Phys 4190 lec (3)

Lec. (3)
The particle-wave duality
1

Photons and Matter Waves
Photons Matter
de Broglie wavelength
2

The physics of the nanometer scale tends to become dominated by quantum
physics.
The first laws of quantum physics dealt with energy quantization.
A new universal constant, Planck’s constant h, was introduced in physics in addition to
other constants like the speed of light, c, the gravitation constant, G, and the charge
quantum, e.
3

The photoelectric effect: small energy particles of light, the so-called light quanta,
are able to knock out electrons from metals just like one billiard ball hitting
another.
Fundamental objects in the physical world, electrons, protons, neutrons, photons
and other leptons, hadrons, and field quanta, all have the same dual nature: they
are at the same time both particles and waves.
the nature of light was proposed in the 17th century, it was debated whether light
were particles (corpuscles), as claimend by Newton, or waves, as claimed by
Huygens.
4

Young’s famous double-slit interference experiments that demonstrated that light
were waves.
In 1913 Niels Bohr published his theory of the hydrogen atom, explaining its
stability in terms of stationary states. Bohr did not explain why stationary states
exist.
Planck’s formula for the energy distribution in black-body radiation demonstrated
that light possesses some element of particle nature.
This particle aspect became more evident with Einstein’s explanation of the
photoelectric effect: small energy parcels of light, the so-called light quanta, are
able to knock out electrons from metals just like one billiard ball hitting another.
5

Two strong evidences for the existence of electron waves. (a) An electron diffraction
from the quasicrystal Al70Co11Ni19 (b) Electron waves on the surface of copper
detected by scanning tunnel microscopy (STM). The waves are trapped inside a ring of
iron atoms. The ring is created by pushing the iron atoms around on the copper
surface using an atomic force microscope. 6

de Broglie’s famous relation between the momentum p of a particle and the
wave length (or wave number ) of its associated wave
Existence of diffraction patterns when electrons are shot through a thin metal
film.
In the beginning of 1926 Schroodinger published his wave equation providing a
firm mathematical foundation for de Broglie’s ideas.
7

de Broglie waves
• First a little basics about waves. Waves are disturbances through a
medium (air, water, empty vacuum), that usually transfer energy.
• Here is one:
8

de Broglie waves
• The distance between each bump is called a wavelength
(λ), and how many bumps there are per second is called
the frequency (f). The velocity at which the wave crest
moves is jointly proportional to λ and f:
V = λ f
• Now there are two velocities associated with the wave:
the group velocity (v) and the phase velocity (V).
When dealing with waves going in oscillations (cycles of
periodic movements), we use notations of angular
frequency (ω) and the wavenumber (k) – which is
inversely proportional to the wavelength. The equations
for both are:
ω = 2πf and k = 2π/ λ
9

de Broglie waves
• The phase velocity of the wave (V) is directly
proportional to the angular frequency, but inversely
proportional to the wavenumber, or:
V = ω / k
The phase velocity is the velocity of the oscillation
(phase) of the wave.
• The group velocity is equal to the derivative of the
angular frequency with respect to the wavenumber, or:
v = d ω / d k
The group velocity is the velocity at which the energy of
the wave propagates. Since the group velocity is the
derivative of the phase velocity, it is often the case that
the phase velocity will be greater than the group
velocity.
10

Derivation for De Broglie Equation
• De Broglie, in his research, decided to look at Einstein’s
research on photons – or particles of light – and how it
was possible for light to be considered both a wave and
a particle. Let us look at how there is a relationship
between them.
We get from Einstein (and Planck) two equations for
energy:
E = h f (photoelectric effect) & E = mc2 (Einstein’s
Special Relativity)
Now let us join the two equations:
E = h f = m c2
11

• From there we get:
h f = p c (where p = mc, for the momentum of a photon)
h / p = c / f
Substituting what we know for wavelengths (λ = v / f, or in this
case c / f ):
h / mc = λ
De Broglie saw that this works perfectly for light waves, but does it
work for particles other than photons, also?
12

• In order to explain his hypothesis, he would have to
associate two wave velocities with the particle. De Broglie
hypothesized that the particle itself was not a wave, but
always had with it a pilot wave, or a wave that helps guide
the particle through space and time. This wave always
accompanies the particle. He postulated that the group
velocity of the wave was equal to the actual velocity of the
particle.
• However, the phase velocity would be very much
different. He saw that the phase velocity was equal to the
angular frequency divided by the wavenumber. Since he
was trying to find a velocity that fit for all particles (not
just photons) he associated the phase velocity with that
velocity. He equated these two equations:
V = ω / k = E / p (from his earlier equation c = (h f) / 13
p )

• From this new equation from the phase velocity we can
derive:
V = m c2 / m v = c2 / v
Applied to Einstein’s energy equation, we have:
E = p V = m v (c2 / v)
This is extremely helpful because if we look at a photon
traveling at the velocity c:
V = c2 / c = c
The phase velocity is equal to the group velocity! This
allows for the equation to be applied to particles, as
well as photons.
14

