Chapter 11 Atomic Physics - Universiti Tunku Abdul …staff.utar.edu.my/limsk/Physics/11 Atomic...
Transcript of Chapter 11 Atomic Physics - Universiti Tunku Abdul …staff.utar.edu.my/limsk/Physics/11 Atomic...
- 263 -
Chapter 11
Atomic Physics
_____________________________________________
11.0 Introduction
Early in twentieth century many prominent scientists doubted about the
existence atom needless to say how the electrons in an atom are arranged, what
are their motions, how atoms emit or absorb light, or even atoms are stable.
Today we can even pick up individual atoms and move them around.
In 1926 all these questions about existence of atom were answered with the
development of quantum physics. Its basic premise is that moving electrons,
protons, and particles of any kind are best viewed as matter waves, whose
motions are governed by Schrödinger’s wave equation. Although quantum
theory also applies to massive particles, there is no point to treat an automotive,
planet etc. as object with quantum theory. For such massive, slow-moving
objects, Newtonian physics and quantum physics yield same answer.
11.1 Some Properties of Atoms
Atoms are stable. Essentially all the atoms that form our tangible worlds have
existed without change for billions of years. Otherwise, we do not know what
the world is like if there is continuous changed into other form.
Atoms combine with each other to form stable molecules and stack up to
form rigid solids. An atom is mostly empty space but we can stand on the floor,
which made up of atoms without falling through it.
Atoms are put together systematically according to their chemical and
physical properties in the periodic table. Figure 11.1 shows examples of
repetitive properties of the element as a function of their positions in periodic
table. The figure also shows the plot of ionization energy of the element, which
is defined as the energy required to remove the most loosely bounded electron
from neutral atom is plotted as a function of the position in periodic table. It
shows remarkable similarity in chemical and physical properties of the elements
in each column show evidences that atoms are constructed according to
systematic rules.
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The elements are arranged in the period table in six horizontal periods.
Except for the first period, each period starts from the left with highly reactive
alkaline metal like lithium, sodium and etc and end with chemical inert noble
gas like neon, argon and etc. Quantum physics accounts for the chemical
properties of these elements. The numbers of elements in six horizontal periods
are 2, 8, 8, 18, 18, and 32.
Figure 11.1: A plot of ionization energies of the elements a function atomic number showing
periodic repetition of properties through six complete horizontal period of the
periodic table
Atoms emit and absorb light. Atom can exist only in discrete quantum states.
Each state has certain energy. An atom makes a transition from one state to
another state by emitting light or absorbing light. The light emitted or absorbed
as photon with energy governed by equation (11.1).
hv = Ehigh – Elow (11.1)
Atoms have angular momentum and magnetism. Figure 9.2 illustrates a
negative charged particle moving in circular orbit around a fixed center. The
orbiting particle has angular momentum L since the path is equivalent to tiny
current loop and magnetic dipole μ . They are perpendicular to the plane but in
opposite direction because the charge is negative.
Model shows in Fig. 11.2 is strictly the classical model. It does not
accurately represent an electron in atom. In quantum physics, the rigid orbit
model has been replaced by the probability model best visualized as a dot plot.
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Figure 11.2: A classical illustration of charged particle has angular momentum and magnetic
dipole moment
In 1915 well before the discovery of quantum physic, Albert Einstein and Dutch
physicist Wander Johannes de Haas carried out an experiment designed to show
that angular momentum and magnetic moment of individual atoms are coupled.
An iron cylinder is suspended using a thin fiber. Initially the magnetic dipole is
random position. Once the current is switched the solenoid around the iron
cylinder, the magnetic field B is created parallel to the axis of cylinder that
causes the magnetic dipole μ of iron aligned along the magnetic field. This
would cause a net angular momentum netL . To maintain zero angular momentum,
the cylinder begins to rotate around its axis to produce an angular
momentum rotL .
(a) (b)
Figure 11.3: Einstein-de Haas Experiment: (a) Initially no magnetic field and random
magnetic dipole and (b) with magnetic field, magnetic dipole lined up parallel
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11.2 Quantization of Angular Momentum
Every quantum state of an electron in an atom has an associated orbital angular
momentum and a corresponding orbital magnetic dipole moment.
The magnitude L of the orbital angular momentum L of an electron in an
atom is quantized, which can have a certain values. These values are governed
by equation (11.2).
)1( llL where l = 0, 1, 2, 3, …… n-1 (11.2)
in which l is the orbital quantum number and is the reduced Planck’s constant,
which is 2
h. l must be a value either zero or a positive value integral no greater
than (n-1), where n is the principal quantum number. Thus, for n = 4, l can have
0, 1, 2, and 3.
The components Lz of the angular momentum are also quantized and they
are given by
lL mz where ml = 0, 1, 2, ….. l (11.3)
Figure 11.4 shows the five quantized components Lz of the orbital angular
momentum for electron with orbital quantum number l = 2. These quantized
components correspond to magnetic quantum number ml=-2, m1=-1, m1= 0,
m1=1, and m1=2.
For every orbital angular momentum vector L shown in Fig. 11.4, there is
a vector pointing in the opposite direction representing the magnitude and
direction of the orbital magnetic dipole momentumorb
μ .
From Fig. 11.4, the minimum angle L between L and Lz is given by
6
2cos 1
z = 35.260 (11.4)
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Figure 11.4: The allowed values of Lz for an electron in a quantum state with orbital
quantum number l = 2
11.3 Magnetic Dipole Moment and Zeeman Effect
A magnetic dipole is associated with the orbital angular momentum L of an
electron. The magnetic dipole has an orbital magnetic dipole momentorb
μ , which
is related with angular momentum by equation (11.5).
Lq
μe
orbm2
(11.5)
The minus sign indicates that orb
μ is directed opposite L . L
μorb is called the
gyromagnetic ratio.
Owing to magnitude of L is quantized, the magnitude of orbμ is also
quantized following equation (11.6).
