7.1

Atomic, Nuclear and
Particle Physics
Topic 7.1 Discrete energy and radioactivity
Image from http://www.onlineopinion.com.au/view.asp?
article=15900

What is an atom?
• Can you explain why you know
that matter is made of atoms?
• How do you know?
• Why is it important?
• How do we collect evidence of
them?
• Why do you need to know about
models of atom that we no
longer “believe”?
TOK

Energy Levels
Thomas Melville was the first to study
the light emitted by various gases.
He used a flame as a heat source, and
passed the light emitted through a
prism.
Melville discovered that the pattern
produced by light from heated gases is
very different from the continuous
rainbow pattern produced when
sunlight passes through a prism.

The new type of spectrum consisted of
a series of bright lines separated by
dark gaps.
This spectrum became known as a line
spectrum.
Melvill also noted the line spectrum
produced by a particular gas was
always the same.

In other words, the spectrum was
characteristic of the type of gas,
a kind of "fingerprint" of the
element or compound.
This was a very important finding
as it opened the door to further
studies, and ultimately led
scientists to a greater
understanding of the atom.

Emission Spectra
Absorption Spectra
What do you notice about these?
TOK

Spectra can be categorised as either
emission or absorption spectra.
An emission spectrum is, as the name
suggests, a spectrum of light emitted
by an element.
It appears as a series of bright lines,
with dark gaps between the lines
where no light is emitted.

An absorption spectrum is just the
opposite, consisting of a bright,
continuous spectrum covering the full
range of visible colours, with dark lines
where the element literally absorbs
light.
The dark lines on an absorption
spectrum will fall in exactly the same
position as the bright lines on an
emission spectrum for a given element,
such as neon or sodium.

Evidence
What causes line spectra?
You always get line spectra from
atoms that have been excited in
some way, either by heating or by
an electrical discharge.
In the atoms, the energy has
been given to the electrons,
which then release it as light.

Line spectra are caused by changes in
the energy of the electrons.
Large, complicated atoms like neon
give very complex line spectra, so
physicists first investigated the line
spectrum of the simplest possible
atom, hydrogen, which has only one
electron.

Planck and Einstein's quantum theory
of light gives us the key to
understanding the regular patterns in
line spectra.
The photons in these line spectra have
certain energy values only, so the
electrons in those atoms can only
have certain energy values.

The electron, has the most
potential energy when it is on the
upper level, or excited state.
When the electron is on the lower
level, or ground state, it has the
least potential energy.

The diagram can show an electron in an
excited atom dropping from the excited
state to the ground state.
This energy jump, or transition, has to be
done as one jump.
It cannot be done in stages.
This transition is the smallest amount of
energy that this atom can lose, and is
called a quantum (plural = quanta).

The potential energy that the electron has
lost is given out as a photon.
This energy jump corresponds to a specific
frequency (or wavelength) giving a specific
line in the line spectrum.
E = hf
This outlines the evidence for the existance
of atomic energy levels.

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.

THE BOHR MODEL OF
THE ATOM
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.

THE BOHR MODEL OF
THE ATOM
0
+
-
free e-
} bound e-
e n e r g y le v e ls
B o h r a t o m

THE BOHR MODEL OF
THE ATOM
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.

THE BOHR MODEL OF
THE ATOM
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.

THE BOHR MODEL OF
THE ATOM
Let’s prove this using
the following
formula…
• LYMAN SERIES
En-Em = hf
(13.6 –0.85)(1.6X10-
19
)=hc/λ
λ=9.7X10-8
m (98nm)
This is UV Light
0
- 2
- 4
- 6
- 8
- 1 0
- 1 2
- 1 4
2
energy(eV)
L y m a n s e r ie s
B a lm e r s e r ie s
P a s c h e n s e r ie s
g r o u n d s ta te
1 s t e x c ite d s ta te
n = 3
n = 1
n = 2
n = 4
n = 5
n = ∞
io n iz e d a to m ( e le c tr o n u n b o u n d
a n d f r e e to ta k e a n y e n e r g y )
( K s h e ll)
( L s h e ll)
( M s h e ll)
( N s h e ll)
- 1 3 .6
- 3 .4
- 1 .5 1
- 0 .8 5

