Quantum physics PHY232 Remco Zegers [email protected] Room W109 – cyclotron building...

Quantum physics

PHY232Remco [email protected] W109 – cyclotron buildinghttp://www.nscl.msu.edu/~zegers/phy232.html

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

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

PHY232 - Remco Zegers - quantum physics 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)

answer: c the speed of light is independent of the motionof the source or observer


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.


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


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.


photo-electric effect

The energy carried by a photon: E=hfh: planck’s constant (h=6.63x10-34

Js)f: frequency with c=f

The energy is localized in the photon-particleThe maximum kinetic energy of a released electron: KEmax=hf-with: : the workfunction (binding of electron to the metal)

So only if hf> will electrons be released from the metalfc=/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


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: it depends!

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 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

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,with =h/(mv) (the so-called ‘de Broglie’ wavelength and thus exhibitwave-phenomena under the right conditions


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.


Schrödinger’s cat

A cat is placed in a closed box. Inside the box a radioactive sourceis placed in which on average once per hour a radioactive decay takes place. If the decay takes place, a bottle of poison breaks, killingthe cat.In quantum-physical sense, the cat is 50% dead and 50 alive after half an hour. Since we can’t 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.


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)

note that the uncertainty principle works for the three dimensions separately:xpxh/(4) ypyh/(4) zpzh/(4)


photons as particles and quanta

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

•Photo-electric effect•Black-body radiation•bremmstrahlung•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 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:maxT=0.2898x10-2 mK

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

(classical)


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: E=nhf f the frequency of the light (Hz) h: planck’s 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


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?

maxT=0.2898x10-2 mKT=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 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


bremmsstrahlung

assume electrons are accelerated in a potential of V Volts. their kinetic energy is E=eV 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 merely deflected, the frequency f is smaller than fmax and its wavelength larger than min.

so eV=hfmax=hc/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 the interacted with the material?

eV=hfmax=hc/min

V=hc/(mine) =6.6x10-34x3x108/(0.35x10-10x1.6x10-19) =3.55x104 V


Braggs law

X-rays scattered off atoms in e.g. will interfere and the interference pattern can be used to identify/studymaterials.


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.e.g.: to observe airplanes by radar, radiation of very longwavelength should be used


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/(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


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 a) 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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