Reference:Nelson
Physics VCE Units 3 & 4Chapters
18 – 21Page 468
In
1911 Robert Millikan devised an experiment to measure small charges, this
was the oil-drop experiment.The
idea of this experiment was that if a mist of very fine oil-droplets was
made, some of them would be charged by static electricity.Those
with positive charge would be attracted to a negative electrode and vice-versa.Millikan
placed two plates one above the other, connected a power source to the
plates, sprayed oil droplets between the plates and observed using a microscope.
With
no potential difference between the plates the droplets will fall towards
the bottom plate.
When
a potential difference is applied between the plates the motion of some
of the droplets will change.Those
that are charged will be attracted to one of the plates
The
behaviour that Millikan was most interested in was when the electric field
force equaled the weight force.
To
obtain this situation a droplet is selected and the voltage adjusted until
the droplet is stationary.
Weight
= Electric field force
W
= mg
Electric
field force F = qE
Somg
= qE
and
wherem
= mass of droplet (kg)
g
= acceleration due to gravity (9.8 kg m s-2)
q
= charge on droplet (Coulomb)
E
= electric field strength (Volts/m)
V
= voltage between plates (Volts)
d
= distance between plates (m)
Thus
the charge on a droplet of known mass can be calculated
Eg.An
oil droplet of mass kg
is held stationary in an electric field.The
lower plate is earthed and the upper plate has a potential of + 750 V.the
spacing of the plates is 5.0 mm.Calculate
the charge on the droplet and state its sign.
mg
= qE
=
mg
= qE
9.8
= q
q
=
q
= C
The
charged particle will be accelerated by the electric force on it and gains
kinetic energy as it moves towards the positive plate.
Since:
and the force on the particle is F
= qE
The
kinetic energy gained is the work done on the particle and is given by:
Ek=qV
Millikan
obtained results from many oil droplets and found that the charge on the
oil droplet was always a multiple of C.Thus
he concluded that the charge on an electron was -
C,
the fundamental charge.This is also
known as quantisation, whole number multiples of a fixed amount.
It
was known that when the cathode of an electric circuit was heated in a
vacuum with a large potential difference applied between that cathode (-ve)
and the anode (+ve), a beam appeared to travel between the two electrodes.Since
they emanated from the cathode they were called cathode rays and have a
negative charge.
J.J.
Thomson (1856-1940) (Nobel 1906, Knighted 1908) investigated
this phenomenon and realized that these cathode rays could be deflected
in their path by magnetic and electric fields. Since they carried a charge,
he concluded that these rays were actually particles, which he called electrons,
and was able to measure their charge-to-mass ratio .
Having
shown that cathode rays consist of moving charged particles which can be
deflected by both electric and magnetic fields, Thomson devised two experiments
to measure the velocity of these particles and the ratio of the mass to
their charge. The first experiment used measurements of the energy and
charge deposited by the beam and the deflection of the beam in a uniform
magnetic field. The second experiment used uniform electric and magnetic
fields to bend the beam a fixed amount. Both methods yielded a value of C/g
for the ratio of charge-to-mass and a velocity of about
m/s.
In
1909, Millikan's oil drop experiment allowed him to determine the charge
of the electron, coulomb.
Earlier, Thompson determined the charge to mass ratio of the electron,
coulomb
/ gram, so this determination of the charge by Millikan allowed the determination
of the mass of the electron,
grams.
i.e.
grams
Thomson
is credited with the discovery of the electron.
Problem
Set #1:TextPage
510Questions 1, 2, 8 – 33
Scientists
always try to come up with a model or theory for the way things behave.Light
has a couple of different models, the particle model and the wave model.
It
can be said that light is made of particles. This
means that a light source is emitting particles.If
this is true, what do we know about these particles.
1.The
particles travel at the speed of light.(3 ´
108 m/s)
2.The
particles are emitted in all directions and the spread out.
3.The
particles are very small.
4.Bright
light emits more particles than does dim light.
Now
lets test the model to see if it holds for all the behaviour of light.
Particles
will reflect in the same way as light.E.g.
Billiard balls
i)the
angle of incidence = the angle of reflection.
ii)The
incident ray, reflected ray and the normal all lie in the same plane.
We
can set up an experiment to simulate refraction.This
involves two horizontal planes with an inclined plane between them.
The
angles i and r can be measured and if we keeps the slope constant the ratio is
a constant.
This
situation is similar to that of light travelling from a low refractive
index to a high refractive index.
Eg.air - glass
Two
Problems
i)The
particles bend towards the normal when they speed up.This
is contrary to the behaviour of light which slows down when it bends towards
the normal.
ii)In
real life light is both reflected and refracted at the same time, this
model does not provide a situation where this phenomenon occurs.
The
behaviour of light is such that the intensity is inversely proportional
to the square of the distance from the light source.
