atomic physics
TRANSCRIPT
ATOMIC PHYSICS
1MIT- MANIPAL
TOPICS
Text BookPHYSICS for Scientists and Engineers
with Modern Physics (6th ed) By Serway & Jewett
• Atomic spectra of gases• Early models of the atom• Bohr’s model of the hydrogen
atom• The quantum model of the
hydrogen atom• The wave functions for
hydrogen• 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
2Halliday, Resnick, Krane
ATOMIC SPECTRA OF GASES
Emission spectra: All objects emit thermal radiation
characterized by a continuous distribution of wavelength
(continuous spectrum).
When a gas at low pressure is subjected to an electric
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
The wavelengths of the Balmer series lines in the hydrogen
spectrum are given by the 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 SPECTRUM
FRAUNHOFER LINES
VISIBLE HYDROGEN SPECTRUM BALMER SERIES LINES
H(656.3 nm) H(656.3 nm)
H(656.3 nm) H(656.3 nm)
22H n1
21
R1
The wavelengths of the other series lines in the
hydrogen spectrum are given by the equation LymanSeries n = 2, 3, 4, . . .
PaschenSeries n = 4, 5, 6, . . .
BrackettSeries n = 5, 6, 7, . . .
Although no theoretical basis existed for these
equations, they are in agreement with the experimental
results.
2H n1
1R1
22H n1
31
R1
22H n1
41
R1
EARLY MODELS OF THE ATOM
EARLY MODELS OF THE ATOM
EARLY MODELS OF THE ATOM
[1] (a) What value of n is associated with the 94.96-nm
spectral line in the Lyman series of Hydrogen ? (b)
Could this wavelength be associated with the Paschen
or Balmer series ?
SOLUTION:
(a)Lyman Series
2H n1
1R1
2
79
1110097.1
1096.94
1
nx
x
5n
(b) Paschen Series
22
1
3
11
nRH
27 1
9
110097.1
1
nx
The shortest wavelength for this series corresponds to n = ∞ for
ionization. For n = ∞, gives λ = 820 nm. This is larger than 94.96
nm, so this wave length cannot be associated with the Paschen
series.
Balmer Series
22
1
2
11
nRH
27 1
4
110097.1
1
nx
with n = ∞ for ionization, λ min = 365 nm. Once again the shorter
given wavelength cannot be associated with the Balmer series.
BOHR’S MODEL OF THE HYDROGEN ATOM
In his semi classical model of the
H-atom Bohr postulated that-
[1] The electron moves in circular
orbits around the proton under the
influence of the electric force of
attraction as shown in the figure.
v+e
me
–e
r
F
[2] Only certain electron orbits are stable (stationary states).
When in one of the stationary state, the atom does not
radiate energy. Hence the total energy of the atom remains
constant in a stationary state.
[3] When the atom makes a transition
from higher energy state (Ei) to lower
energy state (Ef) [ie, the electron makes
a transition from a stable orbit of larger
radius to that of smaller radius],
radiation is emitted. The frequency (f)
of this radiation (photon) is given by
Ei – Ef = h f .
The frequency f of the photon emitted
is independent of the frequency of
electron’s orbital motion.
v+e
me
–e
r
F
[4] The angular momentum of the
electron in any stable orbit is
quantized
L=mev r = n
n = 1, 2, 3, . . .
me = mass of the electronv = speed of the electron in the orbitr = radius of the electron’s orbit
v+e
me
–e
r
F
2h
Electric potential energy of the H-atom is
r
ek
r
qqkU ee
221
v+e
me
–e
r
F
rek
2vm
UKE2
e2
e
The total energy of the H-atom is
Apply Newton’s 2ND law to the electron, the electric force
exerted on the electron must be equal to the product of mass
and its centripetal acceleration (a=v2/r)
rvm
Frek 2
e2
2e
r2ek
2vm
K2
e2
e
ke= Coulomb constant
r2ek
E
rek
r2ek
UKE
2e
2e
2e
The total energy of the H-atom is
+e
–e4ao
ao
9ao
Note that the total energy is negative, indicating
a bound electron-proton system. This means
that energy in the amount of kee2/2r must be
added to the atom to remove the electron and
make the total energy of the system zero.
4. Velocity in different orbits:
rm
ekv
r2
ekvm
2
1K
e
2e2
2e2
e
Speed decreases with increase in radius of the orbits.
