# 6.6

```6. Atomic and
Nuclear Physics
Chapter 6.6 Nuclear Physics
Scattering experiments and distance of closest approach

In scattering experiments such as
Rutherford’s, simple considerations can be
used to calculate the distance of closest
approach of the incoming particle to the
target.

Consider an alpha particle shot head-on
toward a stationary nucleus of charge Q=Ze
v=0
v
d
2e
Ze
Scattering experiments and distance of closest approach


Initially, the total energy of the system
consists of the kinetic energy of the alpha
particle.
v
2e
Ze
The initial distance from the nucleus is so big
that Ep does not exist.
v=0

At the point of closest approach, a distance
d from the centre of the nucleus, the alpha
particle stops and is about to turn back.

The total energy now is the electrical potential energy of the
alpha particle and the nucleus, given by:
Qq
(2e)(Ze)
2Ze 2
Ek
k
k
d
d
d
d
Scattering experiments and distance of closest approach

Assuming that the nucleus does not recoil, its kinetic energy is
zero. Then, by conservation of energy:
2
2Ze
2Ze
Ek  k
d k
d
Ek
2

Assuming a kinetic energy for the alpha particle equal to
2.0MeV directed at a gold nucleus (Z=79) gives d = 1.1x10-13m.

This is outside the range of the nuclear force, which means that
the particle is simply repelled by the electrical force.
Scattering experiments and distance of closest approach

But the distance of closest approach depends on the kinetic
energy of the alpha particle. The bigger the Ek, the smaller the
distance of closest approach.

The smallest it can get is of the same order of the radius of the
nucleus.

By doing experiments of this kind, physicists have estimated

It is found that the nucleus radius R depends on mass number
through:
1
3
15
R  1.2  A 10
m
The mass spectrometer

The existence of isotopes can be demonstrated using a
mass spectrometer.
Ion source
S1
-
+
S2
photographic plate
B into the
page
The mass spectrometer
The mass spectrometer

In a mass spectrometer single ionized ions of an element
(charge e) are made to move through a pair of slits (S1) which
collimates the beam.

The ions enter a region of magnetic and electric fields at right
angles to each other. The B is directed into the page in the region
Ion source


The positive ions are
deflected to the left by the
electric field and to the right
by the magnetic field.
By choosing a suitable value
for the magnetic field, the ions
can continue through
undeflected if the magnetic
and electric forces are equal.
S1
-
+
S2
photographic plate
B into the
page
The mass spectrometer
eV=evB
That is,
E
v
B

If the forces are the same, then

Thus, only ions with this specific velocity will be able to go through
the second slit S2.

The selected ions enter then a second region of magnetic field
and are thus deflected into a circular path, hitting a photographic
plate where they are recorded.
mv
R
eB

The radius of the circular path is given by

If the beam contains atoms of equal mass, all atoms will hit the
plate at the same point. If, however, isotopes are present, the
heavier atoms will follow a longer radius circle and will hit the plate
further to the right.

Measurement of the radius of each isotope's path will allow the
determination of its mass.
Beta decay and the neutrino

The beta decay process originates from a decay of a neutron inside
an atomic nucleus:
1
0
n p e e
1
1
0
1
0
0

The neutron decays into a proton (the Z of the nucleus increases by
1), an electron and an antineutrino.

A free neutron (i.e., outside the nucleus) decays into a proton
according to the equation above with a half-life of 11 minutes

A related decay is that of positron emission, in which a proton inside
the nucleus turns itself into a neutron accompanied by the emission
of a positron (electron’s antiparticle) and a neutrino.
1
1
p n e e
1
0
0
1
0
0
Beta decay and the neutrino
1
0
n p e e
1
1
0
1
0
0
1
1
p n e e
1
0
0
1
0
0

Unlike a free neutron, a free proton cannot decay into a neutron
since the rest energy of a neutron is larger than that of a proton.

Inside the nucleus the reaction is, however, possible because
binding energy is used to make up for the difference.

These reactions must be understood as the disappearance of
one particle and the creation of three particles on the right-hand
side of the decay equation and NOT the splitting of a particle to
form another 3 particles.
Beta decay and the neutrino

The electron antineutrino went undetected until 1953 but its
existence was predicted on theoretical grounds.

If we consider he neutron decay

The mass of the neutron is bigger than the masses of the proton
and electron together by
1
0
n p e e
1
1
0
1
0
0
1.008665u – (1.007276+0.0005486)u = 0.00084u

This mass corresponds to an energy of
0.00084 x 931.5MeV = 0.783MeV

This is the available energy in the decay, which will show up as the
kinetic energy of the products
Beta decay and the neutrino

If only the electron and the proton are produced, than the electron
being lighter of the two will carry most of this energy away as
kinetic energy.

Thus, we should observe electrons with kinetic energies of about
0.783MeV.

In experiments, however, the electron has a range of energies
from zero up to 0.783 MeV.

So, where is the missing energy?

Wolfgang Pauli and Enrico Fermi hypothesized the existence of
a third particle in the products of a beta decay in 1933. This third
very light particle would carry the remainder of the available
energy. Fermi coined the word neutrino, the Italian word for “little
neutral one”
Electron capture

A process related to beta decay is electron capture, in which a
proton inside the nucleus captures an electron and turns into a
neutron and a neutrino:
1
1

p e n e
0
1
1
0
0
0
The creation of a neutron star rests on this process, in which the
huge pressure inside the sat drives electrons into protons in the
nuclei of the star, turning them into neutrons.
Nuclear energy levels

The nucleus, like the atom, exists in discrete energy levels.

The main evidence for this fact is that in alpha and gamma
decays, the energies of alpha particles and gamma ray
photons are discrete.
E/MeV

Like atoms, we can have
diagrams for nuclear
energy levels
8.17
7.60

211
83
The energy of the particle
emitted equals the energy
level difference.
211
84
Po
Bi


0
207
82
Pb
The radioactive decay law

The decay law states that the number of nuclei that will decay
per second is proportional to the number of atoms present
that have not yet decayed,
dN
 N
dt
Here  is a constant, known as the decay constant. Its physical
meaning is that it represents the probability of decay per unit time.

If the number of nuclei originally present (at t=0) is N0, by
integrating the previous equations it can be seen that the number
of nuclei of the decaying element present at time t is
N  N0et
The radioactive decay law

As expected the number of nuclei of the decaying element is
decreasing exponentially as time goes on.

If after a certain time t (lets call it t1/2), the number of decaying
nuclei is reduced by half N=N0/2. So,
N  N 0e

 t
N0
1
  t1 / 2

 N 0e
  e  t1 / 2
2
2
Using the logarithms:


1
1
  t1 / 2
ln   ln e
 ln   t1/ 2 
2
2
 ln 1  ln 2  t1/ 2  ln 2  t1/ 2  t1/ 2 
ln 2

The radioactive decay law
t1/ 2 
ln 2

 t1/ 2 
0.693


This is the relationship between the decay constant and the half-life.

This also means that we can have an equivalent formula for the
decay equation:
1
N  N0  
2

t
T1 / 2
The number of decays per second is called activity where
A0=N0. So,
A  N0et
Why the decay constant is the probability of decay per unit
time

Since
dN
 N
dt
we know that in a short time interval dt the number of nuclei that
will decay is dN=Ndt.

The probability that any one nucleus will decay within the time
interval dt is thus:
probabilit y 
dN
 dt
N
and so the probability of decay per unit time is equal to the decay
constant:
probabilit y

dt
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