Download Atomic physics: Atomic Spectra: Thomson`s plum

Atomic physics:
Atomic Spectra: Thomson’s plum-pudding model,1907:
According to this model, the entire +ve charge is distributed over a sphere of
radius equal to thew atomic radius of the order of 10−10 m. In this sphere of +ve
charge , the -Ve charged point electrons are embedded in such a way , much like the
plums in a pudding, that the system is both stable and electrically neutral.
However, the electrons are not at rest but oscillate with definite frequencies about
their mean positions.
This model could expain the ionization process, but if fail to explain the observed
frequencies of optical spectra and the results of α ray scattering experiment of Geiger
and Marsden.
Rutherford’s nuclear atom model, 1911:
Rutherford’s nuclear model of the atom is an outcome of the scattering experiments of α particles. α- particles are doubly ionised He atom.
according to this model, within the atom there was a small massive core carrying
+ve charge and called it the nucleus. The electrons are assumed to exist outside the
nucleus.
To explain the stability of the atom, Rutherford assumed that electrons revolve
round the nucleus, like planets revolving round sun. The electrostatic attraction
between the nucleus abd the extra-nuclear electrons provided the necessary centripetal
force.
Limitation of Rutherford’s model:
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Drawbacks of Rutherford’s model is that the atom as a whole cannot be stable.
because, according to Rutherford that the electrons are in circular motion, the centripetal force would provided by electrostatic attraction. But uniform rotation is an
accelerated motion and according to the classical electromagnetic theory, an accelerated charge emits electromagnetic radiation. So, as a result the loss of energy would
make the electron spiral into the nucleus. Hence Rutheford model cannot be stable.
Bohr’s quantum model of the atom:
Bohr’s theory contained a combination of ideas from
(i) Planck’s quantum theory
(ii) Einstein’s photon theory of light
(iii) Rutherford model of atom
Bohr’s model can be applied quite successfully to hydrogen atom, hydrogen like
ions like singly ionised He and doubly ionised lithium. But this theory does not
properly describe the spectra of more complex atoms and ions.
Basic assumption or postulates of Bohr’s theory :
The basic assumption or postulates of Bohr’s theory are
(i) An electron in an atom moves in a circular orbit about the nucleus under the
influence of the Coulomb force of attraction between the electron and nucleus.
Ze2
mv 2
=
4π0 r2
r
(ii) An electron moves in astable or stationary orbit in which the electron does not
radiate energy.
(iii) Electromagnetic radiation is emitted or absorbed only when the electron
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jumps from a stationary orbit to other stationary orbit. The radiated energy is given
by Planck-Einstein formula
Ei − Ef = hν
where Ei → the energy of the initial state
and Ef → the energy of the final state. This equation is known as Bohr’s frequency
rule. (iv) The electron moves in certain discrete orbit such that the angular momentum (L) of the electron is an integral multiple of h̄.
L = mvr = nh̄, n = 1, 2, 3, .....
This is known as Bohr’s quantization condition.
Expressions for orbital radius of an hydrogen(Z=1) like atom:
Consider a hydrogen like atom of atomic number Z consisting a nucleus of charge
Ze.
Let a single electron of mass m moves in a circular orbit of radius rn . For the
mechanical stability of the electron
Fc =
Ze2
mvn2
=
4π0 rn2
rn
where vn is the velocity of the electron.
Ze2 = 4π0 mvn2 rn
(1)
Again we have from Bohr’s quantum condition
mvn rn = nh̄ =
nh
2π
Substituting the value of mvn rn in (2), we get
Ze2 = 4π0 mvn2 rn
Ze2 = 4π0 mvn
3
nh
2π
(2)
Ze2
vn =
20 nh
(3)
Hence,
1
n
So, we see that velocity of electron in the inner orbit is greater than that of outer
vn ∝
orbit. Again we have
mvn rn = nh̄ =
nh
2π
(2)
nh 1
2π mvn
nh 20 nh
rn =
2π Ze2
n2 h2 0
rn =
mπZe2
rn =
From this equation we see that
rn ∝ n2
In Bohr’s model n = 0is excluded for if n = 0, rn = 0 and the electron would go
through the nucleus.
Expression for total energy:
Kinetic energy of n-th orbital electron
1
Ze2
EK.E. = mvn2 =
2
8π0 rn
Potential energy of n-th orbital electron
VP.E. = −
Z rn
∞
F drn = −
Ze2
4π0 rn
Total energy of the nth orbital electron
En = EK.E. + VP.E. =
Ze2
Ze2
Ze2
−
=−
8π0 rn 4π0 rn
8π0 rn
Substituting the value of rn , we get the total energy
mZ 2 e4
En = − 2 2 2
80 h n
4
Hence the energy of the electron within the atom is also quantised, i.e. can take only
certain discrete values.
The allowed energy values for a quantised system are called energy levels or energy
states.
The negative sign to the total energy is due to the fact that the potential energy
is taken to be zero when the electron is at infinity. It corresponds to a bound state
implying that energy is required to remove the electron from atom.
