weird experiments schrödinger equation
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Weird experiments Schrödinger equation. Bohr model of an atom 1913. centrifugal is Latin for "center fleeing" It does not exist!. http://regentsprep.org/Regents/physics/phys06/bcentrif/centrif.htm. Bohr model of an atom 1913. Potential energy of the electron. - PowerPoint PPT PresentationTRANSCRIPT
Bohr model of an atom 1913
2e
centrifugalm v
Fr
http://regentsprep.org/Regents/physics/phys06/bcentrif/centrif.htm
Bohr model of an atom 1913
2e
centrifugalm v
Fr
2
Coulomb 20
eF
4 r
Coulomb centrifugalF F
“Introduction to wave phenomena” by Akira Hirose and Karl Lonngren
Potential energy of the electron2
0
(J)4
eU
r
22
204
em ve
r r
Bohr model of an atom 1913
22
0
1 (J)
2 8e
em v
r
Kinetic energy of the electron
22
204
em ve
r r
21
2 eU m vTotal energy of the electron2
0
(J)8
eE
r
2
04e
e
m e rm vr
electron angular momentum
2 22 2 2
04e
e
m e rm v r
r
Bohr model of an atom 1913
electron angular momentum
Niels Bohr postulated that the momentum was quantized
( 1,2,3, )2e
hm vr n n
22 2 110
25.3 10 (m)
e
hr n n
m e
The radius is found to be
2
04e
e
m e rm vr
h is Planck’s constant6.626068 × 10-34 m2 kg / s
2
h
0
2
Bohr model of an atom 1913
http://csep10.phys.utk.edu/astr162/lect/light/bohr.html
The energy then becomes quantized
22 2 110
25.3 10 (m)
e
hr n n
m e
4
2 2 20
2
1
8
1= -13.6 (eV)
en
m eE
h n
n
2
0
8
eE
r
2
22 0
0 2
(J)
8e
e
hn
m e
Photo electric effect - Einstein
http://regentsprep.org/Regents/physics/phys05/catomodel/bohr.htmHoudon
Energy of a photon E = h
2 1E - E h
Einstein’s explanation
19
34
14
2.9 10 J
6.63 10 J sec4.4 10 Hz
cW
vh
Bohr model of an atom 1913
What is the frequency of the light that will be emitted by an electron as it moves from the n = 2 down to n = 1?
2
1 -13.6 (eV)nE n
1 = -13.6 1
4E h
Ionization implies n →
Experiment to understand the photo electric effect.
Experimental conclusions• The frequency must be greater than a “cut off
frequency” that changes with different metals.
• Kinetic energy of the emitted electrons depends upon the frequency of the incident light.
• Kinetic energy of the electrons is independent of the intensity of the incident light.
Sodium has a work function of W = 1.8 eV. Find the cutoff frequency.
c
W
h 144.4 10 Hz 19
34
1.8 1.6 10 J
6.63 10 J sec
cc
c
Å 6900
8
14
m3 10 sec4.4 10 Hz
76.9 10 m
A metal with a work function of 2.3 eV is illuminated with ultraviolet radiation = 3000 Ǻ. Calculate the energy of the
photo electrons that are emitted from the surface.
21
2 em v h W
hch
4.1 eV
34 8
7
6.63 10 3 10
3 10
196.63 10 J
214.1 2.3
2 em v 1.8 eV
Franck-Hertz experiment in mercury vapor. Electrons are accelerated and the current is monitored. 1914 (In 1887, Hertz noted that electrons would be emitted from a metal
that was illuminated with light.)
