Transcript Nuclear Physics

Nuclear Physics
The famous Geiger-Marsden Alpha scattering
experiment (under Rutherford’s guidance)
In 1909, Geiger and Marsden were
studying how alpha particles are scattered
by a thin gold foil.
Thin gold foil
Alpha
source
Geiger-Marsden
As expected, most alpha particles were
detected at very small scattering angles
Thin gold foil
Alpha particles
Small-angle
scattering
Geiger-Marsden
To their great surprise, they found that
some alpha particles (1 in 20 000) had
very large scattering angles
Thin gold foil
Alpha particles
Large-angle
scattering
Small-angle
scattering
Explaining Geiger and Marsdens’ results
The results suggested that the positive (repulsive) charge must be
concentrated at the centre of the atom. Most alpha particles do not pass
close to this so pass undisturbed, only alpha particles passing very close to
this small nucleus get repelled backwards (the nucleus must also be very
massive for this to happen).
Remember on this scale, if the
nucleus is 2 cm wide, the atom
would be 200 m wide!
Rutherford did the calculations!
Rutherford calculated theoretically the
number of alpha particles that should be
scattered at different angles (using
Coulomb’s law). He found agreement with
the experimental results if he assumed the
atomic nucleus was confined to a diameter
of about 10-15 metres.
Closest approach
Using the idea of energy conservation, it is possible to
calculate the closest an alpha particle could get to the
nucleus during a head-on collision.
Alpha particle
nucleus
Closest approach
Initially, the alpha particle has kinetic
energy = ½mu2
K.E. = ½mu2
Closest approach
At the point of closest approach, the
particle reaches a distance b from the
nucleus and comes momentarily to rest.
K.E. = 0
b
Closest approach
All the initial kinetic energy has been
transformed to electrical potential energy.
K.E. = 0
b
Closest approach
Using the formula for electrical potential energy
which is derived from Coulomb’s law
Kinetic energy lost = Electrical potential
K.E. = 0
b
½mu2 =
1
q 1q 2
4πεo
b
Closest approach
Rearranging we get;
b=
1
4πεo
K.E. = 0
b
q1q2
½mu2
Closest approach
For an alpha particle, m = 6.7 x 10-27 kg, q1 = 2 x (1.6 x
10-19 C) and u is around 2 x 107 m.s-1. If the foil is made
of gold, q2 is 79 x (1.6 x 10-19 C).
b=
b=
1
q1q2
4πεo
½mu2
1
4π x 8.854 x 10-12 Fm-1
b = 2.7 x 10-14 m
x (2 x 1.6 x 10-19 C) x (79 x 1.6 x 10-19 C)
½ x 6.7 x 10-27 kg x (2 x 107 m.s-1)2
The mass spectrometer
A VERY useful machine for measuring the
masses of atoms (ions) and their relative
abundances.
What is it and how
does it work Mr
Porter?
The Mass Spectrometer
ions
produced
velocity selector
Region of
magnetic field
ions
accelerated
ion beam
detector
The Mass Spectrometer
Substance to be tested
is turned into a gas by
heating.
Electrons are produced at a
hot cathode and accelerated
by an electric field. Collision
with gas particles produces
ions.
The Mass Spectrometer
Ions acclerated by an electric
field
The Mass Spectrometer
Ions enter the velocity selector which
contains an electric field (up the
page) and magnetic field (into the
page) at right angles to each other.
By choosing a suitable value for the
magnetic field the ions continue in a
straight line. i.e. The force produced
by the electric filed (eE) is equal to
the force produced by the magnetic
field (Bev). eE = Bev
The Mass Spectrometer
eE = Bev
or v = E/B
This means that only ions with a
specific velocity pass through this
region. (Hence ”velocity selector”)
The Mass Spectrometer
The selected ions all with the
same velocity (but different
masses of course) enter the
second region of magnetic
field (also into the page). They
are deflected in a circular
path.
The Mass Spectrometer
Heavier ions continue forward
and hit the sides, as do ions
that are too light. Only ions of
one particluar mass reach the
detector.
The radius of the circle is
given by R = mv/eB so the
mass can be calculated from
m = ReB/v
The Mass Spectrometer
The magnetic field can be
varied so that ions of different
mass can be detected (higher
B would mean that ions of
larger mass could be directed
at the detector).
The Mass Spectrometer
The detector can measure the
numbers of ions detected,
hence giving an idea of
relative abundance of different
ions.
The Mass Spectrometer
The mass spectrometer is
particulary useful for
identifying isotopes of the
same element.
Nuclear energy levels
We have seen
previously that
electrons exist in
specific energy levels
around the atom.
There is evidence that
energy levels exist
inside the nucleus
too.
Wow!
