Here is a semi-log plot of the data with an exponential trendline: The equation of the trendline is: y = 12456e-0.4693x Taking the natural log of both sides: ln(y) = ln(12456) - 0.4693x The slope is -0.4693 Using the equation: t1/2 = 0.693/λ λ = 0.4693 t1/2 = 0.693/0.4693 = 1.5 Therefore, the half-life of the isotope is 1.5 intervals, or 1.5 x 30 s = 45 seconds.

1. Nuclear Physics
AHL 12.2

2. Rutherford’s Experiment
In 1909, Rutherford investigated high
speed alpha particles that passed
through thin, metal foils.
The experiments were actually carried out
by two young physicists named
Hans Geiger and Ernest Marsden.

3. Rutherford’s Experiment
Alpha particles are double charged helium
ions;
 two protons and neutrons.
Emitted spontaneously from heavy
radioactive atoms such as uranium and
plutonium.

4. Rutherford’s Experiment
The beam of alpha particles were directed
at a metal foil,detector picks up the
emerging particles on the other side.
This would give an indication of the nature
of the atom.

5. Rutherford’s Experiment
q
a- source
lead
collimators
vacuum
gold foil
ZnS crystal
microscope

6. Rutherford’s Experiment
A lead block holds the alpha particle source and
emitted through a fine hole in the lead.
It is collimated into a beam by aligning holes in a
series of lead plates.
The beam is then aimed at a gold foil.
Gold is used as it can be beaten into a thin foil,
about 1 m (8000 atoms) thick.

7. Rutherford’s Experiment
The particles are detected by using a
microscope fitted with, a zinc sulfide
crystal.
If an alpha particle hits the crystal, a flash
can be seen and counted.

8. Rutherford’s Experiment
Rutherford made predictions as to what he
would find.
He said there would be very few particles
would suffer deflections those which did
would be deviated, only a few degrees.
Theory suggested that only 1% would be
deflected more than 3o.

9. Rutherford’s Experiment
Rutherford also determined there was
almost no way there could be any
backscattering ie. angles greater than
90o.
It could only happen with a direct collision
with an electron and only 1 particle in
103500 would take this path.

10. Rutherford’s Experiment
The calculations were on
the basis of previous
model of the atom by
Thomson.
Rutherford predicted that
the deflections would be
due to electrostatic
attraction.

11. Rutherford’s Experiment
The actual results came as a great
surprise everyone was expecting
Thomson’s model was correct.
One particle in 10 000 was backscattered
compared to the theory of 1 in 103500.

12. Rutherford’s Atom
He suggested that all the positive matter was
concentrated in
 a very small sphere,
 at the centre of the atom.
As electrons have very little mass
 all the mass must be in this sphere, called the
nucleus.
Around the nucleus is essentially empty space.

13. Rutherford’s Atom
This explains the backscattering found in
the experiment.
The alpha particle would strongly be
repelled when it is directed at the
nucleus, due to electrostatic repulsion.
Those particles that do not hit the nucleus;
would pass through, with little or no
deflection.

14. Rutherford’s Atom
. nucleus

15.  The kinetic energy of the alpha particle is
EK
 This kinetic energy is converted into
potential energy as the alpha particle
overcomes the coulombic force as it
approaches the nucleus
 All the kinetic energy = potential energy
when the approach is the closest

16. The potential energy near a point charge is
Ep = Q1 Q2 / 40 r
EK = Ep = Ze 2e/ 40 d
EK = Ze2/ 20 d
 Calculate the distance of closest
approach for an alpha particle of 5MeV
approaching a gold nucleus

17. Nuclear Physics
Modern scattering techniques show that
the nucleus is not hard like a billiard ball
but has a ‘fuzzy’ surface.
This is due to a variation in the density of
the nuclear material in the outside layer
of the nucleus.

18. Nuclear Physics
The mass of an atom is one of the
characteristic properties that can give an
insight into the structure of an atom.
One device that can be used to determine
the mass is a mass spectrometer.
One particular type is the Bainbridge Mass
Spectrometer.

19. Nuclear Energy Levels
Remember how e/m radiation is given off
when electrons that have been excited,
drop back down to lower energy levels.
The photons given off have specific
energies equal to the difference in
energy levels.

20. Nuclear Energy Levels
Previously, we have studied a, , and 
radiation
Often,  radiation has accompanied the
emission of a or  particles.
The emission of  photons is evidence that
the nucleus has energy levels

21. Nuclear Energy Levels
Radium decays to Radon at different
energy levels.
a-particles are ejected at certain discrete
velocities (energies).
The energy depends on which level the
Radium decays to in the Radon.
 HeRnRa 4
2
224
86
226
88

22. Nuclear Energy Levels
A
B
C
D
Nucleus Shells
Radon 222
Excited State 3
Radon 222
Ground State
a A aB a C a D
Energy
above
the
ground
state of
Radon

23. Nuclear Energy Levels
In the above diagram, 226Ra decays giving
off an aB particle that has a specific K
(aB) when it decays to Rn in the 2nd
excited state.
The 222Rn then might move to the ground
state giving off a photon of energy in the
MeV range called a GAMMA PHOTON
()

24. Radioactive Decay
We will look at the following areas:
 Beta Decay
 The Neutrino
 Half-life
 Decay constant

25. BETA DECAY
Nuclei that have an imbalance of protons
or neutrons can be unstable and also
undergo radioactive decay.
The process involves the change of a
proton into a neutron or more commonly
a neutron into a proton with the ejection
of an electron from the nucleus.
This decay is called beta decay, and the
electron is referred to as a beta particle.