)1(m2 e
orb llq
μ (11.6)
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L and orb
μ cannot be measured directly. However, the components of them can
be measured along a given axis such as z-axis. The component z,orbμ of the
orbital magnetic dipole moment are quantized and given by
Bz,orb m μμ l (11.7)
ml is the orbital quantum number and B is the Bohr magneton, which is defined
as
T/J10x274.9m4
24
e
B
qh
μ (11.8)
me is the mass of electron.
The orbital angular moment and magnetic moment vector are opposite.
Thus, from equation (11.5), the z-component orbital magnetic dipole moment is
z
e
zm2
Lq
μ (11.9)
Substituting equation (11.3) into equation (11.9), one gets lL mz
e
zm2
mq
μ l (11.10)
Zeeman effect is the splitting of atomic energy levels and associated spectrum
lines when the atom is placed in magnetic field. An illustration is shown in Fig.
11.5.
(a) (b)
Figure 11.5: Illustration of Zeeman effect (a) with magnetic field and (b) with magnetic field
The potential energy V associated with the magnetic dipole orb is
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BV orb μ (11.11)
The interaction energy of the z-component of magnetic dipole is
Bm2
BVe
lz q
m where ml = 0, 1, 2, 3, …l (11.12)
In terms of Bohr magneton, which ise
Bm4
qh
μ , the interaction energy V is
BmV Bμl (11.13)
11.4 Electron Spin
Electron has intrinsic spin angular momentum S often called spin. Meaning it is
a basic characteristic of electron. The magnitude of S is quantized and depends
on a spin quantum number s, which is always 2
1 for electron, proton and
neutron. In addition the component S measured along any axis is quantized and
depends on the spin magnetic quantum number ms, which can have value 2
1 or
2
1 .
The magnitude S of the spin angular momentum S of any electron, whether
it is free or trapped, has the single value given by
)1( ssS (11.14)
where s is the spin quantum number that has value 2
1. Thus, the magnitude S of
the spin momentum S of any electron is )1( ssS 2
3 .
An electron has an intrinsic magnetic dipole, which associated with its spin
angular momentum S , whether the electron is confined to an atom or free. The
magnetic dipole has a spin magnetic dipole moments
μ , which follow equation
(11.15).
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Sq
μe
sm
(11.15)
The minus sign mean that s
μ is directly opposite of S . The z-component of the
spin magnetic dipole moment is equal to z
e
zm2
00232.2 Sq
μ , with a small
correction factor.
The magnitude of s
μ is also quantized following equation (11.16).
)1(me
s ss
qμ (11.16)
The components of spin angular momentum S and spin magnetic dipole
moment s
μ along any axis can be measured such as along z-axis. Thus, the
components Sz of the spin angular momentum are quantized and given by
sS mz (11.17)
where ms is the spin magnetic quantum number that can have two values either
2
1 for spin up or
2
1 for spin down i.e.
2
1SZ . The components z,sμ of the
spin magnetic dipole moment are also quantized and they are given by
BBz,s m2 μμμ s (11.18)
Figure 11.6 shows the two quantized components Sz of the spin angular
momentum for an electron and associated orientation for vector S . It also shows
the quantized components z,sμ of the spin magnetic dipole moment and the
associated orientation of sμ .
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Figure 11.6: The allowed values of Sz and and z of an electron
Example 11.1
Calculate the interaction energy for an electron in l = 0 state with no orbital
magnetic moment in magnetic field with magnitude of 2.00T and sketch a graph
to show the splitting of energy levels.
Solution
The interaction energy is equal to BV zμ ; z
e
zm2
00232.2 Sq
μ , and 2
1SZ .
Thus, the interaction energy is
B2
1
m200232.2V
e
q B
m200116.1
e
q.
em2
qis equal to Bohr
magneton.
The interaction energy is then equal to B00116.1V Bμ .
Substitute Bohr magneton value and magnetic field value, the interaction energy
is equal to B00116.1V Bμ 0.2x10x274.900116.1 24 = ±1.86x10-23
J or
±1.16x10-5
eV.
The sketch of energy splitting is as follow.
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11.5 Coulomb Force of Attraction
Two charged particles, which are also called point charges, have charge
magnitude q1 and q2 separated by a distance r. The electrostatic force or
Coulomb force F of attraction or repulsion between these two charged particles
follows equation
2
0
21
r4F
qq (11.19)
where 0 is the permittivity in vacuum, which has value 8.854x10-12
F/m.
Equation (11.18) is called Coulomb’s law, which name after Charles Augustin
Coulomb.
The potential energy of a charge body is equal to r4 0
21
qq. The minus
denotes that its frame of reference is at infinity.
11.6 Principle of Quantum Mechanic
The results of the photoelectric effect experiment performed as shown in Fig.
11.7 with incidence of the photon of various wavelengths on metal surface and
measured the energy of the emitted electron demonstrated the principle of
quantum mechanics. The energy in discrete packet called quanta called photon
was both postulated by Max Planck in 1900 regarding thermal radiation and
Einstein in 1905 regarding energy of light wave.
From the graph an equation can be deduced that is
KEmax = hv- q (11.20)
where e is the work function and the value of gradient of the plot is the value
of Planck's constant, which is 6.626x10-34
Js. KEmax is the maximum kinetic
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energy of electron. This equation is also known as Einstein photoelectric
equation.
Albert Einstein stated that energy of the incident photon hv must be equal
to the energy required to free electron, which is the work function plus the
maximum kinetic energy of the electron.
Figure 11.7: Plot of KEmax of emitted electron versus frequency v
11.7 Electron Wave-Particle Duality
Louis de Broglie, a French physicist, was recognized as the discoverer of the
wave nature of electron. The wave nature of electron is clearly shown in the
development of modern electronic instrumentations and experimental methods
at the turn of twentieth century such as electron microscope, where electron
behaves as wave rather than as particle of small sphere. In 1923 Arthur
Compton demonstrated the particle nature of electron, whereby the scattered
photon had larger wavelength than the wavelength ' of incidence x-ray,
which is termed Compton scattering that has relationship )cos1(cm
e
h
,
where is the scattering angle. Investigations related to the blackbody radiation
i.e., the characteristic radiation that a body emits when heated and the
photoelectric effect i.e., electron emission from matter by electromagnetic
radiation of certain energy revealed the quantum nature of light, whereas the
electron diffraction in crystals demonstrated the wave nature of particles.
