Quantum Physics AHL 13.1. THE PHOTOELECTRIC EFFECT Early last century, several people had noticed that light was capable of ejecting electrons from.

Quantum Physics

AHL 13.1

THE PHOTOELECTRIC EFFECT

Early last century, several people had noticed that light was capable of ejecting electrons from various metal surfaces.

This effect, known as the photoelectric effect, is used in photographer’s light meters, sound tracks of motion pictures and electric eyes used to automatically open doors.

Photoelectric Effect

http://ie.lbl.gov/education/glossary/AnimatedDecays/PhotoElectric.html


Remember that metals have “free” electrons that are not tightly bound. It is these electrons that allow current and heat to flow in a metal, as the electron’s move.


Examples of the freeing of electrons from light energy… Hertz, in 1887, noticed that a spark would

jump between electrodes if exposed to UV light (electrons being released from the electrode).

Other scientists noticed that the leaves of a negatively charged electroscope diverged less over time while a positively charged electroscope did not.


This was due to the electrons “leaking” away from the electroscope as the light energy struck the top plate, allowing the electrons to escape.

CLASSICAL THEORY

We will now look at the predictions made for light using classical (light behaving as a wave) physics and the actual observations that were made in experiments.

CLASSICAL THEORY

CLASSICAL PREDICTION:

The more intense (brighter) the light, the greater the kinetic energy of ejection of the electron. Bright light would eject electrons at high speed

ACTUAL OBSERVATION:

Intensity (brightness) did not lead to high velocity electrons. Instead it led to a greater numbers of electrons being ejected from the metal.

CLASSICAL THEORY

CLASSICAL PREDICTION:More photoelectrons should be ejected by

low frequency radiation (i.e. red) than by high frequency radiation.

e-m waves are oscillating electric and magnetic fields.

Low frequency waves allow more time for the electron to move in one direction before the field reverses and the electron moves in the opposite direction.

High frequency waves would move so fast the electron would hardly begin to move in one direction before it was forced to reverse direction - not ideal for ejection.

ACTUAL OBSERVATION:

Experiments showed that high frequency (UV) radiation ejected photoelectrons more readily than low frequency.

There was a minimum frequency below which no photoelectrons were ejected. This was called the threshold frequency which is different for different materials.

CLASSICAL THEORY


There should be a time delay between when a radiation is incident on a surface and when the photoelectrons are ejected (See point 2)

ACTUAL OBSERVATION:

Photoelectrons were ejected instantaneously.

CLASSICAL THEORY


The radiation’s wavefront falls over the whole surface, billions of photoelectrons should be simultaneously ejected.

ACTUAL OBSERVATION:

By limiting the amount of light on a surface, a single electron could be ejected.

CLASSICAL THEORY


One velocity of ejection should be possible for radiation of one frequency.

ACTUAL OBSERVATION:

Emitted photoelectrons have a range of ejection velocities and energies.

CLASSICAL THEORY

The observations made on the previous slides do NOT AGREE with the predictions made by the Classical Theory.

Photoelectric Effect - Changing VariablesHow can we resolve this?

http://ionaphysics.org/lab/Fendt/ph11e/ph11e/photoeffect.htm



EINSTEIN’S EXPLANATION -PHOTOELECTRIC EFFECT

In 1905 Einstein adopted quantum theory to explain the photoelectric effect and was awarded a Nobel Prize for Physics in 1921.

Planck also used quantum theory to explain the photoelectric effect. He said that the quantum effect occurred at the point where the radiation struck the electrons. The electron would only accept a discrete amount of energy from the incident radiation.


Too little energy and the electron would accept none.

Too much energy, the difference would be emitted as radiation.


Einstein also said that not only was the energy absorbed and emitted by atoms in bursts but the incoming radiation was in the form of discrete entities and not a continuous wave.

He named these discrete entities light quanta. This was renamed photons (a quantum of radiant energy) later as they do behave like particles and in keeping with other particles, electrons, protons and neutrons.

EINSTEIN’S EXPLANATION -PHOTOECLECTRIC EFFECTEinstein’s explanation depends on the relationship

E = hf

A UV photon would have more energy than a blue light photon as it has a higher frequency.

The key to his explanation is that each photon on striking an atom and being absorbed may release only one electron. It never shares its energy amongst electrons.

Any excess energy will be given as kinetic energy to the electron.

EINSTEIN’S EXPLANATION -PHOTOECLECTRIC EFFECT

According to Einstein, only high frequency light would have enough energy

( E = hf ) to eject an electron from a metal surface.