I.e.
Consider
a point source of particles emitting in all directions.Also
consider two spheres of radii r and 2r around this source.
Intensity
= No. particles
Area
Surface
area of sphere = 4 p
r2
The
number of particles emitted from the source each second is given by N.
Intensity
on smaller sphere =
Intensity
on larger sphere =
=
Ratio
of intensities = 4
Ratio
of distances = 0.5
This
means an inverse square relationship exists.
I.e.
However
as we could see in the refraction example the particle model is limited.The
particle model also has trouble in explaining diffraction through narrow
slits.So another model needs to
be used.This does not mean that
the particle model is wrong, just that it cannot be used to explain all
of the behaviour of light.
In
this model we think of light as waves spreading out from a lamp in all
directions.This is similar to a
stone being dropped into water except in three dimensions.
Waves
are reflected from a barrier in such a way that the angle if incidence
is equal to the angle of reflection.
Waves
from two different sources pass through each other, in the same way that
light beams cross over.
Water
waves can easily be made to refract.
As
waves move into the shallow region they slow down and are bent towards
the normal.Thus deep water could
be air and shallow glass.
Note:
This prediction is opposite to the particle model, but is the correct one.
Water
waves can be focused by using various shaped shallow sections or barriers.
If
straight waves are generated in a ripple tank and allowed to pass through
a gap they appear to spread out.
The
amount of spreading out, it is called diffraction, is greatest when the
gap is small and the wavelength large.Diffraction
is also seen when waves (light) pass close to an object, the amount of
diffraction depending on the size of the object.
If
waves are incident on a barrier with two slits in it then an interesting
pattern can be seen.
It
consists of regions of calm water alternating with regions of wave moving
away from the barrier.
The
number and separation of these regions can be altered by adjusting the
distance between the slits, the width of the slits and the wavelength.
Prac
#1:EXPT
4.2Diffraction and Interference
of Light
Demo:Laser
and slit
If
we look at a point source of light through a narrow slit, we see a fringe
pattern formed of light and dark bands.
If
we look at the intensity of the light across the pattern we get a graph
that looks like.
If
we look closely at the middle of the pattern it looks like.
Analysing
the pattern we notice that the central fringe is brighter than the others
and the brightness diminishes the further away from the center.We
also notice thatthe central fringe
is twice the width of the others.The
intensity of the light is zero at certain points along the pattern.
Mathematically
it can be shown that the width of the central maximum of the diffraction
pattern is given by
where
is the width of the central maximum
l
is the wavelength of the light
L
is the distance from the slit to the screen
is
the width of the slit
Note:The
narrower the slit, the greater the diffraction (for constant wavelength)
The
longer the wavelength (towards red), the greater the diffraction (for constant
slit width)
The
dark bands (nodes) will occur for a path difference of
Example
:
A sodium vapour lamp emits yellow light of wavelength 580 nm.If the diffraction pattern of the lamp has a central maximum 11.6 cm wide when viewed on a screen 2.0 m from the slit, what is the width of the slit?
Demo:Laser
and double slit
An
interference pattern is also seen when two slits are used.The
first experiment done in this area was performed by Thomas Young, he was
the first person to show the wave nature of light.The
fringe pattern formed by a double slit looks like.
If
we look at the intensity of the light across the pattern we get a graph
that looks like.
Let
us consider the bright fringe which occurs at point P on the diagram.
The
distances S2 P and Q P are equal, so the path difference for
light travelling from the two slits is the length of S1 P.That
is, d sin q
Constructive
interference will produce a bright band at P when this path difference
is a whole number of wavelengths.That
is, P will be located on a bright band when
If
we let the distance from the centre to the bright fringe equal x.
For
small values of q
we can use the approximation sin q»
tan q.Then:
whereq
is the angle made with the central line
l
is the wavelength used
d
is the distance between slits
n
= 0, ±1, ±2, ±3,
……. (the fringe being looked at)
L
is the distance from the slits to the screen
Xn
is the distance from the central line to the nth fringe
The
bright bands on the screen are lines of constructive interference.I.e.
a path difference of nl;
n = 1, 2, 3,….
The
dark bands are lines of destructive interference.I.e.
a path difference of ;
n = 1, 2, 3,….
These
"light stripes" are excellent evidence that light travels as a wave."Light
stripes" cannot be explained in terms of particles.
ExampleA
screen containing two slits 0.100mm apart is 1.20 m from the viewing screen.Light
of wavelength l
= 500 nm falls on the slits from a distant source.How
far is it between bright fringes on the screen.
m
Problem
Set #2:TextPage
489Questions 1 – 23
Early
this century the theory postulated by Niels Bohr, about the structure of
the atom was generally accepted.Bohr's
theory said that electrons in an atom have fixed energy levels and can
move between energy levels provided they are given the correct amount of
energy.