Consider an electron particle moving with velocity v and mass m, its
momentum p shall be p = mv. The wave nature of electron is shown in
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Einstein's famous E = mc2 equation where E also equals to hv. Combination the
two equations it gives
m =2c
h (11.21)
The momentum p of the photon is p = mc. Substitute this relation into equation
(11.21), it gives
p = c
hv (11.22)
However, v
c then
p =
h (11.23)
which is the Louis de Broglie equation. The equation is important because it
shows the linkage between particle in terms of momentum and wave in terms of
wavelength. Rewrite equation (11.23) as de Broglie wavelength dBr, it becomes
vmr
dBr
h
p
h (11.24)
where mr is the relativistic mass, which is mr = m
v c
e
1 2 2 / and also equal to me
for the case of electron. If 1
2
2m Ve is the kinetic energy E of the electron moving
with velocity v, de Broglie wavelength dBr is Em2 e
dBr
h
p
h .
Example 11.2
What is the de Broglie wavelength dBr of an electron with kinetic energy of
120eV?
Solution
The de Broglie wavelength dBr follows equation
Em2 e
dBr
h
1931
-34
10x602.1x120x10x11.9x2
Js6.63x10
= 1.1x10-10
m.
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According to Louis de Broglie hypothesis, various particles such as electrons,
protons, or macroscopic object could exhibit wave characteristics in certain
circumstances. Whenever, the value of de Broglie wavelength of a given
particle or a macroscopic object is much smaller than the dimensions of the
apparatus i.e., its components, classical mechanics should be applied. For an
example, a ball having a mass of 0.1kg moving with the speed of 100ms-1
, the
de Broglie wavelength is 100x1.0
10x63.6 34
dBr
p
h= 6.63x10
-35m, which is very small
number as compared to the objects that ball interacts with. Thus, classical
mechanics is applied to the description of the motion of the ball. On the other
hand, if a particle, such as electron has a speed of about 8.0x103ms
-1. Its de
Broglie wavelength is 9.1x10-8
m, which is larger than the spacing between
atoms and the size of the atom. Thus, quantum mechanical description is
appropriately applied.
11.8 Heisenberg Uncertainty Principle
According to quantum-mechanical principles, there is inherent limitation to the
accuracy of a measurement of a quantum-mechanical entity. This limitation is
the result of (i) the particle-wave duality and (ii) inevitable interaction between
the instrument of observer and the entity observed.
Heisenberg uncertainty principle discovered by Werner Heisenberg in 1927.
Heisenberg uncertainty principle states that one cannot measure values with
arbitrary precision of certain conjugate quantities such as momentum and
position that share uncertainty relation. Mathematically the product of the
uncertainties of these two measurements follows equation (11.25).
2
xph
(11.25)
In quantum mechanics, a particle may be described in terms of a wave packet of
length x, which is the resultant of individual waves with different amplitudes
and nearly equal wavelengths. The position of the particle is defined so that the
probability of finding the particle at the center of the packet is highest. For
smaller x, the position of the particle is more accurately defined. However,
since the range of wavelengths in the packet now must be wider, there is
greater uncertainty in the momentum p. The particle’s uncertainties are related
by Heisenberg’s uncertainty principle which is shown in equation (11.25). This
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implies that on the atomic scale it is impossible to measure simultaneously the
precise position and momentum of a particle.
The description above can be interpreted as a continuous distribution of
wavelengths can produce localized wave packet, whereby each wave has
different momentum according to Louis de Broglie. Thus, there is a widespread
of wavelength contributing x in the wave packet. If the momentum of the
wave can be more precisely determined, then there is an increase in uncertainty
of its position x.
Example 11.3
An electron moving in x-direction with measured speed of 2.05x106m/s of
0.50%. What is the minimum uncertainty with which one can simultaneously
measure the position of the electron along x-direction?
Solution
The uncertainty equation is
2
xph
.
The delta momentum p of the electron is 0.5/100x9.11x10-31
x2.05x106 =
9.35x10-27
kgms-1
.
The uncertainty of the position x is equal to 6.63x10-34
/2/9.35x10-27
=
1.13x10-8
m = 11.3nm.
11.9 Schrödinger's Wave Equation
In 1926 Erwin Rudolf Josef Alexander Schrödinger provided a formulation
called wave mechanics, whereby it incorporated the principles of quanta
introduced by Plank and wave-particle duality principle by Louis de Broglie.
Based on de Broglie's principle, the motion electron in crystal is explained by
wave theory. This wave theory is described by Schrödinger's wave equation,
which is shown as one-dimensional equation (11.26).
2 2
22m
x t
xV x x t
x t
t
( , )( ) ( , )
( , )j (11.26)
where (x, t) is the wave function which is position and time dependent and j is
the imaginary constant equal to 1 .
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(x, t) can be separated using technique of separation variables into time-
dependent of wave function, position independent, or time-independent, portion
of the wave function, which shall be (x, t) = (x)(t), where (x) is a function
of position only and (t) is the function of time only. Equation (11.27) shall then
be equal to
2 2
22mt
x
xV x x t x
t
t
( )
( )( ) ( ) ( ) ( )
( )j (11.27)
or
2 2
22
1 1
m x
x
xV x
t
t
t
( )
( )( )
( )
( )j (11.28)
Left side of equation (11.28) is position dependent and right side is time
dependent and each side of this equation must be a constant, which can be
denoted by a variable constant . Thus, the time dependent portion can be
written as
j1
( )
( )
t
t
t (11.29)
The solution of equation (11.29) shall be ( ) exp /t t j . This is a form of
classical sinusoidal wave where = E = hv, the total energy of the particle. The
left hand side of equation (11.28) will be equal to
2 2
220
m
x
xE V x x
( )( ) ( ) (11.30)
where V(x) is the potential experienced by the particle.