• Now we can get to an actual derivation of the De Broglie equation:
p = E / V
p = (h f) / V
p = h / λ
With a little algebra, we can switch this to:
λ = h / m v
This is the equation De Broglie discovered in his 1924 doctoral
thesis! It accounts for both waves and particles, mentioning the
momentum (particle aspect) and the wavelength (wave aspect).
This simple equation proves to be one of the most useful, and
famous, equations in quantum mechanics!
15

De Broglie and Bohr
• De Broglie’s equation brought relief to many people,
especially Niels Bohr. Niels Bohr had postulated in his
quantum theory that the angular momentum of an
electron in orbit around the nucleus of the atom is equal
to an integer multiplied with h / 2π, or:
n h / 2π = m v r
We get the equation now for standing waves:
n λ = 2π r
Using De Broglie’s equation, we get:
n h / m v = 2π r
This is exactly in relation to Niels Bohr’s postulate!
16

De Broglie and Relativity
• Not only is De Broglie’s equation useful for small particles, such as
electrons and protons, but can also be applied to larger particles,
such as our everyday objects. Let us try using De Broglie’s equation
in relation to Einstein’s equations for relativity. Einstein proposed this
about Energy:
E = M c2 where M = m / (1 – v2 / c2) ½ and m is rest mass.
Using what we have with De Broglie:
E = p V = (h V) / λ
Another note, we know that mass changes as the velocity of the
object goes faster, so:
p = (M v)
Substituting with the other wave equations, we can see:
p = m v / (1 – v / V) ½ = m v / (1 – k x / ω t ) ½
One can see how wave mechanics can be applied to even Einstein’s
theory of relativity. It is much bigger than we all can imagine!
17

Quantum Theory
Schrodinger started with the idea of
Conservation of Energy: KE + PE = Etotal .
He noted that
• KE = (1/2)mv2
= p2/2m, and that =h/p,
so that p = h/
= (h/2)*(2/)
= k = p, so
KE = 2k2/2m
• Etotal = hf = (h/2)*(2f) = .

Example (1): What are the de Broglie wavelengths of
electrons with the following values of kinetic energy?
(a) 1.0 eV and (b) 1.0 keV.
(a) The momentum of the electron is
p  2
mK
  31   
31

  
2 9.11 10 kg 1.0 eV 1.60 10 J/eV
25

 
5.4 10 kg m/s
and
34
6.626 10 Js 9
1.23 10 m 1.23 nm.
25
5.4 10 kg m/s
  


h
  


p

19

Example continued:
(b) The momentum of the electron is
p  2
mK
  31  3  
31

   
2 9.11 10 kg 1.0 10 eV 1.60 10 J/eV
23

 
1.7 10 kg m/s
and
34
6.626 10 Js 11
3.88 10 m 38.8 pm.
23
1.7 10 kg m/s
  


h
  


p

20

21
Problem
 What is the de Broglie wavelength of a 50 kg person
traveling at 15 m/s?
(h = 6.6 x 10-34 J s) comparable to spacing
34
h
h
6.626 10 Js 37
8.835 10 m

50kg 15 m/s


 


 
mv
p


Emission Energetics - I
Problem: A sodium vapor light street light emits bright yellow light of
wavelength = 589 nm. What is the energy change for a sodium atom
involved in this emission? How much energy is emitted per mole of
sodium atoms?
Plan: Calculate the energy of the photon from the wavelength, then
calculate the energy per mole of photons.
Solution:
h x c
Ephoton = hf = =
wavelength
( 6.626 x 10 -34J s)( 3.00 x 10 8m/s)
589 x 10 -9m
Ephoton = 3.37 x 10 -19J
Energy per mole requires that we multiply by Avogadro’s number.
Emole = 3.37 x 10 -19J/atom x 6.022 x 1023 atoms/mole = 2.03 x 105 J/mol
Emole = 203 kJ / mol 22

Emission Energetics - II
Problem: A compact disc player uses light with a frequency of
3.85 x 1014 per second. What is this light’s wavelength? What portion of
the electromagnetic spectrum does this wavelength fall? What is the
energy of one mole of photons of this frequency?
Plan: Calculate the energy of a photon of the light using E=hf, and
wavelength  C = f . Then compare the frequency with the
electromagnetic spectrum to see what kind of light we have. To get the
energy per mole, multiply by Avogadro’s number.
Solution:
3.00 x 108m/s
3.85 x 1014/s
wavelength  = c / f = = 7.78 x 10 -7 m = 778 nm
778 nm is in the Infrared region of the electromagnetic spectrum
Ephoton = hf = (6.626 x 10 -34Js) x ( 3.85 x 1014 /s) = 2.55 x 10 -19 J
Emole = (2.55 x 10 -19J) x (6.022 x 1023 / mole) = 1.54 x 105 J/m23ole

Phys 4190 lec (3)

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