As h 12 Quantumphenomena

8/12/2019 As h 12 Quantumphenomena

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The photoelectric effect The photoelectric effect is the

emission of electrons from thesurface of a material due to theexposure of the material toelectromagnetic radiation.

For example zinc emits electrons

when exposed to ultraviolet radiation. If the zinc was initially negatively

charged and placed on a gold leafelectroscope, the electroscopes goldleaf would be initially deflected.

However, when exposed to the uvradiation the zinc loses electrons andtherefore negative charge.

This causes the gold-leaf to fall.


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Experimental observations

Threshold frequencyThe photoelectric effect only occurs if the frequency of theelectromagnetic radiation is above a certain threshold value, f0

Variation of threshold frequency

The threshold frequency varied with different materials.

Affect of radiation intensityThe greater the intensity the greater the number of electrons emitted,but only if the radiation was above the threshold frequency.

Time of emission

Electrons were emitted as soon as the material was exposed.

Maximum kinetic energy of photoelectrons

This depends only on the frequency of the electromagnetic radiationand the material exposed, not on its intensity.


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Problems with the wave theory

Up to the time the photoelectric effect was firstinvestigated it was believed that electromagneticradiation behaved like normal waves.

The wave theory could not be used to explain theobservations of the photoelectric effectin particularwave theory predicted: that there would not be any threshold frequencyall frequencies

of radiation should eventually cause electron emission

that increasing intensity would increase the rate of emission atall frequenciesnot just those above a certain minimum

frequency that emission would not take place immediately upon exposure

the weaker radiations would take longer to produce electrons.


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Einsteinsexplanation

Electromagnetic radiation consisted ofpackets or quanta of energy called photons

The energy of these photons:

depended on the frequency of the radiation only

was proportional to this frequency

Photons interact one-to-one with electrons in the material

If the photon energy was above a certain minimum amount(depending on the material)

the electron was emitted

any excess energy was available for electron kinetic energyEinstein won his only Nobel Prize in 1921 for this explanation. This explanation also beganthe field of Physics called Quantum Theory, an attempt to explain the behaviour of verysmall (sub-atomic) particles.
http://en.wikipedia.org/wiki/Einsteinhttp://en.wikipedia.org/wiki/Einstein


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Photon energy (revision)

photon energy (E) = h x f

where h= the Planckconstant = 6.63 x 10-34Js

also as f = c / ;

E = hc /

Calculate the energy of a photon of ultraviolet light(f = 9.0 x 1014Hz) (h= 6.63 x 10-34Js)

E = h f= (6.63 x 10-34Js) x (9.0 x 1014Hz)

= 5.37 x 10-19J
http://en.wikipedia.org/wiki/Planckhttp://en.wikipedia.org/wiki/Planck


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The photoelectric equation

hf = + EKmax

where:

hf= energy of the photons of electromagnetic radiation

= work function of the exposed material

EKmax= maximum kinetic energy of the photoelectrons

Work function,This is the minimum energy required for anelectron to escape from the surface of a material


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Threshold frequency f0

As: hf = + EKmax

If the incoming photons are of the thresholdfrequency f0, the electrons will have the minimumenergy required for emission

and EKmaxwill be zero

therefore: hf0=

and so: f0= / h


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Question 1

Calculate the threshold frequency of a metalif the metals work function is 1.2 x 10 -19J.

(h= 6.63 x 10-34Js)

f0= / h

= (1.2 x 10-19J) / (6.63 x 10-34Js)

threshold frequency = 1.81 x 1014Hz


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Question 2

Calculate the maximum kinetic energy of thephotoelectrons emitted from a metal of workfunction 1.5 x 10 -19J when exposed with photonsof frequency 3.0 x 1014Hz.

(h= 6.63 x 10-34Js)

hf = + EKmax(6.63 x 10-34Js) x (3.0 x 1014Hz) = (1.5 x 10-19J) + EKmaxEKmax = 1.989 x 10

-19- 1.5 x 10-19

= 0.489 x 10-19

J

maximum kinetic energy = 4.89 x 10 -20J


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The vacuum photocell

Light is incident on a metalplate called thephotocathode.

