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Atomic and Nuclear Physics
Photoelectricity
This is the current produced by the freeing ore escaping of electrons from a negatively charged
metal by exposing that metal to ultra-violet light. (Example: ultra-violet radiation ejects electron
from zinc while white light ejects electrons from sodium)
This process is called the photoelectric effect and the light (radiation) gives energy to the
electron in the surface atoms of the metal and enables them to break through the surface.
An experiment which demonstrated the nature of photoelectricity
The surface of the zinc plate is rubbed with emery paper until clean and bright. One of the zinc
plates is insulated and connected to the cap of a gold leaf electroscope and it is given a positive
charge by the induction (the process by which electric or magnetic state is obtained by exposure
to an electric or magnetic field)
Some of this charge spreads to the leaf which opens. In a dark room this positively charged zinc
plate is then exposed to ultra-violet light from a small lamp placed near it. The leaf stays open
which indicates that there is no loss of charge. The free electrons need much more energy to
leave the zinc plate because it is positively charged. The radiation cannot supply enough energy
and therefore no emission of electrons occurs, since the electrons are attracted back to the plate.
However, in the case of the negatively charged zinc plate: the leaf slowly falls because the
electroscope loses it charge. Since both the electrons and the zinc plate now carry a negative
charge, the electrons are repelled away from the zinc plate and are therefore easily emitted.
These electrons are called photoelectrons which indicates that the electron has been emitted
when light fell on the surface of the metal (N.B. These electrons are identical to any other
electrons)
The Phenomenon of Photoelectric Emission
The measurements and investigations have shown the following conclusions:
The kinetic energy (velocity) of the electrons emitted from an illuminated metal is
independent of the intensity of the light (i.e. The intensity of the radiation has no effect
on the kinetic energy of the emitted electrons)
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For a given metal, no photoelectrons are emitted if the light frequency (radiation
frequency) is below a certain value called the threshold frequency. This frequency is
defined as the minimum frequency of electromagnetic radiation for which photoelectric
emission occurs.
No electrons are emitted when the metal is illuminated by light (radiation) of a
wavelength longer than a specific wavelength called the threshold wavelength. Electrons
are emitted when the metal is being illuminated by light (radiation) which has a
wavelength lower than the threshold wavelength
The emission commences as soon as the surface starts to be irradiated.
The number of photoelectrons emitted per second from any given metal is proportional to
the intensity of the incident radiation. (The more intense the radiation the greater the
number of photoelectrons leaving the metal each second).
The emitted electrons have different kinetic energies (ranging from zero up to a
maximum value). Increasing the frequency of the incident radiation will increase the
energies of the electrons emitted and the also the maximum kinetic energy.
The Inability of Classical Physics (Wave Theory) to Explain Aspect of Photoelectric Effect
The photoelectric effect is due to electrons absorbing energy from the radiation and therefore
having the ability to overcome the attractive forces of the nuclei. The wave theory of light states
that the energy of the radiation is distributed evenly over the wavefront.
According to the wave theory, each electron on the surface of the metal will absorb an equal
share of the radiation. Hence the expectation is that the intensity of the radiation would be very
low and no electrons would gain the required amount of energy to escape the metal, or that a
significant amount of time would have lapsed before any electrons escape.
Neither of these two predictions are consistent with the observations seen. Furthermore,
increasing the intensity, increases the energy fall on the surface of the metal and an increase in
the energies of the emitted electrons would be expected. However this is also inconsistent with
the observations made.
In addition, the wave theory does not offer an explanation of the frequency being depend on the
kinetic energy of the electrons being emitted nor why there should be a minimum frequency at
which emission occurs.
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Planckβs Equation
Max Planck showed that the laws could be explained by assuming light and all other forms of
electromagnetic radiation are emitted in wave packets, with each wave packet being a short burst
of light energy from an atom (quantum theory).
He showed that when an atom emits light, its energy changes by specific allowed amounts.
When referring to the light energy, these packets of energy are called photons. (packets of
electromagnetic energy emitted by an atom).
The energy of a single photon is proportional to its frequency and therefore the following
equation can be stated:
πΈ = βπ
where E = photon energy
h = Planckβs constant (6.626 Γ 10-34 Js)
f = the frequency of the radiation
Note that :
π = π
π
where c = speed of light (3Γ 108 m/s)
π = wave length
Work Function (W0)
The least or the minimum amount of work or energy necessary to take a free electron out of a
metal against the attractive forces of surrounding positive ions is called the work function of the
metal.