Low frequency light (like red light) might not have enough energy to pull the electron away from the atom’s nucleus.


Imagine that electrons in an atom are at the bottom of a potential energy well that has a sloping base as shown: -

--

W hf o (work function)

en ergy o f e lec tro n s

in an a to m

-v e

E = 0

photon

E = hf


As each electron is at a different depth, they are bound to the atom by a different amount. Each electron will then be emitted with different energies. If an electron absorbs a photon with sufficient energy, the electron can be freed.

--

-



in an a to m

-v e

E = 0

photon

E = hf


The minimum amount of photon energy required to remove the least bound electron is called the work function (W) and has the units joules but eV are more commonly used.

--

-



in an a to m

-v e

E = 0

photon

E = hf


The “least bound electron” is the electron in the outermost electron shell of the atom. This will be the easiest electron to pull away from the atom.

It is also called the “most energetic electron” because all of the remaining energy given to it will be in the form of kinetic energy which will give it the highest speed of any of the released electrons.


The work function is given by: W = hfo

fo = threshold frequency

The threshold frequency is the minimum frequency required to free the “least bound” electron.


When the photon falls on an electron with more energy than is needed to remove the bound electron, the difference in energy is transformed into kinetic energy of the electron.

The least bound electron is also known as the most energetic electron.

--

-

-v e

E = 0

E = hf

W hf o

( 12mv

2 ) m ax


Eincoming photon = K + Energy required for electron to escape

For the least bound electron,

Eincoming photon = K(max)+ W

As E = hf

EKmax = hf – hf0 --

-

-v e

E = 0

E = hf

W hf o

( 12mv

2 ) m ax

EXAMPLE 1

A certain metal has a work function of 2.0eV. Will light of wavelength 4.0 x 10-7 m cause the ejection of photoelectrons and if so what will be their maximum velocity of ejection?

EXAMPLE 1 SOLUTION

QUANTUM (MODERN) PHYSICS

The photon concept was used by Einstein to explain the experimental observations of the photoelectric effect.

We will now go back to the observations made in the photoelectric effect experiment and look at Einstein’s explanation using quantum physics.


ACTUAL OBSERVATION:

The (kinetic) energy of ejected photoelectrons is independent of the intensity of radiation.

QUANTUM EXPLANATION:

A greater intensity means that more photons will fall on the surface.

This will simply eject more electrons but NOT at a faster speed.


ACTUAL OBSERVATION:

Photoelectrons are more likely to be ejected by high frequency than low frequency radiation.


The energy of a photon depends on the frequency of radiation (E = hf).

A high-frequency photon has more energy and so gives more energy to the photoelectron.

A high frequency photon is more likely to have greater energy than the work function.


ACTUAL OBSERVATION:

Photoelectrons are ejected instantly.


All of the energy of the photon is given up to the electron instantly. Experimental results show that the maximum time delay for the photoelectric effect is about 10-8s.


ACTUAL OBSERVATION:

A range of electron velocities of ejection are possible.


Once the work function is subtracted, the remaining energy exists as kinetic energy.

Depending on which electron absorbs the photon, varying amounts of kinetic energy may be left over.


SUMMARY – the photoelectric effect can be best explained using the Quantum Theory (light behaving as a particle) as opposed to the Classical Theory (light behaving as a wave).

You would use the photoelectric effect in any question that asks you to prove that light behaves as particles (photons).

Experiment to test Einstein’s model

This is a diagram of an apparatus used to investigate the characteristics of photoelectric emission.

It is used to try to determine Plank’s Constant (h) from E =hf

Milikan used it to verify Einstein’s model in 1916


The cathode (negative) and anode (positive) are sealed in an evacuated glass tube to reduce the impedance (number of collisions) of the photoelectrons reaching the anode.

When the light strikes the cathode it causes photoelectrons to be emitted.


If they cross the gap then they will create a current that will be read by a microammeter.

The anode is made progressively more positive attracting more photoelectrons until the saturation current is reached.


This means that there cannot be more electrons given out from the cathode.

It is attracting all of the electrons being given off at the cathode.


Note that we DO NOT vary the frequency or the intensity during the time that we are making the anode more positive.

During this time the current will get stronger, proof that the electrons are being emitted with different kinetic energies.


Only when you make the anode very positive do you finally attract the electrons that have very little kinetic energy (they are drifting around) due to the fact that they required a large amount of energy just to free them (their Work Function).