Nuclear energy levels
When a nucleus decays by
emitting an alpha particle or
a gamma ray, the particles
or photons emitted are only
at specific energies (there is
not a complete range of
energies emitted, only
certain specific levels).
Nuclear energy levels
An alpha particle or photon thus has an
energy equal to the difference between
energy levels of the nucleus.
51.57
energy
levels in
235U
(MeV)
0.051
0.013
0.000
Nuclear energy levels
In the alpha decay of 239Pu to 235U, the plutonium
nucleus with an energy of 51.57 MeV can decay into
Uranium at 3 different energy levels.
51.57
energy
levels in
235U
(MeV)
0.051
0.013
0.000
Plutonium-239
Nuclear energy levels
If the 239Pu (51.57 MeV) decays to the ground state of
235U (0 MeV), an alpha particle of energy 51.57 MeV is
emitted.
51.57
energy
levels in
235U
(MeV)
Plutonium-239
Alpha emission (51.57
MeV)
0.051
0.013
0.000
Nuclear energy levels
If the 239Pu (51.57 MeV) decays to the 2nd excited state
of 235U (0.051 MeV), an alpha particle of energy 51.570.051 = 51.52 MeV is emitted. The uranium nucleus is
now in an excited state so can decay further by gamma
emission to the ground state.
51.57
energy
levels in
235U
(MeV)
Plutonium-239
Alpha emission
(51.52 MeV)
0.051
0.013
0.000
Gamma
emission (0.051
MeV)
Nuclear energy levels
In fact the nucleus could decay first to the 0.013 level,
and then the ground state, thus emitting two gamma
photons.
51.57
energy
levels in
235U
(MeV)
Plutonium-239
Alpha emission
(51.52 MeV)
0.051
0.013
0.000
Gamma
emission (0.038
MeV)
Gamma
emission
(0.013MeV)
Radioactive decay
Positron? That sounds interesting,
what is it?
Beta (β) and
positron decay
(β+) decay
Radioactive decay
In beta decay, a neutron in the nucleus
decays into a proton, an electron and an
antineutrino.
1
0
n
1
1
p+
0
-1
e
+
0
0
ve
Radioactive decay
In positron decay, a proton in the nucleus
decays into a neutron, a positron (the
antiparticle of the electron) and a neutrino.
1
1
p
1
0
n+
0
+1
e
+
0
0
ve
Radioactive decay
We must not think that the decaying
particle actually consists of the three
particles in which it splits.
The antineutrino
The antineutrino in beta decay was not
detected until 1953, although its presence
had been predicted theoretically.
1
0
n
1
1
p+
0
-1
e
+
0
0
ve
The antineutrino
The mass of the neutron is bigger than
that of the proton and electron together.
1
0
n
1.008665 u
1
1
p+
1.007276 u
0
-1
e
+
0
0
ve
0.0005486 u
1.008665u – (1.007276 + 0.0005486)u = 0.00084u
The antineutrino
This corresponds (using E = mc2) to an
energy of 0.783 MeV.
1
0
n
1
1
p+
0
-1
e
+
0
0
ve
The antineutrino
This extra energy should show up as kinetic energy of
the products (proton and electron). Since the electron
should carry most of the kinetic energy away, so we
should observe electrons with an energy of about 0.783
MeV.
1
0
n
1
1
p+
0
-1
e
+
0
0
ve
In fact we observe electrons with a range
of energies from zero up to 0.783 MeV.
The antineutrino
Where is the missing energy? In 1933 Wolfgang Pauli and
Enrico Fermi hypothesized the existence of a third very light
particle produced during the decay. Enrico Fermi coined the
term neutrino for the ”little neutral one”
1n
0
1p
1
+
0 e
-1
+
0v
0 e
Ahhhhh! The little neutral one!
The radioactive decay law
If the number of nuclei present in a sample at t =
0 is N0, the number N still present at time t later
is given by
N = Noe-λt
where λ is the decay constant (the probability
that a nucleus will decay in unit time)
The radioactive decay law
N = Noe-λt
No
Number of original
nuclei present
time
Half-life
After one half-life, the number of original
nuclei present is equal to N0/2. Putting this
into the radioactive decay law;
N0/2 = N0e(-λt½) where t½ is the half-life
Half-life
N0/2 = N0e(-λt½)
taking logarithms we find
λt
λt
½
½
= ln2
= 0.693
This is the relationship between the decay
constant and the half-life.
Measuring half-life
For short half-lives, the half life can usually
be measured directly.
Measuring half-life
For longer half lifes, values of activity can
be measured and the decay law can be
used to calculate λ and thus t½.
Measure the activity A and chemically find
the number of atoms of the isotope.
Use A = λN and then λt½ = ln2
Questions!
• Page 412 Questions 1, 2, 3.
• Page 413 Questions 6, 7, 8, 10.