Combining equation (x, t) = (x)(t) and equation ( ) exp /t t j
shall yield
(x, t) = (x)(t) = t/exp)x( j (11.31)
In 1962, Max Born postulated that |(x, t)|2 is the probability of finding the
particle between position x and x+dx at a given time or |(x, t)|2 is the
probability density function. Thus,
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|(x, t)|2 = (x, t)
*(x, t) (11.32)
where *(x, t) is the complex conjugate function. Therefore,
*(x, t) =
*(x) exp ( / )j E t (11.33)
The product of the total wave function and its complex conjugate is given by
(x, t)*(x, t) = t)/E(exp)x(t)/E(exp)x( * jj- (11.34)
= (x)*(x)
Therefore, the probability density function |(x, t)|2 is |(x, t)|
2 = (x, t)
*(x, t)
= |(x)|2. This implies that probability density is independent of time.
Since |(x, t)|2 represents the probability density function, then for a
single particle ( )x x2
1-
d
and (x) must be finite and single valued.
We shall discuss a few applications of Schrödinger's wave equation. We
shall see how Schrödinger's wave equation is applied to derive the motion of
electron in free space, the wave solution for particle in infinite potential well,
tunneling of particle through a barrier, and harmonic oscillator.
11.9.1 Motion of Electron in Free Space
If there is no force acting on it, the potential energy V(x) is constant and the
total energy is such that E>V(x). For simplicity, the potential energy V(x) = 0
for all x then the time-independent wave equation (11.30) becomes
2
2 2
20
( )( )
x
x
m Ex
e
(11.35)
The solution of equation (11.35) is
Em2xexpB
Em2xexpA)x(
ee j-j (11.36)
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Taking into consideration of time-dependent portion of wave function, which
is ( ) exp /t E t j , the total solution for the wave function shall be
( , ) exp expx t A x m E Et B x m E Ete e
j - j
2 2 (11.37)
Equation (11.37) represents the traveling wave whereby the first term represents
the wave traveling in the +x direction and second term represents the wave
traveling in the -x direction.
The value of coefficients A and B shall be determined by the boundary
conditions. If +x direction is considered than the value of B shall be zero, then
the traveling-wave equation (11.37) shall be
( , ) expx t A x m E Ete
j
2 (11.38)
If we set k 2
, the wave vector k and wavelength
Em2 e
h , then equation
(11.38) shall become the traveling-wave equation as
txexpA)t,x( kj (11.39)
11.9.2 Wave Solution for a Particle in Infinite Potential Well
Let's discuss the wave solution for particle in infinite potential well. The
potential energy V(x) of the particle is a function of position as shown in Fig.
11.8.
If E is finite, then the wave function or (x) must be zero at both regions I
and II. The particle cannot penetrate these potential barriers. Thus, the
probability of finding it is zero at region I and III. The time-independent
Schrödinger's wave equation in region II where V(x) = 0 is
0)x(mE2
x
)x(22
2
(11.40)
A particular solution to this equation is
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xmE2
sinAxmE2
cosA)x(2221
(11.41)
One boundary condition is that the wave function (x) must be continuous so
that (x = 0) = (x = a) = 0. This will give rise A1 = 0 for x = 0 and (x = a) =
A2 sin amE2
2 = 0 for x = a. The equation is valid only if
a
mE22
n
, where n =
1, 2, 3,..., n, which is the quantum number. If n is negative value, it will give
rise to redundant solution and therefore, it is not considered.
Figure 11.8: Potential function of the particle in the infinite potential well
The coefficient A2 can be found from normalization boundary condition that is
( )x x2
1-
d
= ( ) ( )*x x x-
d
1 . If the wave solution (x) is real function then
(x) = *(x). Substitute into x
mE2sinA)x(
22
into ( ) ( )*x x x-
d
1 for
condition x = 0 to x = a, it becomes
1xxmE2
sinA2
22
2
d-
(11.42)
or
1xxa
sinA 22
2
dn
-
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The solution of the equation (11.43) will give rise to coefficient A2 equal to A2
= 2
a. Finally the time independent wave solution is given by
( ) sinxa
x
a
2 n, where n = 1, 2, 3, ....., n (11.43)
Thus, the probability density function is
a
xsin
a
2)x( 22 n
. Figure 11.9 shows
the probability density function plot for energy level n = 1, 2, and 3.
(a) (b) (c)
Figure 11.9: Probability density plot energy level n = 1, 2, and 3
For energy level n = 1, it is clearly shown that maximum probability density for
a/2, which at 50pm and minimum probability density functions are found at x =
0 or x = a. For energy level n = 3, the probability density function is
a
x3sin
a
2)x( 22 . For minimum probability density function,
a
x3 should be
equal to 0, , 2, and 3. This implies that minimum density function occurs at
x = 0, x = a/3, x = 2a/3, and x = a.
From the earlier analysis a
mE22
n
then the total energy E shall follow
equation (11.44).
2
222
n2ma
E
n
2
22
8ma
nh where n = 1, 2, 3, ...., n (11.44)
11.9.3 Wave Solution for Particle in Finite Potential Well
Let's discuss the wave solution for particle in finite potential well. The potential
energy V(x) of the particle is a function of position as shown in Fig. 11.10. If E
is finite then the wave functions or (x) is not zero at both regions I and II. The
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particle can penetrate these potential barriers. Thus, the probability of finding it
is not zero at region I and III. At region x < 0 and x > a, the potential energy is
equal to V0, while in between them the potential energy is equal to zero.
Figure 11.10: Potential function of the particle in the finite potential well
Let the wave functions associated in region I, II, and III to be I, II, and III
respectively, the one dimensional Schrödinger's wave equation for region I and
III is
)x(mE2
)x(mV2
x
)x(22
0
2
2
(11.45)
or )x()EV(m2
x
)x(2
0
2
2
.