If the lights frequency is

above the metals thresholdfrequency electrons areemitted.

These electrons passingacross the vacuum to the

anode constitute andelectric current which canbe measured by themicroammeter.

The photocell is an applicationof the photoelectric effect


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Obtaining Plancks constant

By attaching a variable voltagepower supply it is possible tomeasure the maximum kineticenergy of the photoelectronsproduced in the photocell.

The graph opposite shows howthis energy varies with photon

frequency.hf = + EKmax

becomes:

EKmax= hf

which has the form y = mx +c

with gradient, m = hHence Plancks constant can befound.


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Ionisation

An ion is a charged atom Ions are created by adding or

removing electrons from atoms

The diagram shows thecreation of a positive ion fromthe collision of an incoming

electron. Ionisation can also be caused

by:

nuclear radiationalpha, beta,gamma

heating passing an electric current

through a gas (as in afluorescent tube)

incoming electron


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Ionisation energy

Ionisation energy is the energy

required to remove one electron from

an atom.

Ionisation energy is often expressed in eV.

The above defines the FIRST ionisation

energythere are also 2nd, 3rdetc

ionisation energies.


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Excitation

Excitation is the promotionof electrons from lower tohigher energy levels withinan atom.

In the diagram someof the

incoming electrons kineticenergy has been used to movethe electron to a higher energylevel.

The electron is now said to be

in an excited state. Atoms have multiple excitation

states and energies.

incoming

electron


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Question

An electron with 6 x 10-19J of kinetic energy can cause

(a) ionisation or (b) excitation in an atom.

If after each event the electron is left with

(a) 4 x 10-19

J and (b) 5 x 10-19

J kinetic energycalculate in eVthe ionisation and excitation energy of the

atom.


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Electron energy levels in atoms

Electrons are bound to thenucleus of an atom byelectromagnetic attraction.

A particular electron will occupythe nearest possible position to

the nucleus. This energy level or shell is called

the ground state.

It is also the lowest possibleenergy level for that electron.

Only two electrons can exist inthe lowest possible energy levelat the same time. Furtherelectrons have to occupy higherenergy levels.


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Electron energy levels in atoms

Energy levels are measured withrespect to the ionisation energylevel, which is assigned 0 eV.

All other energy levels aretherefore negative. The ground

state in the diagram opposite is- 10.4 eV.

Energy levels above the groundstate but below the ionisation levelare called excited states.

Different types of atom havedifferent energy levels.


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De-excitation

Excited states are usually

very unstable.

Within about 10 - 6 s the

electron will fall back to a

lower energy level.

With each fall in energy

level (level E1down to

level E2) a photon of

electromagnetic radiationis emitted.

em it ted pho ton energy

= hf = E1E2


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Energy level question

Calculate the frequencies of

the photons emitted when anelectron falls to the ground

state (at 10.4 eV) from

excited states (a) 5.4 eV and(b) 1.8 eV.

(h= 6.63 x 10-34Js)


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Complete:

transition photon

E1/ eV E2/ eV f/ PHz / nm

5.4 10.4 1.20 250

1.8 10.4 2.08 144

1.8 5.4 0.871 344

1.8 4.5 0.653 459


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Excitation using photons

An incoming photon may nothave enough energy to causephotoelectric emission but itmay have enough to causeexcitation.

However, excitation will onlyoccur if the photons energy isexactly equalto the differencein energy of the initial and finalenergy level.

If this is the case the photon willcease to exist once its energy isabsorbed.


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FluorescenceThe diagram shows an incoming

photon of ultraviolet light of energy5.7eV causing excitation.

This excited electron then de-excites intwo steps producing two photons.

The first has energy 0.8eV and will beof visible light. The second of energy4.9eV is of invisible ultraviolet of slightlylower energy and frequency than theoriginal excitating photon.

This overall process explains why

certain substances fluoresce withvisible light when they absorb ultravioletradiation. Applications include thefluorescent chemicals are added aswhiteners to toothpaste and washingpowder.

Electrons can fall back totheir ground states in steps.