We can calculate the threshold frequency (f0) by using the following equation:
π0 = π0
β
And the corresponding maximum wavelength (π0) is given by:
π0 = βπ
π0
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The Electron-volt (eV)
This is the unit of energy. It is defined as the kinetic energy gained by an electron which is
accelerated by a potential difference of one volt.
1 ππ = 1.6 Γ 1019 π½
Einsteinβs Photoelectric Equation
Einsteinβs Theory comprises of the following statements
light of frequency (f) contains quanta of energy (hf)
light consists of particles called photons
the number of photons per unit area of cross section of a beam of light per second is
proportional to its intensity
the energy of a photon is proportional to its frequency but independent of its light
intensity
A simple equation can be obtained for the maximum kinetic energy of electrons liberated from
an illuminated metal, and we can state that:
The maximum kinetic energy of a photoelectron is equal to the energy gained by absorbing a
photon (hf) minus the work done (W0) to escape from the material.
πΎ. πΈ.πππ₯ = βπ β π0
1
2ππ£2 = βπ β π0
Therefore Einsteinβs Photoelectric Equation is summed up by:
βπ = πΎ. πΈ.πππ₯ + π0
OR
βπ = 1
2ππ£2 + π0
These equations show that the maximum kinetic energy depends only on the light frequency and
that K.E. max is unaffected by making the light brighter (light intensity)
πΎ. πΈ.πππ₯ = 1
2πππ£2
where ππ = 9.1 Γ 10β31 ππ (mass of an electron)
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The Stopping Potential (Vs)
The stopping potential of a surface is the minimum positive potential which must be applied to a
surface to stop photoelectric emission at a particular light frequency.
πππ = βπ β π0
Where e = photon energy
W0 = Planckβs constant (6.626 Γ 10-34 Js)
Vs = the frequency of the radiation
If we make Vs the subject of the equation, we get that
ππ = βπ
πβ
π
π
Since this equation is in the format of a straight line (y = mx + c), we can see that
y = Vs, x = f, c (y- intercept) = W/e (depends on the type of material used)
m (gradient) = h/e (not dependent on the type of material since h and e are fundamental
constants),
NOTE: πΎ. πΈ.πππ₯ = πππ
Determining the Stopping Potential
X - photoelectric cell
C - photosensitive metal of large area
A - collecter of electrons in a vacuum
Y - potential divider arrangement for varying P.D. (V) between the anode, A and the cathode, C
d.c. ammeter - ammeter measuring the small current
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What happens?
As the incident light reaches C, electrons are released from the surface of C. Although these
electrons are negative in charge, as so is A, these electrons make their way to A because the
energy they possess upon being released allows them to surpass the repulsive force which they
experience from A.
This occurs for the initial value of V and since electrons are able to move some current in noted
at the d.c ammeter. As the p.d at A rises by the potential divider, A becomes more and more
negative and eventually the electrons are no longer able to surpass the repulsive force (their
energy is less than the force) and A is able to keep the electrons from flowing through out the
circuit. As a result of this no current flows through the circuit. As a result of this no current is
read at the d.c. ammeter. The value of V at which this phenomenon occurs is called the stopping
potential.
Wave-Particle Duality
The particle theory describes light as being made up of small particles moving linearly. This
theory is used to explain reflection and refraction. It basically states that light was made up of
tiny streams of particles, traveling at very high speeds, in straight line. However, could not be
used to explain diffraction of light.
Therefore another model, the wave theory was developed to explain the behavior and properties
of light which couldnβt be explained by the particle theory. This theory suggested that light could
also be diffracted and produce interference effects.
De Borglie suggested that matter could also exhibit a dual nature. He proposed that there was a
relationship between the momentum of any particle and its wavelength.
Further research by Davission and Germer confirmed this relationship when they were able to
succeed in diffracting electrons.
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Production of X-Rays
What are X- Rays?
X- Rays are the electromagnetic waves with short wavelengths, usually 10-10m or less.
How are they produced?
Diagram Shows a Coolidge Type X-Ray Tube
The filament is a very hot wire (usually made of tungsten). When this wire is heated, the kinetic
energy of the electrons in the wire also increases causing them to vibrate more quickly and
vigorously until they reach a point where they are able to escape the wire.
The electrons from the filament travel at high speeds and collide with the target.