For hydrogen atom (Z=1),
En = −
13.6
eV
n2
For
13.6
eV = −13.6eV
12
13.6
n = 2, E2 = − 2 eV = −3.4eV
2
13.6
n = 3, E3 = − 2 eV = −1.5eV
3
13.6
n = ∞, E1 = − 2 eV = 0
∞
The level n = ∞ corresponds to the series limit and a state in which the electron
n = 1, E1 = −
completely removed. The atom is said to be ionised.
Energy required to remove the electron from the ground state n = 1 of the atom
is called the ionization energy or the binding energy.
Hydrogen spectrum:
At room temperature, almost all the H-atoms will be in the ground state. At
higher temperature or during electric discharge, electrons may be in excited states.
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But their life time in the excited states is short so that they make transitions to lower
energy states emitting electromagnetic radiations of definite frequency. There is no
restriction on the change in n during the transition.
Let the electron make a transition from the initial state ni (upper state) of energy
Ei to a final state nf (lower state) of energy Ef , then we get from Bohr’s frequency
condition
hν = Ei − Ef
Ei − Ef
mZ 2 e4 1
1
ν=
=
− 2
2
2
3
h
8 h
nf
ni
!
In terms of wave number
1
ν
mZ 2 e4 1
1
ν̄ = = = 2 3
− 2
2
λ
c
8 h c nf
ni
1
ν
1
1
ν̄ = = = R∞ Z 2 2 − 2
λ
c
nf
ni
where R∞ =
me4
82 h3 c
!
!
is called Rydberg constant for infinite nuclear mass.
Origin of spectral series:
The origin of spectral series of H-atom can be understood from Bohr’s theory. We
have the spectral lines of wave number
1
ν
1
1
ν̄ = = = R∞ Z 2 2 − 2
λ
c
nf
ni
!
For H-atom Z=1,
1
1
1
ν̄ = = R∞ 2 − 2
λ
nf
ni
where R∞ =
me4
82 h3 c
is called Rydberg constant for infinite nuclear mass. Hence, wave
number
1
1
1
ν̄ = = RH 2 − 2
λ
nf
ni
and R∞ = RH =
!
me4
82 h3 c
!
is called Rydberg constant for H-atom.
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1. Lyman series: When an electron jumps from one of the ni = 2, 3, 4, 5, .... etc
quantum states to nf = 1 quantum state , we get Lyman series of wave number
!
1
1
1
ν̄ = = RH 2 − 2 , ni = 2, 3, 4, 5, .....
λ
1
ni
The Lyman series is in the ultraviolet region of e.m. spectrum.
Balmer series: When an electron jumps from one of the ni = 3, 4, 5, .... etc
quantum states to nf = 2 quantum state , we get Balmer series of wave number
!
1
1
1
ν̄ = = RH 2 − 2 , ni = 3, 4, 5, 6, .....
λ
2
ni
The Balmer series is in the visible region of e.m. spectrum. For ni = 3, we get Hα
line
For ni = 4, we get Hβ line
For ni = 5, we get Hγ line
Paschen series: When an electron jumps from one of the ni = 4, 5, .... etc
quantum states to nf = 3 quantum state , we get Paschen series of wave number
!
1
1
1
ν̄ = = RH 2 − 2 , ni = 4, 5, 6.....
λ
3
ni
The Paschen series is in the infrared region of e.m. spectrum.
Brackett series: When an electron jumps from one of the ni = 5, 6, 7.... etc
quantum states to nf = 4 quantum state , we get Brackett series of wave number
!
1
1
1
ν̄ = = RH 2 − 2 , ni = 5, 6, 7.....
λ
4
ni
The Brackett series is also in the infrared region of e.m. spectrum.
Pfund series: When an electron jumps from one of the ni = 6, 7.... etc quantum
states to nf = 5 quantum state , we get Pfund series of wave number
!
1
1
1
ν̄ = = RH 2 − 2 , ni = 6, 7.....
λ
4
ni
7
The Pfund series is also in the infrared region of e.m. spectrum.
Note: If the transition is caused from a lower energy level to a higher one by e.m.
radiation of light energy, it will give to absorption spectrum, observed as dark lines
of the same frequencies as emission lines.
Energy levels:
The allowed energy values for a quantised system is called energy levels or energy
states. According to Bohr’s theory, the electron has a definite energy in a stationary
orbit given by
mZ 2 e4
R∞ ch
=− 2
2 2 2
80 h n
n
= RH , we get
En = −
For H-atom Z = 1, R∞
RH ch
n2
= −RH ch
En = −
Case I: when n = 1 (ground state), En=1
Case II: when n ≥ 2 (excited states)
Case III: As n → ∞, En → 0, the electron is completely removed. This corresponds
to ionisation of the atom.
Case IV: When E > 0, we get continuum , the region of unrestricted +ve energy
values of the free electron.
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Ritz combination principle:
The Ritz combination principle states that certain frequencies in the emission
spectrum can be summed to give other frequencies.
This is an empirical formula.
Let there are three quantum states i, j, k(i > j > k).
Let νij be the frequency of photon emitted when an electron jumps from i−th state
to j−th state.
Similarly, νik , frequency of transition from i → k.
νjk , frequency of transition from j → k.
