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Modern Physics
NCEA AS 3.5
Text Chapters:20,22
The Photoelectric Effect
The photoelectric effect occurs when
shining light (usually UV) onto a piece of
metal causes electrons to be given off.
 This effect can be used in a photoelectric
cell to produce small electric currents.
 Photoelectric cells are used in

Lightmeters
 Burglar alarms
 TV cameras etc

Photoelectric Cells
Thin glass tube (evacuated)
Emitter – curved
metal plate
LIGHT
Collector
A
Photoelectric Experiments
When the photoelectric effect was studied
in detail, the experimental results were
very different to what was expected. A new
theory about the nature of light was
needed to explain what happened.
 Scientists at the time considered light to
behave like a wave……

Photoelectric Experiments

What was expected:


Brighter light would cause electrons with more
kinetic energy to be emitted
What actually happened:

Brighter light caused more electrons to be
emitted, but there was no change in the
amount of energy they had
Photoelectric Experiments

What was expected:


If very dim light was used, it would take some
time before any electrons had absorbed
enough energy to escape from the metal
What actually happened:

With UV light, even the faintest light caused
some electrons to be emitted instantly
Photoelectric Experiments

What was expected:


The frequency (or colour) of the light used
would not affect the energy of the emitted
electrons.
What actually happened:

The higher the frequency, the higher the
energy of the electrons. Below a certain
frequency, no electrons were emitted.
Photoelectric Experiments
Einstein explained these results, using an
idea suggested by Max Planck, that said
electromagnetic radiation comes in fixed
“packets” or quanta of energy called
photons
 The amount of energy each photon has
depends on the frequency of the radiation.

Photoelectric Experiments

Each photon has a fixed amount of energy
given by:
E  hf
h=Planck’s Constant = 6.63x10-34Js
 This suggested that light behaved like a
moving particle, rather than a wave

Photoelectric Experiments
The power supply
provides an
opposing voltage
to the p.e. cell.
 The variable
voltage is
adjusted until the
current in the
circuit is zero

LIGHT
V
Photoelectric Experiments
When the current was zero, the supply
voltage was equal to the cut-off voltage of
the cell
 Different frequencies of light were tried,
and the cut-off voltages measured:

Photoelectric Experiments

These were the
results:

Below a certain
threshold
frequency f0 no
electrons were
emitted
V
f
f0
Photoelectric Experiments

The maximum Ek of the electrons can be
found from the voltage:
energy
Voltage 
charge
Ek  eV

Where e= electron charge = -1.6x10-19
Photoelectric Experiments

Another way
of looking at
that last
graph:
Ek
Gradient = h
f0
Intercept= Work
function Φ (or B)
f
Photoelectric Experiments

By equating to y=mx+c:
Ek  hf  B



Ek = max kinetic energy of emitted electrons
hf = energy of incoming photons
Φ = The work function of the metal – the
minimum amount of energy required for the
electron to escape from the metal surface.
Photoelectric Experiments

Different
metals have
different f0’s
and work
functions
depending
on how
tightly they
hold onto
their
electrons
Ek
Cu
f0
f0
Pb
f
The Conclusion
So the photoelectric effect could be
explained by thinking of light as a stream
of incoming particles that collided with
electrons in the metal. If the photon had
enough energy, it could knock the electron
free of the metal and send it across the
cell to the collector.
 If photon was too small, it couldn’t hit
electrons hard enough (overcome work
function) so no electrons emitted.