In region II, the one dimensional Schrödinger's wave equation is
)x(mE2
x
)x(22
2
(11.46)
The wave function for region I and II are respectively equal to
)xkexp(B)xkexp(A)x( 00I (11.47)
and
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)xkexp(G)xkexp(F)x( 00III (11.48)
where 2
00
)EV(m2k
. From equation (11.47), which is
)xkexp(B)xkexp(A)x( 00I , as x approaching -, I approaches to zero.
This shall mean constant A is equal to zero. As the result the wave function
becomes
)xkexp(B)x( 0I (11.49)
From equation (11.48), which is )xkexp(G)xkexp(F)x( 00III , as x
approaching , III approaches to zero. This shall mean constant G is equal to
zero. As the result the wave function becomes
)xkexp(F)x( 0III (11.50)
The wave function of region II is
)xkcos(D)xksin(C)x( 11II (11.51)
where 21
mE2k
.
At x = 0, )0kexp(B)0( 0I = )0kcos(D)0ksin(C)0( 11II , this implies that
constant B = D. This shall mean that wave function at region I is also equal to
)xkexp(D)x( 0I .
Differentiating wave equations of all regions, which are )xkexp(D)x( 0I ,
)xkcos(D)xksin(C)x( 11II , and )xkexp(F)x( 0III with respect to x, it yields
)xkexp(Dkdx
)x(00
I
(11.52)
)xksin(Dk)xkcos(Ckdx
)x(1111
II
(11.53)
)xkexp(Fkdx
)x(00
III
(11.54)
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At x = 0, dx
)0(
dx
)0( III
, this implies that Dk0 = Ck1. Substituting C = Dk0/k1
into wave function of region II, which is )xkcos(D)xksin(C)x( 11II , it
yields )xkcos(D)xksin(k
kD)x( 11
1
0II .
At x = a, II(a) = III(a) and dx
)a(
dx
)a( IIIII
. This shall mean that
)akcos(D)aksin(k
kD 11
1
0 )akexp(F 0 (11.55)
and
)aksin(Dk)akcos(Dk 1110 )akexp(Fk 00 (11.56)
With the finite well, the wave function is not zero outside the well. So with
xfinite
> a and from Heisenberg uncertainty principle pfinite
< xx , this suggests
that the average value of momentum is less in the finite well. Therefore, the
kinetic energy inside the finite well is less than inside the infinite well. In
addition, the number of allowed energy levels is finite, there is a possibility that
a finite well may be sufficiently narrow or sufficiently shallow that no energy
levels are allowed.
Let’s now attempt to obtain the energy levels in the finite well. Taking
equation (11.55) to divide with equation (11.56), it yields
)aksin(Dk)akcos(Dk
)akcos(D)aksin(k
kD
1110
11
1
0
)akexp(Fk
)akexp(F
00
0
(11.57)
This will yield )Lktan(kkkk2 1
2
0
2
110 . Substituting equation 2
0
2
0
)EV(m8k
h
and 2
2
1
mE8k
h
into this equation to eliminate k0 and k1, it yields a
transcendental equation for the energy level inside finite potential well, which is
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2
22
00
mEa8tan)VE2(EEV2
h (11.58)
The solution of equation (11.52) can be solved by numerical method. However,
one may use try and error method to fine the values of total energy by equating
left hand side of the equation with right hand side of the equation, which is
2
22
0
0 mEa8tan
VE2
EEV2
h. The results of energy of an electron calculated
using trigonometric function plot for V0 = 25eV and a = 0.5nm are shown in Fig.
11.11. The intersection of negative
2
22mEa8tan
hand negative
0
0
VE2
EEV2
are
the E values for level n = 1, n = 2, n = 3, while intersection of positive
2
22mEa8tan
hand positive
0
0
VE2
EEV2
are the E value for level n = 4, and n =
5.
Figure 11.11: Energy levels in finite potential well for a = 0.5nm and V0 = 25eV
11 Atomic Physics
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Using equation (11.44), which is 2
e
22
na8m
Enh
, the energies in infinite potential
well for a = 0.5nm are tabulated. Together with the energies of finite potential
well are shown in Fig. 11.12.
Energy Level Infinite Well (eV) Finite Well (eV)
1 1.504 1.123
2 6.015 4.461
3 13.533 9.905
4 24.059 17.162
5 37.593 24.782
6 54.133 Non quantized
Figure 11.12: Energies in infinite and finite potential wells for a = 0.5nm and V0 = 25eV
The wave functions of three energy levels of a finite potential well with a =
0.5nm and V0 = 25eV are shown in Fig. 11.13.
Figure 11.13: The wave functions of three energy levels of a finite potential well with a =
0.5nm and V0 = 25eV
11.9.4 Wave Solution for a Particle Tunneling
Figure 11.13 shows an electron of total energy E moving parallel along x-axis
and probability density function )x(2 of matter wave showing tunneling of
particle through barrier. The potential energy of the particle is zero except when
it is in the region 0 < x < a, where the potential energy has a constant value V0.
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Figure 11.14: An energy diagram showing potential energy barrier of V0 and the probability
density function )x(2 of electron wave showing tunneling of electron
through barrier
In classical theory, if the energy of particle E is less than the potential energy V0,
the particle will be bounded back at the boundary. In quantum mechanics, there
is a chance of leak through or tunnel through.
The Schrödinger's wave equation for left side and right of the barrier is
0)x(mE2
x
)x(22
2
and the solutions of this equation are
mE2xexpB
mE2xexpA)x(
j-jfor left side and
mE2xexpD
mE2xexpC)x(
j-jfor right side. Wave
function
mE2xexpB)x(
j-denotes the reflected back wave function. Wave
function
mE2xexpC)ax(
jdenotes the transmitted wave function, which
has lower and constant amplitude.
Within the barrier, the Schrödinger's wave equation is
0)x()EV(m2
x
)x(022
2
(11.59)
The solution of the wave function is
)xexp(F)xexp(E)a xand 0x( kk (11.60)
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where k is wave number and is equal to 2
0
2 )EV(m8
hk
.