The target (anode) is set a large positive potential (with reference to the filament lamp) by using
a high voltage unit. By doing this the negatively charged electrons experience a strong pull from
the positive target (anode) and the electrons accelerate towards the target with high speeds. This
is aided by the evacuation of the tube (the vacuum).
Upon collision with the target, the electrons are decelerated (lose speed and hence lose energy).
This type of energy loss is what produces the X- rays. The rest of the energy which was not
βlostβ (usually 99%) goes towards the production of heat.
Since 99% of the energy which the electrons possess is converted into heat energy, the tube is
designed with a cooling system to stop it from melting. In the diagram above this is shown by the
cooled copper rod which conducts the emitted heat by the fast moving electrons away from the
target. The rod is cooled by circulating oil through it or by using cooling fins.
X- rays are produced with the interaction of the electrons with the atoms of the target material
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X-rays produced by the Interaction with Atoms of Target Material
The electrons which bombard the target are very energetic and are capable of knocking electrons
out of the deep lying energy levels of that atom. They are able to do this because the electrons in
the lower energy levels are quite stable and do not move around as the one in the higher level do
and are easier to target.
When one of the electrons from the lower energy is knocked out it leaves a space or a hole. The
electrons in the higher level are seeking to become more stable and seek to occupy the space left
by the electrons which was knock out. The electron from the higher level βfallsβ into the space.
In order to do this, it must lose some of its energy since all the electrons at a particular level must
have the same energy. The energy at which that electron loses in order to occupy that stable
place in a lower level goes towards the production of X-rays.
When an electrons moves from a level with high energy (E2) to one of lower energy (E1), the
frequency (f) of the emitted radiation is calculated by:
πΈ2 β πΈ1 = βπ
where h = Planckβs constant (6.626 Γ 10-34 Js)
(N.B. the more energy emitted means an increase in frequency but this also means a decrease in
wavelengths since f Ξ± 1/Κ).
Energy Levels
The energies of electrons in atoms can have only specific values which are called the energy
levels of the atoms. All the atoms of a named element have the same set of energy levels which
gives the characteristics of the atom and they differ from every other element.
These energy levels are expressed in electronvolts and are usually represented as a series of
horizontal lines. When an electron occupies the lowest level in the atom it is said to be in its
ground state. If it absorbs energy, the atom may be promoted into a higher energy level. The
atom is unstable and is said to be in an excited state.
After a short and random interval the electron βfallsβ back into the lowest level so that atom can
return to its ground state. The energy that was originally absorbed is emitted as electromagnetic
waves.
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Each energy level is described by a quantum number (n). The lowest level is n = 1, then n = 2,
etc. The highest energy level is n = β, and at this level the energy is zero. When an electron
reaches this level, the electron becomes free of the atom and the atom is said to be ionized,
Energy Levels of
Hydrogen atom
X-Ray Emission Spectra
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The diagram above shows a typical X-ray spectrum which has two distinct components:
(i) a background of continuous radiation, the minimum wavelength of which depends on
the operating voltage of the tube. (the energy of the electrons bombarding the target)
(ii) very intense emission at few discrete wavelengths (an X-ray spectrum). These
wavelengths are characteristic of the target material and are independent of the
operating voltage.
The Continuous Background
This is produced by electrons colliding with the target and being decelerated. It is called
continuous because there is large range of energy due to the fact that the electrons do not lose
any one specific value of energy. The energy that each electron lose may be different from other
electrons because of the speed that the electron was originally travelling or the angle at which the
electron collide with the target. All of the factors mention determines what fraction of the
electronβs original energy value will be lost resulting in the large range of values.
The Line Spectrum
This is produced by the electrons knocking out a stable electron from the atom of the target
material. It can be seen that specific values of energy needs to be lost in order for the higher level
electrons to fill the space in the lower level. Therefore a particular energy value needs to be lost.
The total X- ray product is a combination of that produced by the continuous spectrum and the
line spectrum.
Properties of X- Rays
i. They travel in straight lines at the velocity at the speed of light
ii. They cannot be deflected by electric of magnetic fields hence they are not charged
particles
iii. They penetrate matter. Penetrating is the least with materials of high density
iv. They can be reflected but only at large angles of incidences
v. They can be diffracted
vi. The intensity of the X-ray increases with the number of electrons hitting the target and
the filament current. It is also increased by increasing the PD across the tubes.