Atomic Spectra

2 types
Emission – certain frequencies of light given
off by low pressure gases excited by heat or
electricity
 Absorption – certain frequencies absorbed
from a continuous spectrum by low pressure
gases


Spectra are unique to each element and
can be used to identify unknown elements
The Hydrogen Spectrum
Balmer studied the emission spectrum
lines of Hydrogen, as it is the simplest
atom.
 He was limited by the fact that he could
only observe visible frequencies – we now
know there are UV and IR spectral lines
 The spectral lines are caused by the
movement of electrons between different
energy shells in the atom

The Hydrogen Spectrum
In Balmer’s case he was looking at
spectral lines caused by electrons jumping
from higher energy level (shells) down into
the 2nd shell.
 They would release their extra energy as a
photon of light.
 Other Scientists later found series of
spectral lines corresponding to jumps into
the 1st, 3rd, 4th, 5th etc

The Hydrogen Spectrum
Paschen Series S=3 (IR)
Bracket Series S=4 (IR)
Balmer Series S=2
(visible)
Pfund Series
S=5 (IR)
Lyman Series
S=1 (UV)
Nucleus
1
2
3 4 5
∞
Shell no. / Energy level
The Hydrogen Spectrum

A formula was worked out to calculate the
wavelengths of these lines:
1 
 1
 R 2  2 

L 
S
1
R=Rydberg’s Constant=1.097x10-7
 S=Series no. (the shell jumped into)
 L=Line no. (the shell jumped from)

The Hydrogen Spectrum
The formula worked perfectly for
Hydrogen, but started to get more
inaccurate the bigger and more complex
the atom got
 Absorption spectra are produced by
electrons absorbing photons of energy
which allows them to jump up energy
levels

Bohr’s Model of the Atom

Rutherford’s student Niels Bohr proposed
that:
Electrons in H could only exist in stable orbits
with certain fixed amounts of energy, called
energy levels
 An electron moves from one energy level to
another by either emitting or absorbing a
photon of light equal in energy to the
difference between the two energy levels

Bohr’s Model of the Atom

The energy levels in the Hydrogen atom are
given by :
 hcR
En 
2
n




h=Plancks constant = 6.63x10-34
c=speed of light = 3x108
R=Rydbergs constant = 1.097x107
n=energy level = 1,2,3,4…… (quantum number)
Bohr’s Model of the Atom
All energy values are negative – this
represents the fact that it is an energy
which binds the electron to the nucleus
 The lowest energy state n=1 is called the
ground state
 As n∞, E0. This represents the
energy required to ionise the atom by
removing the electron completely.

Bohr’s Model of the Atom
n=∞
n=4
n=3
0
Energy
n=2
(x10-18J)
-1
-2
6.63 10 34  3 108 1.097 107
E1 
12
E1  2.18 10 18
n=1
Electron Volts
Sometimes an alternative unit for energy is
used called the electron volt
 1eV is the energy gained by 1 electron
when accelerated by a potential of 1 Volt
 1eV=1.6x10-19J
 Using this unit:

 13.6
En 
eV
2
n
Nuclear Reactions
3 types:
Radioactive Decay – the spontaneous
emission of particles from the nucleus of
an atom
Nuclear Fission – splitting one large
nuclei into two smaller ones
Nuclear Fusion – combining two small
nuclei into one large one.
Radioactivity

3 types:
Alpha a
 Beta b
 Gamma g

Named in order of their discovery.
 Alpha and beta decay don’t usually occur by
themselves, there is usually some gamma
that occurs with them.

The Nucleus
In small atoms, the number of protons and
neutrons are usually the same (roughly)
 In larger atoms, there are usually many
more neutrons than protons, in order to
keep the nucleus stable.
 If a nucleus is unstable, it may
spontaneously decay to something more
stable by emitting alpha, beta or gamma
radiation

Alpha Particles
Helium nucleus
 Charge of +2
 Mass of 4 (a.m.u)
 Travel slowly ie. 10% of light speed
 Don’t travel very far ie. A few cms in air
 Low penetration power – can be stopped
by a piece of paper
 Very good ionising power – because
they’re big and slow.