If the anode is made negative, electrons are repelled until there is no anode current. When the current is zero, the voltage applied is called the stopping voltage (Vs).


At this point even the most energetic electron (with the smallest work function and hence the most kinetic energy) will not be able to make it to the anode (due to repulsion).


This graph shows what happens as we change the frequency (colour) of the light and the voltage required to stop the most energetic electron for that particular frequency.

Vs e = hf – hf0

The Photoelectric Effect

1.The more intense the incident light, the greater the energy of ejection of the electron.

The voltage required to repel the photoelectron from a current measuring device,should also be greater.


Intensity increased the number of ejected photoelectrons but, the energy is independent of the intensity of the radiation.

The number of electrons emitted per second is,proportional to the intensity of the emitted light.


2.More photoelectrons should be ejected by low frequency radiation (i.e. red),than by high frequency radiation.

Classical theory considers e-m waves to be,oscillating electric and magnetic fields.


Low frequency waves allow more time for the electron to move in one direction,before the field reverses and,the electron moves in the opposite direction.


High frequency waves would move so fast the electron would hardly begin to move in one direction, before it was forced to reverse direction, not ideal for ejection.

Experiments showed that high frequency (UV) radiation ejected photoelectrons more readily, than low frequency.


There was a minimum frequency below which,no photoelectrons were ejected.

This was called the threshold frequency,different for different materials.


3.There should be a time delay between when a radiation is incident on a surface and,when the photoelectrons are ejected.

Photoelectrons were ejected instantaneously.


4.The radiation’s wavefront falls over the whole surface,billions of photoelectrons should be simultaneously ejected.

By limiting the amount of light on a surface, a single electron could be ejected.


5.One velocity of ejection should be possible for radiation of one frequency.

The electric current in the detector should immediately drop to zero once a critical voltage for repulsion of the single energy photoelectron is reached.Emitted photoelectrons have a range of

ejection velocities and energies.


Complete the homework assignment On the wiki

http://pac-physics.wikispaces.com/Topic+13

WAVE NATURE OF MATTER

It has been shown that in some circumstances, light exhibits certain behaviours characteristic of waves.

In other circumstances, light behaves as particles.

Could the reverse be true, namely that particles can behave as waves? This topic investigates this question.


DE BROGLIE’S HYPOTHESIS

Count Louis de Broglie (1892 - 1970) believed in the symmetry of nature. In 1923 he reasoned that if a photon could behave like a particle, then a particle could behave as a wave.


Special Relativity has given us the relationship

E = mc2

But, we also know E = hf

This gives

hf = hc/λ = mc2

mc = h/ λ

p = h/ λ


He turned Compton’s relationship to make wavelength the subject of the equation.

Compton- “a photon has momentum”

De Broglie- “An electron has a wavelength”

h

p

p

h


This is called the de Broglie wavelength of a particle.

All particles (electrons, protons, bullets, even humans) have a wavelength.

They must be moving.

They are called “matter waves”.


We cannot see light. We can only make inferences about the nature of light by looking at its properties.

Its properties indicate that it is both wave like and particle like in nature.


We also cannot see atoms. We often think of them as exhibiting the properties of particles.

But, because we have never seen them, could they be waves pretending to be particles?

De Broglie suggested that particles, in some instances could be wave like.

EXAMPLE 1

Calculate the de Broglie wavelength associated with a 1.0 kg mass fired through the air at 100 km/hr.

EXAMPLE 1 SOLUTION

EXAMPLE 1 SOLUTION

Note the wavelength is so small that it cannot be detected and measured.

We cannot create slits capable of diffracting such small wavelengths.

Can a microscopic object give a more realistic wavelength?

EXAMPLE 2

Calculate the de Broglie wavelength that would be associated with an electron accelerated from rest by a P.D. of 9.0V

EXAMPLE 2 SOLUTION

EXAMPLE 2 SOLUTION

EXAMPLE 3

Calculate the de Broglie of a H atom moving at 158 m s-1 (interstellar space)

EXAMPLE 3 SOLUTION

h

mv

6 626. x 10

1.672 x 10 x 158

-34

-27

= 2.50 x 10-9 m

These are X Rays which do not penetrate the atmosphere

An EXPERIMENT to verify de Broglie

C.J. Davisson and L.H. Germer performed an experiment in 1927 to verify de Broglie’s hypothesis.

DAVISSON-GERMER EXPERIMENT

Electrons were allowed to strike a nickel crystal. The intensity of the scattered electrons is measured for various angles for a range of accelerating voltages.

