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ELEMENTARY MODERN PHYSICS (PHY 201) LECTURE NOTE
PREPARE BY
M.O.OBENDE
DEPARTMENT OF PHYSICAL AND MATHEMATICAL SCIENCES
COLLEGE OF SCIENCES
AFE BABALOLA UNIVERSITY, ADO EKITI (ABUAD)
TABLE OF CONTENT
 REVIEW OF MODERN PHYSICS
 THE ORIGIN OF QUANTUM THEORY
 BOHR’S THEORY OF ATOMIC STRUCTURE
 REVIEW OF SCHRODINGER’S WAVE EQUATION AND APPLICATIONS
 INTRODUCTION TO THE NUCLEUS
 WAVES AND PARTICLES
 ELECTRON AND QUANTA – THE FREE ELECTRON
 X-RAYS
 SPECIAL THEORY OF RELATIVITY
CHAPTER ONE:
REVIEW OF MODERN PHYSICS
1.1: DEFINITION OF MODERN PHYSICS
Gunion and Davis reported that the study of modern physics is the study of the enormous
revolution in our view of the physical universe that began just prior to 1900. At that time,
most physicists believed that everything in physics was completely understood. Normal
intuition and all experiments fit into the context of two basic theories:
1.
Newtonian mechanics for massive bodies
2.
Maxwell’s theory for light (electromagnetic radiation).
Consistency of the two required that there be a propagating medium (and, therefore, a
preferred reference frame) for light.
However, even a little thought made it clear that there was trouble on the horizon and then
came many new experimental results that made clear that the then-existing theoretical
framework was woefully inadequate describe nature.
In a relatively short period of time, physicists were compelled to adopt:
i.
The theory of special relativity based on the idea that there was no propagating
medium for light (so that light travelled with the same speed regardless of the “frame”
from which the light was viewed).
ii.
The theory of quantum mechanics, according to which the precise position and precise
momentum of a particle cannot both be determined simultaneously.
In fact, one must think of particle not as particles, but as waves, much like light.
iii.
At the same time, experiments made it clear that light comes in little quantum particles
like packets called photons.
iv.
In short, both particles and light have both a particle-like and wave-like nature.
1.2: REVIEW OF CLASSICAL MECHANICS
Bums (2012) reported that before developing the transformation equations of classical
relativity, it will prove prudent to review of few of the fundamental principles and defining
equations of classical mechanics, we are primarily concerned with the motion and path of a
particle represented as a mathematical point. The motion of the particle is normally described
by the position of its representative point in space as a function of time, relative to some
chosen reference frame or coordinate system. Using the usual Cartesian coordinate system,
the position of a particle at time‘t’ in three dimensions is described by its displacement
vector ‘r’;
r = xi +yj + zk
(1)
Thus, equation (1) above to the origin of coordinates.
Assuming we know the spatial coordinates as a function of time,
x = x(t), y = y(t), z = z(t)
(2)
Then, the instantaneous translational velocity of the particles is defined by
V = dr/dt
(3)
With the fundamental units of m/s in the SI system of units. Thus, the three-dimensional
velocity vector can be expressed in terms of its rectangular components as:
V = υxi + υyj + υzk
---------------------------------------(4)
Where the components of velocity are defined by:
(5a)
(5b)
(5c)
Although these equations for the instantaneous translational components of velocity will be
utilized in Einsteinian relativity, the defining equations for average translational velocity and
its components, given by:
---------------------------(6)
= axi + ayj + azk
(7)
Having components given by;
(7a)
(7b)
(7a)
Likewise, average translational acceleration is defined by;
(8)
With Cartesian components
(9a)
(9b)
(9c)
The basic units of acceleration in the SI system are m/s 2, which should be obvious from
second equality in equation (7)
(10a)
(10b)
(10c)
(10d)
Will be primarily used in the derivations of classical relativity, as is customary, the Greek
letter ‘delta’ ( ) in these equations is used to denote the change in a quantity. For example,
x=x2-x1 indicates the displacement of the particle along the x-axis from its initial position
x1 to its final position x2.
To continue with our review of kinematics, recall that the definition of acceleration is the
time rate of change of velocity. Thus, instantaneous translational acceleration can be defined
mathematically by the equation.
CHAPTER TWO
2.0: THE ORIGIN OF QUANTUM THEORY
2.1:
BLACK BODY RADIATION
2.1.1: INTRODUCTION
The first indication of inadequacy of classical ideas to explain the properties of matter occur
in black body radiation.
2.1.2: DEFINATION OF BLACKBODY
a.
A black body radiation is an ideal body, which allows the whole of the incident
radiation to pass through itself (without reflecting the energy) and absorbs within itself
the whole of the incident radiation (without passing on the energy).
In other words, black body radiation is an ideal absorber of incident radiation.
b.
A black body is a surface that completely absorbs all incident radiation and emits
radiation at the maximum possible monochromatic intensity in all directions and at all
wave lengths.
In general, the theory of energy distribution of black body radiation was developed by
Planck theory and first appear in 1901. Planck postulated that , energy can be absorbed or
emitted in discrete units of photons with energy :
E =hv
....................................................................... (1)
Where; E = energy
h = Planck's constant
v = frequency
E = ħw
ħ = reduced Planck's constant
w = angular velocity
.......................................................... (2)
However Planck showed that the intensity of radiation emitted by a black body is given by:
Where c1 and c2 are constants
i.e.
c1 = 2πhc2 = 3.74
c2 =
= 1.44
10-6Wm-2
and
10-2 mk
c = speed of light
k = Boltzmann constant
2.2: WIEN'S LAW/WIEN'S DISPLACEMENT LAW
Wien’s displacement law states that the monochromatic wavelength (λm) is inversely
proportional to the absolute temperature (T) of the body. Mathematically;
…………………………………(i)
…………………………………(ii)
Where b is the constant of proportionality
b = 2.898
10-3mk (experimentally)
…………………………………(iii)
Therefore;
Equation (iii) is known as mathematical expression of Wien's displacement law.
