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8/14/2019 Atomic Physics Official Slide
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ATOMIC PHYSICS
1BE-PHYSICS-ATOMIC PHYSICS-2010-11MIT- MANIPAL
TOPICS
Text BookPHYSICS for Scientists and
Engineers with ModernPhysics (6thed)
By Serway &Jewett
Atomic spectra of gases
Early models of the atom
Bohrs model of the
hydrogen atom
The quantum model of
the hydrogen atom
The wave functions forhydrogen
Physical interpretation of
the quantum numbers
The X-ray spectrum of
atoms X-rays and the
numbering of the
elements
Lasers and laser light
TOPICS
Text BookPHYSICS, 5TH Edition Vol 2
Halliday, Resnick, Krane
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ATOMIC SPECTRA OF GASES
Emission spectra: All objects emit thermal radiation
characterized by a continuous distribution ofwavelength (continuous spectrum).
When a gas at low pressure is subjected to anelectric discharge it emits radiations of discrete
wavelengths (line spectrum).
No two elements have the same line spectrum. This
principle is used in identifying the element by
analyzing its line spectrum. H
Hg
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The wavelengths of the Balmer series lines in the
hydrogen spectrum are given by the (empirical) equation
n = 3, 4, 5, . . .
Rydberg constant RH= 1.097 x 107/m
Absorption spectra: An absorption spectrum is obtained
by passing white light from a continuous source through
a gas or a dilute solution of the element being analyzed.
The absorption spectrum consists of a series of dark
lines superimposed on the continuous spectrum of the
light source.
SOLAR SPECTRUMFRAUNHOFER LINES
VISIBLE HYDROGEN SPECTRUMBALMER SERIES LINESH(656.3 nm) H(486.1 nm)H(434.1 nm) H(410.2 nm)
22
1
2
11
nRH
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The wavelengths of the other series lines in the
hydrogen spectrum are given by the equation
LymanSeries n = 2, 3, 4, . . .
Paschen
Series n = 4, 5, 6, . . .
Brackett
Series n = 5, 6, 7, . . .
Although no theoretical basis existed for these
equations, they are in agreement with the
experimental results.
2111n
RH
22
1
3
11
nRH
22
1
4
11
nRH
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EARLY MODELS OF THE ATOM
J. J. Thomson
1897
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EARLY MODELS OF THE ATOM
Ernest Rutherford
1911
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EARLY MODELS OF THE ATOM
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BOHRS MODEL OF THE HYDROGEN ATOM
In his semi classical model of the H-
atom Niels Bohr (1913) postulated that:
[1] The electron moves in circular
orbits around the proton under theinfluence of the electric force of
attraction as shown in the figure.
v+e
mee
r
F
[2] Only certain electron orbits are stable (stationary
states). When in one of these stationary states, the atom
does not radiate energy. Hence the total energy of the
atom remains constant in a stationary state.
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[3]When the atom makes a transition
from higher energy state (Ei) to lower
energy state (Ef) [i.e, the electronmakes a transition from a stable orbit
of larger radius to that of smaller
radius], radiation is emitted. Thefrequency (f) of this radiation (photon)
is given by
EiEf= hf
The frequency f of the photon
emitted is independent of the
frequency of electrons orbital motion.
v+e
mee
r
F
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[4]The angular momentum of the
electron in any stable orbit isquantized
----- (1)
me = mass of the electron
v = speed of the electron in theorbit
r = radius of the electrons orbit
v+e
mee
r
F
2
h
...,3,2,1 nnvrme
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Electric potential energy of the H-
atom is
r
ekU e
2
v+e
mee
r
F
Apply Newtons 2nd law to the electron, the electric
force exerted on the electron must be equal to the
product of mass and its centripetal acceleration (ac=v2/r)
-------(2)
r
vmamF
r
ek ece
e
2
2
2
r
ekvmK ee
22
22
ke= Coulomb constant
229
0
/.1099.84
1CmNke
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r
ekE
r
ek
r
ekUKE
e
ee
2
2
2
22
The total energy of the H-atom is
2
22
222
rm
ek
rm
nv
e
e
e
,....3,2,1
2
22
n
ekm
nr
ee
n
Thus the electron orbit radii are quantized
pmekm
aee
o
9.522
2
Bohr radius
(n = 1)
+e
e4a
o
ao
9ao
Negative energy indicates bound electron-proton system.
From eq(1) and (2) nvrme rekvm ee
22
22
rn= n2 ao
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Energy quantization
Substitute rn = n2 ao in the total energy equation
2
22 1
22 na
ek
r
ekE
o
een
n = 1, 2, 3, . . .
...,3,2,1606.132
neVn
En
E1 =13.606 eV
2
1
n
EEn
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I i ti i i i d t i i
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Ionization energy = minimum energy required to ionize
the atom in its ground state
= 13.6 eV for H-atom
From the equation EiEf= hf
Frequency of the photon emitted during transition of theatom from state i to state f is
22
2 11
2 ifo
efi
nnha
ek
h
EEf
Use c = f
22
111
if
Hnn
R
22
2 11
2
1
ifo
e
nncha
ek
c
f
cha
ekR
o
eH
2
2
RH= 1.097 x 107
/m
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Bohrs correspondence principle:
Quantum physics agrees with classical physics
when the difference between quantized levels
becomes vanishingly small.
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PROBLEMS
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Formulae:
...,3,2,12 2
22
n
n
Z
a
ekE
o
en
22
111
if
H
nnR
pmnanrn 9.522
0
2
rm
ekv
e
e
22
r
ekE e
2
2
r
ekrU e
2
)(
ke= 8.99 x 109N.m2/C2
is Coulomb constant
RH= 1.097 x 107/m
04
1
ekwhere
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[42.1 P-1] (a) What value of n is associated with the
94.96nm spectral line in the Lyman series of
Hydrogen ? (b) Could this wavelength be associated
with the Paschen or Balmer series ?