Taking the exponential decrease function, the probability density function
)a x and 0x(2 is
)x2exp(E)a x0x( 22 k (11.61)
The transmission coefficient T is the measure of fractional amount of particle is
tunneled through. It is approximately equal to
)a2exp(V
E1
V
E16T
00
k
(11.62)
Since transmission coefficient is an exponential function, it depends on mass (in
this case is electron mass), thickness of the barrier a, and (V0 – E).
Tunneling is important in nuclear fusion. A fusion reaction can occur when
two nuclei tunnel through the barrier caused by their electrical repulsion and
approach each other closely enough for attractive nuclear force to cause them to
fuse. Fusion reaction occurs in the core of star including the Sun. Emission of
alpha particle from unstable nuclei is also through tunneling. An alpha particle
of energy E at the surface of a nucleus encounters a potential barrier that results
from the combined effect of the attractive nuclear force and electrical repulsion
of the remaining part of nucleus as shown in Fig. 11.15.
Figure 11.15: The potential energy function for an alpha particle interacting with nucleus of
radius R
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When the distance is such that rR, the radius of nucleus, the alpha particle
encounters square potential well, whereas rR, it encounters potential due to
electrical repulsive force.
11.9.4.1 Application of Particle Tunneling
One of the applications of electron tunneling is the scanning tunneling
microscope. Scanning tunneling microscope STM uses electron tunneling to
create image of surface down to the scale of individual atom. An extreme sharp
needle is brought very close to the surface within 1.0nm. When the needle is at a
positive potential with respect to the surface, electron can tunnel through the
surface potential energy barrier and reach the needle. The width of the barrier is
the distance between tip of needle and surface. The needle or probe scans across
the surface and at the same time move vertical upward or downward to maintain
constant current being registered due to tunneling. In this manner, the surface
topology of the sample can be constructed based on the motion of the probe.
Figure 11.16 illustrates the probe of a scanning tunneling microscope.
Figure 11.16: Probe of scanning tunneling microscope
Other examples of applications are tunneling diode used in microwave
application and Joseph junction. Joseph junction is consisting of two
superconductors separated by an oxide layer of 1.0 to 2.0nm thick. Electron
pairs in superconductors can tunnel through the barrier layer giving such a
device unusual circuit properties. Joseph junction is useful for establishing
precise voltage standard and measuring tiny magnetic field.
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Example 11.4
Suppose that the electron having energy E of 5.1eV approaches a barrier of
height V0 = 6.8eV and thickness a = 750pm. What is the probability that the
electron will be transmitted through the barrier?
Solution
The wave number is 2
0
2 )EV(m8
hk
=
234
19312
)10x63(6.
10x602.1x)1.58.6(10x11.9x8
=
6.67x109m
-1.
The transmission coefficient is
)a2exp(V
E1
V
E16T
00
k
= 3 )10x750x10x67.6x2exp( 129 = 3exp(-10.0) = 135x10
-6.
The probability is 135 electrons will tunnel through for every one million
electrons striking the barrier.
11.9.5 Wave Solution for Harmonic Oscillator
In Newtonian mechanic, a harmonic oscillator is a particle with mass m acted on
by a conservation force component Fx = - k’x. k’ is the force constant. The
corresponding potential energy function is 2
0 x2
1V 'k . The angular frequency is
m
k '
. The energy of the photon is hE . The harmonic oscillator has
the characteristic of Newtonian mechanics. Thus, the quantum-mechanical
analysis of the energy levels of harmonic oscillator would be multiples of
quantity shown in equation (11.63).
m
k '
(11.63)
The energy levels are based on Plank’s radiation law, which is 1e
c2)(
t/c
2
kh5
hI .
It has a good assumption that the energy levels are half integer multiple of .
Thus, equation (11.63) will be rewritten as
m
knn
'
2
1
2
1En n = 0, 1, 2, 3, 4,…, n (11.64)
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The Schrödinger's wave equation for the harmonic oscillator is
)x(Ex2
1m2
x
)x( 2'
22
2
k
(11.65)
The wave function approaches zero when x approaches infinity i.e. (x) 0 as
|| . The solution for the Ex2
1 2' k is
2/x'mexpC)x( 2k (11.66)
The constant C is chosen to normalize the function. Knowing that 1dx)x(2
.
Constant C can be found from integral table, which
a)xaexp( 22 . Thus, C
is found to be 4
8 '
mk. This implies that
2/x'mexp
m)x( 2
4
8 '
kk
.
The illustration of the energy levels of harmonic oscillator is shown in Fig.
11.17.
Figure 11.17: Energy levels of harmonic oscillator
As shown in Equation (11.64), the energy level difference between adjacent
energy level is .
Based on equation (11.64), the Schrödinger's wave equation for the
harmonic oscillator is
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)x(2
1x
2
1m2
x
)x( '2'
22
2
m
knk
(11.67)
The general formula for the normalized wave functions (x) is
2
'
4
'
n4
8 '
x2
mexpx
mH
!2
1m)x(
kk
n
k
n (11.68)
x
mH 4
'
n
kis the Hermite polynomial, after the French mathematician Charles
Hermite. The polynomial states Hn(y) is
22 xx e
dx
de1)y(H
n
nn
n (11.69)
For n = 0, 1, 2, 3, 4, and 5, the Hermite polynomials are 1)y(H 0 , y2)y(H 1 ,
2y4)y(H 2 2 , y12y8)y(H 3 3 , 12y48y16)y(H 24 4 , and
y120y160y32)y(H 35 5 respectively.
Thus, from equation (11.68), the first six wave functions for the harmonic
oscillator will be equal to
2
'
4
8 '
x2
mexp
m)0(
kk
2
'
42
'
4
8 '
x2
mexpx
m2
m)1(
kkk
2
'2
2
'
4
8 '
x2
mexp1x
m2
2
1m)2(
kkk
2
'
42
'34/3
2
'
4
8 '
x2
mexpx
m3x
m2
3
1m)3(
kkkk
2
'2
2
'4
2
'
4
8 '
x2
mexp3x
m12x
m4
24
1m)4(
kkkk
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x2
mexp
xm
30xm
40xm
8240
1k)5(
'
42
'34/3
2
'54/5
2
'
4
8 '
k
kkkm
(11.70)
From the above equations, the first four wave functions for the harmonic
oscillator are shown in Fig. 11.18. A is the amplitude of Newtonian oscillator.