CLASSWORK
Use mathematical expression of Wien's displacement law to compute the "colour"
temperature of the sun ( take
m
= 0.475 m, b = 2.898 103mk)
SOLUTION
Given
λm = 0.475 10-6m,
b = 2.898 103mk
6.10105 x 109k
,T=?
= 6.1x109k
T
= 6.1 X 109K
2.2.1: APPLICATION OF WIEN'S DISPLACEMENT LAW
a.
STEFAN - BOLTZMANN LAW
Stefan - Boltzmann law states that the blackbody flux density (f) obtained by integrating the
Planck function Bλ over all wave lengths is given by ;
F
b.
T4
.............................................. (1)
F=
T4 (where
i.e.
= 5.67 10-8 wm-2k-4
is the constant of proportionality)
KIRCHHOFF'S LAW
Kirchhoff's law states that a body which is a good absorber of energy at a particular
wavelength is also a good emitter at that wavelength. In order to understand this law
properly, let us consider the two building blocks that validate the above law and they are;
Emissivity and Absorptivity.
i.
EMISSIVITY (EΛ )
Emissivity is the ratio of the monochromatic intensity of the radiation emitted by the
body to the corresponding black body radiation.
Mathematically;
Eλ =
Iλ
............................ (i)
B λ(T)
ii.
ABSORPTIVITY (AΛ)
Absorptivity is the fraction of incident monochromatic intensity that is absorbed.
Mathematically;
A λ= I λ (absorbed)
Tλ
..............................(ii)
(Incident)
2.3: RAYLEIGH-JEANS THEORY/FORMULA.
Rayleigh-Jeans formula for the distribution of energy in the blackbody spectrum is based on
the principle of equipartition of energy for all the possible modes of free vibration which
might be assigned to radiation (i.e. per mode of vibration) as;
T
= KT
.................................(i) (classical result)
The number of independent standing waves per unit volume in a cavity in the frequency
range, v and v + dv is given by;
v2 dv
.........................(ii)
C3
The energy U (v) per unit volume in the cavity within frequency range v and v + dv is;
G (v) dv =
u(v)dv =
i.e.
……....................(*)
G (v) dv
= KT, G (v) dv
=
v2 dv
C
3
v2
….....................(iii)
C
2
3 dv
u(v)dv = v KT
..........................(iv)
C
3
In general, Rayleigh -Jeans law agrees
well with the experimental results at low frequencies
u(v)dv = KT
but near the maximum in the spectrum and at higher frequencies, it is in violent
disagreement. According to this law, the energy density will continuously increase with
increase in frequency.
2.4: PLANCK RADIATION FORMULAR
We shall study the application of Bose - Einstein (B - E) distribution to the photons and
obtain Planck radiation formula. Consider a box containing electromagnetic radiation (i.e.
photons) at a temperature T, these photons would absorb and re - emitted by the container
walls. Under thermal equilibrium, the energy distribution can be determined by the Bose Einstein's statistics.
However, there is an important difference between the photon and other particles namely; the
photon number is not conserved. This is because the walls can absorb a photon energy (E)
and emit two photons of energy E1 and E2, such that E1 and E2 = E. Thus we have only the
energy restriction and no restriction on the photon number.
Therefore, the B – E distribution function is:
………………………..(1)
The non- conservation of photons means that
=0
………………………..(2)
Thus, the number of independent standing waves per unit in the frequency interval from “v”
to “v + dv” is
………………………..(3)
The energy of radiation in the frequency range “v” to “v + dv” is
u(v)dv = hv G(v)f(v)dv
u(v)dv =
Equation (4) is know as Planck Radiation Formula
………………………..(4)
CHAPTER THREE: BOHR’S THEORY OF ATOMIC STRUCTURE
3.1: BOHR MODEL OF ATOM
3.1.1: INTRODUCTION
Hydrogen, the simplest of all element was investigated most extensively both experimentally
and theoretically. As long ago as 1885, Balmer succeeded in obtaining a simple relationship
among the wave numbers of the lines in the visible region of the hydrogen spectrum. The
first quantitative correct derivation of the Balmer formula on the basis of an atomic model
was given by Bohr (1913), in his theory of the hydrogen atom. This theory has played such
an important role in the development of atomic physics that even though it has been
modified and extended by the later developments in quantum mechanics, it will be
worthwhile to present the original simplified theory.
In 1913, Niel Bohr proposed a model for the hydrogen atom which retained the earlier
nuclear model of Rutherford but made further stipulations as to the behavior of the electron.
A dramatic explanation of the Rydberg spectral expression resulted. Many modern ideas
about atomic and molecular structure stem from this model.
Bohr suggested the following postulate to explain the electron motion in an atom and the
observed spectral lines:
i.
An electron in an atom moves in a circular orbit around the nucleus which undergoes
the influence of Coulomb force of attraction between the electron and nucleus. The
Columbian force of attraction is balanced by Newtonian Centrifugal force. Thus, we
have that:
(Ze)(e)
ii.
/4π£or2 = mv2/r
(i)
An electron cannot revolve round the nucleus in all possible orbits as suggested by the
classical theory. It can revolve only in a few widely separated permitted orbits. While
moving along these orbits round the nucleus, the electron does not radiate energy.
These non-radiating orbits are called stationary orbits.
iii.
The permissible orbits of an electron revolving round a nucleus are those for which the
angular momentum of the electron is an integral multiple of
h/
2π,
where “h” is Planck’s
constant.
Thus, for any permitted orbit,
Iw = n(h/2π)
i.e. I = mr2n,
(ii)
w = Vn/rn
(mr2n) ( Vn/rn) = n (h/2π)
(iii)
mrnVn = n(h/2π)
Where m and “vn” are the mass and velocity of the electron, “rn” the radius of the
orbits and “n” is a positive integer, called the quantum number. The above equation is
called Bohr’s quantum condition.
iv.