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vii. The penetrating power of an X-ray increase with the PD across the tube. An X-ray with a
low penetrating power is called a soft X-ray while one with a high penetrating power is
called a hard X-ray.
viii. The most energetic X-rays (those whose wave lengths are Κmin) are the result of the
bombarding losing all their energy at once. Since the energy of the electrons depends on
the operating voltage, so does Κmin. Hence the higher the voltage, the smaller the value of
Κmin.
ππππ = βπ
ππ
where h - Planckβs constant
c - speed light
e - electron charge
V - tube voltage
Addition Properties (used to detect X-rays)
i. They ionize gases through which they pass
ii. They blacken photographic film
iii. They can produce fluorescence
iv. They can produce photo-electric emission
Uses of X-Rays
i. Used in medicine to locate bone fractures and to destroy cancer cells. (Bones absorb X-
rays more than flesh)
ii. To detect cracks in metals
iii. To investigate the structure of crystals
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X- Ray Absorption Spectra
X-Rays passing through an Absorber
We can calculate the emerging intensity (I1) of a beam of X-ray of original intensity (I0) after it
has passed through a slab of material of thickness (x1) and absorption coefficient (Β΅1) by using
the equation
πΌ1 = πΌ0πβπ1π₯1
It follows from the equation that
ln πΌ1 = ln(πΌ0πβπ1π₯1)
ln πΌ1 = ln(πΌ0) + ln( πβπ1π₯1)
β΄ ln πΌ1 = βπ1π₯1 + ln(πΌ0)
where y = ln I1, x = x1, m = -Β΅1, c = ln (I0)
If a second slab of thickness (x2) and absorption coefficient (Β΅2) is placed directly behind the first
slab. We can calculate the value for the emerging intensity (I2) of the beam by using
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πΌ1 = πΌ0πβπ1π₯1 and πΌ2 = πΌ1πβπ2π₯2
Hence πΌ2 = πΌ0πβπ1π₯1 Γ πβπ2π₯2
πΌ2 = πΌ0π(βπ1π₯1)+( βπ2π₯2)
β΄ πΌ2 = πΌ0πβ(π1π₯1+ π2π₯2)
We can use a graph to represent the way that Β΅ varies with the wavelength (Κ) for any absorber.
As Κ increases, Β΅ and the absorption rapidly increases until Κ = Κk. This point is called K
absorption edge. Up to this point, absorption is mainly due to the ejections of electrons from the
K shell of the absorber.
The sudden drop in Β΅ at the point Κk occurs because the X-rays of longer wavelengths do not
have enough energy to eject K shell electrons. As Κ increase beyond Κk, the absorption increases
again and is due mainly to the ejection of electrons from the higher lying L shell. The L shell
consists of three energy levels (LI, LII and LIII) which are very close to each other.
The Difference between an Emission and Absorption Spectrum
Emission spectrum is a range of radiation produced by the change in energy level of electrons
which results in the energy being emitted, but the absorption spectrum is caused by an object
absorbing radiation of a particular wavelength from an incident beam and leaving dark bands or
darks spaces in the emergent beam. Absorption emission spectra can be used to determine the
materials because certain materials absorb and emit specific wavelengths of radiation.
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De Broglieβs Equation
In 1924 Louis de Broglie presented a thesis stating that matter possesses wave-particle duality
nature. He proposed that any particle of momentum (p) has an associated wavelength (Κ) called
the de Broglie wavelength.
The de Broglie Equation states that
π = β
π=
β
ππ£
β΄ π = β
π
where m - relative mass
v - velocity of particle
h - Planckβs constant
p - momentum of particles
Table of Electromagnetic Waves showing Frequency and Wave Length Ranges
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The Structure of the Atom
The Atom
Scientists such as Rutherford, Geiger, Thomson, Mardsen, Bohr and Chadwick helped
established modern views of the atom.
Before 1897
Atoms were thought to be small invisible particles
1902 -Thomson
Thomson showed that negative charges called electrons existed within matter. He suggested that
the atom resembled a plum pudding where the electrons were represented by the seeds and the
body of the plum pudding was the positive charge.
1906 - Rutherford
During this time Rutherford observed that alpha particles passing through a thin sheet of mica
without making holes in it.
1911- Geiger and Marsden
A narrow beam of alpha particles from a radon sourced was fired at a thin metal foil. A glass
screen coated with zinc sulphide was used to detect the scattered alpha particles. The experiment
was carried out in a darkened room under a microscope and whenever an alpha particle strike the
screen a faint flash of light was observed.