In
1925 Frank and Hertz designed an experiment to test the predictions of
the Bohr Theory.Their results coincided
with the theory and they received the Nobel prize for their efforts.
Bohr
also postulated that electrons could jump from one stable energy level
to another of lower energy.This
would follow with the emission of a discrete package of energy he called
QUANTA.The value of this quantum
of energy was given by the difference in energy of the two levels.It
was also noted that these quanta of energy often appeared as light.The
quanta of energy emitted in these situations are called PHOTONS.
where
h is Plank's constant = 6.6 ´
10-34 J s or
Example
1:
Calculate
the energy, in joule, of a photon of ultraviolet light with frequency 7.4 ´
1014 Hz.
Example
2:
What
is the energy, in electron volt, of a photon of red light of wavelength
640 nm.
If
photons are released from atoms when electrons drop down energy levels
the opposite could also occur.However
this will only occur if the energy of the incoming photon equals the difference
in the atoms energy levels.Then
the energy of the photon is absorbed, the electron jumps up to its new
energy level and the photon ceases to exist.If
the energy of the incoming photon does not exactly equal the energy difference
of any of the atoms levels, it will not be absorbed.
If
a photon with a large enough energy strikes an atom, an electron may be
completely removed from the atom.This
is called the photoelectric effect.(It
is the principle by which photoelectric solar cells generate electricity).
Light
radiation falling on the cathode causes emission of electrons, photoelectrons,
which are then attracted to the positively charged anode.Thus
a current flows.However a current
is recorded even when V = 0.
If
the photon incident on the plate has sufficient energy, it will raise the
electron up through the energy levels and away from the atom.So
the incident photon must carry enough energy to ionise the atom and give
the electron some energy so that it can move away.
It
was found that even if V is reversed, so that there is a retarding potential
across the plates, a current still flowed.I.e.
some of the photoelectrons leave the cathode with sufficient kinetic energy
to reach, the now negative, anode against the retarding potential.
For
a given frequency of light there will be a range of kinetic energies of
the photoelectrons.This will depend
on which energy level the electron comes from.If
we adjust the retarding potential until it is just large enough to stop
any current flowing in the circuit then we will have stopped the most energetic
photoelectrons.If this happens at
Vmax for electrons of charge q then we have a measure of the
maximum kinetic energy of the ejected electrons, for that frequency of
light.
i.e.KEmax=work
done to stop photoelectron
=q
Vmax
soK.E.=q
Vunits electron volt (or joule)
The
size of the current I is independent of the intensity of the light.I.e.
increasing the intensity of the light source increases the number of electrons
emitted but not their energy.This
is supported by two pieces of theory.
1.Higher
intensity means more photons not more energy for an individual photon.
2.A
photon must give up all of its energy to one electron it cannot be shared.
This
graph shows that increasing V does increase I but the current soon reaches
a maximum value called the saturation current.
Plotting
maximum K.E. of photoelectrons against frequency of incident light yields
- (from three different cathode materials).
Summarizing
these results:
1.The
stronger the beam of light of a given color (f) the greater the photoelectric
current.
2.For
each type of material used for the cathode, photoelectrons are not ejected
if light has a frequency below a certain value called the threshold frequency
(f0).This threshold frequency
is a characteristic of the metal.
3.The
maximum kinetic energy of the emitted electrons increases in direct proportion
to the frequency of the incident light and does not depend on the intensity
of the light source.
4.The
emission of electrons is immediate (<10-9 sec), no matter
how weak the light source, provided the frequency is above the threshold
frequency of the cathode material.
1.The
most puzzling feature of these results is the immediate ejection of electrons
even for very weak light.According
to the classical theory a wave delivers energy continuously. An
electron cannot be ejected until it absorbs enough energy to free it from
the atom.So according to classical
theory the radiation would need to deliver energy for a finite time until
the electron received enough energy to escape.
2.According
to classical theory of Electro Magnetic radiation the energy of a wave
is proportional to its intensity.Why
is there then a threshold frequency below which electrons cannot be ejected,
no matter how bright the light?
3.A
more intense light should increase the energy of the ejected electrons
but their K.E. depends not on intensity but upon frequency.
Einstein
wrote a paper to explain the photoelectric effect for which he received
the 1921 Nobel Prize.
Einstein's
explanation of these results was that the energy of light is not spread
evenly across a uniform wave front but concentrated in separate 'lumps'
or packets of energy.The energy
of these packets has a definite value that depends on the frequency of
light radiation.A more intense
light source contains more packets but each packet still has the same amount
of energy.
As
we have already seen Plank called these packets quanta and later they were
called photons.
This
explains the immediate ejection of electrons even for very weak light,
because if the photon has enough energy it can be absorbed and the electron
is ejected.However if the frequency
is too low, each photon will not possess enough energy for the electron
to be ejected, no matter how many photons are present (i.e. how bright
the light is made).