An atom radiates energy only when an electron jumps from a stationary orbit of higher
energy to another of lower energy. Thus, if the electron jumps from an initial orbit of
energy “Ei” to a final orbit of energy “Ef” (Ei > Ef), the frequency “v” of the radiation
emitted is given by the relationship:
hv = Ei - Ef
Where “h” is Planck’s constant. This equation is called Bohr’s frequency condition.
3.2: BOHR’S THEORY OF HYDROGEN ATOM
Let us now apply these postulates to the classical model of the atom with a nucleus of charge
(Ze). From Bohr’s first postulate, we have:
MVn2
/rn = (Ze)(e)/4π£orn2
(iv)
Cross multiply, we have that:
Mrnvn2 = Ze2/4π£o
(v)
Where z = 1 for hydrogen atom.
From Bohr’s third postulate, the angular momentum of the electron in a permitted orbit must
be an integral multiple of h/2π. From equation (iii), we have that:
Mrnvn2 = n ( h/2π)
vn = n( h/2π)(1/mrn)
(vi)
(vii)
Substitute this value of Vn in equation (vi), we get:
(Mrn)[nh/2πmrn]2 = Ze2/4π£o
/4π2mrn = Ze2/4π£o
(viii)
n2h2
(ix)
Thus, the radius of nth permissible orbit for electron in hydrogen is given by:
rn = (n2h2£o/πmZe2)
(x)
Hence, the radii of different stationary orbits are directly to the square of “n” (called
principal quantum number). Thus, velocity of electron in the stationary orbits can be
obtained by substituting this value of “rn” in equation (vii):
Vn = (nh/2πm)[πmZe2/n2h2£o]
Vn = Ze2/2nh£o
(xi)
(xii)
It is now clear that velocity of the electron is inversely proportional to the principal quantum
number “n”. Thus, electron moves at a lower speed in higher orbits and vice-versa.
The orbital frequency of an electron in the stationary orbits is calculated as follows:
wn = 2π/Tn = 2πFn = Vn/rn
Thus,
Fn = [1/2π][Ze2/2nh£o][mZe2/ n2h2£o]
Fn = mZ2e2/4£o2n3h3
(xiii)
Equation (xiii) is known as orbital frequency of an electron in stationary orbits.
3.3
TOTAL ENERGY OF ELECTRON IN THE STATIONARY ORBITS
The electron revolving round the nucleus has both potential energy (due to its position
with respect to the nucleus) and kinetic energy (due to motion). The potential energy of the
electron is considered to be zero when it is at infinite distance from the nucleus. Potential
energy of an electron in an orbit is given by the work done in taking the electron from the
distance “r” to infinity against the electrostatic attracting between the nucleus and the
electron.
This is obtained by integrating the electrostatic force of attraction between the nucleus and
the electron from the limit as to “rn”
Thus, P.E of the electron = - ∫∞rn f(r)dr
(xiv)
F(r) = Ze2/4π£or2
For anti-symmetric position,
P.E = ∫∞rn [Ze2 dr/4π£orn2]
That is,
(xv)
P.E = -Ze2/4π£orn
Similarly, K.E of the electron = 1/2mV2n.
Substituting the value of “1/2mV2n” from equation (v), we get:
i.e.
1
/2mV2n = ½ x Ze2/4π£orn = Ze2/8π£orn
K.E = Ze2/8π£orn
(xvi)
Thus, the total energy of the electron in the nth orbit is:
En = P.E + K.E
(xvii)
En = Ze2/4π£orn + Ze2/8π£orn
En = - Ze2/8π£orn
(xviii)
Substituting the value of “rn” from equation (x), we get:
En = - [Ze2/8π£orn][πmZe2/n2h2£o]
En = - [mZ2e4/8£on2h2]
(xix)
The negative sign of the energy expression shows that the electron is bound to the nucleus
and some work much be done to pull it away
3.4: CALCULATION OF rn AND En FOR HYDROGEN ATOM
Let us recall that:
rn = (n2h2£o/πmZe2)
Put:
(x)
n=1, h=6.626x10-34, £o=8.854x10-12,
π=3.142, m=9.109x10-31kg, e=1.602x10-19c
If we substitute the parameters above into equation (x), we have that:
rn =
12(6.626x10-34)2(8.854x10-12)
3.142(9.109x10-31)(1.602x10-19)2
rn =
0.529x10-10metre or 0.0529nm
Hence, the radius of the first orbit of hydrogen:
r1 = 0.529x10-10m or 0.0529nm
This is called the Bohr radius (rB)
However, if r1 = rB = 0.0529nm, n = 2 and rn =n2rB
Therefore;
r2 = 22rB = 4rB
Therefore, the energy of the electron in the nth orbit is:
En = - [mZ2e4/8£on2h2]
Put:
Z = 1, B = En =
En = -B/n2
me4
/8£o2h2,
(xx)
then equation (xx) becomes:
(xxx)
By inspection,
En = (9.109x10-31)(1.602x10-19)4
n2 8(8.854x10-12)2(6.626x10-34)2
Thus,
En = - 2.179x10-18/n2 joule
En = - 2.179x10-18
2
-19
ev
n (1.602x10 )
En = -13.6/n2 ev
(ii)
When n=1, this corresponds to the ground state energy of the atom, and is called the
ionization potential of hydrogen atom. Similarly, the energy of the atom with the electron in
the second orbit is:
E2 = -13.6/4 = -3.4ev
And that with the electron in the third orbit:
E3 = -13.6/9 = -1.511ev and so on.