It was found that the majority of the alpha particles deviated through small angles and that only a
small number deviated more than 900 and even less deflected back towards the source. These
observations apposed Thomsonβs βplum puddingβ model of the atom.
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From this experiment it was therefore concluded that the atom has a small positively charged
core which contained most of the mass of the atom and which was surrounded by orbiting
electrons.
1913 - Rutherford and Bohr
These two scientists came up with a new model that suggested that the atom had a dense central
core, which was positive, and a very small negative part compared to the rest of the atom. They
suggested that the most of the volume occupied by the atom was an empty space and negative
electrons orbited around the nucleus
1932 - Chadwick
Chadwick identified the neutron, a neutral particle found within the nucleus of the atom.
Chadwickβs experiment supported the new model proposed by Rutherford and Bohr.
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Today
We still picture the atom today to have a very small and dense nucleus that consists of protons
and neutrons, surrounding by a cloud of negative electrons. The electrons cannot be pinpointed
since it is thought to behave as a wave and a particle.
Atomic Structure Atoms are composed of neutrons, protons and electrons
Standard Notation
πππ΄
where X is the element symbol
Z is the atomic number (proton number)
A is the mass number (nucleon number)
PARTICLE CHARGE MASS
PROTON + 1 1
ELECTRON - 1 1/ 1840
NEUTRON None 1
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Atomic Number (Z)
The atomic number of an element is the number of protons found in the nucleus of that element.
E.g. Carbon has an atomic number of 6; hence there are 6 protons in its nucleus.
Mass Number (A)
The mass number of an element is the number of protons plus the number of neutrons.
(A = Z + N)
Atoms are neutral particles and have the same number of protons as electrons.
E.g. Sodium has 11 protons; therefore it must also have 11 electrons.
Examples:
PARTICLE NO. OF
PROTONS
NO. OF
NEUTRONS
ATOMIC
NO.
MASS NO.
πΆπ1737 17 20 17 37
πΆ612 6 6 6 12
πΊπ3170 31 38 31 70
ππ84210 84 126 84 210
Isotopes
Isotopes are different atoms of the same element. They have the same atomic number but
different mass numbers. Isotopes of the same elements have the same number of protons and
electrons but different neutrons.
Examples
π―ππ π―π
π π―ππ
πΆ612 πΆ6
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Millikanβs Determination of Electron Charge (e)
The principle of Millikanβs experiment is to measure the terminal velocity of a small charged
drop of oil under gravity; then to oppose its motion with an electric field in such a manner that it
remains stationary.
Terminal velocity: An object falling air experiences a force which opposes the motion of the
object (viscous drag). The initial downward force is greater than the drag forces and the object
accelerates. However, as the drag forces increase with the velocity of the object, the gravitational
force and the drag forces eventually become equal. When this occurs there is no more
acceleration. The object has reached its maximum speed and this is called terminal velocity.
When there is no electric field: the forces acting on the oil drop are shown in the diagram
above. Once the oils drop has reached the terminal velocity, there is no more acceleration and
therefore we can state that
π€πππβπ‘ = π’ππ‘βππ’π π‘ ππ’π π‘π πππ + π£ππ πππ’π ππππ β¦ β
We know that the weight of the drop of oil can be calculated by
π€πππβπ‘ = π£πππ’ππ ππ ππππ Γ πππππ π‘π¦ ππ πππ Γ ππππ£ππ‘π¦
Therefore π€πππβπ‘ = 4
3ππ3π0π β¦β‘
where r = radius of the oil drop
p0 = density of the oil drop
Also π’ππ‘βππ’π π‘ = π€πππβπ‘ ππ πππ πππ ππππππ ππ¦ πππ ππππ
= π£πππ’ππ ππ πππ ππππ Γ ππππ ππ‘π¦ ππ πππ Γ ππππ£ππ‘π¦
β΄ π’ππ‘βππ’π π‘ = 4
3ππ3πππ β¦ β’
where pa = density of the oil drop
According to Stokesβ Law, we can calculate the viscous drag by using the equation
π£ππ πππ’π ππππ = 6ππππ£ β¦ β£
where Ζ = coefficient of viscosity of air
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Substituting β‘, β’ and β£ into β ; we get that
4
3ππ3π0π =
4
3ππ3πππ + 6ππππ£ β¦ β€
When an electric field is applied such that the oil drop is stationary: the forces acting on the
oil drop (as shown in the diagram above) are such that the oil drop has no velocity and no
acceleration.