DAVISSON-GERMER EXPERIMENT

DAVIDSON-GERMER EXPERIMENT

It was found that a strong ‘reflection’ was found at θ = 50° when V = 54V.

This appeared to be a place of constructive interference, suggesting that the “matter waves” from the electrons were striking the crystal lattice and diffracting into an interference pattern.


The interatomic spacing of Nickel is close to the ‘wavelength’ of an electron. Therefore it would seem possible that electron matter waves could be diffracted.

Davidson and Germer set out to verify that the electrons were behaving like a wave using the following calculations.


Theoretical Result (according to de Broglie’s calculation)

The kinetic energy of the electrons is

1/2 mv2 = Ve


The de Broglie wavelength is given by:

For this experiment:Vem

h

mv

h

2

m10 x 67.1

)10 x (9.11 x )10 x (1.6 x 54 x 2

10 x 625.6

10-

31-19-

-34

DAVIDSON-GERMER EXPERIMENTExperimental Result (according to Davidson-

Germer)

X-ray diffraction had already shown the interatomic distance was 0.215 nm for nickel.

Since θ = 50°, the angle of incidence to the reflecting crystal planes in the nickel crystal is 25°as shown below:



dsin θ = mλ

For the first order reinforcement…

λ = dsinθ

= (.215 x 10-9)(sin50°)

= 1.65 x 10-10 m


The close correspondence between the theoretical prediction for the wavelength by de Broglie (1.67 x 10-10 m) and the experimental results of Davidson-Germer (1.65 x 10-10 m) provided a strong argument for the de Broglie hypothesis.

SPECTRA

Spectra provide a “picture or model” of what the atom looks like.

About midway through the 19th century, it was discovered that if a gas could be stimulated to produce light, the colour could indicate the nature of the gas.

The light was passed through a prism or diffraction grating to produce a spectrum.

SPECTRA

There are 3 types of spectrum:

CONTINUOUS SPECTRUMHeating an object such as a filament in an

incandescent globe produces this. When viewed through a spectroscope, all spectral colours are seen.

An object’s spectrum can extend beyond the visible part of the spectrum.

CONTINUOUS SPECTRA

A hotter object will produce more energy at all wavelengths than does a cooler object.

The hotter the object, the more energy is emitted at shorter wavelengths. The distribution of wavelengths depends on the temperature.

This is why the colour of objects change as the object is heated.

CONTINUOUS SPECTRA

LINE EMISSION SPECTRA

Produced from light emitted by excited, or

energised gases.

When gaseous atoms become excited, they emit characteristic colours (sodium - yellow and mercury - mauve).


Looking through a spectrometer, the light is made up of specific frequencies and is referred to as spectral lines and the full spectrum seen is known as a line emission spectrum.

Unlike solids and dense liquids, gases do not produce a continuous spectrum.


They are not vibrating back and forth at a variety of frequencies because they are not bonded together.

Instead, they emit SPECIFIC frequencies of light as seen in the spectroscope.


These specific frequencies are produced by electrons falling from higher energy levels to lower energy levels.

A photon of a specific frequency is emitted in the process.


ABSORPTION SPECTRA

This is a continuous spectrum with dark lines. This means colours are removed due to their absorption by matter.

In 1814, a German scientist, Fraunhofer, focused the spectrum of light from the sun and found over 700 dark lines or gaps in the spectrum.

ABSORPTION SPECTRA

ABSORPTION SPECTRA

If we can match the absorption spectrum of the sun with the absorption spectra of an element like calcium, then we can say that calcium is in the sun’s outer atmosphere.

ABSORPTION SPECTRA

The absorbing material used in a lab is generally a gas or liquid but any state could be used. The dark lines depend on the nature of the absorbing material.

The absorption spectrum is as much a fingerprint of the element giving off the E-M energy as is the emission spectrum.

ABSORPTION SPECTRA

In all cases the absorption and the emission spectra will match perfectly.

ABSORPTION SPECTRA

THE BOHR MODEL OF THE ATOM

In 1911 Bohr ignored all the previous descriptions of the electronic structure as they were based on classical physics.

This allowed the electron to have any amount of energy.

Planck and Einstein used the idea of quanta for the energy carried by light.


Bohr assumed that the energy carried by an electron was also quantized.

From this assumption, he formed three postulates (good intelligent guesses) from which he developed a mathematical description.


0

+

-

free e-

} bound e -

energy levels

B o h r a to m


In summary, if the atom had electrons that varied in their energy levels, you would expect to get random frequencies emitted.