SOLUTION:
(a)Lyman Series
2
11
1
nRH
27
9
1110097.1
1096.94
1
n
5n
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(b) Paschen Series
22
1
3
11
nRH
2
7 1
9
110097.1
1
n
The shortest wavelength for this series corresponds to n =
for ionization. For n = ,gives = 820 nm. This is larger than
94.96 nm, so this wavelength cannot be associated with the
Paschen series
Balmer Series
22
1
2
11
nRH
2
7
1
4
110097.1
1
n
with n = for ionization, =365 nm. Once again the shorter;
given wavelength cannot be associated with the Balmer
series
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[SP 42.1] Spectral lines from the star -Puppis :
Some mysterious lines in 1896 in the emission
spectrum of the star -Puppis fit the empiricalequation
22
2
1
2
11
if
H
nnR
Show that these lines can be explained by the
Bohrs theory as originating from He+.
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SOLUTION: The ion He+ has Z = 2, Thus allowed energy
levels are given by
...,3,2,12 2
22
nn
Z
a
ekE
o
en
2
2
42 naekEo
en
22
2 44
2 ifo
efi
nnha
ek
h
EEf
22
2
2
1
2
1
2ifo
e
nnha
ekf
222
1
2
11
if
H
nnR
c
f
cha
ek
Rwhereo
e
H 2
2
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[SP 42.2] (A)The electron in a H-atom makes a transition
from the n=2 energy level to the ground level (n=1). Find
the wavelength and the frequency of the emitted photon.
(B) In interstellar space highly excited hydrogen atoms
called Rydberg atoms have been observed. Find the
wavelength to which radio-astronomers must tune to detect
signals from electrons dropping from n=273 level to n=272.
(C) What is the radius of the electron orbit for a Rydberg
atom for which n=273 ?
(D) How fast is the electron moving in a Rydberg atom for
which n=273?
(E) What is the wavelength of the radiation from the
Rydberg atom in part (B) if treated classically ?
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SOLUTION(A)
22
111
if
Hnn
R
22 2
1
1
11HR
4
3 HR
HR34
)(5.12110215.1 7 tultraviolenmm
Hzc
fFrequency 151047.2
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SOLUTION(B)
22
111
if
Hnn
R
22 273
1
272
11HR
m992.0
SOLUTION(C)
rn = n2 ao = 273
2x(0.0529nm)
r273 = 3.94m
pmekm
aee
o 9.522
2
k 2 2
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SOLUTION(D)
SOLUTION(E)
We have speed vand radius rfrom(C) and (D)
r
v
Tf
2
1
rm
ekv
e
e
22
)1094.3)(1011.9(
)1060.1)(1099.8(631
2199
v
rm
ekv
e
e
2
smv /1001.8 3
Hzxr
v
Tf 81024.3
2
1
mf
c
926.0
rTrv
2
[42 2 P 3] A di t l i l h i h
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[42.2 P-3] According to classical physics, a charge e
moving with an acceleration a radiates at a rate
(a) Show that an electron in a classical hydrogen
atom spirals into the nucleus at a rate
(b) Find the time interval over which the electron
will reach r = 0, starting from ro= 2 x 1010m.
3
22
61
cae
dtdE
o
32222
4
12 cmr
e
dt
dr
eo
SOL A Th t t l i i b
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SOL: A The total energy is given by,
32222
4
12Therefore
cmr
e
dt
dr
eo
r
ekE e
2
2
04
1
ekwhere
r
eE
o8
2
The centripetal acceleration ais given by
8
61
2
2
3
22
er
cae
dtdr o
o
3
22
2
2
6
1
8 c
ae
dt
dr
r
e
dt
dE
oo
3
22
68
car
dtdr
rmekve
e
2
2
SOL B
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SOL:B
32
e
22
o
2
4
cmr12
e
dt
dr
T
x
eo dtedrcmr0
4
0
1000.2
32222
10
12
Tr
e
cmx
eo
101000.2
0
3
4
3222
3
12
[ ]
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[42.3 P-7]A hydrogen atom is in the first excited state
(n = 2). Using the Bohr theory of the atom, calculate
(a) the radius of the orbit(b) the linear momentum of the electron
(c) the angular momentum of the electron
(d) the kinetic energy of the electron(e) the potential energy of the system and
(f) the total energy of the system.
SOLUTION:
a) rn = n2 ao
r2 = 22x(0.0529nm) = 0.212 nm
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[42 3 P 9]
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[42.3 P-9] A photon is emitted as a hydrogen atom
undergoes a transition from the n = 6 state to the n = 2
state. Calculate
(a) the energy
(b) the wavelength
(c) the frequency of the emitted photon.Solution b:
S l ti
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Solution a:
Solution c:
[42 3 P 13] ( ) C t t l l di f
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[42.3 P-13] (a) Construct an energy-level diagram for
the He+ ion (Z = 2). (b) What is the ionization energy for
He+?Solution a:The energy levels of a hydrogen-like ion whose
charge number is Z are given by
Thus for Helium (Z = 2), the energy
levels are
(b) Wh t i th i i ti f H + ?
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(b) What is the ionization energy for He+?