According to Ritz combination principle
νik = νij + νjk
Proof: The frequency of transition from i → k
νik =
Ei − Ek
h
Ei − Ej + Ej − Ek
h
Ei − Ej Ej − Ek
νik =
+
h
h
νik = νij + νjk
νik =
Hydrogenic atoms:
Bohr’s theory for H-atom can be used for any atom with a single electron such as
singly ionisez helium atom He+ , doubly ionised lithium Li++ etc. These hydrogen
like atoms are called hydrogenic atoms.
Simply, in hydrogenic atoms, the nuclear charge = Ze
Z is the atomic number.
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So, in hydrogenic atoms Ze2 should be in place of e2 in H-atom.
With this change, the expressions for the energy and radius of H-like atoms become
En = −
Z2
mZ 2 e4
=
−13.6
eV
n2
32π 2 20 h̄2 n2
n2 a0
Z
This shows that the orbits of atoms with higher Z value are closer to the nucleus.
rn =
Bohr’s correspondence principle:
Bohr’s correspondence principle states that the behaviour of an atomic system
tends asymptotically to that expected in classical theory in transitions involving states
of large quantum number.
or
It states that for large quantum number, the classical orbital frequency of moving
electron equals to the quantum theory frequency of electron or frequency of radiation.
i.e. ν = f , as n be large.
Here, f classical frequency
ν quantum theory frequency or frequency of spectral line.
Proof: Let the classical frequency of revolution f of the electron in an orbit of
radius r is
f=
v
2πr
Again we have from Bohr’s theory
v=
and
rn =
Ze2
20 nh
n2 h2 0
πmZe2
10
(1)
Substituting, the values of r and v in (1) , we get
f=
1 Ze2 πmZe2
2π 20 nh n2 h2 0
mZ 2 e4 2
.
820 h3 n3
We have a radiated frequency in quantum theory
f=
(2)
1
mZ 2 e4 1
− 2
ν= 2 3
2
8 h c nf
ni
!
mZ 2 e4 n2i − n2f
ν= 2 3
8 h c
n2f n2i
!
mZ 2 e4 (ni − nf )(ni + nf )
ν= 2 3
8 h c
n2f n2i
!
As nf and ni be very large, then we can write ni − nf = 1 and ni + nf = 2n we get
for ni = nf = n
!
mZ 2 e4 1.2n
ν= 2 3
8 h c n2 n2
mZ 2 e4 2
ν= 2 3
8 h c n3
!
ν=f
Merits of Bohr’s theory:
The merits of Bohr’s theory are
(i) It gives a convincing explanation and a very simple and elegant picture of the
origin of spectral lines.
(ii) Rydberg constant in terms of fundamental constants offers an excellent proof
of the soundness of Bohr’s theory.
(iii) The reduced mass concept of Bohr’s theory is excellent
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(iv) This theory has been instrumental to the discovery of heavy hydrogen (deuteron)
by A. C. Urey.
(v) The general principle used by Bohr has also been successfully applied to a geat
number of phenomena such as excitation and ionisation of atoms , X-ray spectra etc.
Demerits of Bohr’s theory:
The demerits of Bohr’s theory are
(i) The quantum idea of the statinary orbits is mixed up with the classical idea
of Coulomb force.
Ze2
mv 2
=
4π0 r2
r
(ii) The assumption of only circular orbit is utterly unjustified.
(iii) It can not explain the spectral series of multi-electron systems.
(iv) It can not explain the fine structure of spectral lines.
(v) It can not explain the multiple structure of spectral lines.
(vi) It can not make any calculation about the transitions or the selection rules
which apply to them.
Sommerfeld’s model:
To explain fine structure, Sommerfeld extended Bohr’s theory by introducing
(i) General quantization rule
(ii) the idea of elliptical orbits for the electron.
(i) General quantization rule:
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For any physical system in which the coordinates are periodic functions of time,
there exists a quantum condition for each coordinate given by
Z
pq dq = nq h
where q is one of the coordinates, pq the corresponding momentum, nq a quantum
number that takes integral values.
(ii) the idea of elliptical orbits for the electron:
(a) The electron moves in a stable stationary non-radiating circular as well as in
an elliptical orbit with the nucleus at one of the foci.
(b) Infinite number of elliptical orbit which are possible according to Newtonian
mechanics, only certain discrete orbit satisfy the phase integral
Jr =
Z
Jθ =
pr dr = nr h, nr = 0, 1, 2, 3, 4......
Z
pθ dθ = kh, k = 1, 2, 3, .....
Total angular momentum is given by
J~ = J~r + J~θ
nh = nr h + kh
n = nr + k
where n=Principal quantum number
nr =radial quantum number
k = orbital or azimuthal quantum number= 1, 2, 3, ....
Note: k = 0 corresponds to the motion of electron along a straight line through
the nucleus, but it is impossible. (c) The radiation is emitted or absorbed when an
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electron undergoes a transition from one allowed orbit to another.
|Ei − Ef | = hν
A complete analysis gives
n = nr + k, n = 1, 2, 3, 4........
and energy
1
mZ 2 e4
En = −
2 2 n2
2
32π 0 h̄
!
This energy expression is identical to one for circular orbits so that the mere introduction of elliptic orbits adds no new energy levels and hence does not explain the
fine structure.
From analysis we get
√
b
k
1 − 2 = =
a
n
where b=semi-minor axis and a is semi-major axis of the ellipse.