16
....................3,2,1nekm
nr
rm
ek
rm
nv
rm
ekv,Also
rm
nv
nvrmL:momentumangularknowWe
2ee
22
n
e
2e
22
e
222
e
2e2
22
e
222
e
This equation shows that the radii of the allowed orbits have discrete values-they are quantized. The result is based on the assumption that the electron can exist only in certain allowed orbits determined by the integer n.
5. Radius of the allowed orbits:
a0
-e
+e
4a0
9a0 The first three circular orbits predicted by the Bohr model of the hydrogen atom.
The orbit with the smallest radius, called the Bohr radius a0, corresponds to n=1 and has the value.
)nm0529.0(nanr
:atomhydrogen
theinorbitanyofradiustheforexressionGeneral
nm0529.0ekm
a
20
2n
2ee
2
0
Energy quantization
Substitute rn= n2 ao in the total energy equation
2
o
2e
2e
n n1
a2ek
r2ek
E n = 1, 2, 3, . . .
...,3,2,1n,n
eV606.13E 2n
E1= –13.606 eV
21
n nE
E
Ionization energy = minimum energy required to ionize the atom in its ground state = 13.6 eV for H-atom
From the equation Ei – Ef = h fFrequency of the photon emitted during transition of the atom from state i to state f is
2i
2fo
2efi
n1
n1
ha2ek
hEE
f
Use c = f λ
2i
2f
H n1
n1
R1
2i
2fo
2e
n1
n1
cha2ek
cf1
cha2ek
Ro
2e
H
RH = 1.097 x 107 /m
Energy diagram of Hydrogen Atom
Extension of Bohr’s theory to other one-electron atoms
- Nuclear charge = + Z e
radius
Energy
Za
nr o2n
...,3,2,1nnZ
a2ek
E2
2
o
2e
n
Limitations of Bohr’s theory:
When spectroscopic techniques improved, it was found
that many of the lines in the H-spectrum were not
single lines but closely spaced groups of lines. The
lines appear split when the H-vapour was kept in
magnetic field.
Bohr’s correspondence principle:
Quantum physics agrees with classical physics
when the difference between quantized levels
becomes vanishingly small.
PROBLEMS
[1] Spectral lines from the star ξ-Puppis: Some
mysterious lines in 1896 in the emission spectrum of
the star ξ-Puppis fit the empirical equation
2
i
2
f
H
2n
1
2n
1R
1
Show that these lines can be explained by the Bohr’s
theory as originating from He+.
SOLUTION: The ion He+ has z=2, Thus allowed energy
levels are given by
...,3,2,12 2
22
n
n
Z
a
ekE
o
en
2
2 4
2 na
ekE
o
en
22
2 44
2 ifo
efi
nnha
ek
h
EEf
22
2
2
1
2
1
2ifo
e
nnha
ekf
22
2
2
1
2
1
2
1
ifo
e
nnhca
ek
c
f
cha
ekwhereR
o
eH 2
2
[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 the wavelength of the radiation from the
Rydberg atom in part (B) if treated classically ?
SOLUTION(A)
2i
2f
H n1
n1
R1
22 2
1
1
11HR
4
3 HR
HR3
4
)( 5.12110215.1 7 tultraviolenmmx
Hzxc
fFrequency 151047.2
SOLUTION(B)
2i
2f
H n1
n1
R1
22 273
1
272
11HR
m992.0
SOLUTION(C)
rn= n2 ao= 2732 (0.0529nm)
r273=3.94μm
pm9.52ekm
a 2ee
2
o
SOLUTION(D)
SOLUTION(E)
We have speed v and radius r from (C) and (D)
r
v
Tf
21
rm
ekv
e
e2
2
)1094.3)(1011.9(
)1060.1)(1099.8(631
2199
xx
xxv
rm
ekv
e
e2
smxv /1001.8 3
Hzxr
v
Tf 81024.3
2
1
mf
c926.0
[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 = 2.00 x 10–10 m
3
22
o c
ae
6
1
dt
dE
32e
22o
2
4
cmr12
e
dt
dr
SOL:A The total energy is given by
32222
4
12 cmr
e
dt
drTherefore
eo
r
ekE e
2
2
04
1
ekwhere
r
eE
o8
2
r
vmF
r
ekce ee
2
2
2
sin
The centripetal acceleration a is given by
2
2
3
22 8
6
1
e
r
c
ae
dt
dr o
o
3
22
2
2
6
1
8 c
ae
dt
dr
r
e
dt
dE
oo
3
22
6
8
c
ar
dt
dr
3
22
6
8
c
ar
dt
dr
SOL:B32
e22
o2
4
cmr12
e
dt
dr
T
x
eo dtedrcmr0
40
1000.2
32222
10
12
Tr
e
cmx
eo
101000.2
0
3
4
3222
3
12
[4] 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.
