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Quantum physics

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... of a light source, an electron gun firing electrons at high speeds is used ... In this case, electrons behave like waves and an interference pattern is produced. ... – PowerPoint PPT presentation

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Title: Quantum physics


1
Quantum physics
Anyone who is not shocked by the quantum theory
has not understood it. Niels Bohr, Nobel Price
in 1922 (1885-1962)
I can safely say that nobody understand
quantum physics Richard Feynman Nobel Lecture,
1966, (1918-1988)
  • PHY232
  • Remco Zegers
  • zegers_at_nscl.msu.edu
  • Room W109 cyclotron building
  • http//www.nscl.msu.edu/zegers/phy232.html

2
quiz (extra credit)
  • a distant star moves at a velocity of 0.5 times
    the speed of light away from us. It emits light
    that can be detected in earth-based telescopes.
    What is the speed of the light of the radiation
    we receive from these quasars? Ignore the fact
    that the telescope is in air and not in vacuum.
  • a) it is less than the speed of light in vacuum
    (3x108 m/s) by a factor 1/??(1-v2/c2)?(1-(0.5c)2
    /c2)0.87
  • b) it is larger than the speed of light in vacuum
    (3x108 m/s) by a factor ?1/?(1-v2/c2)1/?(1-(0.5c
    )2/c2)1.15
  • c) it is equal to the speed of light in vacuum
    (3x108 m/s)

3
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 designed by Einstein must be used.
  • Now, we will see that the description of light in
    terms of waves breaks down when looking at very
    small scales. In addition, we will see that
    objects that we usually refer to as particles
    (like electrons) exhibit wave-phenomena.

4
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.
  • if light is described in terms of waves one would
    expect that (classical description)
  • independent of the frequency of the light,
    electrons should be emitted if one waits long
    enough for sufficient energy to be absorbed by
    the metal
  • the maximum kinetic energy depends on the
    intensity (more energy absorbed)
  • the kinetic energy of the electrons is
    independent on the frequency (wavelength of the
    light) and only depends on the intensity
  • electrons take a little time to be released since
    sufficient energy needs to be absorbed

5
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
    well-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.

6
photo-electric effect
The energy carried by a photon Ehf h plancks
constant (h6.63x10-34 Js) f frequency with
cf? The energy is localized in the
photon-particle The maximum kinetic energy of a
released electron KEmaxhf-? with ? the
workfunction (binding of electron to the
metal) So only if hfgt? will electrons be
released from the metal fc?/h fc is the
cut-off frequency ?cc/fc(hc)/? the cut-off
wavelength
see table 27.1 for work functions for various
metals
7
example
  • light with a wavelength of 400 nm is projected on
    a sodium metal
  • surface (?2.46 eV).
  • what is the energy carried by a single photon?
  • what is the maximum kinetic energy of the
    released electrons?
  • what is the cut-off wave length for sodium?
  • what happens if light with a wavelength of 600 nm
    is used?

a) Ehfhc/?6.63x10-34Js x 3x108/(400x10-9)4.97x
10-19 J in eV (1 eV1.6x10-19 J)
3.11 eV b)
KEmaxhf-?3.11-2.46 eV 0.65 eV c)
?cc/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 fltfc no
electrons emitted if ?gt ?c no electrons
emitted
8
particle-wave dualism
  • So, is light a wave or particle phenomenon?
  • answer it depends!

9
question
  • light from a far-away star is used to perform a
    double slit experiment. Approximately once per 10
    minutes will a single photon from the star arrive
    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

10
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.

