Quantum physics PHY232 – Spring 2007 Jon Pumplin pumplin/PHY232 (Ppt courtesy of Remco Zegers)...

Quantum physics

PHY232 – Spring 2007

Jon Pumplin

http://www.pa.msu.edu/~pumplin/PHY232

(Ppt courtesy of Remco Zegers)

“I can safely say that nobody understands quantumphysics” – Richard Feynman Nobel Lecture, 1966, (1918-1988)

“Anyone who is not shocked by the quantum theory has not understood it.” – Niels Bohr, Nobel Prize in 1922 (1885-1962)

PHY232 - Remco Zegers - quantum physics 2

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.


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.


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.


photo-electric effect

The energy carried by a photon: E = h fh: 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 metalfc=/h : fc is the cut-off frequency c=c/fc=(hc)/: the cut-off wavelengthsee table 27.1 for work functions for various metals


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 eVb) KEmax=hf-=3.11-2.46 eV = 0.65 eVc) 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 emittedNote: if f<fc no electrons emitted if > c no electrons emitted


particle-wave dualism So, is light a wave or particle phenomenon?

answer: both!

experiment can be described by light as waves

can be described by light as particles

reflection X Xrefraction X Xinterference Xdiffraction Xpolarization Xphoto-electric effect

X


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!


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.


interference pattern

If one of the slits in a double slit experiment is closed, one sees onlya diffraction pattern from a single slit (P1). If the other slit is opened andthe 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=E0

2cos2. 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.

P1P1

P2

P2

A B


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.


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 usinga 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.xph/(4) with x: precision of position measurement

with p: precision of momentum (mv) measurementthis can also be expressed in terms of energy and time measurementsEth/(4) with E: precision of energy measurement

with p: precision of time measurement

note ħ = h/(2)


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:xph/(4) with x=1x10-9 m, h=6.63x10-34 Jsso 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)


photons as particles and quanta

Some other examples of where the particle nature oflight plays a role:

•Photo-electric effect•Black-body radiation•bremsstrahlung•Compton effect


black-body radiationA 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

Until 1900, the intensity distribution,predicted using classical equations,predicted a steep rise at small wavelengths. However, the opposite was determined experimentally…

(classical)


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.


example

Hot lava can be considered as a blackbody emitting radiation at a variety oftemperatures. 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 KT=1200+273=1473 Kmax= 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 colorsred/orange in the visible spectrum


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


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


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


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 probemust have a wavelength of the same scale or smaller.


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:


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.

0


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 mb) minimum shift occurs if cos=1 (=00) in which case

essentially no collision takes place: =0c) gain in kinetic energy by electron is loss in energy of x-ray:case b) no kinetic energy gained by electroncase 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)

=-0=h/(mec) x (1-cos)


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