[5] 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:
Solution a:
Solution c:
[6] (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) 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
39
THE QUANTUM MODEL OF THE HYDROGEN ATOM
Bohr’s model views the electron as a particle orbiting the nucleus in nonradiating, quantized energy levels (n=1,2,3…... principal quantum number).
Bohr orbits are quantized only with respect to their size, shape and energy.
Bohr’s model demonstrates excellent agreement with the experimental values, it can not explain others such as the ‘splitting’ of spectral lines.
The quantum model involving the Schrödinger equation in polar coordinates (r,θ,φ) explains the hydrogen spectrum completely.
THE QUANTUM MODEL OF THE HYDROGEN ATOM
40
Bohr’s theory :The potential energy function for the H-atom is
rek
)r(U2
eke = 8.99 x 109 N.m2/C2 is Coulomb constant
r = radial distance of electron from proton.
The time-independent schrodinger equation in 3-dimensional space is
Ezyxm2 2
2
2
2
2
22
41
Quantum theory: Since U has spherical symmetry, it is easier to solve the schrodinger equation in spherical polar coordinates (r, θ, φ):r is the radial coordinate is the radial distance from the origin.
where
Angular coordinate θ specifies the angular position relative to the z-axis. θ is the angle between z-axis and
222 zyxr
φ
P
y
x
z
θr
φ- is the angle between the x-axis and the projection of onto the xy-plane.
r
It is possible to separate the variables r, θ, φ and their
corresponding wave functions can be written as
(r, θ, φ) = R(r) f(θ) g(φ)
By solving the three separate ordinary differential equations
for R(r), f(θ), g(φ), with conditions that the normalized and
its first derivative are continuous and finite everywhere, one
gets three different quantum numbers for each allowed
state of the H-atom.
Principal quantum number (n):The radial function R(r) of is
associated with orbits: the principal quantum number n. The
quantum numbers are integers and correspond to the three
independent degrees of freedom.
From this theory the energies of the allowed states for the
H-atom are
2o
2e
n n
1
a2
ekE
...,3,2,1n,
n
eV606.132
Orbital quantum number l The polar function f(θ) is associated with the orbital quantum number l. Magnetic quantum number ml.The azimuthal function g(φ) is associated with the orbital magnetic quantum number ml.
which is in agreement with Bohr theory.
44
The application of boundary conditions on the three
parts of leads to important relationships among
the three quantum numbers:
[1] n can range from 1 to .
[2] l can range from 0 to n–1 ; [n allowed values].
[3] ml can range from –l to +l ; [(2l+1) allowed
values].
Hence, in quantum model wave function of H atom, can be written as
),,r(),,r(lm,l,n
All states having the same principal quantum are
said to form a shell. All states having the same
values of n and l are said to form a subshell:
n = 1 K shell l = 0 s subshell
n = 2 L shell l = 1 p subshell
n = 3 M shell l = 2 d subshell
n = 4 N shell l = 3 f subshell
n = 5 O shell l = 4 g subshell
n = 6 P shell l = 5 h subshell
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. .
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 be represented
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, l = 0):
ao = Bohr radius.
|1s|2 is the probability
density for H-atom
in 1s-state.
o
3o
s1ar
ea
1)r(
o3o
2
s1ar2
ea
1
THE WAVE FUNCTIONS FOR HYDROGEN
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.
o3o
2
s1ar2
ear4
)r(P
P(r) dr is the probability of finding the
electron in this shell.