3.5: BOHR’S INTERPRETATION OF HYDROGEN SPECTRUM
In general, if an electron jumps from an outer initial orbits “n 2” of higher energy to an inner
final orbit “n1” of lower energy, the frequency of the radiation emitted is given by:
Hυ = (Ei - Ef) = [-mZ2e4/8£o2h2n22] - [-mZ2e4/8£oh2n12]
(i)
Hence,
Hυ = mZ2e4/8£o2h2 [1/n12 - 1/n22]
(ii)
Divide both sides by “h”, we have that:
υ = mZ2e4/8£o2h3 [1/n12 - 1/n22]
But v = c/λ and 1/λ = v/c
v = c/λ, c/λ = mZ2e4/8£o2h3 [1/n12 - 1/n22] x 1/c2
(iii)
Therefore;
1
/λ = mZ2e4/8£o2ch3 [1/n12 - 1/n22]
(iv)
Hence, the reciprocal of the wavelength is called the “wave number”
Thus,
ύ = 1/λ = mZ2e4/8£o2ch3 [1/n12 - 1/n22]
(v)
[mZ2e4/8£o2ch3] with z=1, is called Rydberg constant for hydrogen (RH).
Thus,
RH = (9.109x10-31)(1.602x10-19)4
8(8.854x10-12)2(3x108)(6.626x10-34)3
RH = 1.096x107m-1
More elegantly, an atom is said to be excited, if the electron is raised to an orbit of higher
energy. The electron can be completely free from the influence of the nucleus by supplying
sufficient energy and the minimum energy needed for this is called ionization energy.
For hydrogen:
EI = E∞ – E1 = me4Z2/8£o2h2 [1/1- 1/∞], with z=1
Thus,
E1 = 13.6ev
3.6: SPECTRAL SERIES OF HYDROGEN
A.
LYMAN SERIES
When electron jumps from second, third, etc. orbits to the first orbit, we get the Lyman
series which lies in the ultraviolet region. Here n1 = 1, n2 = 2, 3, 4, ……….
ύ = 1/λ = RH [1/n12 - 1/n22],
Put n1 = 1, n2 = 2, we have that:
ύ1 = RH [1/12 - 1/22] = 3/4RH
Therefore;
ύ2 = 8/9RH and ύ3 = RH(15/16) and so on.
This is identified as Lyman Series
B.
BALMER SERIES
When an electron jumps from outer orbits to the second orbit (n1 = 2, n2 =3, 4, 5…),
this series is called Ballmer series and lies in the visible region of the spectrum.
i.e.
ύ = RH [1/n12 - 1/n22]
RH [1/22 - 1/32]
ύ1 = RH [9-4/36]
RH [5/36] and so on.
Thus, the first line in the series (n2=3) is called H∞ line, the second line (n2=4) is called
Hβ and so on.
C.
PACHEN SERIES
Paschen series lies in the infrared region which is due to the transition of electrons
from outer orbits to the third orbit i.e. n1= 3, n2= 4, 5, 6,……….
Thus,
ύ = RH [1/n12 - 1/n22]
ύ = RH [1/32 - 1/42] with n2= 4, 5, 6,
ύ1 = RH [1/9 – 1/16] = RH (16 – 9/44) = RH (7/144)
D.
BRACKETT SERIES
If n1= 4, n2= 5, 6, 7,………., we get Bracket series lines
ύ = RH[1/42 - 1/n22] with n2=5, 6, 7,
E.
PFUND SERIES
If n1= 5, n2= 6, 7, 8,……….,we get the pfund series lines. For this series,
ύ = RH[1/52 - 1/n22] with n2=6, 7, 8,
In general, Brackett and Pfund series lie in the far infrared region of hydrogen
spectrum. By putting n2 = (n1 + 1) in each one of the series, we get the longest
wavelength of the series. Similarly by putting “n2 = ” in each of the series, we get the
shortest wavelength of the series (or series limit).
3.7: SHORTCOMINGS OF BOHR THEORY
Bohr’s theory is based on circular electron orbits and was able to explain successfully a
number of experimentally observed facts and has correctly predicted the spectral lines of
neutral hydrogen atom and strongly ionized helium atom in terms of the principal number,
“n”. However, the theory fails to explain the following facts:
i.
It could not account the spectra of atoms more complex than hydrogen.
ii.
It fails to give any information regarding the distribution and arrangement of electrons
in atoms.
iii.
It does not explain the experimentally observed variations in intensity of the spectral
lines of an element.
iv.
It fails to explain the transitions of electrons from one level to another and also the rate
at which they occur or the selection rules which apply to them.
v.
It fails to account the structure of spectral line.
vi.
The theory cannot be used for the quantitative study of chemical bonding.
vii.
It was found that when electric or magnetic field is applied to the atom, each spectral
line is splitted into several lines. The former one is called Stark Effect while the later
as Zeeman Effect. Bohr’s theory fails to explain these effects.
3.8: ALPHA PARTICLE SCATTERING EXPERIMENT
In 1911, E. Rutherford and his student H. Geiger and E. Marsden, through a series of
scattering experiments, used the most direct way to find out what is inside an atom, by
sending fast alpha particles as probes into it.
Rutherford’s experiments considered in projecting “ ” particles into thin foils of metal; and
by observing the trajectories in which the projected particles were deflected, conclusions
were drawn about the distribution of charge within the “target” atoms. Rutherford used “∞”
particles produced in natural radioactivity. These “ ” particles are now known to be identical
with the nuclei of helium atoms, each consisting of two protons and two Newton bound
together, but without the two electrons normally present in a neutral helium atom.
3.9: RUTHERFORD’S MODEL OF ATOM
The Rutherford’s scattering experiment led to the nuclear atom model, the main features of
which are:
a.
An atom consists of a positively charged and massive core called the nucleus.
b.
The nucleus occupies a very small space (about 10-14m) as compared with the size of
the atom (about 10-10m).
c.
The nucleus is surrounded by an adequate number of electrons so that the atom as a
whole is electrically neutral.
d.
To keep the electrons in place against the electric force attracting them to the nucleus,
they revolve around the nucleus in orbits, like the planets around the sun.
3.9.1: DRAWBACKS OF RUTHERFORD ATOMIC MODEL
a.
According to electromagnetic theory, an accelerated charge must radiate energy in the
form of electromagnetic waves. In the Rutherford model, an orbiting electron
undergoes a centripetal acceleration and hence would emit electromagnetic radiation
continuously, losing energy and spirally into the nucleus. This leads to an unstable
atom. Contrary to the observed stability of atom.
b.