Solution b: For He+ , Z = 2 , so we see that the ionization
energy (the energy required to take the electron from the n =1 to the n = state) is
THE QUANTUM MODEL OF THE HYDROGEN ATOM
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THE QUANTUM MODEL OF THE HYDROGEN ATOM
37
The potential energy function for the H-atom is
r
ek
rU e
2
)(
ke= 8.99 x 109N.m2/C2
is Coulomb constant
r = radial distance of electron from proton (at r = 0)
The time-independent schrodinger equation in three
dimensional space is
EUzyxm
2 2
2
2
2
2
22
Since U has spherical symmetry, it is easier to solvethe schrodinger equation in spherical polarcoordinates (r, , ):
where
is the angle between z-axis and
222 zyxr
r
P
x
z
r
i th l b t th i d th j ti f
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is the angle between the x-axis and the projection of
onto the xy-plane.
It is possible to separate the variables r, , as follows:
(r, , ) = R(r) f() g()By solving the three separate ordinary differential
equations for R(r), f(), g(), with conditions that thenormalized and its first derivative are continuous andfinite everywhere, one gets three different quantum
numbers for each allowed state of the H-atom.The quantum numbers are integers and
correspond to the three independent
degrees of freedom.
r
P
y
z
r
The radial function R(r) of is associated with the principal
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The radial function R(r) of is associated with the principal
quantum number n. From this theory the energies of the
allowed states for the H-atom are
2
21
2 na
ekE
o
e
n
...,3,2,1,
606.132
nn
eV
The polar function f() is associated with the orbitalquantum number l.The azimuthal function g() is associated with the orbital
magnetic quantum number ml.
The application of boundary conditions on the three parts of leads to important relationships among the three quantumnumbers:[1] n can range from 1 to .[2] l can range from 0 to n1; [n allowed values].
[3] ml can range froml to +l ; [(2l+1) allowed values].
which is in agreement with Bohr theory.
All states having the same principal quantum number
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All states having the same principal quantum number
are said to form a shell. All states having the same
values of n and are said to form a subshell:
n = 1 K shell = 0 s subshelln = 2 L shell = 1 p subshelln = 3 M shell = 2 d subshelln = 4 N shell = 3 f subshelln = 5 O shell = 4 g subshelln = 6 P shell = 5 h subshell. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
[SP 42 3]: For a H-atom determine the number of allowed
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Solution:When n = 2, can have the values 0 and 1.
If = 0, m can only be 0.
If = 1, m can be -1, 0, or +1.Hence, we have one 2s state with quantum numbers
n = 2, = 0, m = 0
and three 2p states for which the quantum numbers are
n= 2, =1, m =-1
n= 2, =1, m =0
n= 2, =1, m =+1
All these states have the same principal
quantum number, n=2, they also have the
same energy, En=(-13.66eV) Z2/n2
E2=-(13.66eV)/22= -3.401eV
[SP 42.3]: For a H-atom, determine the number of allowed
states corresponding to the principal quantum number n
= 2, and calculate the energies of these states.
[42 4 P 16]: A general expression for the energy
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[42.4 P-16]: A general expression for the energy
levels of one-electron atoms and ions is
where ke is the Coulomb constant, q1and q2are the
charges of the electron and the nucleus, and is
the reduced mass, given by
The wavelength for n = 3 to n = 2 transition of the
hydrogen atom is 656.3 nm (visible red light). Whatare the wavelengths for this same transition in (a)
positronium, which consists of an electron and a
positron, and (b) singly ionized helium ?
22
2
2
2
1
2
2 n
qqk
E e
n
21
21
mm
mm
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so the energy of each level is one half as large as in
hydrogen. The photon energy is inversely proportional to its
wavelength, so for positronium,
so the transition energy is 22= 4 times larger than hydrogen.
22
2
2
2
1
2
2 n
qqkE en
[42.4 P-17]: An electron of momentum p is at a distance
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[42.4 P 17]:An electron of momentum p is at a distance
r from a stationary proton. The electron has a kinetic
energy
The atom has a potential energy and total
energy E = K + U. If the electron is bound to the proton
to form a H-atom, its average position is at the proton,
but the uncertainty in its position is approximately equal
to the radius r of its orbit. The electrons average
vector momentum is zero, but its average squared
momentum is equal to the squared uncertainty in its
momentum, as given by the uncertainty principle.
em
pK
2
2
r
ekU e
2
Treating the atom as one dimensional system
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Treating the atom as one-dimensional system,
(a) estimate the uncertainty in the electrons momentum
in terms of r.
(b) Estimate the electrons kinetic, potential, and total
energies in terms of r.
(c) The actual value of r is the one that minimizes thetotal energy, resulting in a stable atom. Find that
value of r and the resulting total energy. Compare
your answer with the predictions of the Bohr theory.
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THE WAVE FUNCTIONS FOR HYDROGEN
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The potential energy for H-atom depends only on the
radial distance r between nucleus and electron.
Some of the allowed states for the H-atom can berepresented by wave functions that depend only on r
(spherically symmetric function).
The simplest wave function for H-atom is the 1s-state
(ground state) wave function (n = 1, = 0):
ao= Bohr radius.
|1s|2 is the probabilitydensity for H-atom in 1s-state.
o
o
s
ar
ea
r
3
1
1)(
o
o
s
ar
ea
2
3
2
1
1
The radial probability density P(r) is the probability
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The radial probability density P(r) is the probability
per unit radial length of finding the electron in a
spherical shell of radius r and thickness dr.
o
o
s
ar
ea
rrP
2
3
2
1
4)(
P(r) dr is the probability of finding
the electron in this shell.
P(r) dr = ||2 dV = ||2 4r2 drP(r) = 4r2||2
Radial probability density for H-atom in its ground
state:
Plot of the probability of finding the electron as a
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p y g
function of distance from the nucleus for H-atom in the
1s (ground) state. P1s(r) is maximum when r = ao (Bohr
radius).