P(r) dr = ||2 dv = ||2 4r2 dr
P(r) = 4r2 ||2
Radial probability density for H-atom in its ground state:
Plot of the probability of finding the electron as a function of
distance from the nucleus for H-atom in the 1s (ground)
state. Probability (P1s(r)) is maximum when r equals to (Bohr
radius). r= ao
Cross-section of
the spherical
electronic charge
distribution of H-
atom in 1s-state.
rMOST PROBABLE = ao
rAVERAGE = 3ao/2
rMOST PROBABLE = 5ao
The next simplest wave function for the H-atom is the
2s-state wave function (n = 2, l = 0):
o
o
23
os2
ar
ear
2a1
24
1)r(
2s is spherically symmetric (depends only on r). E2 = E1/4 = –3.401 eV (1ST excited state).
MIT-MANIPAL 50
PROBLEMS
[1] - SJ-Example-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.Solution:
When n= 2, l can have the values 0 and 1.
If l=0, ml can only be 0.
If l=1, ml can be -1, 0, or +1.
So, we have a 2s state with quantum numbers n= 2, l=0, ml =0
and three 2p states for which the quantum numbers are
n= 2, l=1, ml =-1
n= 2, l=1, ml =0
n= 2, l=1, ml =+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
[3] SJ-Example-42.5 Probabilities for the electron in H-atom:
Calculate the probability that the electron in the ground state
of H-atom will be found outside the Bohr radius.
Solution:
The probability is found by integrating the radial probability density
for this state, P1s(r), from the Bohr radius a0 to ∞ .
o3o
2
s1ar2
ear4
)r(P
We can put the integral in dimensionless form by changing
variables from r to z = 2r/a0. Noting that z=2 when r=a0, and that
dr=(a0/2)dz, we get
o3o
2
s1ar2
ear4
)r(P
This is about 0.677, or 67.7%.
[2] SJ-Example-42.4.Calculate the most probable value of
r (= distance from nucleus) for an electron in the ground
state of the H-atom.
Solution:
The most probable distance is the value of r that makes the radial
probability P(r) a maximum. The most probable value of r is
obtained by setting dP/dr= 0 and solving for r.
04)(
2
3
21
o
o
s ar
ea
r
dr
d
dr
rdPo3o
2
s1ar2
ear4
)r(P
04)(
2
3
21
o
o
s ar
ea
r
dr
d
dr
rdP
0)()(
22 2
2
dr
dr
dr
rd oar
oar ee
0)2(22 22 oaroar ere oar
0]1[2 2
oaroarre
01 oa
r
oar
The expression is satisfied if
The most probabale value of r is the Bohr radius
[5] SJ-Problem-42.16: A general expression for the
energy levels of one-electron atoms and ions is
where ke is the the Coulomb constant, q1 and q2 are the
charges of the electron and the nucleus, and μ 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). What are
the wavelengths for this same transition in (a)
positronium, which consists of an electron and a
positron, and (b) singly ionized helium ?
22
22
21
2
2 n
qqkE e
n
21
21
mm
mm
Solution:
The reduced mass of positronium is less than hydrogen, so the
photon energy will be less for positronium than for hydrogen. This
means that the wavelength of the emitted photon will be longer
than 656.3 nm.
On the other hand, helium has about the same reduced mass but
more charge than hydrogen, so its transition energy will be larger,
corresponding to a wavelength shorter than 656.3 nm.
All the factors in the given equation are constant for this problem
except for the reduced mass and the nuclear charge. Therefore,
the wavelength corresponding to the energy difference for the
transition can be found simply from the ratio of mass and charge
variables.
so the energy of each level is one half as large as in
hydrogen, which we could call “protonium”. The photon
energy is inversely proportional to its wavelength , so for
positronium,
so the transition energy is 22 = 4 times larger than hydrogen.
[6] SJ-Problem-42.17: 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 electron’s average vector
momentum is zero, but its averaged squared momentum is
equal to the squared uncertainty in its momentum, as given
by the uncertainty principle.
e
2
m2p
K
rek
U2
e
An electron of momentum p is at a distance r from a
stationary proton. Treating the atom as
one-dimensional system,
(a) estimate the uncertainty in the electron’s momentum in
terms of r.
(b) Estimate the electron’s kinetic, potential, and total
energies in terms of r.
