Loading...

PPT – Quantum physics PowerPoint presentation | free to view - id: a4ef5-Y2M0Y

The Adobe Flash plugin is needed to view this content

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

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)

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 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

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

particle-wave dualism

- So, is light a wave or particle phenomenon?
- answer it depends!

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

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.

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.

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.

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.

(No Transcript)

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

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?)

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

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)

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

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?

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 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

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

Braggs law

X-rays scattered off atoms in e.g. will interfere

and the interference pattern can be used to

identify/study materials.

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

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

- ???-?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

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)

applications