Cross-section of the spherical electronic charge
distribution of H-atom in 1s-state
rMOST PROBABLE= ao
rAVERAGE= 3ao/2
The next simplest wave function for the H-atom is the
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rMOST PROBABLE = 5ao
p
2s-state wave function (n = 2, = 0):
o
oo
s
a
r
ea
r
ar
21
24
1
)(
2
3
2
2s is spherically symmetric.(depends only on r).
E2= E1/4 =3.401 eV
(1ST
excited state).
[SP 42.4]. Calculate the most probable value of r (=
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[SP 42.4]. Calculate the most probable value of r (
distance from nucleus) for an electron in the ground
state of the H-atom. Also calculate the average value r
for the electron in the ground state.
Solution:
The most probable distance is the value of r that makes the
radial probability P(r) a maximum. The slope of the curve (P
v/s r) at this point is zero, so the most probable value of r is
obtained by setting dP/dr= 0 and solving forr.
04)(
2
3
2
1
o
o
s ar
e
a
r
dr
d
dr
rdPo
o
s
ar
ea
rrP
2
3
2
1
4)(
4)(2
2 r
ddP
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04)(
3
2
1
o
o
s aea
r
dr
d
dr
rdP
022 22
o
aroar eedr
drrdr
d
0222 2
2
oaroar
ere )a(r
o
0]1[22
o
aroar
re
01 oar
oar
The expression is satisfied if
The most probable value of r is the Bohr radius
dxxx The expectation value is given by
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0
3
2
0
24
)( dra
rrdrrrPrr
o
o
av
a
r
e
The average value of r is the expectation value of r
0
3
3
24
drra
o
o
ar
e
ooo
aaa 2
3
/2
!3443
oav
ar2
3
*)( rPHere
dxxx The expectation value is given by
dxrr
0
[SP 42 ]
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Solution:
The probability is found by integrating the radial probability
density for this state, P1s(r), from the Bohr radius a0to .
o
o
s a
r
earrP
2
3
2
1 4)(
[SP 42.5]Calculate the probability that the electron in
the ground state of H-atom will be found outside the
Bohr radius.
r224
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We can put the integral in dimensionless form by changing
variables from rto z = 2r/a0. Noting that z = 2 when r = a0, and
that dr = (a0/2)dz, we get
o
o
s
aea
rrP
3
2
1
4)(
This is about 0.677, or 67.7%.
[42.5 P-21]: For a spherically symmetric state of a
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[42.5 P 21]: For a spherically symmetric state of a
H-atom the schrodinger equation in spherical
coordinates is
Show that the 1s wave function for an electron in
H-atom
satisfies the schrodinger equation.
o
o
s
ar
ea
r
3
1
1)(
Er
ek
rrrm
e
e
2
2
22 2
2
Solution:oar
er
1
)(
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o
se
a
r 3
1 )(
This is true , so the schrodinger equation is satisfied
Er
ek
rrrmhavewe e
e
2
2
22 2
2
o
o
oaadr
d ar
e 11
5
By Substituting the above values
PHYSICAL INTERPRETATION OF THE QUANTUM
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The orbital quantum numberAccording to quantum mechanics, an atom in a
state whose principal quantum number n can take
the following discrete values of the magnitude of
the orbital angular momentum:
NUMBERS
1,...,2,1,0)1( nL lll
The orbital magnetic quantum number m
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The energy U of the electron with a magnetic moment
in a magnetic field is According to
quantum mechanics, there are discrete directions allowed for
the magnetic moment vector with respect to magnetic field
vector
Since
one finds that the direction of is quantized. This means
that LZ the projection of along the z-axis [direction of ]
can have only discrete values. The orbital magnetic quantum
number m specifies the allowed values of the z-component
of the orbital angular momentum.
.B
B
.B-U
Lm
e
e
2
L
L
B
l
mLz
The quantization of the possible orientations of withL
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respect to an external magnetic field is called
space quantization. Following vector model describes
the space quantization for = 2.
B
THE ALLOWED VALUES OF LZ
LIES ON THESURFACE OF A CONEAND PRECESSES ABOUT
THE DIRECTION OF
L
B
is quantized 0)1(
mLcos Z
ll
l
L
The Zeeman effect:
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Splitting of energy levels and hence spectral lines
in magnetic field
ENERGY
n=1, =0
n=2, =1
hfohfo
h(fof)
h(fo+f)
m=0
m=0m=1
m=+1NO MAG-FIELD
MAG-FIELD PRESENT
fo fo (fo+f)(fof)
SPECTRUMWITHOUT
MAG-FIELD
SPECTRUM WITHMAG-FIELD
PRESENT
The spin magnetic quantum numberms
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p g q s
The quantum numbers n, l, ml are generated by applying
boundary conditions to solutions of the schrodinger
equation. The electron spin does not come from the
schrodinger equation. The experimental evidence showed the
necessity of the spin magnetic quantum number ms which
describes the electron to have some intrinsic angular
momentum. This originates from the relativistic properties of
the electron. There can be only two
directions for the spin angular
momentum vector spin-up and
spin-down as shown in the figure:
,S
Spin is an intrinsic property of a particle, like mass andh Th i l t it d S f th
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65
2
31 ssS
S
is quantized in space as
described in the figure:
It can have two orientations
relative to a z-axis, specified by
the spin magnetic quantum
number ms= .