Hence, only those elliptical orbits are permitted for the electron for which the
ratio of the major and minor axis is the ratio of two integer. This is the condition of
quantization of the orbits. As
b≤a
b
≤1
a
Then
b
k
= ≤1
a
n
k≤n
Circular orbit:
when b = a, then k = n, = 0, Hence nr = 0 , as n = k + nr
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k = 0, This means that the electron executes S.H. M. through the nucleus. But
this is impossible.
Hence k 6= 0. So, the permissible values of k are
k = 1, 2, 3, 4, .......
The values of
nr = n − k
nr = 0, 1, 2, 3, 4, .....(n − 1)
Orbit:
Case I: If n = 1, then nr = 0, k = 1, we get circular orbit.
Case II: If n = 2, then,
nr = 0, k = 2, → circular orbit.
nr = 1, k = 1, → elliptical orbit.
nr = 2, k = 0, → not possible
Case III: If n = 3, then,
nr = 0, k = 3, → circular orbit.
nr = 1, k = 2, → elliptical orbit.
nr = 2, k = 1, → elliptical orbit
Degeneracy: Except for the ground state, all other states of H-atom have multiple orbits and shows a degeneracy. For the higher energy states, the larger will be
degeneracy in the state of higher energy. e.g. For n=2, we get two fold degeneracy.
for n=3, three fold degeneracy. Because of this degeneracy Sommerfeld theory fails
to explain the fine structure of H-atom. Because , the total energy is found to be
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independent of azimuthal quantum number K. i.e. obtained total energy
mZ 2 e4
1
En = −
2
2
32π 2 0 h̄ n2
!
Sommerfeld considered the relativistic effects to remove degeneracy and explain fine
structure.
Relativistic correction and fine structure:
This correction should be done due to following reasons.
The speed of the electron in elliptic orbit changes with its position, speeding up
when closer to the nucleus and slowing down when far away from it. So there is a
variation of mass of electron according to the relativistic relation
m0
m= q
1 − v 2 /c2
This variation of mass is sufficiently to demand the relativistic correction.
Taking relativistic effects into consideration, the energy of hydrogen -like atom is
given by
mZ 2 e4
Z 2 α2 1
3
En = −
1
+
−
2 2 2
2
n
k 4n
32π 0 h̄ n
where α is the fine structure constant defined by
α=
!!
e2
1
=
4π0 ch
137
a dimensionless universal constant.
For a given principal quantum number n, k can have n different values leading
to different states with slightly different energies. So, we shall have a number of
transitions in place of a single one of Bohr’s theory. Thus, the lines generated are
the fine structure. But all possible transition are not allowed. Only those for which
k changes by unity are allowed. i.e.
∆k = ±1
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which is the selection rule for a allowed transition. There is no restriction on the
change in n.
Short coming of Bohr-Sommerfeld model:
(i) It applies only to H-atom
(ii) It provides no method for calculation of intensity of spectral line.
(iii) Even the Sommerfeld relativistic theory is partially successful and in order to
obtain this partial success, a selection rule ∆k = ±1 had to be imposed.
Problem: Show that in a Bohr atom if the electron is considered as a wave travelling along the circular path, then the n-th orbit will contain n complete de Broglie
waves.
We have the radius rn of the n-th Bohr’s circular orbit
rn =
n2 h2 πme2
So, circumference of the n-th Bohr orbit is
2πrn = 2π
n2 h2 πme2
Velocity of electron in n-th orbit is
v = vn =
e2
2nh0
1
2nh0
=
v
e2
The de Brogli wavelength is given by
λ=
h
h 2nh0
=
mv
m e2
17
Number of complete de Broglie waves in n-th orbit is
2 2
h 2π nπme
2
h 2nh0
m e2
=n
Vector Atom model:
Introduction:
(i) Bohr’s theory was able to explain only the series spectra of the H2 atom. It
could not explain the multiple structure of spectral lines in H2 atom.
(2) Sommerfeld’s theory could not predict the correct number of the fine structure
lines. It cannot give any information about the relative intensities of the lines. It can
not explain the complex spectra of alkali metals like sodium.
(3) These theories can not explain zeeman effect, Stark effect.
(4) Bohr model can not explain how the orbital electrons in an atom were distributed around the nucleus.
Vector atom model: In order to explain the complex spectra of atoms and their
relation to atomic structure, the vector atom model was introduced.
The two distinct features of the vector atom model are
(a) The conception of space quantization: The motion of an electron in
elliptical orbits which are two-dimensional and the electron have possesses two degree
of freedom. Hence, only two quantum number are sufficient to define electron orbit
or energy state of an atom.
But in general, the motion of the electron in atom is three dimensional and therefore possesses three degree of freedom. Hence an additional quantum number and
a corresponding quantum condition is necessary to describe the true state of affairs.
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The third quantum condition quantises the orientation of the elliptical orbit in three
dimensional space and does not alter the original Sommerfeld orbits in regard to their
size and shape. When an electron moves in three-dimensional orbit it possesses the all
orientation w.r.t a fixed direction. But according to space quantisation, only certain
discrete orientation are allowed . Thus we observe the need of a fixed direction of
space.
For this , to specify the orientation of the orbit in space, we must need a fixed
reference axis. This reference line is chosen as the direction of an external magnetic
field that is applied to the atom. The different permitted orientations of an electron
orbit are determined by the projections of the quantised orbits on the field direction
must be quantised.