The model can also not explain why an atom emits only certain characteristic
frequencies of electromagnetic radiation rather than a continuous spectrum.
WORKED EXAMPLE
Find (a) the longest and (b) the shortest wavelengths in the Lyman series of the hydrogen
atom (take λ = 1.09737x107m-1).
SOLUTION
The wavelengths of the lines in the Lyman series are given by:
1
a.
/λ = RH [1/n12 - 1/n22]
The longest wavelength occurs for n2 = 2 i.e. n1 = 1.
1
/λ = (1.09737x107)[1/12 - 1/22]
λ = 1.215x10-7m.
b.
The shortest wavelength, known as a series limit, occurs as n
1
/λ = RH [1/n12 - 1/n22]
i.e. n1 = 1, n2 = ∞, R = 1.09737x107m-1
1
/λ
= RH [1/12 - 1/∞2]
= (1.09737x107)[12]
λ = 9.1x10-8m
∞
CHAPTER FOUR: REVIEW OF SCHRODINGER’S WAVE EQUATION AND
APPLICATIONS
4.1: INTRODUCTION
Schrodinger equation was formulated in late 1925, and published in 1926, by the Austrian
physicist ERWIN SCHRODINGER.
Schrodinger equation is a partial differential equation that describes how the quantum state
of physical system changes with time.
At the beginning of the twentieth century, experimental evidence suggested that atomic
particles were also wave-like in nature. For example, electrons were found to give diffraction
patterns when passed through a double slit in a similar way to light waves. Therefore, it was
reasonable to assume that a wave equation could explain the behavior of atomic particles.
Schrodinger was the first person to write down such a wave equation. Much discussion then
centered on what the equation is meant for: The eigenvalues of the wave equation were
shown to be equal to the energy levels of the quantum mechanical system, and the best test
of the equation was when it was used to solve for the energy levels of the Hydrogen atom
and the energy levels were found to be in accord with Rydberg’s law.
In general, Schrodinger’s wave equation can be define as a linear, homogenous partial
differential equation that determines the evolution with time of a quantum – mechanical
wave function.
4.2: DERIVATION
OF
ONE-DIMENSION
SCHRODINGER’S
TIME-
DEPENDENT WAVE EQUATION
Let us consider a wave function, “ψ” for a particle moving freely in the positive x-direction
which has the same form as the solution for undamped harmonic waves in the positive xdirection. i.e.
(i)
Ψ(x,t) = Aexp[-iw(t – x/v)]
i.e. w = 2πυ
Substitute the value of “w” into equation (i), we have:
(ii)
Ψ(x,t) = Aexp[-i2πυ(t – x/v)]
(iii)
Ψ = Aexp[-2πi(υt – υx/v)]
i.e. v = υλ
Substitute the value of “v” into equation (iii), we have that:
Ψ = Aexp[-2πi(υt – υx/υλ)]
(iv)
Ψ = Aexp[-2πi(υt – x/λ)]
By inspection, υ = E/h (from E=hυ) and λ = h/p (from p=h/λ). If we insert the value of “υ and
λ” into equation (iv) we have that:
Ψ = Aexp[-2πi(E/ht – x
Ψ = Aexp[Also, by inspection
h
Aexp[-2πi(Et/h – px/h)]
/p)]
(v)
(Et/h – px/h)]
= (from
)
Therefore, equation (vi) becomes:
(vi)
Ψ = Aexp[- (Et – px)]
Differentiating equation (vi) twice with respect to “x” and once with respect to “t”.
i.e.
= Aexp[- (Et – px)]( )
= Aexp[- (Et – px)](
)
(vii)
Put Ψ = Aexp[- (Et – px)], i2 = -1 (from MAT101 i.e. complex number)
Equation (vii) becomes
= Ψ(
)
(viii)
If we multiply both side of equation (viii) by “
”, we have that:
(ix)
Also,
= Aexp [- (Et – px)]( ) ( ) {Reference to equation (vi)}
(x)
Put Ψ = Aexp[- (Et – px)], thus equation (x) becomes:
= Ψ (- )
(xi)
=
If we multiply equation (xi) both side by “
”, we have that:
(xii)
= Eψ
Hence, the total energy of a particle is the same of its K.E and P.E
i.e.
E = K.E + P.E
E = 1/2mv2 + v
But,
P = mv, P2 = m2v2
E = 1/2(
E=
)+v
(xiii)
+v
If we multiply both side of equation (xiii) by wave function (ψ), we have that;
Eψ =
(xiv)
+ vψ
If we extract the statement of “Eψ” and P2ψ from equation (ix) and (xii) and substitute it into
equation (xiv), we have that:
=
=
+ vψ
+ vψ
(xv)
Multiply both side of equation (xv) by “-1”, we have that:
(xvi)
=
- vψ
Equation (xvi) is known as one dimensional Schrodinger’s time-dependent wave equation.
PROBLEM SET
Write down the generalized two and three dimensional Schrodinger’s time-dependent
wave equation
CHAPTER FIVE: INTRODUCTION TO THE NUCLEUS
5.1: INTRODUCTION
The atomic nucleus was discovered in 1911 by Rutherford. Rutherford’s “α” particle
scattering experiments showed that the atom consists of a very small nucleus (10 -14m in
diameter) surrounded by orbiting electrons. Thus, all atomic nuclei are made up of
elementary particles called protons and neutrons. A proton has a positive charge of same
magnitude as that of an electron. A neutron is electrically neutral. The proton and the neutron
are considered to be two different charge states of the same particle which is called a
nucleon. A species of nucleus, known as a nuclide, is represented schematically by ZXA
where “z”, the atomic number indicates the number of protons, “A”, the mass number
indicates that total number of proton plus neutrons and “X” is the chemical symbol of the
species.
Thus, N = Number of neutrons = A – Z
As an example, the chlorine nucleus 17C35 has Z=17 protons, A=35 nucleons and N=35-17 is
equal to 18 neutrons.
