Transcript ppsx

About these slides
• These slides are used as part of my lessons and
shouldn’t be considered comprehensive
– There’s no excuse for not turning up to lessons!
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Recap
Wave-particle duality
• You should be familiar with the idea that
light can be thought of as a wave or a
particle
• The particle is called a __________
• The energy of this particle is given by the
equation
Rewrite the equation in
terms of wavelength, l
E=hf
• Where h is ________ constant and f is
the frequency of the light
Einstein
Evidence for wave-particle
duality
• One piece of evidence that light can
behave as a wave or a particle is the
photoelectric effect
• This was discovered by Hertz as a side
effect of his work on radio waves
• He noticed that metals emit electrons
when electromagnetic radiation of a
certain frequency is directed at a metal
The photoelectric effect in
action
Incident Radiation
zinc plate
gold leaf
electrode
Photoelectric effect
• The photoelectric effect gives rise to the
following observations:
– Emission of electrons only takes place above
a certain threshold frequency of incident
electromagnetic radiation. This minimum
frequency is dependent on the metal.
If light was a wave then electrons would be
emitted at any frequency, as electrons would
gradually gather energy from the waves regardless
of their frequency
Photoelectric effect continued
• The photoelectric effect gives rise to the
following observations:
– Electrons are emitted as soon as the source
of electromagnetic radiation is switched on,
no matter what the intensity of the radiation.
If light was a wave then electrons would gather
energy more quickly, and be emitted sooner, by
high intensity light sources.
Photoelectric effect continued
• The photoelectric effect gives rise to the
following observations:
– The number of electrons emitted per second
is proportional to the intensity of the incident
radiation, unless its frequency is less than
the threshold frequency, in which case
emission isn’t observed
Particles as waves
Electron diffraction
Electron diffraction
• If electrons were acting as particles we
would see them either pass straight
through the crystal, or bounce off
• Instead we see an interference pattern
– Exactly the same as if electrons are acting as
waves
• It is possible to do the Young’s slits
experiment and see an interference
pattern even if the electrons are fired one
at a time
de Broglie
• In 1923 de Broglie was the first to suggest
that matter could be a wave
• He said that the de Broglie wavelength, l,
of a particle is related to its momentum,
p, by the equation h
l
p
• We know that momentum = mass x
velocity so
h
l
mv
Question
• Calculate the de Broglie wavelength of
– An electron moving at 2.0 x 107 ms-1
– A proton moving at the same speed
• I weigh 85kg and can run at 6 ms-1 what
would my de Broglie wavelength be?
?
Question
?
• Calculate the momentum and speed of
– An electron that has a de Broglie wavelength
of 500nm
– A proton that has the same de Broglie
wavelength
✔
The photoelectric effect
• We saw this last lesson
• As long as light is above a certain frequency it
can eject electrons from a metal surface
• This can be explained as the photon having a
certain energy, E = hf
• It therefore follows that there is a minimum
energy needed to eject an electron from a
metal surface at zero potential
– This is the work function, f
Explaining the photoelectric
effect
• In order to explain photoelectricity
Einstein said that:
– Electrons must be absorbing a single photon
in order to escape from the metal
– This photon transfers energy given by E=hf
– If the photon has less energy than the work
function, f, of the metal the electron cannot
escape
– The work function is the minimum energy
required for an electron to escape the metal
surface
Photoelectric effect
• We can calculate the maximum kinetic
energy of an electron ejected by a
photon with energy hf
Ekmax = hf – f
• So emission will take place as long as hf > f
• Which means there is a threshold frequency
above which an electron is emitted:
f min 
f
h
Plotting Ekmax against frequency
• Since EKmax = hf – f we can plot a graph
of the form y = mx + c
• So if we plot EKmax against f we get a
straight line
What is the gradient?
What is the intercept on
the x axis?
What would the intercept
on the y axis be?
EKmax /J
f / Hz
Questions
• Why does photoelectric emission only
take place above a certain frequency of
incident radiation?
• Calculate the frequency and energy of a
photon of wavelength
– 450nm
– 1500nm
?
Questions
• The work function of a metal plate is
1.1 x 10-19J
• Calculate:
– The threshold frequency
– The maximum kinetic energy of electrons
emitted from this plate when light of
wavelength 520nm is directed at the metal
surface
?