The results also shown that for x < -A and x> A are not forbidden region. The
maxima and minima for each function is n+1, an extra than the quantum number.
(a) (b)
(c) (d)
Figure 11.18: First wave functions for harmonic oscillator (a) n = 0, (b) n = 1, (c) n = 2, and
(d) n = 3
The probability density function 2(x) for n = 0, 1, 2, and 3 and its
corresponding Newtonian probability density function are shown in Fig. 11.19.
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(a) (b)
(c) (d)
Figure 11.19: Probability density function of harmonic oscillator for (a) n = 0, (b) n = 1, (c)
n = 2, and (d) n = 3
11.10 Atomic Model
In 1911, Ernest Rutherford suggested that atom is consisted of positively
charged nucleus with negatively charged electrons orbiting the nucleus at
constant speed. This was a great difference from the earlier model, which stated
that electrons were stuck into the spherical surface of the nucleus.
From the glow discharge experiment of hydrogen gas, Johannes Robert
Rydberg came out with an empirical formula, which stated that the line spectra
obeyed.
22
11R
1
Pjpj
(11.71)
where j<p and they are integrals. R c8
m32
0
0
4
h
q
is the Rydberg constant =
1.097x107m
-1.
Niels Bohr knew the results of hydrogen glow discharge and he thought
that the electron stayed at fixed stable orbit and did not radiate continuous range
of energies. He revised Rutherford hydrogen model and suggested that electron
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orbiting the nucleus in a circular orbit at constant velocity v. He came out with
three postulations. The first one states that there are certain orbits in which
electron is stable and does not radiate energy. The second one states that when
electron falls from orbit p to j, the energy, which it loses, is (Ep - Ej) and it is
radiated as a quantum of light, which follows equation (11.72).
Epj = hv (11.72)
where h is Planck's constant = 6.626x10-34
Js and v is the frequency of photon.
Frequency v can also be expressed as v =c
, where c is the speed of light in
vacuum which is 2.998x108ms
-1. Thus, the energy lost by the electron as it falls
from orbit p to orbit j shall be
Ep - Ej = hvpj =pj
c
h (11.73)
The total energy En of an electron orbiting in nth
orbit is equal to
En = n0
2
r8
q (11.74)
It comprises of kinetic energy n
2
0
2
er8
1Vm
2
1 q
and potential energy due to
coulomb force of attractionn
2
0 r4
1 q
, where the coulomb force of attraction is
2
n
2
0 r4
1 q
= m
v
re
n
2
and mv
re
n
2
is the centripetal force.
However, from Bohr's third postulation, which states that the angular
momentum is
mevrn = 2
nh= n (11.75)
where
2
h is the reduced Planck's constant.
The relation of rn and n can be obtained
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rn = 2
2
e
2
0
mn
q
h
(11.76)
For n =1, r1 = ao is called Bohr's radius and is equal to 0.529o
A . Thus, the radius
of n orbit rn 2
o
n A529.0r n . Substituting equation (11.76) into equation (11.74), it
yields equation (11.77).
222
0
e
4
n
1
8
mE
nh
q
(11.77)
where n is the subscript to denote energy level and eV6.138
m22
0
e
4
h
q. Thus,
energy of the electron at orbit n is E eVn 136
2
.
n.
For hydrogen atom whereby it has one electron occupying first orbit
closed to nucleus would have highest ionization energy. Ionization energy is
defined as energy required removing an electron from its stable state to infinity,
which is not in control of nucleus.
The shortcoming of Bohr's atomic model that it could not explain many
electron atom, was overcome by Arnold Johannes Wilhelm Sommerfeld
suggesting two more quantum numbers associated with the principal quantum
number n. They are: l denotes orbital quantum number for angular momentum
of electron and m is magnetic quantum number for behavior of electron in
magnetic field. The additional two quantum numbers meant that the orbit
denoted by n could be split into more orbits, which need not be circular. They
can be elliptical. The rough representation of an atom with circular and elliptical
orbits is shown in Fig. 11.20.
Figure 11.20: A representation of an atom with circular and elliptical orbits
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Later in 1925, George Eugene Uhlenbeck and Samuel Abraham Goudsmit
proposed a fourth quantum number called spin quantum number s, which is
responsible for the magnetic property of the element. Electron can have one of
the two different spin which either clockwise or anti-clockwise spin. Therefore,
the spin s can be s = + 1
2 or s =
1
2.
The summary of four quantum numbers is:
1. principal quantum number n;
2. orbital quantum number l;
3. magnetic quantum number ml;
4. spin quantum number s.
The maximum number of electron in an atom that can have the same principal
quantum number n is 2n2. l may be zero or a positive number such as l n -1. ml
l so ml can be negative, zero or positive integral.
l can also be shorthand written such that l = 0 means s; l = 1 means p, l = 2
means d and l = 3 means f.
Beside the way how to fill the quantum numbers, Friedrich Hund - Hund
rule - deduced from the measurement of magnetic moment of atom that same l
level is filled first by electron having the same value of spin i.e. 1
2 to be filled
first and followed by 1
2. This is also the reason why iron has magnetic
property because its outermost shells have four electrons having s = 1
2 and
they are not paired which termed spin pair. This gives rise to magnetic moment
of four Bohr magneton B, whereby one Bohr magneton e
Bm4
qh
= 9.27x10-
24Am
2.
Pauli exclusive principle states that no two electrons in atom can have the
same set of four quantum numbers i.e. a set of these four quantum numbers
describes an individual electron in a single atom.
As an illustration of how the four quantum numbers are applied, the
electronic configuration of chlorine that has atomic number Z = 17, is shown in
Fig. 11.21.