This is not the case.Electrons give off photons of SPECIFIC

frequencies. More evidence for the Quantum Theory!


An electron can be moved to a higher energy level by…

1. INCOMING PHOTON- Must be of exactly the same energy as E2 – E1

2. INCOMING ELECTRON- remaining energy stays with the incoming electron.

3. HEAT- gives the electron vibrational energy.


IONIZATION- energy required to remove an electron from the atom.

Example: the ionization energy required to remove an electron from its ground state (K=1) for Hydrogen is 13.6 eV.


Let’s prove this using the following formula…

LYMAN SERIES

En-Em = hf(13.6 –0.85)(1.6X10-19)=hc/λλ=9.7X10-8 m (98nm)This is UV Light

0

-2

-4

-6

-8

-1 0

-1 2

-1 4

2

L y m a n se rie s

B a lm e r se rie s

P a sc h e n se rie s

gro u n d sta te

1 st e x c ite d sta te

n = 3

n = 1

n = 2

n = 4

n = 5n =

io n ize d a to m (e le c tro n u n b o u n d a n d fre e to ta k e a n y e n e rgy )

(K sh e ll)

(L sh e ll)

(M sh e ll)(N sh e ll)

-1 3 .6

-3 .4

-1 .5 1-0 .8 5


BALMER SERIES

En-Em = hf

(3.4 –0.85)(1.6X10-19)=hc/λ

λ=4.87X10-7 m

= 487 nm

This is Blue Light.

0

-2

-4

-6

-8

-1 0

-1 2

-1 4

2

ener

gy (

e V)

L y m a n se rie s

B a lm e r se rie s

P a sc h e n se rie s

gro u n d sta te

1 st e x c ite d sta te

n = 3

n = 1

n = 2

n = 4

n = 5n =

io n ize d a to m (e le c tro n u n b o u n d a n d fre e to ta k e a n y e n e rgy )

(K sh e ll)

(L sh e ll)

(M sh e ll)(N sh e ll)

-1 3 .6

-3 .4

-1 .5 1-0 .8 5

EXAMPLE 1Use the first three energy levels for the

electron in hydrogen to determine the energy and hence wavelength of the lines in its line emission spectrum.

n = 3

n = 2

n = 1-1 3 .6 e V

-3 .4 e V

-1 .5 1 e V

EXAMPLE 1 SOLUTION

From the diagram, the atom can be excited to the first (n = 2) and second

(n = 3) excited states. From these, it will return to the ground state emitting a photon. The electron can make the following transitions:

EXAMPLE 1 SOLUTION

n = 3 n = 1, Ephoton = E3 - El

= -1.51 - (-13.6) = 12.09eV

n = 2 n = l, Ephoton = E2 - E1

= -3.4 - (-13.6) = 10.2eV

EXAMPLE 1 SOLUTION

n = 3 n = 2, Ephoton = E3 - E2

= -1.51 - (-3.4) = 1.89 eV

n = 3

n = 2

n = 1

EXAMPLE 1 SOLUTION

To find the wavelengths of the three photons, use

Note: convert eV to J

E = ie. =E

hfhc hc

EXAMPLE 1 SOLUTION

n=3 n=l,

= 1.024 x 10 -7 m = 102 nm (ultraviolet)

6.6 x 10 x 3 x 10

12.09 x 1.6 x 10

-34 -8

-19

EXAMPLE 1 SOLUTION

n=2 n=l, λ=

= 1.21 x 10-7 m = 121nm (ultraviolet)

n=3 n=2, λ=

= 6.55 X 10-7m

= 655 nm (visible-red)

19-

834

10 x 1.6 x 10.210 x 3 x 10 x 6.6

19-

834

10 x 1.6 x 1.8910 x 3 x 10 x 6.6

EXAMPLE 1 SOLUTION

Note that we have two Lyman series lines (those ending in the ground state) and one Balmer line (ending in the first excited state).

n = 3

n = 2

n = 1

EXAMPLE 1 SOLUTION

Those from the Lyman series produce lines in the ultra-violet part of the spectrum while the line in the Balmer series produces a line in the visible part of the spectrum.

n = 3

n = 2

n = 1

EXAMPLE 2

The valence (outer) electron of sodium is involved in the emission of two photons that have almost the same energy. The wavelengths are 589.0 nm and 589.6 nm. It is known that the valence electron falling to the ground state from one or two higher levels will produce these photons. Determine the excitation energies of these two energy levels, and find the difference in their energy. (Work to 4 sig. Figs.)

