6. Atomic and Nuclear Physics
Chapter 6.6 Nuclear Physics
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
Scattering experiments and distance of closest approach
v
2e
Ze
v=0d
Initially, the total energy of the system consists of the kinetic energy of the alpha particle.
The initial distance from the nucleus is so big that Ep does not exist.
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.
Scattering experiments and distance of closest approach
v
2e
Ze
v=0d
The total energy now is the electrical potential energy of the alpha particle and the nucleus, given by:
d
Zek
d
Zeek
d
QqkE
22))(2(
Assuming that the nucleus does not recoil, its kinetic energy is zero. Then, by conservation of energy:
Scattering experiments and distance of closest approach
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.
kk E
Zekd
d
ZekE
22 22
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 the nuclear radii.
It is found that the nucleus radius R depends on mass number through:
Scattering experiments and distance of closest approach
mAR 1531
102.1
The existence of isotopes can be demonstrated using a mass spectrometer.
The mass spectrometer
Ion source
S1
+-
S2photographic plate
B into the page
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 shaded grey.
The mass spectrometer
Ion source
S1
+-
S2photographic plate
B into the page
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.
If the forces are the same, then eV=evB That is,
The mass spectrometer
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.
The radius of the circular path is given by
B
Ev
eB
mvR
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.
The beta decay process originates from a decay of a neutron inside an atomic nucleus:
Beta decay and the neutrino
eepn 00
01
11
10
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.
eenp 00
01
10
11
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
eepn 00
01
11
10
eenp 00
01
10
11
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
1.008665u – (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
eepn 00
01
11
10
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?
Beta decay and the neutrino
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”
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:
Electron capture
enep 00
10
01
11
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.
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.
Nuclear energy levels
Like atoms, we can have diagrams for nuclear energy levels
The energy of the particle emitted equals the energy level difference.
E/MeV
8.17
7.60
0
Bi21183
Po21184
Pb20782
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,
The radioactive decay law
Ndt
dN
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
teNN 0
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,
The radioactive decay law
2/12/1 0
00 2
1
2ttt eeN
NeNN
Using the logarithms:
2ln2ln2ln1ln
2
1lnln
2
1ln
2/12/12/1
2/1 2/1
ttt
te t
The radioactive decay law
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:
693.02ln
2/12/1 tt
2/1
2
10
Tt
NN
The number of decays per second is called activity where
A0=N0. So,
teNA 0
Why the decay constant is the probability of decay per unit time
Since
and so the probability of decay per unit time is equal to the decay
constant:
Ndt
dN
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:
dtN
dN yprobabilit
dt
yprobabilit