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Fluorescent tubes A fluorescent tube consists of a

glass tube filled with lowpressure mercury vapour and aninner coating of a fluorescentchemical.

Ionisation and excitation of the

mercury atoms occurs as thecollide with each other and withelectrons in the tube.

The mercury atoms emitultraviolet photons.

The ultraviolet photons areabsorbed by the atoms of thefluorescent coating, causingexcitation of the atoms.

The coating atoms de-excite andemit visible photons.

See pages 37 and 38 for further

details of the operation of a

fluorescent tube


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Line spectra

A line spectrum is produced from the excitation of a low

pressure gas.

The frequencies of the lines of the spectrum are

characteristic of the element in gaseous form.

Such spectra can be used to identify elements.

Each spectral line corresponds to a particular energy level

transition.


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Question

Calculate the energy level transitions (ineV) responsible for (a) a yellow line of

frequency 5.0 x 1014 Hz and (b) a blue line

of wavelength 480 nm.


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The hydrogen atom

- 21.8


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With only one electron, hydrogenhas the simplest set of energylevels and corresponding linespectrum.

Transitions down to the loweststate, n=1 in the diagram, give riseto a series of ultraviolet lines calledthe Lyman Series.

Transitions down to the n=2 stategive rise to a series of visible lightlines called the Balmer Series.

Transitions down to the n=3, n=4etc states give rise to sets of infra-red spectral lines.

- 21.8


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The discovery of helium

Helium was discovered in the Sun before it wasdiscovered on Earth. Its name comes from theGreek word for the Sun helios.

A pattern of lines was observed in the Suns

spectrum that did not correspond to any knownelement of the time.

In the Sun helium has been produced as theresult of the nuclear fusion of hydrogen.

Subsequently helium was discovered on Earthwhere it has been produced as the result ofalpha particle emission from radioactiveelements such as uranium.


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The wave like nature of light

Light undergoes diffraction(shown opposite) and displaysother wave properties such aspolarisation and interference.

By the late 19thcentury mostscientists considered light andother electromagnetic

radiations to be like waterwaves


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The particle like nature of light

Light also producesphotoelectric emissionwhich can only beexplained by treating

light as a stream ofparticles.

These particles withwave properties arecalled photons.


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The dual nature

of electromagnetic radiation Light and other forms of electromagnetic radiation

behave like waves and particles.

On most occasions one set of properties is the mostsignificant

The longer the wavelength of the electromagnetic wavethe more significant are the wave properties.

Radio waves, the longest wavelength, is the mostwavelike.

Gamma radiation is the most particle like Light, of intermediate wavelength, is best considered tobe equally significant in both


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Matter waves

In 1923 de Broglie proposed that particles such aselectrons, protons and atoms also displayed wave likeproperties.

The de Broglie wavelengthof such a particle depended

on its momentum, paccording to the de Broglie relation:= h / p

As momentum = mass x velocity= mv

= h / mv

This shows that the wavelength of a particle can be alteredby changing its velocity.


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Question 1

Calculate the de Broglie wavelength of an

electron moving at 10% of the speed of light.

[me= 9.1 x 10-31kg]

(h= 6.63 x 10-34Js; c= 3.0 x 108ms-1)


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Question 2

Calculate the de Broglie wavelength of a person

of mass 70 kg moving at 2 ms-1.

(h= 6.63 x 10-34Js)


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Question 3

Calculate the effective mass of a photon of

red light of wavelength 700 nm.

(h= 6.63 x 10-34Js)


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Evidence for de Broglies hypothesis

A narrow beam of electrons in a vacuum tube is directed at a thinmetal foil. On the far side of the foil a circular diffraction pattern isformed on a fluorescent screen. A pattern that is similar to that formedby X-rays with the same metal foil.

Electrons forming a diffraction pattern like that formed by X-raysshows that electrons have wave properties.

The radii of the circles can be decreased by increasing the speed ofthe electrons. This is achieved by increasing the potential differenceof the tube.


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Energy levels and electron waves

An electron in an atom hasa fixed amount of energythat depends on the shell itoccupies.

Its de Broglie wavelengthhas to fit the shape and sizeof the shell.

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As h 12 Quantumphenomena