As a result of applied field, the plane of the orbit precess about the magnetic
field and orbital angular momentum pl generates a cone with fixed angle θ. The
precessional frequency depends upon the magnetic field and given by
ν=
eH
4πmc
(b) The spining electron hypothesis: According this hypothesis the electron spins
about an axis of its own, while it moves round the nucleus of the atom in its orbits.
According to the quantum theory, the spin of the electron also should be quantised.
Hence, a new quantum number called the spin quantum number (s) is introduced.
As the orbital and spin motions are both quantised in magnetic field and direction
according to the idea of spectral quantization, they are known as quantised vector.
Hence, the atomic model based on these quantised vectors is called the vector atom
model.
Quantum numbers in vector atom model:
(i) The Principal quantum number (n):
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The serial number of the shells starting from innermost is leveled as its principal
quantum no. (n). It can take only integral values. i.e.
n = 1(K), 2(L), 3(M ), 4(N )....
(ii) The orbital quantum number (l): According to Bohr’s theory, the orbital angular
quantum number is given by
~ = ~lh̄
Pl = L
According to wave mechanics
Pl ψ =
q
l(l + 1)h̄ψ
where l is the orbital quantum number, which takes up the values l = 0, 1, 2, 3, ......(n−
1).
~ drawn perpenThe orbital angular momentum of an electron can be represented by L
dicular to the plan of the electron orbit. If an atom contains more than one electron,
the total orbital angular momentum is the vector sum of the angular momentum
vectors of the individual electrons. i.e.
~ = ~l1 + ~l2 + ~l3 + ........
L
(ii) The spin quantum number (s):
The certain features of the atomic spectra such as fine structure of spectral lines
and the splitting of spectral lines in a magnetic field can be explained if it is assumed
that the electron spins about an axis passing through its centre of mass.
Electron spin gives rise to angular momentum ~s. The spin angular momentum of
the electron is intrinsic and it is independent of the orbital angular momentum. The
spin quantum number has only one value which is equal to 21 . The spining electron
has spining angular momentum
1
ps = sh̄ =
2
20
According to wave mechanics
ps ψ =
q
s(s + 1)h̄ψ
f an atom contains more than one electron, the total spin angular momentum is the
vector sum of the spin angular momentum vectors of the individual electrons. i.e.
~ = ~s1 + ~s2 + ~s3 + ........
S
Rules of addition:
~ is an odd multiple of 1 i.e. 1/2, 3/2, 5/2, 7/2, ......
1. For an odd number of electrons, S
2
~
2.For an even number of electrons S is an integeri.e. 0, 1, 2, 3...
These rules show that the spin vectors must be parallel or antiparallel. i.e. the
spins must point either in the same direction or in opposite directions. Examples:
(iii) Total angular momentum(~j):
Here we consider the spectral characteristics of an one electron atom. The orbital
angular momentum ~l and the spin angular momentum ~s of the electron in an atom
combine to give a total angular momentum ~j . According to the vector atom model
~j is the vector sum of ~l and ~s. i.e.
~j = ~l + ~s
The magnitude of J~ is given by
|~j| =
q
j(j + 1)h̄
21
The total angular momentum quantum number for single electron can have values
j = l + s; j = l − s
as s = 12 , then
1
1
j = l + ;j = l −
2
2
In case, l = 0, then j = 1/2
If l = 1, j = 1 ± 21 , i.e. 3/2,1/2
If l = 2, j = 2 ± 12 i.e. 5/2,3/2
(iv) Magnetic orbital quantum number (ml :
The projection of the orbital quantum no. l on the magnetic field direction (z-axis)
is called the magnetic orbital quantum no. ml . i.e.
Lz = ml h̄
where
θ = cos
−1
lz
l
ml = lcosθ
ml h̄
−1
= cos
q
= cos
l(l + 1)h̄
−1
ml
q
l(l + 1)
Here, ml can take up (2l + 1) values from −l to +l in steps of unity.
For example: for l = 3, ml has 2.3 + 1 = 7 values.
These are ml = −3, −2, −1, 0, +1, +2, +3 Here the angle θ between l and applied
magnetic field (B) is given by
θ = cos−1
ml
l
(v) Magnetic spin quantum number (ms ):
22
This is the projection of the spin vector ~s along the direction of the applied
magnetic field. i.e.
Sz = ms h̄
ms has (2s + 1) values, i.e. ± 21 .
θ = cos
−1
sz
s
= cos
−1
ms h̄
q
= cos
−1
ms
q
s(s + 1)h̄
s(s + 1)
Hence the energy of the two atomic states corresponding to
s
values ± 12 will be
degenerate. The atom when placed in a magnetic field (z-axis), the states for ms = + 12
and ms = − 21 will split into two distinct states.
The ms = + 21 state is known as the spin ”up” state
and the ms = − 12 is the spin ”down” state.
