Transcript Today: Quantum mechanics

Exam 3 is Tuesday Nov. 25
5:30-7 pm, 2103 Ch (here)
Students w / scheduled academic conflict please stay after class
Tues. Nov. 18 (TODAY) to arrange alternate time.
Covers: all material since exam 2.
Bring: Calculator
One (double-sided) 8 1/2 x 11 note sheet
Exam review: Thursday, Nov. 20, in class
Tues. Nov. 18, 2008
Phy208 Lect. 23
1
From Last Time…
Photoelectric effect
and light quantization
Tues. Nov. 18, 2008
Phy208 Lect. 23
2
Summary of Photoelectric effect

Light comes in photons - particles of light

h=Planck’s constant

Red photon is low frequency, low energy.

(Ultra)violet is high frequency, high energy.


E photon  hf  hc/ 
Electron in metal absorbs one photon

Can escape metal if photon energy large enough

Ephoton>Work function Eo

Excess energy Ephoton-Eo shows up as kinetic energy
Tues. Nov. 18, 2008
Phy208 Lect. 23
3
Photon properties of light




Photon of frequency f has energy hf
 E
photon  hf  hc/ 

h  6.626 1034 J  s  4.14 1015 eV  s

hc 1240eV  nm
Red light made of ONLY red photons

The intensity of the beam can be increased by
increasing the number of photons/second.

Photons/second = energy/second = power
Tues. Nov. 18, 2008
Phy208 Lect. 23
4
How many photons can you see?
In a test of eye sensitivity, experimenters used 1 milli-second
(0.001 s) flashes of green light. The lowest power light that
could be seen was 4x10-14 Watt.
How many green (500 nm, 2.5 eV) photons is this?
A. 10 photons
B. 100 photons
C. 1,000 photons
D. 10,000 photons
Tues. Nov. 18, 2008
4 10 J /s0.001s  4 10 J
4 10 J1eV /1.6 10 J 250eV
14
17
17
19
250eV 1photon/2.5eV  100photons

Phy208 Lect. 23
5
Quantization of light
Quantum mechanically, brightness can only be
changed in steps, with energy differences of hf.
Possible energies for green light (=500 nm)




One quantum of energy:
one photon
Two quanta of energy
two photons
etc
Think about light as a
particle rather than wave.
Tues. Nov. 18, 2008
Phy208 Lect. 23
E=4hf
E=3hf
Energy

E=2hf
E=hf
6
Thompson’s model of atom

J.J. Thomson’s model of atom




A volume of positive charge
Electrons embedded throughout
the volume
A change from Newton’s model
of the atom as a tiny, hard,
indestructible sphere
This model is not correct!
Tues. Nov. 18, 2008
Phy208 Lect. 23
7
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Tues. Nov. 18, 2008
Phy208 Lect. 23
8
Resulted in new model




Planetary model
Based on results of thin
foil experiments
Positive charge is
concentrated in the
center of the atom,
called the nucleus
Electrons orbit the
nucleus like planets orbit
the sun
Tues. Nov. 18, 2008
Phy208 Lect. 23
9
Difference between atoms

Simplest is Hydrogen:


Other atoms






1 electron orbiting 1 proton
number of orbiting negative electrons same as number of
positive protons in nucleus
Different elements have different number of
orbiting electrons
Helium: 2 electrons
Copper: 29 electrons
Uranium: 92 electrons!
Organized into periodic table of elements
First concentrate on hydrogen atom
Tues. Nov. 18, 2008
Phy208 Lect. 23
10
Planetary model and radiation

Circular motion of orbiting electrons:

electrons emit EM radiation at orbital frequency.

Similar to radio waves emitted by accelerating
electrons in a antenna.

In an atom, emitted EM wave carries away energy



Electron predicted to continually lose energy.
The electron would eventually spiral into the nucleus
However most atoms are stable!
Tues. Nov. 18, 2008
Phy208 Lect. 23
11
Line spectra from atoms

Atoms do emit radiation,
but only at certain discrete frequencies.

Emission pattern unique to different atoms

Spectrum is an atomic ‘fingerprint’,
used to identify atoms (e.g. in space).
Hydrogen
Mercury
Wavelength (nm)
Tues. Nov. 18, 2008
Phy208 Lect. 23
12
The Bohr atom

Retained ‘planetary’ picture
with circular orbits

Only certain orbits are stable

Radiation emitted only when
electron jumps from one
stable orbit to another.

Einitial
Photon
Efinal
Here, the emitted photon has
an energy of
Einitial-Efinal
Stable orbit
Stable orbit
Tues. Nov. 18, 2008
Phy208 Lect. 23
13
Energy levels
Instead of drawing orbits, just indicate energy an
electron would have if it were in that orbit.
Zero energy
n=4
n=3
E3  
13.6
eV
32
n=2
E2  
13.6
eV
22
E1  
13.6
eV
12

Energy axis


n=1
Tues. Nov. 18, 2008
Phy208 Lect. 23
14
Hydrogen atom energies


Quantized energy levels:
Each corresponds to
different
 Orbit radius
 Velocity
 Particle wavefunction
 Energy
Each described by a
quantum number n
Zero energy
n=4
n=3
E3  
13.6
eV
32
n=2
E2  
13.6
eV
22
E1  
13.6
eV
12


n=1
13.6
E n   2 eV
n
Tues. Nov. 18, 2008
Energy


Phy208 Lect. 23
15
Emitting and absorbing light
Zero energy
n=4
n=3
13.6
E 3   2 eV
3
n=2
13.6
E 2   2 eV
2
Photon
emitted
hf=E2-E1
n=1


E3  
13.6
eV
32
n=2
E2  
13.6
eV
22
E1  
13.6
eV
12
Photon
absorbed
hf=E2-E1
13.6
E1   2 eV
1
Photon is emitted when electron
drops fromone quantum
state to another
Tues. Nov. 18, 2008
n=4
n=3
Phy208 Lect. 23
n=1


Absorbing a photon of correct

energy makes
electron jump
to higher quantum state.
16
Hydrogen emission

This says hydrogen emits only
photons of a particular wavelength, frequency

Photon energy = hf,
so this means a particular energy.