26. BETA+ DECAY
BETA+ DECAY (too
many protons)-
When a nucleus has to
increase its neutron
number to become
more stable, a proton
can spontaneously
change into a
neutron.
B+

27. BETA+ DECAY
An electron (positively
charged) is ejected
with a neutrino.
The positive electron
is called a positron
and is an example
of antimatter. The
atomic number is
reduced by one but
the mass number is
unaffected.

28. BETA+ DECAY
On the line stability
on the graph, any
atom below the
line would decay
this way.
+eY+X 0
1+1
A
Z
A
Z  B+

29. BETA+ DECAY
In the nucleus, the
reaction is:
An example of this
is:
+e0
1+
1
0
1
1  np
+e+CN 0
1+
13
6
13
7


30. BETA+ DECAY
Notice that both mass and charge are conserved.
A ‘positron’, a positively charged electron (the
same mass as an electron) is ejected.
The positron is an example of antimatter
(“opposite of”).
This is known as ‘proton decay”.
+e+CN 0
1+
13
6
13
7


31. BETA+ DECAY
The positron is known as the B+ particle.
A Neutrino (v) is also released.
Note a new element is formed. There are no
natural positron emitters since positron half-
lives are very small.
Note- as the 13N might decay into a metastable
form of 13C, the 13C could then drop down to a
more stable state, giving off a GAMMA RAY.

32. NEUTRINOS AND ANTINEUTRINOS
Beta particles are emitted with a range of
energies up to a maximum of a few MeV.
It seemed strange that the electrons with
the maximum kinetic energy carried
away all the available energy, yet those
with less than he maximum kinetic
energy appeared to have energy
missing.

33. NEUTRINOS AND ANTINEUTRINOS
This did not obey the law of conservation
of energy.
Other experiments with momentum
confirmed that linear momentum was not
conserved.
7N14
6C14 e- Speed and direction
of the electron if
momentum was
conserved.
e- Actual speed and
direction of the electron.

34. NEUTRINOS AND ANTINEUTRINOS
In 1934 Enrico Fermi developed the theory
of beta decay and that the conservation
laws did hold because there was a
particle that had yet to be detected
carrying the lost energy and momentum.
He called this particle a neutrino (Italian for
‘little neutral one’).
The antimatter of the neutrino ( ) is the
antineutrino ( ).
_


35. NEUTRINOS AND ANTINEUTRINOS

36. NEUTRINOS AND ANTINEUTRINOS
Using the conservation laws, he postulated the
properties for the neutrino.
 Neutrinos are uncharged. This is because
charge is already conserved. A neutron
decays into a proton and an electron.
 Neutrinos have zero rest mass but carry
energy and momentum. The conservation
laws would not hold otherwise.

37. NEUTRINOS AND ANTINEUTRINOS
Neutrinos react very weakly with matter. It
took 25 years to detect them and there are
millions of neutrinos that pass through the
Earth from the sun as if the Earth was not
there. This is because they have no real
mass or charge.
Neutrinos travel at the speed of light. As
they have no mass but have energy, they
must travel at the maximum speed
possible - the speed of light.

38. NEUTRINOS AND ANTINEUTRINOS
The neutrino was accepted readily as it
solved awkward problems but was not
discovered until 1956.
It is given the symbol (the Greek letter
nu) and has zero atomic number and
mass number.


39. 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.

40. 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.

41. 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.

42. Radioactive Decay
An example would be the half-life of tritium
(3H), 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) and so on.

43. Radioactive Decay
The decay curve
is
exponential.The
only difference
from one
sample to
another is the
value for the
half-life.

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

45. 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.

46. 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:

47. Radioactive Decay
in which , 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

t
oeNN 

You have to
derive this

48. Radioactive Decay
-dN/dt = N
Collect like terms
dN/N = -dt
Integrate
ln N = -t + c
But c = ln N0
So, ln N = -t + ln N0
ln N - ln N0 = -t
N/ N0 = e-t

49. Radioactive Decay
 Solving for t½ yields, that is when N = N0/2
 t1/2 = 0.693
 t1/2 = 0.693/ 

2ln
2
1 t or
2
1
2ln
t


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

51. 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

52. Radioactive Decay

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

54. 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.

55. 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

56. 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

57. 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 transfer
time problem

58. 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?

59. 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

60. 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

61. 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.

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

63. EXAMPLE 2 SOLUTION
The data are plotted
on a graph with
the point placed
at the end of the
time interval since
the count reaches
this value after the
full 30 seconds.

64. 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.

65. 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.

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

67. methods
 Short half life
 Long half life