We can therefore state the equation
π€πππβπ‘ = π’ππ‘βππ’π π‘ + πππππ‘πππ πππππ
We can state
4
3ππ3π0π =
4
3ππ3πππ + ππΈ β¦ β₯
where Q = the charge of the oil drop
E = electric field strength
When we subtract β€ from β₯, we get that
0 = ππΈ β 6ππππ£
β΄ π = 6ππππ£
πΈ β¦ β¦
However, Millikan measured E and v and did and did another experiment to find Ζ. He was not
able to measure r directly but rearranging β€ to get that
4
3ππ3(π0 β ππ)π = 6ππππ£
β΄ π = (9 ππ£
2(π0βππ)π)
1
2 β¦ β§
Substituting β§ into β¦, we get that
π = 6ππ
πΈ(
9 ππ£
2(π0 β ππ)π)
12 π£
But since the density of air at RTP is extremely small compared to a drop of oil, we can use the
equation
π = 6ππ
πΈ(
9 ππ£
2π0π
)
12 π£
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NUCLEAR STABILITY, FUSION AND FISSION
Einsteinβs Mass β Energy Equation
Einstein stated that the relationship between the mass (m) and the speed of light (c) is equivalent
to the amount of Energy (E), and can be calculated by the equation
πΈ = ππ2
Therefore as there is a change in energy there is also a change in mass.
The unified atomic mass unit (u) is defined as 1
12 the mass of the Carbon β 12 atom
Mass of πΆ612 = 12π
So 12
6 Γ1023 π = 12
6 Γ1026 ππ = 12 π’
12 u is the mass of one particle in 1 mole of Carbon 12 atoms, therefore
1π’ = 12
12 Γ (6.0 Γ 1026)
β΄ 1 π’ = 1.661 Γ 10β27 ππ
And we can state that
1 π’ = 932 πππ
Mass Defect
The mass of nucleus is always less than the total mass of it individual nucleons. The reduction in
mass occurs because combining the nucleons causes some of their mass to be released as energy
(in the form of Ξ³ β rays). The difference in mass is called the mass defect.
πππ π ππππππ‘ = πππ π ππ’ππππππ β πππ π ππ ππ’ππππ’π
or
πππ π ππππππ‘ = ππ ππ’ππππππ πππ ππππ‘ππππ β πππ π ππ ππ‘ππ
Binding Energy and Nuclear Forces
The nucleus consists of protons and neutrons. The protons repel each other, yet the nucleons are
held together in the nucleus. In order for the nucleons to be stable there must be strong force
between them. This force is called the nuclear force.
Any attempt to separate the nucleons requires energy. When the nucleons come together in the
nucleus, there is loss of energy which is equal to the binding energy. The binding energy,
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therefore is the energy required to completely split up the nucleons into its neutrons, protons and
electrons which exists in the nucleus of the atom.
πππππππ ππππππ¦ (π½) = πππ π ππππππ‘ (ππ) Γ π 2 (ππ β1)2
or
πππππππ ππππππ¦ (πππ) = 932 Γ πππ π ππππππ‘ (π’)
Worked Example
Calculate the mass defect (in both u and kg) and the r=energy released (in both J and MeV)
πΏπ37 + π»1
1 β π»π24 + π»π2
4 + π
Atomic masses: Li = 7.016 u H = 1.008 u He = 4.004 u
π‘ππ‘ππ πππ π ππππππ = ( 7.016 + 1.008) = 8.024 π’
π‘ππ‘ππ πππ π πππ‘ππ = 2(4.004) = 8.008 π’
πππ π ππππππ‘ (ππ π’) = (8.024 β 8.008) = 0.016 π’
πππ π ππππππ‘ (ππ ππ) = 0.016 Γ (1.66 Γ 10β27) = 2.66 Γ 10β29 ππ
ππ πππ πΈ = ππ2
πΈ (ππ π½ππ’πππ ) = (2.66 Γ 10β29) Γ (3.0 Γ 108)2 = 2.39 Γ 10β12π½
πΈ (ππ πππ) = 2.39 Γ 10β12
1.6 Γ 10β19 = 1.49 Γ 107ππ = 14.9 πππ
Variation of Binding Energy per Nucleon with Mass Number
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Nuclear Fission
Nuclear Fission is the disintegration of a heavy nucleus of a heavy nucleus into two lighter
nuclei. The energy is released because the average binding energy per nucleon of the fission
products is greater than the average binding energy per nucleon of the fission products is greater
than that of the parent.