Conservation of energy:

Energy carried away by photon is lost by the orbiting
electron.
Tues. Nov. 18, 2008
Phy208 Lect. 23
17
Hydrogen atom
An electron drops from an -1.5 eV energy level to one with
energy of -3.4 eV. What is the wavelength of the photon
emitted?
Zero energy
A. 827 nm
B. 653 nm
C. 476 nm
D. 365 nm
E. 243 nm
n=4
n=3
Photon
emitted
hf=E2-E1
n=2
hf = hc/
= 1240 eV-nm/ 
E 3  1.5 eV


E1  13.6 eV
n=1
Tues. Nov. 18, 2008
Phy208 Lect. 23
E 2  3.4 eV

18
Energy conservation for Bohr atom




Each orbit has a specific energy
En=-13.6/n2
Photon emitted when electron
jumps from high energy to low
energy orbit.
Ei – Ef = h f
Photon absorption induces
electron jump from
low to high energy orbit.
Ef – Ei = h f
Agrees with experiment!
Tues. Nov. 18, 2008
Phy208 Lect. 23
19
Hydrogen emission spectrum



n=4
Hydrogen is simplest atom
n=3
One electron orbiting around one
proton.
The Balmer Series of emission
lines given empirically
 1 1 
 RH  2  2 
2 n 
m
1
RH=0.01097nm-1
n = 3,  = 656.3 nm
n = 4,  = 486.1 nm
Hydrogen
Tues. Nov. 18, 2008
Phy208 Lect. 23
20
Balmer series

Transitions terminate at n=2

Each energy level has energy
En=-13.6 / n2 eV

E.g. n to 2 transition

Emitted photon has energy
E photon


 13.6eV   13.6eV 
 1 1 
 
 
13.6 eV  2  2 
2
2
2 n 
 n   2 
Emitted wavelength

Tues. Nov. 18, 2008
hc
E photon
1


1240 eV  nm 1 1
91.18nm



 2
2 

13.6 eV
2 n  1/22 1/n 2 
Phy208 Lect. 23
21
Why stable orbits?
Bohr argued that the stable orbits
are those for which the electron’s
orbital angular momentum L is quantized as
L  me v r  n
Electron
velocity

Electron
orbit
radius

h 
 

 2 
Integer:
n=1,2,3…

Bohr combined this with the Coulomb force to find allowed
orbital radii and energies.
Tues. Nov. 18, 2008
Phy208 Lect. 23
22
Including more physics
Circular orbit, electron is accelerating
(centripetal acceleration = v2/r = Force/mass)
Force causing this accel. is Coulomb force ke2/r2
between pos. nucleus and neg. electron
v2
FCoulomb
=
r
m
Also gives a condition for angular momentum.

Tues. Nov. 18, 2008
Phy208 Lect. 23
23
Bohr model of H-atom
v2
e2
= FCoulomb /m  k 2 / m
r
r
Orbital motion:

p  mv

centripetal
acceleration
mvr  L2  mke2r
2


Quantization:
Tues. Nov. 18, 2008

Coulomb force / mass

L2  mvr  n 2
Phy208 Lect. 23
2
2
24
Radius of H-atom states
L2  n 2
2
and
Quantization

L2  mke 2 r
Orbital motion
Quantized orbital radius
n
1
2
3
orbit radius
ao
4ao
9ao
Tues. Nov. 18, 2008
 m ke2 r
 2  2
2
r  n 
 n ao
2 
m ke 
n2

2
ao  Bohr radius
 0.529Å

Phy208 Lect. 23
25
Energy of H-atom states
Total Energy = kinetic +
 p 2 
  
2m 
 e 2 
k  
 2r 
potential
rn  n 2 ao
 e 2 
k 
 r 
ke 2  1
 e 2 
e2
  k    2
k 
2r
 r 
2ao n
9 10 N  m /C 1.6 10 C

ke
18


2.1810
J
10

2ao
20.52910 m
2
9
2
2
19
2
Quantized energy
Tues. Nov. 18, 2008
Phy208 Lect. 23
En  
13.6 eV
n2
26
Energy quantization in a pendulum
Swinging pendulum.
Larger amplitude, larger energy
Small energy
Large energy
Quantum mechanics:
Not every swing amplitude is possible
energy cannot change by arbitrarily small steps
Tues. Nov. 18, 2008
Phy208 Lect. 23
27
Energy quantization

Energy can have only certain discrete values
Energy states are separated by E = hf.
E = hf=3.3x10-34 J for pendulum
f = frequency
= spacing between energy levels
-34
h = Planck’s constant= 6.626 x 10 J-s
Suppose the pendulum has
Period = 2 sec
Freq = 0.5 cycles/sec

d


Tues. Nov. 18, 2008
Phy208 Lect. 23
E=mgd=(1 kg)(9.8 m/s2)(0.2 m)
~ 2 Joules
Emin=hf=3.3x10-34 J << 2 J
Quantization not noticeable
28
Question

This quantum system has
equally-spaced energy
levels as shown. Which
photon could possibly
 be
absorbed by this system?
A. 1240 nm
E photon
hc 1240 eV  nm




E3=7 eV
E3=5 eV
E2=3 eV
B. 413 nm
E1=1 eV
C. 310 nm
D. 248 nm
Tues. Nov. 18, 2008
Phy208 Lect. 23
29