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