The z-component of is :
SZ= ms= /2
S
charge. The spin angular momentum magnitude S for the
electron is expressed in terms of a single quantum number
(spin quantum number), s = (for electron) :
The value m = + is for spin up case and m =
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The value ms= + is for spin-up case and ms=is for spin-down case.The spin magnetic moment of the electron is
related to its spin angular momentum
Z-component of thespin magnetic moment:
Bohr magneton
Sm
e
e
SPIN
S
SPIN
e
SPIN,Zm
e
2
J/T.m
e
e
B
24102792
[SP 42.6]: Calculate the magnitude of the orbital
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[ ] g
angular momentum of an electron in a p-state of
hydrogen.
ll )1( L
2)11(1
sJ.1049.1 34
Solution:
with = 1 for a p state
[SP 42.7] Consider the H-atom in the = 3 state.C l l t th it d f th ll d l f L
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Calculate the magnitude of the allowed values of LZ,and the corresponding angles that makes withthe z-axis. For an arbitrary value of , how many
values of m are allowed.Solution:with = 3
,L||L
ll 32)13(3)1( L
32)1(cos ll
ll mmLZ
L
The allowed values of LZ is given by
LZ= m -3,-2,-,0, 1,2,3
SJ-Example-42.8For a H-atom, determine the quantumb i t d ith th ibl t t th t
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numbers associated with the possible states thatcorrespond to the principal quantum number n = 2.
n m ms subshell shell No of statesin subshell
--------------------------------------------------------------------
2 0 0
2 0 0 - 2s L 2
2 1 1
2 1 1 -
2 1 0 2p L 6
2 1 0 -
2 1 -1
2 1 -1 -
[42.6 P-27] How many sets of quantum numbers are
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possible for an electron for which (a) n=1, (b) n=2, (c)
n=3, (d) n=4, and (e) n=5 ? Check your results to show
that they agree with the general rule that the number of
sets of quantum numbers for a shell is equal to 2n2.
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THE X-RAY SPECTRUM OF ATOMS
History
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1895 : Wihelm ConradRoentgen discovered X-Rayswhile experimenting withdischarge tubes
X-unknown => X - radiation or
X rays
When a beam of fast moving
electron strikes on solid target
an invisible and high
penetrating radiation isproduced. These radiations are
called X rays.
y
X-ray spectrum
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X-ray spectrum
The X rays have wide range ofwavelengths (from 0.01 nm to10 nm) with the intensitydistributed over the entirerange.
Based on their characteristics& their origin, X-ray spectramay be classified as
a) Continuous X-ray spectrum
b) Characteristic X-ray spectrum minminmin
I
maxmaxmax
KK
kV40VkV30VkV20V
kV50V
min
Braking
lungBremsstrah
raysXContinuous
raysXsticCharacteri
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"Bremsstrahlung" means "braking radiation" and is
retained from the original German to describe the
radiation which is emitted when high energetic
electrons are decelerated or "braked" when they are
fired at a metal target..
Characteristic X-ray spectrum
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In an X-ray tube, an electron emittedfrom cathode strikes the target withtremendous velocity it may penetrateswell inside the atoms of the target andknockout one of the electrons frominner shell.
Immediately the transition of electron
from outer shell n2 to inner shell n1take place and the energy ( En2-En1)difference appears as X-ray photon offrequency
A K series of lines results from thetransition of electron from the highershell to K shell.
Ex: LK transitionK, M K transitionK
M L transitionL N L transition L
h
EE 1n2n
Similarly L seriesconsists of L L lines O
E0En
O5n
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consists of L, L lineswhen electrons jumpsfrom M, N shell to L
shell. The K, L, M, N
series constitute the X-rays spectrum which isthe characteristic of
particular material.
KK LL L
KE
O
LE
NE
ME
K1n
L2n M3n N4n O5
I
min K K LLL
Energy level diagramShell to shell transitions
Summary of continuous and characteristic X-rays
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Summary of continuous and characteristic X rays
To examine the motions of electrons that lie deep withinmulti-electron atoms one needs to consider the x-ray
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82
multi electron atoms, one needs to consider the x ray
spectrum of atoms, shown in the figure below:
The x-rays are emitted byatoms in a target when the
atoms are bombarded with high
energy electrons. The x-rayspectrum has two parts:
Continuous spectrum and
characteristic spectrum.
Sharply defined cutoff wavelength
(MIN) is a prominent feature of
the continuous x-ray spectrum.
TARGET: MOLYBDENUMX-RAY TUBE VOLTAGE:
V = 35 kV
MIN= 35.5 pm
Consider an electron accelerated through a potential
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difference of V (x-ray tube voltage), hitting a target atom.
The electrons initial kinetic energy is K = e V. The electron
loses its kinetic energy by an amount K = hf, which
appears in the form of x-ray photon energy (Bremsstrahlung).
K can have any value from 0 to K.
Thus the emitted x-rays can have any value for the
wavelength above MIN in the continuous x-ray spectrum.
Thus
MINMAX
ch
hfVe
Ve
chMIN
MIN depends only on V
The peaks in the x-ray spectrum have wavelengths
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characteristic of the target element in the x-ray tube and
hence they form the characteristic x-ray spectrum.
When a high energy (K = e V, V= x-ray tube voltage)
electron strikes a target atom and knocks out one of its
electrons from the inner shells with energy Em(| Em | K,
m = integer), the vacancy in the inner shell is filled up
by an electron from the outer shell (energy = En, n =
integer).
The characteristic x-ray photon emitted has the energy:
mn EEch
hf
X RAY ENERGY
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A Kx-ray results due to the transition of the electronfrom L-shell to K-shell.
A Kx-ray results due to the transition of the electronfrom M-shell to K-shell.
When the vacancy arises in the L-shell, an L-series (L,L, L) of x-rays results. Similarly, the origin of M-seriesof x-rays can be explained.