(vi) Total magnetic quantum number (mj ):
The projection of the total angular momentum j along the direction of the applied
magnetic field is also quantized and have (2j +1) values between +j and −j excluding
zero.
i.e.mj = −j, −j + 1, ..... + j − 1, +j Here mj = jcosθ
θ = cos−1
jz
j
= cos−1 q
mj h̄
j(j + 1)h̄
= cos−1 q
mj
j(j + 1)
We have
jz = Lz + Sz
mj h̄ = ml h̄ + ms h̄
mj = ml + ms
As the largest possible values of ml is l and that of ms is 1/2. So the largest possible
value of mj is
!
mj
max
1
=l+ .
2
23
We vector inequality
|~l| + ~s| ≥ |~l| − ~s|
|~j| ≥ |~l| − ~s|
q
j(j + 1)h̄ ≥
q
l(l + 1)h̄ −
q
s(s + 1)h̄
when l = 0, j = 1/2. When l 6= 0, there are only two members of the series satisfying
the inequality. i.e. j = l +
1
2
and j = l −
1
2
Notation: The notation used to describe the different atomic states is
n2s+1 Lj
Here n is the principal quantum number
j the total angular momentum quantum number
L the orbital angular momentum quantum number
Different L-values are represented by different symbols denoting energy states.
L = 0(S), 1(P ), 2(D), 3(F ), 4(G), 5(H).........
Example: The ground state of the H-atom.
n = 1, l = n − 1 = 0, s = 1/2andj = l + 1/2 = 1/2................1S1/2 State
For the first excited state
n = 2, l = 0, s = 1/2, j = 1/2...............................2S1/2
n = 2, l = 1, s = 1/2, j = 3/2, 1/2...............................2P3/2 , 2P1/2
~ and S
~
Angles between L
24
We have
~ +S
~
J~ = L
~ J~ = (L
~ + S).(
~ L
~ + S)
~
J.
~ S
~
J 2 = L2 + S 2 + 2.L.
J 2 = L2 + S 2 + 2LScosθ
J 2 − L2 − S 2
cos(L, S) = cosθ =
2LS
j(j + 1) − l(l + 1) − s(s + 1)
q
q
cos(L, S) = cosθ =
2 l(l + 1) s(s + 1)
∗∗∗ The capital letters S, P, D... describe the orbital angular momentum of atomic
states.
For individual electrons are described by small letters s, p, d, ......
Selection rules:
Spectral transitions:
The transition of a system from one state to another
under the influence of electromagnetic radiation depends on the interaction of the
electronic field radiation with the dipole moment of the atom (or molecule).
The transition from state m to state n is determined by the transition dipole
moment
µmn =
Z
ψ ∗ µ̂ψdv
where ψ are the wave functions of the corresponding state and µ̂ the dipole moment
operator. If µmn is finite value then the transition is allowed. otherwise, it is forbidden.
So allowed transitions follow some selection rules. i.e.
∆n = any
value
25
∆l = ±1
∆ml = 0, ±1
Transitions in violation of the selection rules are forbidden as they are less probable.
Change in the state of the atom during transition implies that the photon must
possess energy, linear momentum and angular momentum. e.g. ∆l = ±1 suggests
that the photon carries on unit, h̄, of angular momentum.
An electron cannot jump from one energy level to all other energy levels. A
transition of an electron between two levels is possible only if certain rules called
selection rules are satisfied. For the vector atom model , three selection rules are
there.
(i) Selection rules for L:
Seletion rule for L is ∆L = ±1
e.g. L = 0(S) → 1(P ), ∆L = 1 possible
L = 1(P ) → 0(S), ∆L = −1 possible
L = 0(S) → 2(D), ∆L = 2 not possible
L = 2(D) → 0(S), ∆L = −2 not possible
(ii) Selection rule for J:
The selection rule for J is ∆L = ±1 and ∆J = 0. But transition from ∆J = 0 →
∆J = 0 excluded.
(iii) Selection rule for S:
26
The selection rule for S is ∆S = 0 which forbids transition between the single and
the triplet states.
Hund rules:
When the spin-orbit interaction splits the energy level corresponding to different
J values, the lowest J value will be the lowest in energy.
Magnetic moment of an orbital electron:
According to Sommerfeld’s theory of atomic spectrum, the electrons move in elliptical orbits of various eccentricity,
with the nucleus at one of the foci. The angular
q
momentum of the electron L =
l(l + 1)h̄ at any instant of time is given by
L = mr2 ω = mr2
dθ
dt
L
dθ
= r2
m
dt
1 dθ
1L
= r2
2m
2 dt
We know the areal velocity
dA
1 dθ
= r.r
dt
2 dt
Z T
dθ
1
r.r dt
A=
2 0
dt
Z T
1
L
A=
dt
2 0 m
LT
A=
2m
Let the electron is moving in a closed plane orbit of area A. So current in this loop
is given by
i=
e
T
27
Here e is the charge of the electron. According to the Ampere’s laq, this current gives
rise to a magnetic dipole of moment
µl = iA =
eA
T
e LT
T 2m
e ~
L
µ
~l =
2m
~ = ~lh̄, so the magnetic dipole moment
As the angular momentum is quantised, L
µl =
µ
~l =
where µB =
eh̄
.
2m
~l
eh̄ ~l
= gl µB
2m h̄
h̄
As charge e is negative, the magnetic dipole moment µl is directed
opposite to that of angular momentum. The ratio of magnetic dipole moment and
angular momentum is called the gyromagnetic ratio. i.e. gyromagnetic ratiogl = 1
This is a numerical factor.
