Wave Properties of Particles


In 1924, Louis de Broglie postulated
that because photons have wave and
particle characteristics, perhaps all
forms of matter have both properties
Furthermore, the frequency and
wavelength of matter waves can be
determined
de Broglie Wavelength and
Frequency

The de Broglie wavelength of a
particle is
h h
 
mv p

The frequency of matter waves is
E
ƒ
h
The Davisson-Germer
Experiment





They scattered low-energy electrons from
a nickel target
They followed this with extensive
diffraction measurements from various
materials
The wavelength of the electrons calculated
from the diffraction data agreed with the
expected de Broglie wavelength
This confirmed the wave nature of
electrons
Other experimenters have confirmed the
wave nature of other particles
Importance of Hydrogen Atom


Hydrogen is the simplest atom
The quantum numbers used to
characterize the allowed states of
hydrogen can also be used to
describe (approximately) the allowed
states of more complex atoms
• This enables us to understand the
periodic table
More Reasons the Hydrogen
Atom is so Important

The hydrogen atom is an ideal system for
performing precise comparisons of theory
and experiment
• Also for improving our understanding of atomic
structure

Much of what we know about the
hydrogen atom can be extended to other
single-electron ions
• For example, He+ and Li2+
Early Models of the Atom

J.J. Thomson’s model
of the 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
Early Models of the Atom, 2

Rutherford
• 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
Difficulties with the Rutherford
Model

Atoms emit certain discrete characteristic
frequencies of electromagnetic radiation
• The Rutherford model is unable to explain this
phenomena

Rutherford’s electrons are undergoing a
centripetal acceleration and so should
radiate electromagnetic waves of the
same frequency
• The radius should steadily decrease as this
radiation is given off
• The electron should eventually spiral into the
nucleus

It doesn’t
Bohr’s Hydrogen Atom
(1 proton and 1 electron)
centripetal force = Coulomb force
mv2
e2
k 2
r
r
+
-
-
-
electron as a wave
(de Broglie)
2pr = , 2, 3, …
 = h/mv
This is the necessary condition for electron to maintain an orbit.
de Broglie Waves in the
Hydrogen Atom


In this example, three
complete wavelengths
are contained in the
circumference of the
orbit
In general, the
circumference must
equal some integer
number of
wavelengths
• 2 p r = n λ n = 1, 2, …
2
2pr = , 2, 3, …
 = h/mv
2
mv
e
k 2
rn
rn
2
2prn = nh/mv
2
nh
rn 
4p 2 ke2 m
2pke
vn 
nh
2
rn = (5.3 x 10-11)n2 (m); vn = 2 x106/n (m/s)
n
rn (nm)
vn (m/s)
2
0.212
1 x 106
3
0.477
5 x 105
1
0.053
2 x 106
Now, let’s think about the total energy of the electron in the nth orbit.
Etot
2
2
(1/2)mv
= Potential
+ Kinetic Energy
-ke /rEnergy
n
n
2
2pke
vn 
nh
2
2
nh
rn 
4p 2 ke2 m
2 2 2
2
p
k em
13.6
n
Etot  
  2 eV
2 2
nh
n
n
1
En (eV)
-13.6
2
-3.4
3
-1.5
∞
0
Etot
-
0
n=∞
-1.5 eV
n=3
-3.4 eV
n=2
+
-
-
-
Ionized state of Hydrogen: proton
-13.6 eV
n=1
2 2 2
2
p
k em
13.6
n
Etot  
  2 eV
2 2
nh
n
Electron Energy diagram of Hydrogen atom
Etot
n=∞
0
-1.5 eV
n=3
Electron has to absorb 12.1 eV energy for this
-3.4 eV
n=2
Electron has to absorb 10.2 eV energy for this
-13.6 eV
n=1
-
2p k e m
13.6
E 
  2 eV
2 2
nh
n
2
n
tot
2 2
lowest energy:
ground state
2 2 2
2
p
k em
13.6
n
Etot  
  2 eV
2 2
nh
E = hc/ n
1 
 1
E (n  m)  En  Em  13.6 2  2 
n 
m
in eV
1 
1
 1
 R 2  2 
 
n 
   nm
m
R  1.097 10 m
7
1
(Rydberg const.)
E(nm) or  < 0
Absorb photons of a given 
E (nm) or  > 0 Emit photons of a given 
Etot
E = hc/
n=∞
0
-1.5 eV
n=3
-3.4 eV
n=2
E (12) = 10.2 eV
= 10.2 x (1.6 x 10-19 J/eV)
= 1.63 x 10-18 J
  = 121 nm
Ultraviolet range
-13.6 eV
-
n=1
Ex. 29.6 What is the longest wavelength em radiation that can ionize
unexcited hydrogen atom?
ground state (n = 1)
smallest energy
n=1m=∞
1 
1
 1
 R 2  2 
 
n 
   nm
m
R  1.097 10 7 m 1
1/ = (1.097 x 107) x (0 – 1)
= - 1.097 x 107 (m-1)
 = -9.12 x 10-8 (m)
= -91.2 (nm) (- means absorption)
1 
1
1
 R 2  2 
 
   m3
3 m 
1 
1
1

R

 
 2
2 
   m 2
2 m 
1 
1

 R1  2 
 
   m1
 m 
Q. What is the shortest wavelength for the Balmer series?
Largest energy difference in Balmer series
From n = ∞ to m = 2 (Balmer)
1 
1
 1
 R 2  2 
 
n 
   nm
m
R  1.097 10 7 m 1
 = (R x (1/4 – 0))-1
= 365 nm
The Balmer series are in the visible range.
Hydrogen Spectrum
ABSORPTION SPECTRUM
ABSORPTION SPECTRUM
Q. What is the de Broglie wavelength of an electron that has a
kinetic energy of 100 eV?
After an electron is accelerated in 100 V potential difference, its
kinetic energy is 100 eV.
eV unit has to be converted into SI unit, Joule.
1 eV = 1.6 x 10-19 J
Ek = (1/2)mov2 = 1.6 x 10-17 J
v2 = 2Ek/mo = 2(1.6 x 10-17 J)/(9.1 x 10-31 kg)
= 3.52 x 1013 m2/s2
v = 5.93 x 106 m/s
low speed: no need to use relativistic
 = h/p
= h/mov = (6.6 x 10-34 Js)/(9.1 x 10-31 kg x 5.93 x 106 m/s)
= 1.23 x 10-10 m = 0.123 nm