EXAMPLE 2 SOLUTION

The diagram represents the two energy levels whose energies are El and E2

above the ground state.

gro u n dsta te

E1

E2

EXAMPLE 2 SOLUTION

E1 hfhc

1

1

6.625 x 10 x 3.00 x 10

5.896 x 10

-34 8

-7

= 3.371 X 10-19 J

= 3.374 X 10 -19 J

E2 hc

26.625 x 10 x 3.00 x 10

5.890 x 10

-34 8

-7

EXAMPLE 2 SOLUTION

Now it is more convenient to work in electron volts. E1 = 2.107eV, and E2 =

2.109eV, so the energy difference is 0.002 eV (3.2 x 10-22 J). The energy level diagram can now be drawn.

gro u n dsta te

E1

E2

= 2 .1 0 9 e V= 2 .1 0 7 e V

5 8 9 .0 n m 5 8 9 .6 n m

Electron in a box modelThe electron is bound to the nucleus by the Coulombic

attraction

Consider, the electron to be confined in a one dimensional box whose edges are defined by 1/r

Classical wave theory states that such a confined wave would be a standing wave and hence

λn =2 L /n …where n = 1,2,3…

But pn = h / λn = n h / 2 L

Also EK = p2/2m … En = n2h2/ 8mL2

Schrödinger Model

The Bohr Model was a landmark in the history of Physics

Limitations show that a new model was needed

Less than 2 years after de Broglie gave us the matter wave

Erwin Schrödinger (1887-1961) an Austrian Physicist developed a new comprehensive theory

Schrödinger Model

Schrödinger Model

De Broglie determined the wavelength and momentum of a matter wave

What about amplitude?

The amplitude of a matter wave is given the symbol Greek letter psi

Schrödinger Model

Schrödinger developed an equation to determine the wavefunction,.

The wavefunction represents the amplitude of a matter wave as a function of time and position

It is a differential equation – the solution of which exactly predicts the line spectra of a hydrogen atom

may vary in magnitude from point to point in space and time

Schrödinger Model

Consider Young’s Double Slit Experiment If slits are in the order of the wavelength

of lightAn interference pattern would be seen

Reduce the flow of photons (or electrons) to one at a time

Schrödinger Model

Initially electrons appears to be random

Schrödinger Model

If time is allowed to elapse, the pattern starts to become visible

Schrödinger ModelEventually the pattern follows that

expected by wave theory

Schrödinger Model

We actually see:

Schrödinger Model

Where 2 is zero; A minimum in the pattern would be seen

Where 2 is a maximum A maximum in the pattern would be seen

Schrödinger Model

To get an interference pattern; Electrons must pass through both slits at

the same time

This is possible as an electron is as much a wave as a particle

What would happen if we covered up one slit?

Schrödinger Model

Electron would pass through one slit only; A diffraction pattern would be seen

If we then covered up the other slit? A different diffraction pattern would be

seen

There would be no interference pattern

Schrödinger Model

If both slits were open Electron passes through both slits As if it were a wave Forming an interference pattern

Yet each electron would make a tiny spot on the screen as if it were a particle

Schrödinger Model

The main point: Treat electrons as waves;

represents the wave amplitude Treat electrons as particles

Must treat them on probabilistic basis 2 gives the probability of finding a given

electron at a point

We cannot predict or follow the path of an electron precisely through space and time

Schrödinger Model

Can use Schrödinger’s equation to determine the probability of finding an electron at any given place or time around a nucleus

2 tells us an electron is more likely to be found close to the nucleus than far away

This allowed the development of an electron cloud

Heisenberg Uncertainty Principle

In 1927 Werner Heisenberg proposed a principle that helped understand the interpretation of the wavefunction.

Let the uncertainty in position be Δx Let the uncertainty in momentum be Δp

The uncertainty principle states Δx Δp h

(More accurately h/4)

If p is known precisely then = h/p is also known

But a completely defined means that a wave must be infinite in space and time

Therefore if p is known then we cant know the position and vice versa

Uncertainty principle also applies to energy

ΔE Δt hThis explains why spectral lines are of

finite width

Quantum Physics AHL 13.1. THE PHOTOELECTRIC EFFECT Early last century, several people had noticed that light was capable of ejecting electrons from.

Documents

classical light

uv light electrons

high frequency radiation

high velocity electrons

free electrons

freeing of electrons

high frequency waves

low frequency radiation