~ is given by
The magnetic moment associated with spin angular momentum S
µ
~s = −
~
e~
eh̄ S
S = −2
m
2m h̄
µ
~ s = −2µB
~
~
S
S
= −gs µB
h̄
h̄
For this case gs = 2.
Magnetic dipole in a magnetic field:
~ So, the energy
Let a magnetic dipole of moment µ
~ be placed in a magnetic field B.
of interaction between the magnetic moment and the magnetic field, given by
~
U = −~µ.B
~ is in the z-direction, then
If B
~ = µl Bcos(π − θ)
U = −~µ.B
28
U=
where ω =
eB
2m
eB
eB
h̄lcosθ =
h̄ml
2m
2m
U = ωml h̄
is called Larmor frequency.
Larmor theorem:
The effect of a magnetic field on an electron moving in an orbit is to superimpose
on the orbital motion , a precessional motion of the entire orbit about the direction
eB
of the magnetic field take place with angular frequency ω = 2m
.
Larmor precession:
~ it
If a dipole of magnetic moment µl be placed in an external magnetic field B,
experiences a torque
~ = µl Bsinθ
τ =µ
~l × B
We have from mechanics
~
dL
dt
Hence the change in angular momentum is in the direction of ~τ . Since the magnitude
~ remains the same, its direction must change to produce the change in L.
~ Thus
of L
~τ =
~ requires the precession of L
~ about the magnetic field with frequency
the change dL
ω=
dφ
dt
From Fig. dl = Lsinθdφ, so we get
ω=
ω=
dφ
1 dL
=
dt
L sin θ dt
1
1
τ=
µl Bsinθ
L sin θ
L sin θ
µl B
eB
ω=
=
L
2m
29
Stern-Gerlach Experiment:
The orbital and spin motions of the electrons in atoms endow the atoms with
magnetic moments. Direct evidence for the existence of magnetic moment of atoms
and their space quantization is provide by this experiment.
Principle: The experiment is based on the behaviour of a magnetic dipole in a
non-uniform magnetic field. In a uniform magnetic field (B), the dipole experiences a
torque that tends ti align the dipole parallel to the applied field. If the dipole moves
in such a field in a direction normal to the field, it will trace a straight line without
any direction.
In an inhomogeneous magnetic field, the dipole experiences, a translatory force.
If the atomic magnet flies across such an inhomogeneous magnetic field normal to the
field direction, it will be deviated away from its path.
Deviation calculation:
Let a magnetic field (B) vary along x-axis. and field gradient=
dB
dx
CD is an atomic magnet of length l and pole strength p. Hence dipole moment
µ = pl. Let θ is the angle between the atomic magnet and field direction.
The magnetic field at c= B
the magnetic field at D= B +
dB
lcosθ
dx
30
!
So, force on the pole at C= Fc = pB and force on the pole at D, FD = p B + dB
lcosθ
dx
So, translatory force on the atomic magnet
!
dB
Fx = FD − Fc = p B +
lcosθ − pB
dx
Fx = p
dB
lcosθ
dx
Now, acceleration of the atom
Fx
m
So, the displacement of the atom along the field direction
ax =
1
1 Fx L
S = ax t 2 =
2
2m v
!2
where time travel t = L/V , L is length of the path of the atom in the field. v is the
velocity of the atomic magnet of mass m. Hence
1 dB plcosθ L
S=
2 dx m
v
1 dB µcosθ L
S=
2 dx m
v
1 dB µx L
S=
2 dx m v
Experimental arrangement is given in Fig.
31
!2
!2
where component of magnetic moment µx = µcosθ
Experimental arrangement :
!2
Procedure:
(i) Silver is boiled in an oven and atoms of silver stream out from an opening in
the oven.
(ii) By the use of slits S1 andS2 , a sharp linear beam of atoms is obtained.
(iii) These atoms then pass through a very inhomogeneous magnetic field between
the sharped poles of a magnet MM.
(iv) The magnetic field is at right angles to the direction of movement of the
atoms. Finally the atoms fall on a photographic plate P.
(v) The whole arrangement is enclosed in an evacuated chamber.
Observation:
(i) With no field, the beam produces a narrow continuous line on the plate.
In terms of vector atom model, these atoms will experiences a force in one direction. Because, electron spin is parallel to the magnetic field.
(ii) When the beam of atoms pass through the inhomogeneous magnetic field, it
was observed that the stream of silver atoms splits into two separate lines on the
plate.
This splitting of the beam into two parts of approximately equal intensity was
observed in this experiment.
According to the vector atom model , the beam of atoms should split into two
beams in its passage through the inhomogeneous field . Because, for this case, the
spin and magnetic field are in opposite direction i.e. antiparallel.
32
(iii) Knowing dB/dx,L, V andS, µ can be calculated. It is found that each silver
atom had a magnetic moment of one Bohr magneton in the direction of the field.
Pauli’s exclusion Principle:
We know that the state of an electron is completely specified by the four quantum
number n, l, ml , ms .
Pauli’s exclusion principle states that no two electrons in any atom can be in the
same quantum state, that is , no two electrons in any atom can have the same four
quantum numbers.
