quantum physics and nuclear physics
Download
Skip this Video
Download Presentation
Quantum Physics and Nuclear Physics

Loading in 2 Seconds...

play fullscreen
1 / 74

Quantum Physics and Nuclear Physics - PowerPoint PPT Presentation


  • 297 Views
  • Uploaded on

Quantum Physics and Nuclear Physics. 13.1 Quantum Physics. Quantum Physics Revision When an electron falls to a lower energy level , the change in energy ( Δ E) is emitted as a photon of e-m radiation. . Δ E = E 2 – E 1

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Quantum Physics and Nuclear Physics' - cormac


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide3
Quantum Physics
  • Revision
  • When an electron falls to a lower energy level, the change in energy (ΔE) is emitted as a photon of e-m radiation.
  • ΔE = E2 – E1
  • The change in energy is proportional to the frequency of the emitted photon:
  • ( E = Photon energy
  • h = Plank’s constant
  • = 6.6 x 10-34 m2kgs-1 )

E = hf

slide4
E.g. Energy levels in the hydrogen atom:

Thus the emission spectrum for Hydrogen (this shows only the visible section)...

slide5
The Photoelectric Effect
  • Demo:
  • u.v. light shone onto a zinc plate will cause a negative charged electroscope to discharge .
  • visible light will not cause it to discharge (even if very intense).
  • if the charge is positive, even the u.v. will not cause it to discharge.
slide6
Conclusion
  • u.v. light causes negative electrons to be emitted from the surface of zinc (this is called photoelectric emission) but visible light does not, hence the first two observations.
  • if u.v. causes negative electrons to be emitted, the positive charge will not be discharged.

Applet link

slide7
Threshold Frequency

Electrons will only be emitted from zinc by photoelectric emission if the electromagnetic radiation incident upon its surface has a frequency of 1 x 1015 Hz or above. This is called the threshold frequency of zinc.

Limitation of Wave Theory of Light

Wave theory would suggest that once enough visible light energy had been absorbed by the zinc, the electron would be able to escape. This is not the case. No matter how intense the incident radiation, if its frequency is below the threshold frequency for a particular material, no photoelectric emission will occur.

slide8
Photons – the Quantum Model
  • In 1900 Max Planck came up with the idea of energy being ‘quantised’ in some situations. i.e. existing in small ‘packets’.
  • In 1905 Einstein suggested that all e-m radiation is emitted in small quanta called photons rather than in a steady wave.
  • Intensity of radiation depends on the number of photons being emitted per second (not amplitude as suggested by the wave model).
  • Energy per photon depends upon its frequency:

E = hf

slide9
Work Function and Photoelectric Emission

When u-v light is incident upon a zinc surface, each photon gives its energy to a single electron on the zinc surface. u-v photons have a high frequency. As a result they give enough energy to the electron to escape from the surface.

The minimum energy needed to just remove an electron from a metal surface is called the work function, .

slide10
Low intensity u-v light will still cause electrons to be emitted. Because there are less photons per second there will be less electrons emitted per second.

If the incident light has a lower frequency, each photon has less energy and so no electrons are emitted, irrespective of the intensity.

slide11
Einstein’s Photoelectric Equation

If the photon energy (E=hf) is greater than the work function (), any remaining energy becomes kinetic energy of the electron (= ½mv2). Thus Einstein stated...

This is one version of Einstein's photoelectric equation.

hf =  + ½ mv2

m = mass of an electron

v = speed of fastest electrons (ms-1)

slide12
Einstein’s theory was confirmed by Robert Millikan in 1916. He realised that if the clean metal emitting surface was given a positive potential, the electron emission could be stopped.

Thus, if the p.d. applied was known, the KE ‘removed’ from the fastest electron can be found:

We know... V = W / q  W = eV

So...

Link - PhET simulation

V = stopping voltage

e = charge on an electron

= 1.6 x 10-19C

 = Work function (Joules)

hf =  + eV

slide13
Testing Einstein's Photoelectric Equation
  • Einstein’s photoelectric equation can be rearranged to give...
  • Thus plotting a graph of V against f enables us to determine Plank’s constant.

V = h f - 

e e

slide14
Incident radiation is shone onto a photoelectric cell with a low work function
  • This causes electrons to be emitted from the larger emitting electrode. If they reach the small receiving electrode a current is detected on the ammeter
  • The stopping voltage V is increased until zero current flows through the ammeter. The p.d. has made it impossible for even the fastest electrons to escape from the large electrode.
  • The experiment is repeated with different frequency incident radiation and a set of values for frequency and stopping voltage are collected.
slide16
Q. Explain why this graph will always have the same gradient, whatever metal is used for the emitting electrode.