THE BOHR MODEL OF
THE ATOM
• 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
energy(eV) L y m a n s e r ie s
B a lm e r s e r ie s
P a s c h e n s e r ie s
g r o u n d s t a t e
1 s t e x c it e d s t a t e
n = 3
n = 1
n = 2
n = 4
n = 5
n = ∞
io n iz e d a t o m ( e le c t r o n u n b o u n d
a n d f r e e t o t a k e a n y e n e r g y )
( K s h e ll)
( L s h e ll)
( M s h e ll)
( N s h e ll)
- 1 3 .6
- 3 .4
- 1 .5 1
- 0 .8 5

EXAMPLE 1
Use 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
hf
hc hc
=
λ
λ

EXAMPLE 1 SOLUTION
n=3 n=l,
• = 1.024 x 10 -7
m
• = 102 nm (ultraviolet)
λ =
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 10-7
m
= 655 nm (visible-red)
19-
834
10x1.6x10.2
10x3x10x6.6 −
19-
834
10x1.6x1.89
10x3x10x6.6 −

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

Mass Number
The total number of protons and
neutrons in the nucleus is called
the mass number (or nucleon
number).

Nucleon
Protons and neutrons are called
nucleons.
Each is about 1800 times more
massive than an electron, so
virtually all of an atom's mass is
in its nucleus.

Atomic Number
All materials are made from about 100
basic substances called elements.
An atom is the smallest `piece' of an
element you can have.
Each element has a different number
of protons in its atoms:
it has a different atomic number
(sometimes called the proton
number).
The atomic number also tells you the
number of electrons in the atom.

Isotopes
Every atom of oxygen has a
proton number of 8. That is, it has
8 protons (and so 8 electrons to
make it a neutral atom).
Most oxygen atoms have a
nucleon number of 16.
This means that these atoms also
have 8 neutrons.This is 16
8O.

Some oxygen atoms have a
nucleon number of 17 or 18.
16
8O and 17
8O are both oxygen
atoms.
They are called isotopes of
oxygen.
Isotopes are atoms with the
same proton number, but
different nucleon numbers.

Since the isotopes of an element
have the same number, of
electrons, they must have the
same chemical properties.
The atoms have different masses,
however, and so their physical
properties are different.

Evidence for Neutrons
The existence of isotopes is evidence for the
existence of neutrons because there is no
other way to explain the mass difference of
two isotopes of the same element.
By definition, two isotopes of the same
element must have the same number of
protons, which means the mass attributed to
those protons must be the same.
Therefore, there must be some other particle
that accounts for the difference in mass, and
that particle is the neutron.
TOK

Interactions in the
Nucleus
Electrons are held in orbit by the force of
attraction between opposite charges.
Protons and neutrons (nucleons) are bound
tightly together in the nucleus by a different
kind of force, called the strong, short-
range nuclear force.
There are also Coulomb interaction
between protons.
Due to the fact that they are charged
particles.

What is radiation?
• What does it do?
• Where does it come from?
• How can we use it?
• Can we stop it?
• How do we protect ourselves?
TOK

Radioactivity
In 1896, Henri Becquerel
discovered, almost by accident,
that uranium can blacken a
photographic plate, even in the
dark.
Uranium emits very energetic
radiation it is radioactive.‑

Then Marie and Pierre Curie
discovered more radioactive elements
including polonium and radium.
Scientists soon realised that there
were three different types of radiation.
These were called alpha (α), beta (β),
and gamma (γ) rays
from the first three letters of the
Greek alphabet.

Properties 2
The diagram on the right shows
how the different types are
affected by a magnetic field.
The alpha beam is a flow of
positively (+) charged particles,
so it is equivalent to an electric
current.
It is deflected in a direction given
by Fleming's left hand rule the‑ ‑
rule used for working out the
direction of the force on a
current carrying wire in a‑
magnetic field.

The beta particles are much lighter than the alpha
particles and have a negative ( ) charge, so they‑
are deflected more, and in the opposite direction.
Being uncharged, the gamma rays are not
deflected by the field.
Alpha and beta particles are also affected by an
electric field in other words, there is a force on‑
them if they pass between oppositely charged
plates.

