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Wave Particle Duality
Quantum Physics Lesson 3
Today’s Objectives

Explain what is meant by wave-particle duality.

Describe the main points of de Broglie’s
hypothesis that matter particles also have a
wave-like nature.

State and use the equation λ = h/p = h/mv

Describe evidence for de Broglie’s hypothesis.
Wave particle duality

We have seen……………..
Photons : Quanta (particles) of light
Electrons: Being diffracted. A property
of waves
Prince Louis de Broglie
1892-1987

Electrons should not be
considered simply as
particles, but that frequency
must be assigned to them
also.
(1929, Nobel Prize Speech)
De Broglie (1924)


Suggested that particles such as electrons might
show wave properties.
He summised that the de Broglie wavelength,
λ was given by:
m = mass
h
h
 
p mv
v = velocity of the particle
Note that:• This is a matter wave equation not
electromagnetic wave
• The de Broglie wavelength can be altered by
changing the velocity of the particle.
In words...
Planck' s constant
de Broglie wavelengt h 
momentum
Planck' s constant
de Broglie wavelengt h 
mass  velocity
The diffraction tube
Summary of Experiment



Beam of electrons directed at a thin
metal foil.
Rows of atoms cause the electron beam
to be diffracted in certain directions
only.
We observe rings due to electrons being
diffracted by the same amount from
grains of different orientations, at the
same angle to the incident beam.
What we should see
Electron diffraction




1927: Davisson & Gerner confirmed this
prediction with experiments using
electron beams.
They actually used a nickel target
instead of a carbon one (we used)
The wavelength they measured agreed
with de Broglie
There is a relationship between the
accelerating voltage V and the k.e. of
the particles
Diffraction effects have been shown for
Hydrogen atoms
Helium atoms
Neutrons
Neutron diffraction is an excellent way of
studying crystal structures.
De Broglie Wavelength

In 1932, De Broglie
discovered that all
particles with
momentum have an
associated
wavelength.
h
h


p
mv
What is the wavelength of a human being, assuming
he/she weighs 70 kg, and is running at 25 m/s?
Practice Questions



1.Find the wavelength of an electron of mass
9.00 × 10-31 kg moving at 3.00 × 107 m s-1
2. Find the wavelength of a cricket ball of mass
0.15 kg moving at 30 m s-1.
3. It is also desirable to be able to calculate the
wavelength associated with an electron when the
accelerating voltage is known. There are 3 steps
in the calculation. Calculate the wavelength of
an electron accelerated through a potential
difference of 10 kV.

Step 1: Kinetic energy
EK = eV = 1.6 × 10-19 × 10000 = 1.6 × 10-15 J
Step 2:
EK = ½ mv2 = ½m (mv) 2 = p2 / 2m,
so momentum
p = √2mEk = √2 × 9.1 × 10-31 × 1.6 × 10-15
= 5.4 × 10-23 kg m s-1

Step 3: Wavelength
λ = h / p = 6.63 × 10-34 / 5.4 × 10-23 = 1.2 × 10-11 m
= 0.012 nm.

Slit spacing, d
Wavelength, 
Distance to screen, L
Fringe spacing, x
Laser d2
d1
L1
Slits
L2
Screen 1
Screen 2
End