“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)

 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

 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

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

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: KE max
=hf with:  : the workfunction (binding energy of electron to the metal) f
So only if h f >  can electrons be released from the metal c
=  /h : f c
 c
=c/f c is the cut-off frequency
=(hc)/  : the cut-off wavelength see table 27.1 for work functions for various metals quantum physics 5

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
-34 Js x 3x10 8 /(400x10 -9 )=4.97x10
-19 J in eV (1 eV=1.6x10
-19 J) =3.11 eV b) KE c)  c max
=hf =3.11-2.46 eV = 0.65 eV
=c/f c
=(hc)/  =6.63x10
-34 x3x10 8 /(2.46x1.6x10
-19 J)=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<f c if  >  no electrons emitted c no electrons emitted quantum physics 6

 So, is light a wave or particle phenomenon?
experiment reflection refraction interference diffraction polarization photo-electric effect can be described by light as waves
X
X
X
X
X can be described by light as particles
X
X
X
 answer: both!
quantum physics 7

 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

 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

A
P
1
P
1
B
P
2
P
2
If one of the slits in a double slit experiment is closed, one sees only a diffraction pattern from a single slit ( P
1
). If the other slit is opened and the first one closed, one sees only the diffraction pattern from the other slit ( P
2
). If both are opened, one does not simply see the sum of P
1 and
P
2
(like in A ), but the double-slit interference pattern (like in
The reason is the following:
B ).
Remember that the intensity (I) is proportional to the E-field squared:
I~E 2 =E
0
2 cos 2  . In A, it is assumed that the intensities add:
=(E
1
+E
2
I sum
=I
1
+I
However, one should add the E-fields (which can be positive or negative) and than squared, like in B: I sum treated as vectors.
) 2 where E
1
2 and E
2
.
are quantum physics 10

A and if you think that you’ve seen it all…
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

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 velocity (momentum) with limited accuracy.
note ħ = h/(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 quantum physics 13

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
 v min
= 5.28x10
-26 /9.11x10
-31 =5.79x10
4 m/s
(use p=mv) quantum physics 14

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

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 T 4 with  =5.67x10
-8 W/m 2 K 4
The peak in the intensity spectrum varies with wavelength using the
Wien displacement law:
 max
T = 0.2898x10
-2 m K
Until 1900, the intensity distribution, predicted using classical equations, predicted a steep rise at small wavelengths. However, the opposite was determined experimentally…
(classical) quantum physics 16

 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

Hot lava can be considered as a black body emitting radiation at a variety of temperatures.
If temperature of molten lava is about 1200 0 C, 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

 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

 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 f max 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 f max and its wavelength  larger than
 min.
 e V = h f ma x
= h c /  min quantum physics 20

 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 f max
= h c /  min
V = h c / (  min
= 3.55x10
4 V e)
=6.6x10
-34 x 3x10 8 / (0.35x10
-10 x1.6x10
-19 ) quantum physics 21

 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

 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

  =  
0
= h /(m e c) x (1-cos  with:  = wavelength of photon after collision
)

0
= wavelength of photon before collision h/(m e c) = “ Compton wavelength” of electron (2.43x10
-3 nm) m e
= mass of electron
 = angle of outgoing X-ray relative to incoming direction


0
This formula comes from energy and momentum conservation.
quantum physics 24

 =  
0
=h/(m e c) x (1-cos  )
 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 (  =180 0 ). This is usually referred to as
Compton backscattering. in that case:
 =2h/(m e c)=2x2.43x10
-12 =4.86x10
-12 m b) minimum shift occurs if cos  =1 (  =0 0 ) 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/  energy of X-ray after collision: hf=hc/( 
0
0
=1.98x10
-13 J
+  ) =3.38x10
-14 J kinetic energy gained by electron: 1.64x10
-13 J=1.0 GeV (giga) quantum physics 25

quantum physics 26