Ionising Properties
α particles,‑ β particles and‑ γ ray photons‑
are all very energetic particles.
We often measure their energy in
electron volts (eV) rather than joules.‑
Typically the kinetic energy of an α particle‑
is about 6 million eV (6 MeV).
We know that radiation ionises molecules by
`knocking' electrons off them.
As it does so, energy is transferred from the
radiation to the material.
The next diagrams show what happens to an
α particle‑

Why do the 3 types of
radiation have different
penetrations?
Since the α-particle is a heavy,
relatively slow moving particle‑
with a charge of +2e, it interacts
strongly with matter.
It produces about 1 x 105
ion pairs
per cm of its path in air.
After passing through just a few
cm of air it has lost its energy.

the β particle‑ is a much lighter
particle than the α particle and it‑
travels much faster.
Since it spends just a short time in the
vicinity of each air molecule and has a
charge of only le, it causes less‑
intense ionisation than the α particle.‑
The β particle produces about 1 x 10‑ 3
ion pairs per cm in air, and so it
travels about 1 m before it is
absorbed.

A γ ray photon‑ interacts weakly with
matter because it is uncharged and
therefore it is difficult to stop.
A γ ray photon often loses all its‑
energy in one event.
However, the chance of such an event
is small and on average a γ photon‑
travels a long way before it is
absorbed.

Stability
If you plot the neutron number N
against the proton number Z for
all the known nuclides, you get
the diagram shown here

Can you see that the stable nuclides of
the lighter elements have approximately
equal numbers of protons and neutrons?
However, as Z increases the `stability
line' curves upwards.
Heavier nuclei need more and more
neutrons to be stable.
Can we explain why?

It is the strong nuclear force that holds
the nucleons together, but this is a very
short range force.
The repulsive electric force between
the protons is a longer range force.
So in a large nucleus all the protons
repel each other, but each nucleon
attracts only its nearest neighbours.

More neutrons are needed to hold the
nucleus together (although adding too
many neutrons can also cause
instability).
There is an upper limit to the size of a
stable nucleus, because all the
nuclides with Z higher than 83 are
unstable.

Alpha Decay
An alpha particle is a helium nucleus‑
and is written 4
2He or 4
2α.
It consists of 2 protons and 2
neutrons.
When an unstable nucleus decays by
emitting an α particle‑
it loses 4 nucleons and so its
nucleon number decreases by 4.
Also, since it loses 2 protons, its
proton number decreases by 2

The nuclear equation is
A
Z X → A-4
Z-2 Y + 4
2α.
Note that the top numbers
balance on each side of the
equation. So do the bottom
numbers.

Beta Decay
Many radioactive nuclides
(radio nuclides) decay by‑
β emission.‑
This is the emission of an
electron from the nucleus.
But there are no electrons in the
nucleus!

What happens is this:
one of the neutrons changes into
a proton (which stays in the
nucleus) and an electron (which
is emitted as a β particle).‑
This means that the proton
number increases by 1,
while the total nucleon number
remains the same.

The nuclear equation is
A
Z X → A
Z+I Y + 0
-1e
Notice again, the top numbers
balance, as do the bottom ones.

A radio nuclide‑ above the
stability line decays by
β emission.‑
Because it loses a neutron and
gains a proton, it moves
diagonally towards the stability
line, as shown on this graph

Gamma Decay
Gamma emission does not‑
change the structure of the
nucleus, but it does make the
nucleus more stable
because it reduces the energy of
the nucleus.

Decay chains
A radio nuclide often produces an unstable‑
daughter nuclide.
The daughter will also decay, and the
process will continue until finally a stable
nuclide is formed.
This is called a decay chain or a decay
series.
Part of one decay chain is shown below

When determining the
products of deacy
series, the same
rules apply as in
determining the
products of alpha and
beta, or artificial
transmutation.
The only difference is
several steps are
involved instead of
just one.