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Atomic
Number Z Electron Number
Quantum Number
n l ml s
17
1 1 0
0 + 1/2
2 0 - 1/2
3
2
0 0 + 1/2
4 0 - 1/2
5
1
-1 + 1/2
6 0 + 1/2
7 +1 + 1/2
8 -1 - 1/2
9 0 - 1/2
10 +1 - 1/2
11
3
0 0 + 1/2
12 0 - 1/2
13
1
-1 + 1/2
14 0 + 1/2
15 +1 + 1/2
16 -1 - 1/2
17 0 - 1/2
Figure 11.21: Electronic configuration of chlorine
11.10.1 Electron Probability Distribution
Instead treating the electron as a point particle moving in a precise circle like
the one shown in Fig. 11.20, Schrödinger’s equation gives a probability
distribution surrounding the nucleus. The probability of finding an electron in a
small volume dV is dV2
. The integration over all space should be unity. i.e.
1dv2
. This shall mean 100% probability of finding electron somewhere in
the universe. The volume of the shell with inner radius r and other radius r+dr is
approximately equal to drr4dV 2 . Thus, the probability P(r) of finding an
electron within the radial range dr is equal to
drr4dV)r(P222 (11.78)
Figure 11.22 shows the probability P(r) of finding an electron for several
hydrogen atom wave functions. The figures show that for each function, the
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number of maxima is (n-1). Take an example n = 3, the number of maxima is
two, which is shown in 3p and n = 4, the number of maxima is three, which is
shown in 4p.
(a) l = 0 (b) l = 1 (c) l = 2 and l = 3
Figure 11.22: Radial probability distribution function P(r) for several hydrogen wave
functions plotted as functions of ration r/a
The graph with l = n-1, there is only one maximum located at r = n2a, where a is
Bohr’s radius. Thus, the largest possible l for each n like 1s, 2p, 3d, and 4f, P(r)
has a single maximum at n2a. Figure 11.22 shows the radial probability
distribution function 22r4)r(P , which indicates the relative probability of
finding an electron within a thin spherical shell of radius.
Figure 11.23 and Fig. 11.24 show three dimensional probability
distribution function 2
, which indicate the relative probability of finding an
electron within a small box at a given position. The darker cloud indicates high
value of 2
. Figure 11.23 shows the cross sections of spherically symmetric
probability clouds for three lowest s sub-shells which 2
depend only on the
radial coordinate r.
Figure 11.23: Three dimensional probability distribution function
2 for spherically
symmetric 1s, 2s, and 3s hydrogen atom wave functions
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Figure 11.24 show cross sections of the clouds for other electron for which 2
depends on both r and . In any stationary state if hydrogen atom, 2
is
independent of .
Figure 11.24: Cross sections of three dimensional probability distribution function
2 for a
few quantum states of the hydrogen atom. Mentally rotate each drawing along
z-axis to get three dimensional representation of 2
Tutorials
11.1. How many l value associated with n = 3?
11.2. How many ml number associated with l = 3?
11.3. For orbital quantum number with l = 3.
(a). What is the magnitude of the orbital angular momentum in a state
with l = 3?
(b). What is the magnitude of its largest projection on an imposed z-axis?
(c). What is the magnitude of its minimum projection on an imposed z-
axis?
11.4. An atom in a state with l = 1 emits a photon with wavelength 600nm as it
decays to a state with l = 0. If the atom is placed in a magnetic field with
magnitude B = 2.00T, determine the shifts in the energy levels and in the
11 Atomic Physics
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wavelength resulting from the interaction of the magnetic field and
atom’s orbital magnetic moment.
11.5. Calculate the value of two equal charges if they repel one another with
force of 0.1N when situated 50cm apart in vacuum.
11.6. The work function for cesium is 2.14eV.
(a). What is the maximum kinetic energy of the electron ejected from the
surface of cesium by light of wavelength = 546nm?
(b). What is the maximum speed of the electron?
11.7. A gain of sand with mass of 1.0x10-3
g appears to be at rest on smooth
surface. We locate its position to within 0.010mm. What velocity limit is
implied by our measurement of its position?
11.8. An electron revolving an orbit n around the nucleus has total energy equal
to En = n0
2
r8
q. Prove this formula using the formula of Coulomb force
of attraction between the electron and nucleus in which it follows
expression 2
n0
2
r4
q.
11.9. Calculate the de Broglie wavelength or an electron of kinetic energy
1.0eV and 100eV.
11.10. The equation is 1xxa
sinA 22
2
dn
-
. Evaluate for the value of constant A2
for limit x = 0 to x = a.
11.11. A ground state electron is trapped in a one-dimensional infinite potential
well with a = 100pm.
(a). What is the probability of that an electron can be detected in left one-
third of the well?
(b). What is the probability that the electron can be detected in middle
one-third of the well?
11.12. Calculate the energies and radii of second, third, and fourth orbit of
hydrogen atom.
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11.13. What is the maximum number of electron that occupied in a shell with
principal quantum number 3?
11.14. Reference to question 11.13, state the electronic configuration of this
shell.
11.15. The home-base level of Lyman series of hydrogen atom is n = 1. What is
the least energetic photon emitted in this series?
11.16. A 2.0eV electron encounters a barrier with height 5.0eV. What is the
probability that it will tunnel through the barrier of width a is 1.0nm?
11.17. A sodium atom of mass 3.82x10-26
kg vibrates with simple harmonic
motion in crystal. The potential energy increases by 0.0075eV when the
atom is displaced by 0.014nm from its equilibrium position.
(a). Find the angular frequency according to Newtonian mechanics.
(b). Find the spacing of adjacent energy levels.
(c). If the atom emits a photon during a transition from one vibrational
level to the next lower level, what is the wavelength of the emitted
photon?
11.18. Iron, transition element has atomic number Z = 26 and its electron
configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d6. Using four quantum
numbers to tabulate the electronic configuration of iron. From the results
discuss and comment why iron has magnetic property.