5.2: CLASSIFICATION OF NUCLEI
Atoms of different elements are classified as follows:
i.
Isotopes are nuclei with the same atomic number Z but different mass numbers, A.
The nuclei
14Si
28
,
14Si
29
,
30
14Si ,
and
14Si
32
are isotopes of silicon. The isotopes of an
element all contain the same number of neutrons since the nuclear charge is what is
ultimately responsible for the characteristics properties of an atom, all the isotopes of
an element have identical chemical behaviour and differ physically only in mass.
ii.
Those nuclei, with the same mass number A, but different atomic number Z, are called
isobars. The nuclei 8O16 and 7N16 are examples of isobars. The isobars are atoms of
different elements and have different physical and chemical properties.
iii.
Nuclei, with equal neutrons, that is, with the same N, are called isotones. Some
isotones are 6C14, 7N15 and 8O16 (N= 8 in each case).
iv.
There are atoms, which have the same Z and same A, but differ from one another in
their nuclear energy states and exhibit difference in their internal structure. These
nuclei are distinguished by their different life time such nuclei are called isomeric
nuclei or isomers.
v.
Nuclei, having the same mass number A, but with the proton and neutron number
interchanged (that is, the number of protons in one is equal to the number of neutrons
in the other) are called mirror nuclei.
Example: 4Be7 (Z=4 and N=3) and 3Li7 (Z=3 and N=4).
5.3: GENERAL PROPERTIES OF NUCLEUS
a.
NUCLEAR SIZE:
Rutherford’s work on the scattering of “α” particles showed that the mean radius of an
atomic nucleus is of the order of 10-14 to 10-15m while that of the atom is about 10-10m.
Thus, the nucleus is about 10000 times smaller in radius than the atom.
The empirical formula for the nuclear radius is:
R= ro A1/3
Where “A” is the mass number and ro= 1.3 x 10-15m = 1.3fm. Nuclei are so small that
the Fermi (fm) is an appropriate unit of length. 1fm = 10-15m
Example: The radius of Ho165 is 7.731 Fermi. Deduce the radius of He4.
SOLUTION
Therefore, R1 = A11/3
R2 = A21/3
R2 = R1 A21/3
A11/3
b.
NUCLEAR MASS:
OR
=
7.731 x 41/3
(165)1/3
= 2.238fm
We know that the nucleus consist of protons and neutrons. Then the mass of the
nucleus should be: (assumed nuclear mass = zmp + Nmn, where mp and mn are the
respective proton and neutron masses and “N” is the neutron number, Nuclear masses
are experimentally measured accurately by mass spectrometers. Measurements by
mass spectrometer, however show that: (real nuclear mass < zmp + Nmn.
The difference in masses: (zmp + Nmn – real nuclear mass =∆m) is called the mass
defect.
c.
NUCLEAR DENSITY:
The nuclear density (ρN) can be calculated from:
ρN = Nuclear mass
Nuclear volume
i.e. Nuclear mass = AmN, where A = mass number and mN = mass of the nucleon =
1.67 x 10-27kg.
d.
/3πR3 = 4/3π (ro A1/3) =
4
ρN
=
AmN
4
/3πro3A
ρN
=
1.815 x 1017kgm-3 (Approximately)
=
mN
4
/3πro3
=
4
/3π ro3A
=
Nuclear volume
1.67 x 10-27
4
/3π(1.3 x 10-15)3
NUCLEAR CHARGE:
The charge of the nucleus is due to the protons contained in it. Each proton has a
positive charge of 1.6 x 10-19C. The nuclear charge is “Ze” where “Z” is the atomic
number of the nucleus. The value of “Z” from x-ray scattering experiments, from the
nuclear scattering of “α” particles, and from the x-ray spectrum.
e.
SPIN ANGULAR MOMENTUM:
Both the proton and neutron, like the electron, have an intrinsic spin. The spin angular
momentum is computed by Ls = √L (L+1) h/2π, where the quantum number L,
commonly called the spin is equal to 1/2. The spin angular momentum, then has a value
Ls = √3/2 h/2π
f.
RESULTANT ANGULAR MOMENTUM:
In addition to the spin angular momentum, the protons and neutrons in the nucleus
have an orbital angular momentum of the nucleus is obtained by adding the spin and
orbital angular momenta of all the nucleons within the nucleus. The total angular
momentum of a nucleus is given by:
LN = √L(L+1) h/2π. This total angular momentum is called nuclear spin.
5.4: BINDING ENERGY
The theoretical explanation for the mass defect is based on Einstein’s Equation, E=mc 2.
When the z protons and N neutrons combine to make a nucleus, some of the mass (∆m)
disappears because it is converted into an amount of energy ∆m=(∆m)c 2. This is called the
Binding energy (B.E) of the nucleus.
Mathematically,
B.E = [(Zmp + Nmn) – M]c2
ILLUSTRATION
Let us illustrate the calculation of B.E by taking the example of the deuteron. A deuteron is
formed by a proton and a neutron.
Mass of proton = 1.007276u.
Mass of neutron = 1.008665u.
Therefore;
Mass of proton + neutron in Free State = 2.0159414u.
Mass of deuteron nucleus = 2.013553u.
Mass of defect = ∆m = 0.002388u
B.E = 0.002388 x 931
B.E = 2.23 MeV (1u = 931ev)
CHAPTER 6
6.0: WAVES AND PARTICLES
6.1: INTRODUCTION
The Bohr's theory of hydrogen atom which we treated before involved the use of the concept
of emission and absorption of radiation in whole quanta. This was really an extension of
Planck's hypothesis. In 1901, Planck in proposing a successive theory of black body
radiation introduced a revolutionary assumption regarding the way in which radiation was
emitted discontinuously as little bursts of energy. The unit energy hf is called quantum
energy of a photom: in referring to electromagnetic radiation of frequency F.
6.2: DEFINATION OF PHOTOELECTRIC EFFECT
Photoelectric effect is a phenomenon in which electrons are librated from the surface of
material when radiation of stable frequency is incident upon it.