X-RAY ENERGY
LEVEL DIAGRAM
FOR MOLYBDENUM
EK= 17.4 keV
K= 71 pm
[HRK 48.1 P-1]: Show that the short-wavelengthcutoff in the continuous x-ray spectrum is given by
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cutoff in the continuous x ray spectrum is given by
where V is the applied potential
difference in kilovolts.
pm
VMIN
1240
Solution: The highest energy x-ray photon will have an
energy equal to the bombarding electrons,
Ve
chMIN
pm
V
1240
HRK-Sample Problem 48-1: Calculate the cutoff
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wavelength for the continuous spectrum of x-rays
emitted when 35-keV electrons fall on a
molybdenum target.
Solution:
Ve
chMIN
nmeVhc .1240
pmnm
eV
nmeVMIN
5.350355.0
1035
.12403
HRK 48.1 P-5: Electrons bombard a molybdenum target,
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producing both continuous and characteristic x-rays. If
the accelerating potential applied to the x-ray tube is
50kV, what values of (a) MIN (b) K (c) K result ? Theenergies of the K-shell, L-shell and M-shell in the
molybdenum atom are 20.0 keV, 2.6 keV and -0.4 keV,
respectively.
pmpmpmVMIN 8.2450
12401240
hc
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pmxxx
xxxEE
hc
hchEE
K
K
K
K
39.71106.110)6.220(
10310625.6193
834
12
12
pmxxx
xxxEE
hc
hchEE
K
K
K
K
37.63106.110)4.020(
10310625.6193
834
12
12
[HRK 48.1 P-9]: X-rays are produced in an x-raytube by a target potential of 50 keV If an electron
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mE
hc
hckeVE
electronincidentkeVE
Photon
Photon
o
12
193
834
1
1
1
1
1068.49106.11025
10310625.6
2
50
)(50
Solution
tube by a target potential of 50 keV. If an electronmakes three collisions in the target before coming torest and loses one-half of its remaining kinetic energy
on each of the first two collisions, determine thewavelengths of the resulting photons. Neglect therecoil of the heavy target atoms.
keVcollisionondthebeforeelectronofEnergy 25sec
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mE
hc
hckeVE
keVcollisionthirdthebeforeelectronofEnergy
Photon
Photon
12
193
834
3
3
3
3
10375.99106.1105.12
10310625.6
5.12
5.12
mE
hc
hckeVE
keVcollisionondthebeforeelectronofEnergy
Photon
Photon
12
193
834
2
2
2
2
10375.99106.1105.12
10310625.62
25
25sec
HRK 48.1 P-12: The binding energies of K-shell andL-shell electrons in copper are 8.979 keV and 0.951 keV,
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L shell electrons in copper are 8.979 keV and 0.951 keV,respectively. If a Kx-ray from copper is incident ona sodium chloride crystal and gives a first-order Bragg
reflection at 15.9 when reflected from the alternatingplanes of the sodium atoms, what is the spacingbetween these planes ?Solution:
K
2nL
1nK
keVBE 951.02
keVBE 979.81
2nL keVBE 951.02
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nmxxx
xxx
EE
hc
hchEE
K
K
K
K
154.0106.110)951.0979.8(
10310625.6193
834
12
12
.282)9.15sin(2
10154.0
sin2
1,,sin29
pmm
d
norderfirstfornd
K
1nK keVBE 979.81
Bohr theory and the Moseley plot: Bohrs formula
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for the frequency of radiation corresponding to a
transition in a one-electron atom between any two
atomic levels differing in energy by E is
2232
42 11
8 ifo nnh
eZm
h
Ef
In a many-electron atom, for a K transition, theeffective nuclear charge felt by an L-electron can be
thought of as equal to +(Zb)e instead of +Ze, where
b is the screening constant due to the screening effect
of the of the only K-electron.
Frequency of the K x ray is
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MOSELEY PLOT OFTHE K X-RAYS
bZhemfand
o
2
1
32
4
323
Frequency of the Kx-ray is
2232
42
21
11
8 hebZmf
o
1since b
2
1
32
4
32
3
h
emCwhere
o
1 ZCf
X-RAYS AND THE NUMBERING OF THE ELEMENTS
Moseleys observation on the characteristic K x rays shows
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Moseley s observation on the characteristic Kx-rays shows
a relation between the frequency (f) of the K x-rays and
the atomic number (Z) of the target element in the x-ray
tube:MOSELEY PLOT OF
THE K X-RAYS
1 ZCf
C is a constant.
Based on this observation,the elements are arrangedaccording to their atomicnumbers in the periodic
table.
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HRK-Sample Problem 48-3: A cobalt (Z=27) target isbombarded with electrons, and the wavelengths of its
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characteristic x-ray spectrum are measured. A second,
fainter characteristic spectrum is also found, due to an
impurity in the target. The wavelengths of the Klines are178.9 pm (cobalt) and 143.5 pm (impurity). What is the
impurity ?
cf
11
Co
X
X
Co
zz
1 ZCf
1 coco
ZCc
1 X
X
ZCc
and
127
1
5.143
9.178
Xz
pm
pm
)(30 ZincZX
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BASICS OF
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L ight
A mplification byS timulated
E mission of
R adiation
Laser-Professionals.com
BASICS OF
LASERS AND LASER LIGHT
CHARACTERISTIC OF THE LASER LIGHT
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1. Monochromaticity
The light emitted by a laser is almost pure in color,almost of a single wavelength or frequency.
2 Coherence
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2. Coherence
3. Directionality
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3. Directionality
The astonishing degree of directionality of a laser
light is due to the geometrical design of the lasercavity and to the monochromaticity and coherentnature of light generated in the cavity.