Q. Explain how you would determine a value for the work function of the metal used to produce the graph above.

Q.

slide17
Photocurrents

If the metal surface and frequency of incident radiation are both kept constant, a graph can be plotted showing how the photoelectric current (photocurrent) in a photocell varies with applied p.d. (voltage).

Consider these situations and explain the photocurrent that will flow in each case:

+

-

-

+

slide18
0 +

Photocurrent

Applied p.d.

Saturation current

Stopping potential, Vs

slide19
0 +

Photocurrent

Applied p.d.

High intensity

Low intensity

slide20
0 +

Photocurrent

Applied p.d.

Low frequency (red)

Higher frequency (blue)

slide22
The Wave Nature of Matter

We have seen that light behaves both like a wave (it diffracts) and like a particle (in explaining photoelectric emission). It has wave particle duality.

Demo 1: Shine a beam of laser light (a wave) through a single diffraction grating (like a very fine gauze) then through many gratings crossed at angles to each other.

Demo 2: Fire a beam of electrons (particles of matter) at a thin piece of graphite using a cathode ray tube.

Wave - particle duality is the ability of something to exhibit both wave and particle behaviour.

slide23
Control grid

Anode

Cathode

Graphite foil

Flourescent screen

VA

slide24
De Broglie’s Equation

Considering a photon of light, in 1924 Prince Louis Victor de Broglie equated Einstein’s mass-energy relation and Planck’s equation:

E = mc2 and E = hf

thus... mc2 = hf

so... mc2 = h c

λ

so... or

But c = f λ

λ = h

mc

λ = h

p

slide25
This is the de Broglie equation.

( Where h = Planck’s constant = 6.6 x 10-34Js )

By analogy, this equation can be applied to any other particle of matter. Thus de Broglie’s Hypothesis:

Q1 A year 13 student runs with joy to his physics lesson. If he runs at 5 ms-1 and has mass 60kg, determine the de Broglie wavelength of his motion. Comment upon your answer.

All matter can behave like a wave, with the wavelength given by Plank’s constant divided by the matter’s momentum.

slide26
Q2 An electron is accelerated in a cathode ray tube through a potential difference of 2kv.
  • Determine the velocity of the electron (me = 9.11 x 10-31 kg)
  • 2.65 x 107 ms-1
  • Determine the de Broglie wavelength of the electron.
  • 2.7 x 10-11 m
slide28
Energy Levels

E.g.

h = Planck constant

= 6.6 x 10-34 m2kgs-1

E = hf

hf = E2 - E1

slide29
Emission Line Spectra

Atoms of a gas emit e-m radiation if they become excited. This means the electrons jump to a higher energy level and then fall back, losing potential energy and emitting it as a photon of e-m radiation. This will have a frequency according to ΔE = hf

  • Experiment: Observing emission spectra
  • Place a slit in front of a hydrogen lamp.
  • View the light through a diffraction grating or by refracting it through a prism.
slide30
Results:

Conclusion

Explain why...

only certain colours are seen for any particular lamp.

coloured fringes are produced

This is the image seen when light from one lamp is diffracted through a grating:

slide34
Xenon

Images from this website

slide35
Hydrogen...

Helium...

Neon...

slide37
Absorption spectrum

If light with a continuous spectrum of frequencies (a filament light bulb approximately emits this) is shone into a gas its photons will interact with electrons in the gas atoms, boosting them to a higher energy level. Thus the gas will absorb only certain frequencies of the light. These frequencies will be missing from the light that passes through.

slide38
Absorption spectrum for Hydrogen.

Q. Why des the intensity not fall to zero for the absorption lines?

Energy is re-emitted as photons in all directions

slide39
Fraunhoffer Lines

This effect is visible in the Fraunhoffer lines seen in spectra of light from the Sun. The Sun emits a virtually continuous spectrum. The absorption lines are due to gases in the Sun’s atmosphere absorbing some frequencies of photon. Hence astronomers can deduce what gases are in a star’s atmosphere as well as their proportions.

slide43
The ‘Electron in a Box’ Model of an Atom

We have seen how the Bohr model of an atom suggests that the atom has discreet energy levels. Thus the energy of the electrons and of the atom is not continuous.

Next we will look at how this idea is compatible with the de Broglie hypothesis.

According to de Broglie, the electron must have wave properties.

So how does this wave fit into an atom...?

slide44
We can create a quantum model of an electron trapped in an atom by considering the electron as a standing wave travelling back and forth across a box of side L:

Clearly, if there are nodes at either end, the electron can only have certain fixed wavelengths and frequencies.

i.e. λ = 2L (fundamental)

λ = L (2nd harmonic)

λ = 2/3 L (3rd harmonic)

So in general λ = 2L

n

Java applet

slide45
According to de Broglie...