Example of a fission reaction
π92235 + π0
1 β π΅π56141 + πΎπ36
92 + 3 π01
Worked Example
Nuclear Fusion
Nuclear Fusion is the combination of two light nuclei to produce a heavier nucleus. The energy
released by this process is less than that which results from fission.
Example of fusion reaction
π»12 + π»1
2 β π»π23 + π0
1
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Worked Example
The Relationship between Binding Energy per Nucleon and Nuclear Fusion and Fission
A nucleus becomes more stable when it releases energy. Therefore the greater the stability the
more energy is needed to break the nucleus up. So the higher the binding energy, the more stable
the element. (mass defect is directly proportional to the binding energy)
Since nuclear reactions occur in order to increase the stability of the nucleus; fusion and fission
reactions will occur in order to produce a nucleus with a higher binding energy.
Note that the rising part of the binding energy curve indicates that elements with lower mass
numbers can produce energy by fusion. The falling portion of the curve indicates that the heavy
elements can produce energy by fission.
Note that Iron does not usually undergo nuclear reactions because it is very stable, hence it has
the highest binding energy.
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RADIOACTIVITY
Radioactivity is the spontaneous emissions from the nucleus of certain atoms, of either alpha,
beta or gamma radiation. These radiations are emitted when the nuclei of the radioactive
substance breaks down to form a new and more stable nuclei.
One of the early workers with this phenomenon was Marie Curie. She experimented with
uranium compounds and discovered new elements such as polonium and radium. These
substances emitted invisible radiation that eventually killed her in 1934 from overexposure.
Nature of Radioactive Emissions
Alpha particles () are helium nuclei. They have 2 protons and 2 neutrons existing together.
Beta particles () are electrons moving at high speeds.
Gamma rays () are electromagnetic waves of very short wavelengths.
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Cloud Chamber
Radioactivity can be detected by using by using a device called the diffusion cloud chamber. The
device consists of a cylinder chamber under which dry ice is placed. Inside the chamber is filled
with alcohol vapour, which is cooled about β65 oC by using dry ice. When the radioactive source
is placed close to the base, the inside is illuminated with white vapour trails, which can be seen
shooting from the source. The vapour trails formed, shows the paths taken by the radiation and
can be used to distinguish the type of radiation present.
The Effect of Magnetic Fields on Radioactive Emissions
Some radioactive emissions are deflected by both electric and magnetic fields; the diagram
below shows the effects of magnetic fields on particles, particles and rays.
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Gamma () rays are not affected by magnetic fields since they have no charge, hence they do not
deflect.
Alpha () particles show a slight deviation to the left as shown in the diagram above. By using
Fleming Left Hand Rule we can determine that the alpha particles are positively charged.
Beta () particles deviate to the right, which shows that they have a negative charge.
Note that alpha particles deviate less than beta particles because they are massive compared to
beta particles.
Radioactive Decay
If an isotope is radioactive it has an unstable arrangement in its nuclei. The emission of alpha or
beta particles can make the isotope more stable. This type of reaction however alters the number
of protons or neutrons in the nucleus making the nucleus of a different element. The original
nucleus is called the parent nucleus and the new nucleus is called the daughter nucleus. (N.B.
Further decay can also produce a granddaughter nucleus.) The daughter nucleus and the
emitting products are called decay products.
Radioactive decay, which occurs naturally, is random and spontaneous and there is no way of
telling how much of the nuclei will decay or when they will decay.
Alpha Particle Decay
When a nucleus decays by emitting alpha () particles, its atomic mass number (Z) decreases by
two and its mass number (A) decreases by four. This type of decay occurs when the nuclei has
too many protons to be stable.
General Equation:
πππ΄ β ππβ2
π΄β4 + π»π24 + πΈπππππ¦
Examples
π π88226 β π π86
222 + πΌ24 + πΈπππππ¦
ππ84212 β ππ82
208 + π»π24 + πΈπππππ¦
π92238 β πβ90
234 + π»π24 + πΈπππππ¦
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Beta Particle
Beta emission causes the proton number (Z) to increase by 1, but the nucleon number (A) does
not change. Beta particles are emitted when the nuclei have too many neutrons to be stable.