6.3: FAILURES OF THE WAVE THEORIES OF LIGHT IN EXPLAINING
PHOTOELECTRIC EFFECT
Some of the important features of photoelectric effect which the wave theory of light fails to
explain are;
1.
It fails to account for threshold frequency at low intensity and light frequency above
threshold frequency. For each metal, there exist a threshold frequency V˳ of the
incident light, below which no electrons are ejected, no matter how intense the light,
or for how long it falls on the surface. Light frequency above the threshold eject
electrons, no matter how low the intensity of the light.
2.
It fail to account for dim light of the same frequency irrespective of the maximum
kinetic energy bright light yields more photoelectrons than a dim one of the same
frequency, but the maximum kinetic energy of the photoelectrons is independent of the
light intensity.
3.
It fails to account for linearity behavior of light frequency at high kinetic energy. The
maximum kinetic energy of the photoelectrons increases linearly with increasing light
frequency.
4.
It fails to account for detectable time delay of photoelectrons when ejected on the
surface. Photoelectrons are ejected from the surface almost instantaneously, with no
detectable time decay (less than 10 -9s after the surface is illuminated), no matter how
weak the light source, if the frequency is above the threshold.
6.4: EINSTEIN'S PHOTOELECTRIC EQUATION -PARTICLE NATURE OF
LIGHT
In 1909, Einstein proposed a photoelectric effect using a concept first put forward by MaxPlanck that light- wave consist of tiny bundles or packets of energy known as photons or
quanta. Each photon of light (or electromagnetic wave) has an energy “hf”, where h is
Planck's constant (= 6.62 10-34Js) and “f” is the frequency of the light.
Einstein's idea can be expressed as;
hf = W + 1/2mv2
..............................................(i)
Where; hf= Total energy
W= Work function
m= Mass of electron
v= Velocity of the photoelectron
From equation (1), if w =hfo,
Ek =1/2mv2, then;
hf = hfo +Ek
………………........................(ii)
Ek = hf – hfo
……...................................... (iii)
E= h(f – fo) or 1/2mv2 = h(f – fo)
………................................ (iv)
Equation (iv) is known as Einstein photoelectric equation.
In terms of potential differences, Einstein photoelectric equation can be written as ;
ev = hf – hfo
………………….....................(v)
…………………,,,...................(vi)
WORKED EXAMPLE
1.
Calculate the energy in Joules of ultraviolet light of wavelength
3ₓ10 -7m. Take the
velocity of light as 3×108ms-1 and Planck's constant as 6.6 × 10-34Js
2.
Compute the frequency of the photon whose energy is required to eject a surface
electron with a kinetic energy of 3.5 ×10-16ev. Take h =6.6 × 10-34Js, 1 ev =1.6 × 1019
J.
SOLUTION
1.
Given that;
E(J) = ?, h = 6.6 × 10-34Js, λ = 3 10-7m, c = 3 108ms-1 ,
Recall that
……………………………….(i)
E = hf =
E = 6.6 × 10-34 =
E = 6.6 × 10-19J
(2) Given that;
F =?, K.E = 3.5
10-16ev , W = 3.0
10-16ev , h = 6.6 × 10-34Js
Recall that;
E = hf – W
…………………………….(i)
…………………………….(ii)
hf = E + W
…………………………….(iii)
f = 0.157Hz
or
1.57x10-1Hz
6.5: COMPTION EFFECT
The Compton Effect further confirmed the particular nature of radiation when x-rays were
scattered by a target with loosely bound electrons e.g. carbon, the scattered radiation was
found to consist of two compounds. One having the same wavelength as the incident beam
(unmodified line).
According to the classical electromagnetic theory, there should be no change in wave length
or frequency. The incident radiation was expected to set the atomic electron vibrating with
the frequency of the incident radiation, and then produce radiation emitted in all directions
with the same frequency (scattered radiation). Compton in 1923 provided the explanation to
the observed effects by treating the incident radiation as a stream of individual photons each
of which could interact with single electron .This is the Compton Effect.
If
the
frequency
of
the
incident
photon
is
F,
the
momentum
is
hf
/c (i.e. from p = E/C = hf/C = h/λ). The photon strikes a relatively free electron of mass in
which recoils with momentum, p =
through angle
, while the incident x- ray photon is scattered
with new energy hf' , the corresponding momentum being
shown in the figure below;
hf’
/C. This is
P
E
Recoil electron
hf
hf
m
/C
hf
hf
/C
FIG 1(A)
Scattered x-rays
hf
/C
P
hf
/C
FIG 1(B)
Figure 1:
(a) The Compton Effect
(b) Momentum triangle for the Compton Effect
In general, the interaction is treated as a simple collision problem in mechanics in which case
initial momentum vector hf/C of the x-ray is equal to the two p and
hf'
/C as shown in figure
1(b). From energy conservation;
hf + mc2= hf' + E
……………………………………(i)
E=
Therefore; h (f -f’) + mc2 =
……………………………………(ii)
Dividing both sides by mc2, squaring and subtracting 1,
………………………(iii)
Multiplying both sides by m2c4,
…..……………………(iv)
……………………….………………(v)
From the vector triangle, figure (1b)
p2c2 = (hf') 2 + (hf) 2 - 2h2 ff’ cos
…………………..……………(vi)
(hf') 2 + (hf) 2 - 2h2 ff’ cos …………………..……………(vii)
Comparing the equation (v) and (vii), and simplifying, we get;
………………………..……………(viii)
mc2(f -f') = hff' (1- cos )
If this is written in terms of wavelength, where λ = c/f and λ' = c, we finally obtain
λ' - λ =
……………………………………..(ix)
Equation (ix) is the Compton formula which conforms with experimental observation.
6.6: WAVE - PARTICLE DUALITY
The wave theory of light is successfully in explaining the variation of the velocity of light
with medium,
Intensity
0
M
λ
FIGURE 2:
λ
Intensity against wave length of scattered x-rays.