11
interference pattern
P1
B
A
P1
P2
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 IE2E02cos2?. In A, it is assumed that
the intensities add IsumI1I2 . However, one
should add the E-fields (which can be positive or
negative) and than squared, like in B
Isum(E1E2)2 where E1 and E2 are treated as
vectors.
12
and if you think that youve seen it all
B
A
lets 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.
13
Schrödingers cat
A cat is placed in a closed box. Inside the box a
radioactive source is placed in which on average
once per hour a radioactive decay takes place.
If the decay takes place, a bottle of poison
breaks, killing the cat. In quantum-physical
sense, the cat is 50 dead and 50 alive after
half an hour. Since we cant see it, it is in a
superposition of those two states and there is a
certain probability of being in one of either
states. Only when we open the box, do we
determine what state the cat is in. The
observation is crucial to determine the state of
the cat.
14
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15
heisenbergs 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. Lets 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 velocity (momentum) with
limited accuracy.
note hh/(2?)
?x?p?h/(4?) 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
16
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 ?x1x10-9 m, h6.63x10-34 Js so ?p?
6.63x10-34/(4? x 1x10-9)5.28x10-26 kgm/s ?vmin
5.28x10-26/9.11x10-315.79x104 m/s (use pmv)
note that the uncertainty principle works for the
three dimensions separately ?x?px?h/(4?)
?y?py?h/(4?)
?z?pz?h/(4?)
17
photons as particles and quanta
  • Some other examples of where the particle nature
    of
  • light plays a role
  • Photo-electric effect
  • Black-body radiation
  • bremmstrahlung
  • Compton effect

18
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 Tlt700 K, almost no visible light
is produced (hence a black body). The energy
emitted from a black body P?T4 with ?5.67x10-8
W/m2K4
The peak in the intensity spectrum varies with
wavelength using the Wien displacement
law ?maxT0.2898x10-2 mK Until 1900, the
intensity distribution, predicted using classical
equations, predicted a steep rise at small
wavelengths. However, the opposite was determined
experimentally
(classical)
19
Planck to the rescue
  • Max Planck devised a theory for a simple black
    body that could describe the measured spectra.
  • He assumed that the walls consisted of little
    radiators that only emitted light at certain
    discrete energies Enhf
  • f the frequency of the light (Hz)
  • h plancks constant (6.63x10-34 Js)
  • n integer.
  • His achievement was really the first success of
    quantum theory
  • In essence, his theory showed that because the
    energy is quantized, it is hard to emit light of
    small wavelengths (high frequency) since a lot of
    energy is required

20
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?
21
X-rays
  • when energetic electrons are shot on a material,
    photons with small wavelengths (0.1 nm) are
    produced.
  • The spectrum consist of two components
  • broad bremsstrahlung spectrum
  • 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

22
bremmsstrahlung
  • assume electrons are accelerated in a potential
    of V Volts.
  • their kinetic energy is EeV with e1.6x10-19 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 merely deflected, the frequency f is
    smaller than fmax and its wavelength ? larger
    than ?min.
  • so eVhfmaxhc/?min

23
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 the
    interacted with the material?

eVhfmaxhc/?min Vhc/(?mine)
6.6x10-34x3x108/(0.35x10-10x1.6x10-19)
3.55x104 V
24
Braggs law
X-rays scattered off atoms in e.g. will interfere
and the interference pattern can be used to
identify/study materials.
25
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

26
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

27
compton scattering
  • ???-?0h/(mec) x (1-cos?)
  • with ? wavelength of photon after collision
  • ?0 wavelength of photon before
    collision
  • h/(mec) Compton wavelength (2.43x10-3
    nm)
  • me mass of electron
  • ? angle of outgoing X-ray relative
  • to incoming direction

?
?0
28
example
???-?0h/(mec) x (1-cos?)
  • A beam of X-rays with ?010-12 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.
  • maximum shift occurs if cos?-1 (?1800). This is
    usually referred to as
  • Compton backscattering. in that case
  • ??2h/(mec)2x2.43x10-124.86x10-12 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 a) energy of X-ray before collision
    hfhc/?01.98x10-13 J
  • energy of X-ray after collision
    hfhc/(?0??) 3.38x10-14 J
  • kinetic energy gained by electron
    1.64x10-13 J1.0 GeV (giga)

29
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