The Photon:
The basic equation for energy of a photon is
E = hν = h̄ω
where ω = 2πν
The modified Planck’s constant h̄ =
h
2π
The velocity of photon is the same as that of light. So, a photon is a relativistic
particle and obys the relativistic energy equation
E 2 = p2 c2 + m20 c4
But rest mass of photon m0 = 0.Hence
E 2 = p 2 c2
E = pc
So, momentum of photon
p=
E
hν
h
=
=
c
c
λ
Again , we have
E = mc2
33
asc =
ν
λ
So, mass of the photon
m=
E
hν
= 2
2
c
c
The Compton effect:
When a beam of monochromatic X-rays or γ-rays is scattered by a light element
such as carbon, the scattered radiation in a particular direction consists not only of the
unmodified incident radiation, but also of another new modified radiation of slightly
shorter frequency or longer wavelength. Scattering with a modified wavelength is
called incoherent scattering. This type of incoherent X-ray or γ-ray scattering is
called the Compton effect.
Importance:
Compton effect provides the most direct evidence for the particle nature of radiation and the fundamental validity of quantum theory.
Compton found that the radiation scattered through a given angle consists of two
components.
(i) one whose wavelength (λ0 ) is same as that of incident radiation.
(ii) other wavelength (λ0 ) shifted relative to the incident wavelength by an amount
that depends on the angle.
Compton measured the dependence of scattered X-ray intensity on wavelength at
three different scattering angle θ = 00 , 450 , 900 , 1350 which are shown in fig.
34
These plots show that there are two peaks.
(i) Peak at λ0 :
The peak at λ0 is called by X-ray scattered from electrons that are tightly bound
to the target atoms.
(ii)Shifted peak at λ0 :
The shifted peak is caused by scattering of X-rays from free electrons in the target.
Compton was able to explain the modified component λ0 by treating the incoming
radiation as a beam of photons of enegy hν. Here, the individual photon scattered
elastically with individual electrons.
Calculation:
We know that for relativistic particle energy
E=
q
p2 c2 + m20 c4
For photo, rest mass m0 = 0
Energy of photon E = pc = hν
and momentum p = hν
c
35
Before collision:
hν
c
Let the momentum of incident photon p =
the energy of incident photon E = hν
the momentum of electron=0
and energy of electron= m0 c2
After collision
the momentum of scattered photon p0 =
the energy of scattered photon E = hν 0
hν 0
c
the momentum of electron=P
q
and energy of electron=
P 2 c2 + m20 c4
In elastic collision, we know that momentum and energy are conserved. Hence, we
get from conservation of momentum
p~ = p~0 + P~
P~ = p~ − p~0
P~ .P~ = (~p − p~0 ).(~p − p~0 )
P 2 = p2 + p02 − 2~p.~p0
P 2 = p2 + p02 − 2pp0 cosθ
hν 2
hν 0
+
c
c
From conservation of energy , we get
P2 =
2
−2
hν + m0 c2 = hν 0 +
q
q
hν
c
2 hν 0
c
2
cosθ
(1)
P 2 c2 + m20 c4
P 2 c2 + m20 c4 = hν − hν 0 + m0 c2
Squaring both sides we get
P 2 c2 + m20 c4 = (hν − hν 0 )2 + m20 c4 + 2(hν − hν 0 )m0 c2
P 2 c2 = (hν − hν 0 )2 + 2(hν − hν 0 )m0 c2
hν
P =
c
2
2
hν 0
+
c
2
hν
−2
c
36
2 hν 0
c
2
+ 2(hν − hν 0 )m0 c2
(2)
From (1) and (2), we get
hν
c
2
hν
c
2
+
hν 0
+
c
hν 0
c
2
−2
hν
c
2 hν 0
c
2
cosθ =
hν 0 2
+ 2(hν − hν 0 )m0
c
h2 νν 0
2hm0 (ν − ν 0 ) = 2 2 (1 − cosθ)
c
0
ν−ν
h
=
(1 − cosθ)
0
νν
m0 c2
h
1
1
− =
(1 − cosθ)
0
ν
ν
m 0 c2
2
hν
−2
c
2 λ0 λ
h
− =
(1 − cosθ)
c
c
m0 c2
h
(1 − cosθ)
λ0 − λ =
m0 c
Hence, compton shift
∆λ = λ0 − λ =
Here,
h
m0 c
h
(1 − cosθ)
m0 c
= λc is known as compton wavelength of the electron, because m0 is the
rest mass of electron.
Energy of scattered Photon:
We have relation
1
h
1
−
=
(1 − cosθ)
ν0 ν
m 0 c2
1
ν0 = 1
h
+ m0 c2 (1 − cosθ)
ν
hν 0 =
E0 =
hν
1+
hν
(1
m0 c2
1+
E
(1
m0 c2
− cosθ)
E
− cosθ)
Here, E 0 is the energy of scattered photon. From this equation , we see that E 0 < E.
37
Kinetic energy of recoil electron:
K.E. of recoil electron
Ek.E. = E − E 0
E
=E−
E
1 + m0 c2 (1 − cosθ)
= hν −
Ek.E.
hν
1+
hν
(1
m0 c2
− cosθ)
1
= hν 1 −
hν
1 + m0 c2 (1 − cosθ)
38
!