General Equation:
πππ΄ β ππ+1
π΄ + πβ10 + πΈπππππ¦
Examples
πΆ614 β π7
14 + π½β10 + πΈπππππ¦
ππ82212 β π΅π83
212 + πβ10 + πΈπππππ¦
π΄π89230 β πβ90
230 + π½β10 + πΈπππππ¦
Gamma Ray Emissions
With some isotopes the emission of alpha particles and/ or beta particles leaves the neutrons and
protons in an excited arrangement. As the protons and neutrons arrange themselves into a more
stable arrangement they lose energy. This energy is emitted as a burst of gamma radiation.
Gamma radiation does not cause changes in the mass number (A) or the atomic number (Z) of
the isotope.
General Equation:
πππ΄ β ππ
π΄ + πΎ πππ¦π
Examples
ππ82212 β ππ82
212 + πΎ πππ¦π
π714 β π7
14 + πΎ πππ¦π
Exponential Law of Radioactive Decay
This law states that for large numbers of any particular nuclei the rate of decay is proportional to
the number of parent nuclei. ππ
ππ‘= βππ β¦ β
where Ξ» = positive constant of proportionality called the decay constant (unit: s-1)
βππ
ππ‘ = the rate of decay and is called the activity of the source. (unit: becquerel, Bq)
NB: 1 Bq = 1 disintegration per second and 1 Ci (curie) = 3.7 Γ1010 s-1
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Rearranging β we get that
ππ
π= βπ ππ‘ β¦ β‘
Integrating β‘ we get that
β«ππ
π= β« βπ ππ‘
β«1
π ππ = βπ β« ππ‘ β¦ β’
Since we know that β«1
π₯ = ln π₯ and β« ππ‘ = π‘, then β’ becomes
ln π = β ππ‘ + π β¦ β£
Given that N0 = the number of original radioactive atoms (i.e. at t = 0), and substituting into β£,
ln π0 = β π (0) + π
β΄ π = ln π0 β¦ β€
Substituting β€ into β£, we get that
ln π = β ππ‘ + ln π0
ln π β ln π0 = β ππ‘
ln (π
π0) = β ππ‘
β΄ π
π0= πβ ππ‘
Hence, we can conclude that
π = π0πβ ππ‘ β¦ β₯
Since activity is proportional to the number of parent nuclei, we can also say that
π΄ = π΄0πβ ππ‘ β¦ β¦
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Graph showing the Exponential Nature of Radioactive Decay
If a radioactive source has a short half-life, apparatus can be used to obtain data to plot a decay
curve. It is not easy to measure the number of nuclei in a sample of radioactive material but the
rate of decay can be used to indicate how much of the nuclei is present.
To plot a radioactive decay curve we need to find the value for the background count, this would
be subtracted from each reading obtained. Readings are then taken at regular intervals and the
corrected count rate is plotted against the time. We can then use the graph to find the half-life of
the substance.
Activity (A)
This is the number of particle emission per second from a radioactive source and hence is simply
the rate of decay. (unit: Becquerel, Bq)
Decay Constant (Ξ»)
This is the positive constant of proportionality. Decay is the change of a property such that the
value decreases with time
π΄ = ππ where A = activity
Ξ» = decay constant
N = number of parent nuclei
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Half-Life (T1/2)
The half-life of a radioactive substance is the time taken for half of the unstable atoms to decay
(decrease by half of its original mass).
If we represented the half -life by t = T1/2 and N = N0 / 2 and substituted these values into β₯, we get that
π0
2= π0πβ ππ1/2
1
2= πβ ππ1/2
ln (1
2) = β ππ1/2
β0.6931 = β ππ1/2
β΄ π1/2 = 0.6931
π ππ π1/2 =
ln 2
π
N.B. Radioactive decay occurs randomly, hence there is no way of predicting when a particular
nucleus will degrade. The radioactive process is not affected by temperature change, pressure
change or chemical change.
Uses of Radioactivity
Cancer cells can be destroyed by Ξ³ β radiation from high β activity source of Cobalt-60
The thickness of metal sheets can be monitored by using Ξ³ βrays and a detector. Thicker
sheets absorb more Ξ³- rays.
A small quantity of short-lived radioactive liquid can be used to detect the exact positions
of pipes and leaks.
Radioactive dating (example Carbon - 14 dating)
Radioisotopes can be used to monitor different abilities in both plants and humans.
Operations of Simple Detectors
(See Muncaster AβLevel Physics Page 831 β 836 β Detectors of Radiation)
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Worked Example 1
Worked Example 2
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