Beta Particles
An electron from the nucleus
 Charge of -1
 Same mass as an electron (effectively 0)
 Travel relatively fast – up to 95% of light
speed
 Travel about 30 cms in air
 Average penetration power – can be
stopped by a few mm of Aluminium
 Average ionising power

Gamma Radiation
A wave of electromagnetic radiation
(energy)
 No charge
 No mass
 Travels at light speed
 Travels several metres in air
 High penetration power – Several cms of
lead needed to stop it
 Low ionising power – because no mass

Radiation

One way that the different types of
radiation can be distinguished is by
observing their behaviour in a magnetic
field:
b
g
a
The Nucleus

Writing nuclei
X = element symbol
A = mass number or
nucleon number (the
number of p+n)
Z = atomic number
(the number of
protons)
A
Z
X
Isotopes
Atoms with the same atomic number but
different mass numbers
 Eg:
12
1
6
1

C
H
2
1
3
1
H (deuterium )
13
6
H (tritium )
14
6
C
C
Alpha Decay

Example: Radium 226 decays to Radon 222 by
alpha decay:
226
88
Ra  Rn  He  g
222
86
4
2
Note: Both mass and charge must be conserved
(ie 226=222+4, 88=86+2

Beta Decay

Cobalt 60 decays by beta decay to Nickel
60
Co Ni e  g
60
27
60
28
0
1
Again, mass and charge are conserved
 NB. the a or b symbols can be used
instead of He or e

Half-life
The time it take for the decay rate to have
halved, or….
 The time taken for half of the original
atoms to have decayed
 Usually shown on a graph

Half-life
Half Life
4500
4000
No. of Atoms
3500
3000
2500
2000
1500
1000
500
0
0
1
2
3
Time in days
4
5
Detecting Radioactivity

Geiger Counter – detects electrical current
caused by the ionisation of atoms in a gas
Geiger-Muller
tube filled with
low pressure Ar
End: thin mica
window
+Cathode: metal cylinder
400V DC
Supply
- Anode: central wire
Counter or
speaker
Uses of Radioactivity
Radiation therapy to treat cancer
 Sterilisation
 Carbon dating
 Nuclear medicine eg tracers
 Smoke detectors

Binding Energy
If we put together a nucleus from
individual protons and neutrons, we would
find that the mass of the resulting nucleus
is less than the total mass of the individual
nucleons.
 This reduction in mass is called a mass
deficit

Binding Energy
In order to break up a nucleus into
separate nucleons the mass deficit must
be restored by adding extra energy.
 This energy changes into mass according
to Einstein’s famous equation:

E  mc
2
Binding Energy
This energy shortage has the effect of
holding the nucleus together so it is called
the binding energy.
 Binding energy represents the amount of
“glue” holding the nucleus together.
 The more binding energy per nucleon, the
more stable an atom will be

Binding Energy
B.E per
nucleon
(MeV)
56Fe
8
238U
4He
6
7Li
4
2
Fusion
Fission
50
100
150
200
Mass
number
Nuclear Fission
Breaking large unstable nuclei into smaller
ones.
 Lots of possible combinations of fragments
from one initial nucleus
 Eg:

1
0
n U  Ba  Kr 3 n
235
92
141
56
92
36
1
0
Nuclear Fission
When a large nucleus is split into smaller
fragments, the fragments have less mass
per nucleon
 The lost mass is released as energy in the
form of kinetic energy of neutrons and
gamma rays

n
Nuclear Fission
U


Only one
neutron is
needed to
start the
reaction, but
several are
produced
This starts a
“chain
reaction”
Kr
Ba
n
Kr
Ba
n
n
U
U
U
n
n
n
Kr
Ba
n
n
n
Kr
Ba
n
n
n
Nuclear Fission
If the chain reaction is controlled it can be
used in a nuclear reactor
 If it is uncontrolled it explodes as a nuclear
or atomic bomb

Nuclear Fusion
The joining of two small nuclei to form one
larger one
 This is the process that powers the sun
 Eg:

2
1
H  H  He  n
3
1
4
2
1
0
Nuclear Fusion
Fusing two light atoms together results in
a nucleus with less mass per nucleon
 The lost mass results in a release of
energy

Nuclear Fusion
Fusion requires extreme temperature (eg
millions of degrees) to occur, and has not
practically and economically been used in
power generation (yet….)
 Hydrogen bombs have been successfully
made, but require a fission reaction to
provide the necessary temp.
