Download Quantum physics

Quantum physics
“Anyone who is not shocked by the quantum theory has not
understood it.” – Niels Bohr, Nobel Prize in 1922 (1885-1962)
“I can safely say that nobody understands quantum
physics” – Richard Feynman Nobel Lecture, 1966, (1918-1988)
so far…
 we have treated light as being waves and used that
formalism to treat optics and interference
 we have seen that under extreme conditions (very high
velocities) the Newtonian description of mechanics
breaks down and the relativistic treatment discovered by
Einstein must be used.
 Now, we will see that the description of light entirely in
terms of waves breaks down at very small scales.
 In addition, we will see that objects that have mass, which
we usually think of as particles (like electrons) also exhibit
wave phenomena.
quantum physics
2
photoelectric effect
 when light hits a metal, electrons are
released. By providing a voltage difference
between the metal and a collector, these
electrons are collected and produce a
current.
quantum physics
3
however…
one observes that:
• if the frequency of the light is too low, no electrons
are emitted
•the maximum kinetic energy of the electrons is
independent of the intensity.
• the maximum kinetic energy increases linearly
with frequency
•the electrons are emitted almost instantaneously,
even at very low light intensities
These observations contradict the classical
description. It suggest that energy is delivered to the
electrons in the metal in terms of localized packets
of energy. The photons in the light beam are thus
seen as ‘particles’ that deliver packets of energy
(so-called energy quanta) to the electron it strikes.
quantum physics
4
photo-electric effect
The energy carried by a photon: E = h f
h: Planck’s constant (h = 6.63x10-34 Js)
f: frequency (where c = f )
The energy is localized in the photon-particle
The maximum kinetic energy of a released
electron: KEmax=hf-
with: : the workfunction (binding energy of
electron to the metal)
So only if h f >  can electrons be released from
the metal
fc=/h : fc is the cut-off frequency
c=c/fc=(hc)/: the cut-off wavelength
see table 27.1 for work functions for various metals
quantum physics
5
example
light with a wavelength of 400 nm is projected on a sodium metal
surface ( = 2.46 eV).
a) what is the energy carried by a single photon?
b) what is the maximum kinetic energy of the released electrons?
c) what is the cut-off wave length for sodium?
d) what happens if light with a wavelength of 600 nm is used?
a) E=hf=hc/=6.63x10-34Js x 3x108/(400x10-9)=4.97x10-19 J
in eV (1 eV=1.6x10-19 J)
=3.11 eV
b) KEmax=hf-=3.11-2.46 eV = 0.65 eV
c) c=c/fc=(hc)/=6.63x10-34x3x108/(2.46x1.6x10-19J)=505x10-9 m
the cut-off wavelength is 505 nm.
d) if light with a wavelength of 600 nm is used: no electrons are emitted
Note: if f<fc no electrons emitted
if > c no electrons emitted
quantum physics
6
particle-wave dualism
 So, is light a wave or particle phenomenon?
experiment
can be described by can be described by
light as waves
light as particles
reflection
X
X
refraction
X
X
interference
X
diffraction
X
polarization
X
photo-electric effect
X
 answer: both!
quantum physics
7
question
 light from a far-away star is used to perform a double slit
experiment. Approximately once per 10 minutes a single
photon from the star arrives at the double slit setup on
earth. Which of the following is true?
 a) since light is a wave-phenomenon, an interference
pattern will be seen on a screen placed behind the
double slits.
 b) since only one photon arrives every 10 minutes,
interference is not possible since one can hardly think of
the light coming in as waves
interference is a pure wave-phenomena; it doesn’t depends on how
many photons are there!
quantum physics
8
question
 instead of a light source, an electron gun firing electrons at high
speeds is used in a double-slit experiment. Which of the following is
true?
a) since electrons are massive particles, an interference pattern is not
produced
 b) electrons are similar to photons; they exhibit both wave and
particle phenomena. In this case, electrons behave like waves and
an interference pattern is produced.
