weird experiments schrödinger equation

Bohr model of an atom 19132

ecentrifugal

m vFr

http://regentsprep.org/Regents/physics/phys06/bcentrif/centrif.htm

Bohr model of an atom 1913

2e

centrifugalm vFr

2

Coulomb 20

eF4 r

Coulomb centrifugalF F

“Introduction to wave phenomena” by Akira Hirose and Karl Lonngren

Potential energy of the electron2

0

(J)4eU

r

22

204

em ver r

Bohr model of an atom 1913

22

0

1 (J)2 8e

em vr

Kinetic energy of the electron

22

204

em ver r

212 eU m vTotal energy of the electron

2

0

(J)8eE

r

2

04e

em e r

m vr

electron angular momentum

2 22 2 2

04e

em e r

m v rr

Bohr model of an atom 1913

electron angular momentum

Niels Bohr postulated that the momentum was quantized

( 1,2,3, )2ehm vr n n

22 2 110

2 5.3 10 (m)e

hr n n

m e

The radius is found to be

2

04e

em e r

m vr

h is Planck’s constant6.626068 × 10-34 m2 kg / s

2h

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

2 5.3 10 (m)e

hr n n

m e

4

2 2 20

2

18

1= -13.6 (eV)

en

m eE

h n

n

2

0

8eE

r

2

22 0

0 2

(J)8

e

ehn

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 J6.63 10 J sec

4.4 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 14

E 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.

cWh

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.

212 em v h W

hch

4.1 eV 34 8

7

6.63 10 3 10

3 10

196.63 10 J

21 4.1 2.32 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

1 m v qV2

0e

e

2qVI n q Am

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 Ehk

1 Ek

2

particleparticle

particle particle

1 mvE 2p (mv )

particlev

( 2 )k

Planckh2

de Broglie wavelengthwave energymomentumwave velocity

2mass velocity

velocity

hpc

h

de Broglie argued that there was a wavelength that could be written from

de Broglie hp

Interpretation of the Davisson-Germer experiment

particle

particle

E 1 Ep k

particlep kh 22

de Broglieparticle

hp

de Broglien 2d sin

Schrödinger equation

energy of particles2p U

2m

energy of photon hh ( 2 )2

2

deBroglie

h

U2m

2 2k U2m

2

deBroglie

2 h2

U2m

Schrödinger equation2 2k U2m

j( t kz )

0e

jt

jkz

22 2

2 jk kz

2 2k U2m

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 aMax Born

2z a

Schrödinger equationa2a1

( z,t ) * ( z,t )dzprobability

( z,t ) * ( z,t )dz

j( t kz )

0e

elsewhere0

1 - 1 z 10

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

2dT( t ) d Z( z )j Z( z ) T( t ) UZ( z )T( t )dt 2m dz

2 2

21 dT( t ) 1 d Z( z )j U

T( t ) dt 2m Z( z ) dz

Schrödinger equationelectron in free space

2 2

21 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 kE U

2m 2m

2 2

21 dT( t ) 1 d Z( z )j

T( t ) dt 2m Z( z ) dz

Schrödinger equation

2 2

21 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 U2m 2m

Schrödinger equation

Ej t

0T( t ) T e

Z( z ) Asin( kz ) Bcos( kz )

Z(0 ) 0 B 0 nkL

Z( L ) 0

( z,t ) Z( z )T( t ) Ej t

0n zAT e sinL

L0

normalization ( 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 z3

n yn x8 n z( x, y,z ) sin sin sin eL L LL

2 22 22yx znn nE U

8mL L L L

Schrödinger equationEj tyx z

3

n yn x8 n z( x, y,z ) sin sin sin eL 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 & -1/2Lithium 2 0 0 +1/2 or -1/2

Heisenberg uncertainty principle

http://www.aip.org/history/heisenberg/

( position ) ( momentum ) hx p h

( energy ) ( time ) hE t h

2 2m mv v v h2 2

mv v h vh

m v h

x(m v ) m v h