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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Nuclear physics
Syllabus
Daniel Hilton
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Nuclear Physics – overview diagram
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
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You Tube links
The Discovery of the Atomic Nucleus - Dr. Brian Cox
3:28
Gold foil experiment and transmutation of Nitrogen in to Oxygen
8:15
Rutherford Gold Foil Experiment - lab
4:06
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford experiment
The Rutherford experiment or Gold foil
experiment or the Geiger–Marsden
experiment was an experiment to probe
the structure of the atom performed by
Hans Geiger and Ernest Marsden in
1909, under the direction of Ernest
Rutherford at the Physical Laboratories
of the University of Manchester.
The unexpected results of the
experiment demonstrated for the first
time the existence of the atomic
nucleus, leading to the downfall of the
plum pudding model of the atom, and
the development of the Rutherford (or
planetary) model.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering
Alpha particles from a
radioactive source
strike a thin gold foil.
If the gold foil was 1
micrometer thick, then
using the diameter of
a gold atom from the
periodic table, the foil
is ~ 2800 atoms thick.
Alpha particles produce
a tiny, but visible flash
of light when they strike
a fluorescent screen.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering
Surprisingly, alpha particles
were found at large deflection
angles and some were even
found to be back-scattered.
This showed that the positive
matter in atoms was
concentrated in an incredibly
small volume and gave birth
to the idea of the nuclear
atom. In so doing, it
represented one of the great
turning points in our
understanding of nature.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering
“It was quite the most incredible event that has ever
happened to me in my life.
It was almost as incredible as if you fired a 15-inch shell at a
piece of tissue paper and it came back and hit you.
On consideration, I realized that this scattering backward
must be the result of a single collision, and when I made
calculations I saw that it was impossible to get anything of
that order of magnitude unless you took a system in which
the greater part of the mass of the atom was concentrated in
a minute nucleus. It was then that I had the idea of an atom
with a minute massive center, carrying a charge.
—Ernest Rutherford
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Alpha scattering - “plum pudding model”
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Alpha scattering - “Rutherford model”
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Rutherford Scattering – Coulomb's Law
The scattering process can be treated in terms of the electrostatic force (Coulomb force) between the alpha particle and the
nucleus, which is considered to be a point charge Ze.
r : distance between charges; ε0 : “permittivity of free space
Work done (or PE) =F r=(
Daniel Hilton
k qQ
k qQ
)r=
2
r
r
A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering – scattering process
The scattering of the alpha particle by the central repulsive
Coulomb force results in a hyperbolic trajectory. From the
scattering angle and momentum, the impact parameter and
closest approach to the target nucleus can be calculated.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering – [Eqn is not A level]
For a detector at a specific angle with respect to the
incident beam, the number of particles per unit area
striking the detector is given by the Rutherford formula:
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford scattering – playing with data
We can explore Rutherfords scattering equation:
2 2 4
Ni n L Z k e
N (θ)= 2
2
4
4 r KE sin (θ/2)
For a fixed radiation source and target material we
could place the following into one value eg Const
N i n L Z 2 k 4 e2
Const =
2
2
4 r KE
Thus:
Const
N (θ)= 4
sin (θ/ 2)
We can use a spreadsheet to look at this curve.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford scattering – playing with data
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For a spreadsheet plot let Const = 1, thus N (θ)= 4
sin (θ/ 2)
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Rutherford scattering – playing with data
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For a spreadsheet plot let Const = 1, thus N (θ)= 4
sin (θ/ 2)
Sin (θ/2) ^4
Sin (θ/2) ^4
800000.00
1000000.00
700000.00
100000.00
600000.00
10000.00
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1000.00
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300000.00
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10.00
100000.00
0.00
1.00
0
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Nuclear radius
Rutherford used the conservation of energy to estimate the
nuclear radius. The Alpha particle KE “becomes” electrostatic
PE at it's closest approach to the nucleus, just before the Alpha
particle rebounds back.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Nuclear radius – conservation of energy
Electrostatic energy of the Alpha/nucleus pair can be seen
as resulting from the work done to move the two charges
together from “infinity” to r [for more details see “Fields”]
Work done (or PE) =F r = (
k q alpha q nucleus
r2
k qalpha qnucleus
)r =
r
Using conservation of energy we can “find”:
KE before + PE before =KE after +PE after
qalpha q nucleus
1/2 m v +0=0+k
d
q alpha q nucleus q alpha q nucleus
2
1/2 m v =k
=
d
4 π ϵo d
2
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Nuclear radius – point of closest approach
Using conservation of energy we can also “find” the
distance of closest approach (ie r = d):
KE before +PE before =KE after +PE after
qalpha q nucleus
1/ 2 m v +0=0+k
d
k qalpha qnucleus k (2 e)(Ze) 2 k Z e 2
d=
=
=
KE before
KE alpha
KE alpha
2
Show that for an Alumiminium nucleus (Z=13) and an
Alpha particle of energy 8x10^-13J that d is ~ 7x10^-15m
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Nuclear radius issues
Some points:
20
2
2k Z e
d=
KE particle
The nucleus was treated as a point charge - at this level it is not.