http://hyperphysics.phy-astr.gsu.edu/hbase/FrHz.html
2e 0
1m v qV
2
0e
e
2qVI n q A
m
eI n qvA
Reflected wave is strong if n = 2d sin
dd sin
Davisson-Germer experiment – electrons incident on nickel 1925
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/davger2.html
Interpretation of the Davisson-Germer experiment
Energy of a photon E = h
gv k
2 E
hk
1 E
k
2particle
particle
particle particle
1mvE 2
p (mv )
particlev
( 2 )
k
Planckh
2
de Broglie wavelengthwave energy
momentumwave velocity
2
mass velocity
velocity
h
pc
h
de Broglie argued that there was a wavelength that could be written from
de Broglie h
p
Interpretation of the Davisson-Germer experiment
particle
particle
E 1 E
p k
particlep k
h 2
2
de Broglieparticle
h
p de Broglien 2d sin
Schrödinger equation
energy of particles2p
U2m
energy of photon h
h ( 2 )2
2
deBroglie
h
U2m
2 2kU
2m
2
deBroglie
2 h
2U
2m
Schrödinger equation2 2k
U2m
j( t kz )
0e
jt
jk
z
2
2 22 jk kz
2 2kU
2m
2 2
2j Ut 2m z
22j U
t 2m
Schrödinger equation2 2
2j Ut 2m z
22j U
t 2m
a2a1
( z,t ) * ( z,t )dzprobability
( z,t ) * ( z,t )dz
1probability of finding a state in a
Max Born
2z a
Schrödinger equationa2a1
( z,t ) * ( z,t )dzprobability
( z,t ) * ( z,t )dz
j( t kz )
0e
elsewhere0
1 - 1 z 1
0
a1 a0 a2a1 a0 a2a1 a0 a2 a1 a0 a2
a0
+20-2
1
Schrödinger equation2 2
2
( z,t ) ( z,t )j U ( z,t )
t 2m z
( z,t ) Z( z )T( t ) 2 2
2
dT( t ) d Z( z )j Z( z ) T( t ) UZ( z )T( t )
dt 2m dz
2 2
2
1 dT( t ) 1 d Z( z )j U
T( t ) dt 2m Z( z ) dz
Schrödinger equationelectron in free space
2 2
2
1 dT( t ) 1 d Z( z )j U
T( t ) dt 2m Z( z ) dz
Ej t
0T( t ) T e
jkz jkzZ( z ) Ae Be2 2 2p k
E U2m 2m
2 2
2
1 dT( t ) 1 d Z( z )j
T( t ) dt 2m Z( z ) dz
Schrödinger equation
2 2
2
1 dT( t ) 1 d Z( z )j U( z )
T( t ) dt 2m Z( z ) dz
Ej t
0T( t ) T e Z( z ) Asin( kz ) Bcos( kz )
2 2 2p kE U
2m 2m
Schrödinger equation
Ej t
0T( t ) T e
Z( z ) Asin( kz ) Bcos( kz )
Z(0 ) 0 B 0 n
kL
Z( L ) 0
( z,t ) Z( z )T( t ) E
j t
0n z
AT e sinL
L
0normalization ( z,t ) * ( z,t )dz 1
Schrödinger equation2
2j Ut 2m
2 2 22
2 2 2x y z
( x, y,z ) X ( x )Y( y )Z( z )
Schrödinger equation2
2j Ut 2m
( x, y,z ) X ( x )Y( y )Z( z )
Ej tyx z
3
n yn x8 n z( x, y,z ) sin sin sin e
L L LL
2 22 22yx z
nn nE U
8mL L L L
Schrödinger equationE
j tyx z3
n yn x8 n z( x, y,z ) sin sin sin e
L L LL
Schrödinger equation
22j U
t 2m
2
sin
sin sin
22
22 2 2 2
r1 1 1r
rr r r
( r , , ) R( r ) ( ) ( )
Schrödinger equation
Schrödinger equation
element n l m s
Hydrogen 1 0 0 +1/2 or -1/2
Helium 1 0 0 +1/2 & -1/2
Beryllium 2 0 0 +1/2 & -1/2
Lithium 2 0 0 +1/2 or -1/2
Heisenberg uncertainty principle
http://www.aip.org/history/heisenberg/
( position ) ( momentum ) h
x p h
( energy ) ( time ) h
E t h
2 2m mv v v h
2 2 mv v h
vh
m v h
x(m v ) m v h