All particles can be associated with a characteristic wavelength
= h/p where p=momentum.
In the case of particles with mass, this wavelength is called the
‘de Broglie’ wavelength. Particles with mass (like the electron) can
exhibit wave phenomena just like like light.
quantum physics
9
interference pattern
A
P1
P2
P1
B
P2
If one of the slits in a double slit experiment is closed, one sees only
a diffraction pattern from a single slit (P1). If the other slit is opened and
the first one closed, one sees only the diffraction pattern from the other
slit (P2). If both are opened, one does not simply see the sum of P1 and
P2 (like in A), but the double-slit interference pattern (like in B).
The reason is the following:
Remember that the intensity (I) is proportional to the E-field squared:
I~E2=E02cos2. In A, it is assumed that the intensities add: Isum=I1+I2 .
However, one should add the E-fields (which can be positive or
negative) and than squared, like in B: Isum=(E1+E2)2 where E1 and E2 are
treated as vectors.
quantum physics
10
and if you think that you’ve seen it all…
A
B
let’s assume I determine through which hole the photon (or electron)
goes by placing a detector before the slits. Would I still observe an
interference pattern like in B?
Answer: no! By measuring the location of the photon, we have ‘turned’
the light-wave into a particle and the interference pattern gets lost.
The interference happens only if the photon or electron can go through
both slits.
quantum physics
11
quantum physics
12
heisenberg’s uncertainty principle
If we want to determine the location and
velocity (momentum) of an electron at a
certain point in time, we can only do that with
limited precision.
Let’s assume we can locate the electron using
a powerful light microscope. Light scatters off
the electron and is detected in the
microscope. However, some of the
momentum is transferred and observing the
electron means we can only determine its
note ħ = h/(2)
velocity (momentum) with limited accuracy.
with x: precision of position measurement
with p: precision of momentum (mv) measurement
this can also be expressed in terms of energy and time measurements
Eth/(4)
with E: precision of energy measurement
with p: precision of time measurement
xph/(4)
quantum physics
13
example
The location of an electron is measured with an uncertainty of 1 nm.
One also tries to measure the velocity of the electron. What is the
(minimum) uncertainty in the velocity measurement? The mass of the
electron is 9.11x10-31 kg.
use the uncertainty principle:
xph/(4) with x=1x10-9 m, h=6.63x10-34 Js
so p 6.63x10-34/(4 x 1x10-9)=5.28x10-26 kgm/s
vmin = 5.28x10-26/9.11x10-31=5.79x104 m/s
(use p=mv)
quantum physics
14
photons as particles and quanta
Some other examples of where the particle nature of
light plays a role:
•Photo-electric effect
•Black-body radiation
•bremsstrahlung
•Compton effect
quantum physics
15
black-body radiation
A black body is an object that absorbs all electromagnetic radiation
that falls onto it. They emit radiation, depending on their temperature. If
T<700 K, almost no visible light is produced (hence a
‘black’ body).
The power emitted from a black body of surface area A at temperature
T is P =  A T4 with =5.67x10-8 W/m2K4
The peak in the intensity spectrum
varies with wavelength using the
Wien displacement law:
maxT = 0.2898x10-2 m K
(classical)
Until 1900, the intensity distribution,
predicted using classical equations,
predicted a steep rise at small
wavelengths. However, the opposite
was determined experimentally…
quantum physics
16
Planck to the rescue
 Max Planck devised a theory for a
simple black body that could describe
the measured spectra.
 The key assumption E = h f for the
photons.
f = frequency of the light (Hz)
h= planck’s constant (6.63x10-34 Js)
In essence, it is hard to emit light of
small wavelengths (high frequency)
since a lot of energy is required to
produce those photons.
quantum physics
17
example
Hot lava can be considered as a black
body emitting radiation at a variety of
temperatures.