Half Life
Suppose you have a sample of 100
identical nuclei.
All the nuclei are equally likely to
decay, but you can never predict
which individual nucleus will be the
next to decay.
The decay process is completely
random.
Also, there is nothing you can do to
`persuade' one nucleus to decay at a
certain time.
The decay process is spontaneous.

Does this mean that we can never
know the rate of decay?
No, because for any particular
radio nuclide there is a certain‑
probability that an individual nucleus
will decay.
This means that if we start with a
large number of identical nuclei we
can predict how many will decay in a
certain time interval.
TOK

Iodine 131 is a radioactive‑
isotope of iodine.
The chart on the next slide
illustrates the decay of a sample
of iodine 131.‑
On average, 1 nucleus
disintegrates every second for
every 1000 000 nuclei present.

To begin with, there are 40 million undecayed nuclei.
8 days later, half of these have disintegrated.
With the number of undecayed nuclei now halved, the
number of disintegrations over the next 8 days is also
halved.
It halves again over the next 8 days... and so on.
Iodine 131 has a‑ half life‑ of 8 days.

Definition
The half life of a radioactive‑
isotope is the time taken for half
the nuclei present in any given
sample to decay.

Activity and half life‑
In a radioactive sample, the
average number of
disintegrations per second is
called the activity.
The SI unit of activity is the
becquerel (Bq).
An activity of, say, 100 Bq means
that 100 nuclei are disintegrating
per second.

The graph on the next slide of the
next page shows how, on
average, the activity of a sample
of iodine 131 varies with time.‑
As the activity is always
proportional to the number of
undecayed nuclei, it too halves
every 8 days.
So `half life' has another meaning‑
as well:

Definition 2
The half life of a radioactive‑
isotope is the time taken for the
activity of any given sample to
fall to half its original value.

Exponential Decay
Any quantity that reduces by the
same fraction in the same period
of time is called an exponential
decay curve.
The half life can be calculated
from decay curves
Take several values and the take
an average

Radioactive Decay
Radioactive decay is a
completely random process.
No one can predict when a
particular parent nucleus will
decay into its daughter.
Statistics, however, allow us to
predict the behaviour of large
samples of radioactive isotopes.

Radioactive Decay
We can define a constant for the
decay of a particular isotope,
which is called the half-life.
This is defined as the time it
takes for the activity of the
isotope to fall to half of its
previous value.

Radioactive Decay
From a nuclear point of view, the
half-life of a radioisotope is the
time it takes half of the atoms
of that isotope in a given
sample to decay.
The unit for activity, Becquerel
(Bq), is the number of decays
per second.

Radioactive Decay
An example would be the half-life
of tritium (3
H), which is 12.5
years.
For a 100g sample, there will be
half left (50g) after 12.5 years.
After 25 years, one quarter (25g)
will be left and after 37.5 years
there will be one eighth (12.5g)

Radioactive Decay
The decay
curve is
exponentia
l.The only
difference
from one
sample to
another is
the value
for the half-

Radioactive Decay
Below is a decay curve for 14
C.
Determine the half-life for 14
C.

Radioactive Decay
The half-life does not indicate
when a particular atom will
decay but how many atoms will
decay in a large sample.
Because of this, there will always
be a ‘bumpy’ decay for small
samples.

Radioactive Decay
If a sample contains N
radioactive nuclei,
we can express the statistical
nature of the decay rate
(-dN/dt)
is proportional to N:

Radioactive Decay
in which λ, the
disintegration or decay
constant, has a
characteristic value for
every radionuclide. This
equation integrates to:
No
is the number of
radioactive nuclei in a
sample at t = 0 and N is
the number remaining
at any subsequent time
t.
N
dt
dN
λ=−
t
oeNN λ−
=
You have to
derive this

Radioactive Decay
-dN/dt = λN
Collect like terms
dN/N = -λdt
Integrate
ln N = -λt + c
But c = ln N0
So, ln N = -λt + ln N0
ln N - ln N0 = -λt
N/ N0 = e-λt

Radioactive Decay
Solving for t½
yields, that is when N =
N0/2
λt1/2 = 0.693
t1/2 = 0.693/ λ
λ
2ln
2
1 =t or
2
1
2ln
t
=λ