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LASERS AND LASER LIGHT
Characteristics of laser light: Laser light is highly
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Characteristics of laser light: Laser light is highlymonochromatic. Laser light is highly coherent. Laserlight is highly directional. Laser light can be sharplyfocused.
Interaction of radiation with matterAbsorption: Absorption of a photon of frequency f
takes place when the energy difference E2
E1
of theallowed energy states of the atomic system equals theenergy hf of the photon. Then the photon disappearsand the atomic system moves to upper energy state
E2 (see figure).
Spontaneous Emission: The average life-time of theatomic system in the excited state is of the order of
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108s. After the life-time of the atomic system in theexcited state, it comes back to the state of lower
energy on its own by emitting a photon of energyhf = E2E1
In an ordinary light source, the radiation of light fromdifferent atoms is not coherent. The radiations are
emitted in different directions in random manner. Suchtype of emission of radiation is called spontaneousemission.
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Spontaneous and Stimulated emission
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1. Emission take place
without external
agency.
2. Independent on
incident light intensity
3. Transition take place
b/n two states
4. Ordinary light radiation
is emitted
1. Emission take place
with external agency
namely photon of right
frequency
2. Dependent on incidentlight intensity.
3. Transition take place
b/n three states
4. Laser radiation is
emitted
Population inversion: Boltzmann statistics gives thepopulation of atoms in various energy states at temp T.
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p p gy p
k = Boltzmann constant. n(E1) = density of atoms withenergy E1, n(E2) = density of atoms with energy E2.
n(E2) < n(E1) if E2> E1 (Figure a).This is the normal condition in which the population of theatoms in upper energy state is less than that in lowerenergy state.
For the stimulated emission rate to exceed the absorptionrate, it is necessary to have higher population of upperenergy state than that of lower energy state. This conditionis called population inversion [n(E2) > n(E1)] (Figure b).This is a non equilibrium condition and is facilitated by the
presence of metastable states.
Tk
EE
En
En 12
1
2
exp
Metastable state: A metastable state is an excitedenergy state of an atomic system from which
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spontaneous transitions to lower states is forbidden(not allowed by quantum mechanical selection rules).
The average life time of the atomic system in themetastable state is of the order of 103s which ismuch longer than that in an ordinary excited state.
Stimulated transitions from the metastable stateare allowed. An excited atomic system goes tometastable state (usually a lower energy state) dueto transfer of its extra energy by collision with
another atomic system.
Thus, it is possible to have population inversionof atomic systems in a metastable state relative to a
lower energy state.
Principle of a Laser: The main parts of a laser arelasing medium, resonant cavity and pumping system.
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In a laser the medium chosen to amplify light iscalled lasing medium (active medium). This medium
has atomic systems (active centers), with special systemof energy levels suitable for laser action (see figure).This medium may be a gas, or a liquid, or a crystalor a semiconductor. The atomic systems in this medium
may have energy levels including a ground state (E1),an excited state (E3) and a metastable state (E2).
For eg., in Ruby laser the lasing medium is a ruby rod.Ruby is Al2O3 doped with Cr2O3. Cr
3+ ions are the activecentres which have approximately similar energy level
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The atoms in the state E3may come down to state E1 byspontaneous emission or they
may come down to metastablestate (E2) by collision.
The atoms in the stateE2come down to state E1 by
stimulated emission.
centres, which have approximately similar energy levelstructure shown above.
The resonant cavity is a pair of parallel mirrorsto reflect the radiation back into the lasing medium.Pumping is a process of exciting more number of
atoms in the ground state to higher energy states,
which is required for attaining the population inversion.In Ruby laser thepumping is done byxenon flash lamp.
These radiations may bereflected due to mirror action ofthe end faces (see figure)
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the end faces (see figure).When population inversion takes
place at E2, a stray photon ofright energy stimulates chainreaction, accumulates morephotons, all coherent.
The reflecting ends turn thecoherent beam back into activeregion so that the regenerativeprocess continues and part of
the light beam comes out fromthe partial mirror as a laserpulse. The output is an intensebeam of coherent light.
The ruby laser gives red light.
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During collisions
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118MIT- MANIPAL
between He- andNe- atoms, the
excitation energy (E3=20.61eV) of He-atom is transferred toNe-atom (level E2=20.66eV). Thus, population inversion occursbetween levels E2 and E1. This population inversionbetweenE2 and E1 is maintained because:
(1) the metastability of level E3 ensures a ready supply ofNe-atoms in level E2 and
(2) level E1 decays rapidly to Eo.Stimulated emission from level E2 to level E1
predominates, and red laser light is generated. The mirrorM1 is fully reflective and the mirror M2 is partially reflectiveto allow the laser beam to come out. The Brewsterswindows W & W are at polarizing angles to the mirrors, tomake the laser light linearly polarized.
BE-PHYSICS-ATOMIC PHYSICS-2011-12
HRK-Sample problem 48-7: A three level laser emits light ofwavelength 550 nm. (a) What is the ratio of population of the
upper level (E ) to that of the lower level (E ) in laser
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upper level (E2) to that of the lower level (E1) in laser
transition, at 300 K? (b) At what temperature the ratio of the
population of E2to that of E1becomes half?
kT
EE
N
Na 12
1
2 exp)
eV
x
Jx
Jx
x
xxxhchEE
26.2
106.1
10616.3
10616.3
10550
10310625.6
19
19
19
9
834
12
923.86exp1
2 N
N
38
1
2 1077.1 xN
N
KT = 0.0259eV
K=1.38 x 10-23/1.6 x 10-19= 8.625 x 10-5eV/K
This is very small number !