The KE of the electron is given by KE = ½ mv2

Substituting the equations for v and λ gives…

The energy of the electron ‘wave’ is clearly quantised (it has discreet values depending upon n), thus fitting scientific observations e.g. work function in the photoelectric effect.

λ = h = h v = h

p mv mλ

n = integer

h, m and L are all constants.

KE = n2 h2

8mL2

slide47
Limitations of the Bohr Model of the Atom

The Bohr model of the atom worked well with basic ideas of quantum theory (as we have seen). However it has limitations.

E.g. It only works well with the Hydrogen atom

It does not predict intensities of different spectral lines.

slide48
Schrödinger's Model of the Atom
  • Schrödinger developed a new model that accounted for all these limitations. He suggested that...
  • electrons exist in the atom with a position determined by the wavefunction (ψ – psi), a function of position and time.
  • the position of the electron at any time is undefined
  • the probability of the finding the electron at any particular position in the atom is determined by the square of the amplitude of the wavefunctioni.e. ψ2
  • Thus for a given energy there are some places where the electron is more likely to exist. This can be represented by ‘probability clouds’.
slide50
Additional note:

This illustrates an important feature of quantum mechanics. Its outcomes are probabilities and not certainties; it is a probabilistic model.

Classical mechanics (e.g. Newtons laws) are deterministic: the future is determined by the past and is therefore predictable with some certainty!

slide51
Heisenberg’s Uncertainty Principle

Consider one electron amongst a beam of electrons moving with identical momentum towards a narrow slit:

If the slit is wide, the electron will pass straight through without diffracting. Hence we can know its direction and thus its momentum beyond the slit. However we cannot be sure where exactly it passed through the slit so the uncertainty in position is large.

slide52
If the slit width is similar to the diameter of the electron, it will diffract as it passes through.

So we are not sure about its direction and thus momentum although we can be quite sure of its position.

This is an example of the Heisenberg uncertainty principle which states that...

Note: This is a fundamental property of the universe and is nothing to do with our ability to measure accurately.

It is not possible to measure the exact position and the exact momentum of a particle at the same time.

slide53
This imposes a minimum uncertainty on the product of the uncertainties of position and momentum:

This can also be written in terms of energy and time:

Clearly the value of h is so small that the effects of this uncertainty are not seen in everyday events.

h = Plank’s constant

Δp Δx ≥ h

ΔE Δt ≥ h

slide54
You tube link
  • E.g. Quantum tunnelling
  • As a result of the uncertainty principle energy can be ‘borrowed’ to overcome potential barriers:
  • Alpha particles should not be able to escape from the nucleus due to the ‘strong force’ between quarks. However they escape by borrowing energy and paying it back within a time limit defined by the equation.
  • Two protons (hydrogen nuclei) should not be able to fuse together in the Sun at its current density and temperature. They do fuse together by borrowing energy and then paying it back later on.
slide55
Q1

Consider a football moving through the air. If the uncertainty in its momentum is 2.25 x 10-2 kgms-1, Determine the uncertainty in its position (Δx).

Δx ≥ +/- 2.33x10-33m

Δx Δp ≥ h

so… Δx ≥ h

4πΔp

Δx ≥ 6.6 x 10-34

4π x (2.25 x 10-2)

slide56
Q2

A proton in the LHC takes 1.0 x 10-10s to collide. Assuming a 1% uncertainty in measurement of time, determine the uncertainly in measurements of proton energy (ΔE).

1% uncertainty in time = 1/100 x (1.0 x 10-10)

= +/- 1.0 x 10-12 s

ΔE Δt ≥ h

so… ΔE ≥ h

4πΔt

ΔE ≥ 6.6 x 10-34

4π x (1.0 x 10-12)

ΔE ≥ +/- 5.3 x 10-23 J

slide57
Q3

Assuming 1% uncertainty in velocity, determine the uncertainty in the position of a proton moving at

2.5 x 107ms-1

Close to the speed of light (3.0 x 108 ms-1)

( mp = 1.673 x 10-27 kg )

slide58
Q4

Outline by reference to position and momentum how the Schrödinger model of the hydrogen atom is consistent with the Heisenberg uncertainty principle.

- the Schrödinger model assigns a wave function to the electron that is a measure of the probability of finding it somewhere;- therefore the position of the electron is uncertain;- resulting in an uncertainty in its momentum;

slide66
Probability Waves

From the de Broglie hypothesis we can see that our picture of the electron as a small ‘ball’ is too simplistic. It

slide74
Subtitle

Text

http://www.physics.uq.edu.au/people/mcintyre/applets/cathoderaytube/crt.html

ad