Extrapolating
the previous graph:
The
equation of this line isK.E.max=h
f - W
hf
is the quantity of energy delivered to the photoelectron by the photon.However
the electron loses some of this energy on the way to the metal's surface
through collisions and some energy is required for the electron to escape
the surface.This work done by the
electron in escaping the surface is W.The
excess energy hf - W will be the kinetic energy that the photoelectron
escapes with.ThusKEmax
= h f - W.
Note:
1.W
is the minimum amount of work done in escaping the metal's surface and
is called the work function.
2.The
work function W is characteristic of the cathode material.W
= h fo
3.Energy
is often measured in electron volts (eV).1
eV = 1.60 ´
10-19 J
Example
:(from
text page 525)
Ultra
violet light of wavelength 200 nm is incident on a clean silver surface.The
work function of the silver is 4.7 eV.What
is
a)The
kinetic energy of the fastest-moving ejected electrons?
b)The
kinetic energy of the slowest-moving electrons?
c)The
cut-off frequency (fo) for silver?
d)The
cut-off wavelength for silver?
e)The
cut-off potential Vo for silver?
Since
the velocity of a photon is that of the speed of light c.We
can write the momentum as
p
= m c
Using
Einstein's famous equation E = mc2 and E = hf.We
get
mc2
= hf
m
= hf
c2
So
the momentum is
(since
c = ln)
Example
:
What
is the momentum of a photon of red light of wavelength 650 nm?
The
photoelectric effect suggests that light is a stream of energy 'packets'
or particles.However if we direct
a stream of these 'particles' at a double slit apparatus such that the
photon intensity is so weak that less than one photon per second arrives
at the slits we find that an interference pattern exactly like that predicted
by the wave model is produced.
So,
is light a particle or a wave?It
displays properties of both and yet it cannot be both a wave and a particle.The
solution to this dilemma lies in a complex and mathematically abstract
subject called Quantum Mechanics.The
details of which were worked out independently by Erwin Schrodinger (wave
mechanics) and Werner Heisenberg (matrix mechanics) in 1933.
These
models do not provide a physical picture in familiar concepts.So
we must accept wave-particle duality and choose the model appropriate to
the problem we are solving.
E.g.Refraction
Diffraction
InterferenceWave
model
Reflection
Intensity
Photoelectric
effectParticle (photon) model
Quantised
energy levels
Problem
Set #3:TextPage
531Questions 1 – 52
Demo:Spectroscopes
and sodium lamp
Back in sections 3.1 and 3.2, it was stated that electrons could move between energy levels in an atom.Each atom has its own series of energy levels.
If an electron moves from a higher energy level to a lower one the energy will be given out in the form of a photon.The energy of the photon given out is given by:
Ephoton = Energy of higher level – Energy of lower level
Example 1:Some energy levels for sodium are shown above.
What is the energy of the photon emitted when a atom of sodium falls from:
i)The 2.10 eV level to the ground state?
ii)The 3.75 eV level to the 2.10 eVlevel?
Example 2:What is the minimum energy required to ionise a hydrogen atom?What is the frequency of the photon associated with this energy?
In
1923 the French Physicist, Prince Louis de Broglie postulated that all
matter has waves associated with it.According
to de Broglie, all particles exhibit both wave and particle properties.The
waves associated with these particles
are called de Broglie waves.
Louis
de Broglie also predicted some properties of these waves, including the
wave length.He looked at the momentum
of a photon and said that this equation would hold for all particles.If
this is the case then we can work out the wavelength.
and
or
Thus
the wavelength for any particle can be calculated.If
this is done then we find that large objects have an extremely small wavelength
that cannot be detected.
i.e.a
1 Kg mass travelling at 30 m/s
l=6.625 ´
10-34
1 ´
30
=2.2 ´
10-35 m
Smaller
objects such as electrons have wavelengths that can be detected.
i.e.an
electron moving at 2 x 106 m/s
l=6.625 ´
10-34
9.11 ´
10-31´
2 ´
106
=3.6 ´
10-10 m
If
de Broglie's theory was correct then electrons should behave as waves.Thus
they should have the same behaviour as waves exhibiting diffraction and
interference effects.
In
1927 Davisson and Gerner in the USA and G.P. Thompson (son of J.J.) in
the UK both performed experiments showing diffraction of electrons.Davisson
and Gerner were also able to prove de Broglie's prediction of the wavelength
to be correct.Davisson and Thompson
shared the 1937 noble prize for their work in this area.
However
interference properties were still not able to be shown.It
was not until 1989 that a team of Japanese Physicists were able to show
that single electrons exhibited an interference pattern.
In
1991 interference patterns for large particles, in particular Helium and
Sodium, were shown.
Problem
Set #4:TextPage
555Questions 1 – 56