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