Radioactive Decay
The half-life of an isotope can be
determined by graphing the
activity of a radioactive
sample,over a period of time

Radioactive Decay
The graph of activity vs time can
be graphed in other ways
As the normal graph is
exponential it does not lead to a
straight line graph
Semi logarithmic graph paper can
solve this problem

Radioactive Decay
If we take the natural log of N =
Noe-λt
we get:
ln N = ln No -λt
The slope of the line will
determine the decay constant λ

Radioactive Decay
The accuracy in determining the
half-life depends on the number
of disintegrations that occur per
unit time.
Measuring the number of
disintegrations for very long or
short half-life isotopes could
cause errors.

Radioactive Decay
For very long half-life isotopes
i.e. millions of years
Only a small number of events will
take place over the period of a
year
Specific activity is used
Activity of sample is measured
against a calibrated standard

Radioactive Decay
Standard is produced by
reputable organisations
i.e. International Atomic Energy
Agency
Calibrated standard measures the
accuracy of the detector
making sure it is accurate
Specific activity and atomic mass
of isotope is then used to
calculate the half-life

Radioactive Decay
With very short half-life isotopes,
the isotope may disintegrate
entirely before it is
measured.Time is therefore of
the essence
As most of these isotopes are
artificial. Produce them in or
near the detector
This eliminates or reduces the

EXAMPLE 1
(a) Radium-226 has a half-life of
1622 years. A sample contains
25g of this radium isotope. How
much will be left after 3244
years?
(b) How many half-lives will it
take before the activity of the
sample falls to below 1% of its
initial activity? How many years
is this?

EXAMPLE 1 SOLUTION
(a) 3244 years is 2 half lives
(2 x 1622)
N= No(1/2)n
= 25 x (1/2)2
= 25 x (1/4)
=6.25

EXAMPLE 1 SOLUTION
(b)The activity of a radioactive
sample is directly proportional
to the number of remaining
atoms of the isotope. After t1/2,
the activity falls to ½ the initial
activity. After 2 t1/2, the activity
is ¼. It is not till 7 half-lives
have elapsed that the activity is
1/128th
of the initial activity.
So, 7 x 1622 = 11354 years

EXAMPLE 2
A Geiger counter is placed near a
source of short lifetime
radioactive atoms, and the
detection count for 30-second
intervals is determined. Plot the
data on a graph, and use it to
find the half-life of the isotope.

EXAMPLE 2
• Interval Count
• 1. 12456
• 2. 7804
• 3. 5150
• 4. 3034
• 5. 2193
• 6. 1278
• 7. 730

EXAMPLE 2 SOLUTION
The data are
plotted on a
graph with
the point
placed at
the end of
the time
interval
since the
count

EXAMPLE 2 SOLUTION
A line of best fit is
drawn through the
points, and the
time is determined
for a count rate of
12 000 in 30
seconds. Then the
time is determined
for a count rate of
6000, and 3000.

EXAMPLE 2 SOLUTION
t(12 000) = 30s
t( 6000) = 72s, so
t1/2
(1) = 42s
t( 3000) = 120s, so
t1/2
(2) = 48s
The time difference
should have be the
half-life of the
sample.

EXAMPLE 2 SOLUTION
Since we have two
values, an
average is taken.
t s1 2
42 48
2
45/ =
+
=

Background radiation
• Can you
overdose on
radiation?

https://www.flickr.com/photos/oregonstateuniversity/8528630449

• The Four Forces in Nature by relative strength
Fundamental Interactions
Type Relative
Strength
(2 p in
nucleus)
Field Particle
Strong nuclear 1 Gluons (was
mesons)
Electromagnetic 10-2
Photon
Weak nuclear 10-6
W±
and Zo
Gravitational 10-38
Graviton (?)

An analogy is used to help understand
how a force can be experienced due
to exchange of a particle….
Imagine Harry & Julius throwing
pillows at each other
Each catch results in the child being
thrown backwards
A repulsive force
Fundamental
Interactions

If they grab the pillow out of the
other’s hand to exchange pillows
They are pulled towards each
other
An attractive
force
Fundamental
Interactions

7.1

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