As well as interference and diffraction effects. These effects give a picture in which there are
alternate regions of brightness (maximum) and darkness (minimum). Thus, it has become a
fundamental part of present day physical theory that the two characters (wave and particle
duality. According to the De Broglie hypothesis (1924), this dual character applies not only
to radiation but to all fundamental entities of physics
On this hypothesis, electrons, protons, neutrons, atoms, and molecules should have some
type of wave motion associated with them. The wave length of the motion is given by;
λ = h/p
Where;
…………………………………(i)
λ = wavelength
h = Planck's constant
p = momentum.
6.7: UNCERTAINITY PRINCIPLE
According to the Heisenberg uncertainty principle, it is impossible to know (or measure)
precisely and simultaneously both the momentum and position of the particle.
Mathematically, the principle is represented as;
Δx . Δp
Where;
h
…………………………………(ii)
/2π
Δx = uncertainty in position
Δp= uncertainty in momentum
WORKED EXAMPLE
Find the uncertainty in the position of an electron that has an uncertainty in its speed of
105ms-1. Take h= 6.6 10-34Js, m= 7.31 10-3kg)
SOLUTION
According to the uncertainty principle
Δx . Δp
i.e.
h
/2π
Δp = mΔv
Therefore,
Δx =
Δx = 1.1x10-9m
CHAPTERSEVEN
7.0: ELECTRON AND QUANTA – THE FREE ELECTRON
7.1: INTRODUCTION
The emission of electron from the surface of metal can be carried out in several ways and
they are:
a.
By heating a metal
b.
By irradiating a metal surface with electromagnetic radiation of appropriate frequency.
c.
By imposing an intense electric field at the surface of metal.
d.
By bombarding a metal surface with primary electron.
7.2: EMISSION OF ELECTRON – RICHARDSON’S HYPOTHESIS
According to Max. Planck in 1902, energies of a said body i.e. metal that emit electron is a
function of frequency of light illuminated on the body.
i.e.
Eαf
(i)
E = hf
(ii)
In general, a scientist whose name is Richardson O.W step into the work of Max Planck and
deduce the following observations and they are:
i.
Current density of a material varies with absolute temperature and work function of
the filament.
ii.
Each material used during absorption of electron on the surface of metal must be
define appropriately.
iii.
Absolute temperature must be noted during the emission of electrons.
iv.
Minimum energy require for the liberation of electron from the surface of each object
must be considered.
7.2.1 RICHARDSON’S THERMODYNAMIC EQUATION
Richardson’s express the equation in terms of thermodynamic consideration as:
i = AT2exp (- ϕ/KT)
N.B:
i = Current density
A = Richardson material constant value
D = work function of the filament
T = Absolute temperature
K = Boltzmann constant
PROBLEM SET
If the work function require to liberate a tungsten material surface is 4.3 x 10 -23J/s and its
absolute temperature is 4K. Find the current density illuminated on the surface of the
material (Take K = 1.38 x 10-23J/k, A = 4.5460).
(Answer: i = 33.8A/m2).
CHAPTER EIGHT
8.0: X-RAYS
8.1: INTRODUCTION
X-rays have been identified as electromagnetic radiations of very high frequency. X-rays
have very short wavelengths in the order of “10 -10m”, indeed much shorter than ordinary
light waves. On the electromagnetic chart, x-rays and -rays overlap.
The main difference between x-rays and Y-rays is in their production mechanism. X-rays are
produced in x-ray tubes by bombarding certain materials, e.g. copper or tungsten, with
electron beams but Y-rays are produced during radioactive decay.
8.2: PRODUCTIONS OF X-RAYS
Briefly, x-rays are produced when an electron beam bombards certain materials at high
velocity. The rapid deceleration of the electrons generates electromagnetic forces which
produce the radiation. The production of x-rays is therefore an inverse process to the
photoelectric effect.
Figure “1” shows the apparatus used in production of x-rays. The apparatus is called an x-ray
tube.
In the x-ray tube, a high potential difference is applied between the hot cathode and the
anode. Electrons are accelerated to an extremely high speed. The sudden deceleration, as
they strike the anode, causes the emission of high energy radiation of short-wavelength.
The rays can be detected by fluorescent screens which glow on being hit by x-rays.
8.3: PROPERTIES OF X-RAYS
i.
X-rays are not deflected by electric and magnetic fields. This indicates that x-rays are
not charged particles but electromagnetic radiations.
ii.
X-rays have great penetrating power. X-rays pass through substances which are
opaque to light. X-rays pass through substances which are opaque to light. X-rays pass
through flesh, bones, metals and indeed through almost all known substances. X-rays
with very short wavelength or very high frequency are referred to as hard x-rays. Xrays with low penetrating power or low frequency are known as soft x-rays. Soft xrays are produced with low voltage.
iii.
When x-rays pass through a gas, they ionize the atoms of the gas by ejecting the
electrons in the gas.
iv.
Using ionization chamber to investigate the intensity of x-rays, it has been found that,
increasing the filament current in an x-ray tube, increases the intensity of x-rays.
v.
If x-rays are directed at a block of lead, or very dense materials, the temperature of the
materials absorb x-rays.
8.4: USES OF X-RAYS

X-rays are used in examining the human body for broken bones.

X-rays are used in mass radiography for the investigation of tuberculosis within

Barium salt, being good absorber of x-rays, improves the quality of the photography.

Hard x-rays are used in the treatment of cancer and other malignant growths in the
human body.

X-rays are used in examining baggage for hidden weapons.

Defects and fractures in metals are detected with x-rays.

X-rays are very important tools in the investigation of crystal structures.
8.5: HEALTH HAZARDS OF X-RAYS AND SAFETY PRECAUTION
X-rays may cause damage to living tissues and bring about the death of small organisms. Xrays can cause damage to cells of organisms including the chromosomes in the nucleus of the
cells.
Unnecessary exposure to x-rays should be avoided. Adequate and well tested safety
precautions must be observed whenever an x-ray apparatus is in use.