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HRK-Exercise 48.9 P-28: A ruby laser emits light atwavelength 694.4nm. If a laser pulse is emitted for 12ps and
the energy release per pulse is 150mJ
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the energy release per pulse is 150mJ
a) What is the length of the pulse and
b) How many photons are there in each pulse?
mxxxx
ctpulsetheofLengtha
3128
106.31012103
)(
17
193
1025.5
1240
.4.694106.110150
,)(
x
eV.nm
nmeV
hc
En
cnhnhEpulseperEnergyb
HRK-Exercise 48.9 P-29. Assume that lasers are available
whose wavelengths can be precisely "tuned" to anywhere in
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the visible range (400 nm to 700 nm). If a television channel
occupies a bandwidth of 10MHz, how many channels could
be accommodated within this wavelength range?
Solution:
The lower frequency is
f1 = c/ 1= 4.29 x 1014Hz
The higher frequency is
f2 = c/ 2= 7.50 x 1014Hz
The number of signals that can be sent in this range is
(f2-f1)/(10 x 106) = 3.21 x 107
That's quite a number of television channels.Hence more number of TV channels can be obtained by
re lacin microwave beam with Laser beam as si nal carrier !
HRK-Exercise 48.9 P-30. A He-Ne laser emits light ofwavelength of 632.8 nm and has an output power of 2.3 mW.
How many photons are emitted each minute by this laser
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How many photons are emitted each minute by this laser
when operating?
HRK-Exercise 48.9 P-33: An atom has two energy levels with atransition wavelength of 582 nm. At 300 K, 4 x 1020 atoms are therein the lower state. (a) How many occupy the upper state under
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( ) y py ppconditions of thermal equilibrium? (b) Suppose, instead, that 7.0 x1020atoms are pumped into upper state, with 4.0 x 1020in the lower
state. How much energy could be released in a single laser pulse?
kT
EE
N
Na 12
1
2 exp)
eV
hc
hEE 13.212
eVkTAlso 026.0,
kT
EE
NN12
12 exp
92.81exp104 202 xN
16
2 1066
xN
nhEb )
JxxxxE1920
106.113.2107
JE 240
02N
That's effectively
none.
HRK-Sample Problem 48-8: A pulsed ruby laser has aruby rod (Al2O3 doped with Cr2O3) as an active medium,which is 6 cm long and 1 cm in diameter. There is one
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which is 6 cm long and 1 cm in diameter. There is onealuminium ion (active centre, with energy levels of the type
shown in the figure) for every 3500 chromium ions. Theruby laser light has a wavelength of 694.4 nm. Supposethat all the chromium ions are in metastable state (E2) andnone are in ground state (E1). How much energy is therein a single laser pulse if all these ions come down to
ground state in a single stimulated emission chain reactionepisode ? Density of Al2O3 is 3700 kg/m
3. Molar mass ofAl2O3 is 0.102 kg/mol.Solution:
HOME WORK
ATOMIC PHYSICS
01 Mention the postulates of Bohrs model of H-atom
QUESTIONS
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01. Mention the postulates of Bohr s model of H atom. [2]
02. Based on the Bohrs model for H-atom, obtain theexpression for (a) the total energy of the H-atom(b) radii of the electron orbits. [5]
03. Sketch the energy level diagram of H-atom
schematically, indicating the energy value for eachlevel and the transition lines for the Lymanseries, Balmer series and Paschen series. [4]
04. Write the expressions for total energy of (a) the H- atom (b) other one-electron atoms. From this, obtain
the expressions for the reciprocal wavelengths H- spectral lines in terms of quantum numbers. [4]
ATOMIC PHYSICS05. Give a brief account of quantum model of H-atom.
[2]
QUESTIONS
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[2]06. The wave function for H-atom
in ground state is
Obtain an expression for the radial probability densityof H-atom in ground state. Sketch schematically theplot of this vs. radial distance. [4]
07. The wave function for H-atom in 2s state is
Write the expression for the radial probability densityof H-atom in 2s state. Sketch schematically the plot
of this vs. radial distance. [2]
o
o
sa
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ATOMIC PHYSICS
08. Sketch schematically the plot of the radial probability
QUESTIONS
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08. Sketch schematically the plot of the radial probabilitydensity vs. radial distance for H-atom in 1s-state
and 2s-state. [2]
09. Give the physical interpretation of the following:(a) Orbital quantum number [1]
(b) Orbital magnetic quantum number m [4](c) Spin magnetic quantum number ms [3]
10. Explain the continuous x-ray spectrum with aschematic plot of the spectrum. [2]
11. Obtain an expression for the cutoff wavelength in thecontinuous x-ray spectrum. [4]
ATOMIC PHYSICS
12. Explain the characteristic x-ray spectrum with a
QUESTIONS
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12. Explain the characteristic x ray spectrum with aschematic plot of the spectrum. [2]
13. Explain the origin of characteristic x-ray spectrum witha sketch of x-ray energy level diagram. [3]
14. Write Moseleys relation for the frequency of
characteristic x-rays. Sketch schematically the Moseleysplot of characteristic x-rays. [2]
15. Obtain Moseleys relation for characteristic x-ray
frequency from Bohr theory. [4]
16. Mention the characteristics of a laser beam. [2]
ATOMIC PHYSICS
17. Explain the following terms with reference to lasers:
QUESTIONS
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17. Explain the following terms with reference to lasers:(a) spontaneous emission [2]
(b) stimulated emission [2](c) metastable state [2](d) population inversion [2](e) pumping [1](f) active medium [2]
(g) resonant cavity. [1]
18. Explain the principle of a laser. [5]
19. Give a brief account of a He-Ne laser. [4]