "T
!T
!t+
i"
!T = 0
!!
2
2m
1
"#2" +V (
"r ) !$ = 0 #2" +
2m
!2
$ !V ("r )( )" = 0
Time dependent equation: !
Space dependent equation: !
!T
!t+
i"
!T = 0
Time equation:
!2" +
2m
!2
# $V ("r )( )" = 0
Space equation:
!T
!t+
i"
!T = 0
Solution to the t dependent equation:
bAebt+
i!
!Ae
bt= 0 b+
i!
!= 0 b = "
i!
!T = Ae
"i!
!t
Assume an exponential solution: bt
T Ae=
A!ebt
!t+
i"
!Ae
bt= 0Insert into the diff. eq. above:"
T = Ae! i"t
" =
#
!
The solution is then
Quantum mechanics relates the energy of a particle and the "
frequency of the associated wave according to: "
2
hE hf ! !
"= = = !
!T
!t+
iE
!T = 0 T = Ae
" iE
!t
Schroedinger’s equation in one dimension:"
( )2
2 2
( ) 2( ) ( ) 0
x mE V x x
x
!!
"+ # =
" !
The space dependent part of Schroedinger’s equation."
!2" +
2m
!2
# $V ("r )( )" = 0
!2" +
2m
!2
E #V ("r )( )" = 0
Assume a simple one dimensional “toy” atom "
in which the electron is free to move between x=0 and x=L,
and outside "=0. This is called the “particle in the box”.
2
2 2
( ) 2( ) 0, ( ) 0
x mV x E x
x
!!
"= + =
" !
Simple example:"
!2Be
cx
!x2
+2m
!2
EBecx
c2+
2m
!2
E = 0 c = ± "2m
!2
E = ±i2m
!2
E
Particle in the Box"Rohlf Sec. 7.2, p193"
!
"2# (x)
"x2
+2m
!2
E# (x) = 0
Inside the box the particle is free; V(x)=0.
The equation is a simple linear equation. "
Assume a solution:"( ) cxx Be! =
Insert into Schroedinger’s equation.
c = ± !2m
!2
E = ±i2m
!2
E
E =1
2mv
2v
2=
p
m
!"#
$%&
2
' E =1
2m
p
m
!"#
$%&
2
=p
2
2m
Assumed solution:" ( ) cxx Be! =
Write it in terms of momentum p .
Then . c = ±i
2m
!2
E = ±i2m
!2
p2
2m= ±i
p
!
The solution is: ! = Fei
p
!x
+Ge" i
p
!x
with"
! ="T = Fei
p
!x+Ge
# ip
!x$
%&'
()e# i
E
!t= Fe
i
!( px#Et )
+Ge# i
!( px+Et )
Exercise: Show these are oscillating functions which can be written:"
! = Acos kx( ) + B sin kx( ) with"
k !
p
!
p = !k =
h
2!k =
h
2!
2!
"=
h
": the deBroglie relationNote:
Superposition of wave moving to right and left: standing wave."
! = 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( )
#(x, t) = Acos kx( ) + B sin kx( )( )e"i$t
= Acos kx( ) + B sin kx( )( ) cos $t( ) + i sin $t( )( )
#(x, t) = Acos kx( )cos $t( ) + B sin kx( )cos $t( ) i i i i
! (x) = Feikx
+Ge" ix
Standing waves"
Apply boundary conditions."
Solution"
! (0) = 0 = Asin(0)+ Bcos0 " B = 0
! (L) = 0 = Asin kL " kL = n# k =n#L
! (x) = Asinn#L
x$%&
'()
! = Asin kx + Bcos kxInside the box "
Outside the box " ! = 0