Electron volts
• We’ve already encountered the electron
volt
• It is defined as the unit of energy
equivalent to the work done when an
electron is moved through a potential
difference of 1 volt and is equal to 1.6 x
10-19J
Ionisation
• An ion is a charged atom
• Ionisation is the process of making an ion
by adding or removing electrons
• The energy to do this can be provided by
electricity, heat, light or by collisions
between alpha, beta or gamma radiation
and an atom
Energy levels in atoms
• Electrons surrounding atoms are like
standing waves
• They have certain allowed energy levels
• It is the movement between these levels
that happens during excitation
• The lowest energy state of an atom is its
ground state
• If an electron moves into a higher energy
level the atom is in an excited state
Excitation
• In excitation an electron can be moved
from an inner shell to an outer shell
• This only happens at specific excitation
energies which are dependent on the
atom and correspond to its excited
This is excitation by
states
collision – one
electron hits
another and
promotes it to a
higher energy level
Excitation continued
• The excitation energy will always be less
than the ionisation energy
• There are a number of excitation energies
for each atom
– These correspond to the difference between
different shells
– Electrons can also ‘jump’ more than one shell
at a time
Energy Levels Example:
Mercury
Ionised 0.00 eV
This is excitation by
absorption of a
photon. If it has
exactly the right
frequency it
promotes the
electron to a higher
energy level
-1.59 ev
-1.60 eV
-2.51 eV
-2.71 eV
-3.74 eV
-4.98 eV
-5.55 eV
-5.77 eV
-10.44 eV
De-excitation
• The excited states of an atom are
unstable
• They have a gap in a low energy electron
shell
• Electrons from a higher shell can drop
into this lower level, emitting a photon in
the process
De-excitation Example: Mercury
• This is fluorescence
• An atom in an excited
state can de-excite by
emitting photons of the
same or lesser energy
than the one that
excited it
• In this case the photons
have energies of 0.79
eV and 4.67 eV
Ionised 0.00 eV
-1.59 ev
-1.60 eV
-2.51 eV
-2.71 eV
-3.74 eV
-4.98 eV
-5.55 eV
-5.77 eV
-10.44 eV
Calculating the frequency of
photons
• Since the photon emitted comes from the
transition between two energy levels it will
have an energy, E given by:
E = E1 – E2
• Where E1 is the energy of the higher level
and E2 the energy of the lower
• We also know that E = hf, so we can
calculate the frequency of the emitted
photon
Fluorescent Tubes
• Fluorescent tubes contain mercury
vapour and have a fluorescent coating on
the inner surface
Filament Electrodes
Mercury Vapour
Fluorescent Tubes
• When a tube is turned on:
– Ionisation and excitation of mercury atoms
occurs as they collide with each other and with
electrons
– Photons are emitted at UV and visible
frequencies
– UV photons are absorbed by atoms in the
fluorescent coating
– Coating atoms de-excite and emit visible
photons
– A range of coatings is used to give a ‘white’ light
Questions
• How much energy is
needed, in Joules, to
excite the atom to its
highest excitation level?
• How many different
energies could photons
released by a mercury
atom in the 5.46 eV
excited state have?
?
Ionised 0.00 eV
-1.59 ev
-1.60 eV
-2.51 eV
-2.71 eV
-3.74 eV
-4.98 eV
-5.55 eV
-5.77 eV
-10.44 eV
Question
• An atom absorbs a photon of energy 3.8
eV and subsequently emits photons of
energy 0.6 eV and 3.2 eV
– Sketch an energy level diagram to represent
these changes
– Describe what is happening to the electrons
in the atom during this process
– What is the frequency of the photons
emitted?
?
Answer
?
Spectral Lines
• Since each element has different energy
levels, and therefore transitions between
levels, each element will emit different
frequencies (wavelengths) of
electromagnetic radiation
• This means the photons emitted are
characteristic of that element
Mercury
Hydrogen
Question
?
• A mercury atom de-excites from its 4.9 eV
energy level to the ground state
• Calculate the wavelength of the photon
released
c = 3.0 x 108 ms-1, e = 1.6 x 10-19 C, h = 6.63 x
10-34 Js