Alpha particles stop some distance away from the nucleus.
It takes higher energy alpha particles to penetrate the nucleus.
The values for the nuclear radius given by other particles such as
protons, neutrons and electrons are slightly different.
Nuclear units
1 electron volt = 1eV = 1.6 x 10^-19 joules
1 MeV = 106 eV; 1 GeV = 109 eV; 1 TeV = 1012 eV
Atomic sizes are on the order of 0.1 nm = 1 Angstrom = 10^-10 m
Nuclear sizes are ~ femtometers (fermis) 1 fm = 10^-15m
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Nuclear radius – estimates from scattering
Various types of scattering experiments suggest that nuclei are
roughly spherical (but not hard spheres) and have essentially
the same density. The scattering data can be summarized in
an expression for the nuclear radius (called the Fermi model):
R=r 0 A1/3
R : nuclear radius
A: mass number
r 0=(1.2±0.2) x 10−15 m=1.2 fm
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
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Nuclear density
Nuclear density is the density of the nucleus of an atom,
and is used when describing objects like neutron stars.
Nuclear Density = Nuclear mass / Nuclear Volume
Mn
ρn =
Vn
Let M n= A u
(u=atomic mass unit =1.66 x 10−27 kg )
3
1/3
Let V n=4/ 3π R n
(nuclear radius R n=r 0 A )
Mn
Au
Au
u
ρn =
=
=
=
3
1/3 3
V n 4 /3 π R n 4 /3 π(r o A ) 4/3 π r 3o
3u
ρn =
3
4 π ro
Daniel Hilton
A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Nuclear density
3u
ρn =
3
4 π ro
This expression indicates that the nuclear density is
independent of A and that the average distance between
nucleons is similar, despite the element
−15
Let r 0=1.05 x 10 m=1.05 fm
−27
u=atomic mass unit =1.66 x 10 kg
−27
3 u 3(1.66 x 10 kg )
ρn =
=
3
−15
3
4 π r o 4 π(1.05 x 10 m)
17
−3
ρn =3.4 x 10 kg m
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Rutherford Scattering – some historical notes
1899+: Discovered the concept of radioactive half-life; proved that
radioactivity involved the transmutation of one chemical element to
another; differentiated and named alpha and beta radiation. Awarded
Nobel Prize in Chemistry in 1908.
1909-13: Experiments on alpha scattering by nucleus, leading to...
1911: Discovery of the nucleus, theorized as a +ve charge
concentrated in a very small space [Rutherford atomic model]
1917: First "splitting of the atom" - nuclear reaction between nitrogen
and alpha particles, discovered (and named) the proton.
1920: Conceptualized the possible existence of the neutron - he
considered that the disparity found between atomic number (Z) and
atomic mass (A) due to existence of a neutrally charged particle
within the atomic nucleus. He considered the neutron to be a neutral
double of an electron orbiting a proton. By early 1930's it was found
that the neutron was is a single particle.
1937: After his death he was interred near Isaac Newton's and J.J.
Thomson in Westminster Abbey.
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Rutherford Scattering
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A2 Physics: Unit5 - Nuclear physics - Scattering, radius and density
Afterword : the alpha particle
A 'diagram' of the helium-4 atom the protons (red) and neutrons
show as separate particles.
In an actual helium atom, the
protons are superimposed in
space and most likely found at
the very center of the nucleus,
and the same is true of the two
neutrons.
All four particles are most likely
found in exactly the same space.
Classical images of separate
particles thus fail to model known
charge distributions in very small
nuclei [Quantum Mechanics!]
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