If temperature of molten lava is about 1200 0C,
what is the peak wave length of the light
emitted?
max T = 0.2898x10-2 m K
T=1200+273=1473 K
max= 0.2898x10-2/1473
=1.96x10-6 m=1970 nm
The peak is in the infrared region (not
visible by eye), but closest to the colors
red/orange in the visible spectrum
quantum physics
18
X-rays
 when energetic electrons are shot at a
material, photons with small
wavelengths (~0.1 nm) are produced.
 The spectrum consist of two
components:
(1) A broad bremsstrahlung spectrum
(2) Peaks at characteristic
wavelengths depending on the
material (see next chapter)
 the bremsstrahlung (braking radiation)
is due to the deflection of the electron
in the field of the charged nucleus.
 a light quantum is produced when the
electron is deflected. It takes away
energy from the electron
quantum physics
19
bremsstrahlung
 assume electrons are accelerated in a potential of V Volts.
 their kinetic energy is E = e V with e =1.6x10-19 C and V the potential
 If the electron is completely stopped in the material, all its kinetic
energy is converted into the photon with maximum frequency fmax
and hence minimum wavelength min
 if it is merely deflected, the electron retains some of its kinetic energy,
so the frequency f is smaller than fmax and its wavelength  larger than
min.
 e V = h fma x= h c / min
quantum physics
20
example
 an X-ray spectrum is analyzed and the
minimum wavelength is found to be
0.35 Angstrom (1 Angstrom = 10-10 m).
What was the potential over which the
electrons were accelerated before
they interacted with the material?
eV = h fmax = h c / min
V = h c / (min e)
=6.6x10-34 x 3x108 / (0.35x10-10x1.6x10-19)
= 3.55x104 V
quantum physics
21
question
 X-rays are sometimes used to identify crystal structures of materials.
This is done by looking at the diffraction pattern of X-rays scattered off
the material (see ch 27.4). Why are X-rays used for this and not for
example visible light?
 a) the wavelength of X-rays is close to the spacing between atoms in
a crystal
 b) since the frequency (and thus energy) of X-rays is much larger
than that of visible light, they are easier to detect
 c) X-rays are much easier to produce than visible light
In order to observe structures of a given scale, the probe
must have a wavelength of the same scale or smaller.
quantum physics
22
compton effect
 When photons (X-rays) of a certain wavelength are directed towards
a material, they can scatter off the electrons in the material
 If we assume the photon and the electron to be classical particles, we
can describe this as a normal collision in which energy and
momentum conservation must hold.
 after taking into account relativistic effects (see previous chapter) one
finds that:
quantum physics
23
compton scattering
  = -0 = h /(me c) x (1-cos)
with:  = wavelength of photon after collision
0 = wavelength of photon before collision
h/(mec) = “ Compton wavelength” of electron (2.43x10-3 nm)
me = mass of electron

 = angle of outgoing X-ray relative
to incoming direction
This formula comes from energy and
momentum conservation.
quantum physics
0
24
=-0=h/(mec) x (1-cos)
example
 A beam of X-rays with 0=10-12 m is used to bombard a material.
 a) What is the maximum shift in wavelength that can be observed due
to Compton scattering?
 b) What is the minimum shift in wavelength that can be observed due
to Compton scattering?
 c) What are the minimum and maximum kinetic energies of the struck
electrons, ignoring binding to the material they are in.
a) maximum shift occurs if cos=-1 (=1800). This is usually referred to as
Compton backscattering. in that case:
=2h/(mec)=2x2.43x10-12=4.86x10-12 m
b) minimum shift occurs if cos=1 (=00) in which case essentially no
collision takes place: =0
c) gain in kinetic energy by electron is loss in energy of x-ray:
case b) no kinetic energy gained by electron
case c) energy of X-ray before collision: hf=hc/0=1.98x10-13 J
energy of X-ray after collision: hf=hc/(0+) =3.38x10-14 J
kinetic energy gained by electron: 1.64x10-13 J=1.0 GeV (giga)
quantum physics
25
applications
quantum physics
26