2e
2
e
2
e
2
rm8m2
p
m2
pK
rek
U2
e
r
ek
rm8UKE
2e
2e
2
[5] SJ-Problem-42.21: 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
3o
s1ar
ea
1)r(
Erek
rr2
rm2
2e
2
22
Solution: o
o
sar
ea
r
31
1)(
This is true , so the schrodinger equation is satisfied
E
r
ek
rrrmwehave e
2
2
22 2
2
[1]The orbital quantum number l
According to quantum mechanics, an atom in a state
whose principal quantum number n can take on the
following discrete values of the magnitude of the
orbital angular momentum:
PHYSICAL INTERPRETATION OF THE QUANTUM NUMBERS
1n,...,2,1,0)1(L
[2] The orbital magnetic quantum number ml
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 ml specifies the allowed values of the z-
component of the orbital angular momentum. LZ = ml Ћ
.B μ
μ
B
.Bμ
-U
Lμ
em2e
L
L
B
The quantization of the possible orientations of
with respect to an external magnetic field is
called space quantization. Following vector
model describes the space quantization for l = 2.
L
B
THE ALLOWED VALUES OF LZ
LIES ON THE SURFACE OF A CONE AND PRECESSES ABOUT THE DIRECTION OF
L
B
θ is quantized θ ≠ 0 )1(
mLcos Z
L
The Zeeman effect:
Splitting of energy levels and hence spectral lines
in magnetic field.
ENERGY
n=1, l=0
n=2, l=1
hfo
hfoh(fo–∆f)
h(fo+∆f)
ml=0
ml=0ml=–1
ml=+1NO MAG-FIELD MAG-FIELD PRESENT
fo fo (fo+∆f)
(fo–∆f)
SPECTRUM WITHOUT MAG-FIELD
SPECTRUM WITH MAG-FIELD
PRESENT
[3]The spin magnetic quantum number ms
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
69
Spin is an intrinsic property of a particle, like mass and charge. The spin angular momentum magnitude S for the electron is expressed in terms of a single quantum number (spin quantum number), s = ½ :
23
1ss S
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 = ±½. The z-
component of is :
SZ = msЋ = ±Ћ/2
S
The value ms = +½ is for spin-up case and ms = –½ is for spin-down case. The spin magnetic moment of the electron is related to its spin angular momentum
Z-component of the spin magnetic moment:
Bohr magneton
SμSPIN
eme
SSPINμ
em2e
ZSPIN,μ
T/J10x27.9m2
e 24
e
Bμ
SJ-Example-42.6 Calculate the magnitude of the
orbital angular momentum of an electron in a p-
state of hydrogen.
Solution:
with l = 1 for a p state
)1( L
2)11(1
sJx .1049.1 34
SJ-Example-42.7Consider the H-atom in the l = 3 state. Calculate the magnitude of the allowed values of LZ, and the corresponding angles θ that makes with the z-axis. For an arbitrary value of l, how many values of ml are allowed.Solution: with l = 3
,L
|| L
32)13(3)1( L
32)1(cos
mmLZ
L
The allowed values of LZ is given by LZ = ml Ћ -3 Ћ, -2 Ћ ,- Ћ, 0, Ћ, 2 Ћ ,3 Ћ
SJ-Example-42.8For a H-atom, determine the quantum numbers associated with the possible states that correspond to the principal quantum number n = 2.n l ml ms subshell shell No of states in subshell--------------------------------------------------------------------
2 0 0 ½
2 0 0 -½ 2s L
2
2 1 1 ½
2 1 1 -½
2 1 0 ½ 2p L
6
2 1 0 -½
2 1 -1 ½
2 1 -1 -½
SJ-Problem-42.27 How many sets of quantum
numbers are possible are 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.
SJ-Problem-42.31 The ρ-meson has a charge of –e, a
spin quantum number of 1, and a mass 1507 times that
of the electron. Imagine that the electrons in an atom
were replaced by ρ-mesons. List the possible sets of
quantum numbers for ρ-mesons in the 3d-subshell.
Solution:
78MIT- MANIPAL
THE X-RAY SPECTRUM OF ATOMS
When a beam of fast moving electron strikes on solid target an
invisible and high penetrating radiation is produced.
These radiations are called X – rays. i.e.The x-rays are
emitted by atoms in a target when the atoms
are bombarded with high energy electrons.
MIT-MANIPAL 79
The x-ray spectrum has two parts: continuous spectrum
and characteristic spectrum. Sharply defined cutoff
wavelength (λMIN) is a prominent feature of the continuous
x-ray spectrum. TARGET: MOLYBDENUMX-RAY TUBE VOLTAGE:
∆V = 35 kVλMIN = 35.5 pm
To examine the motions of electrons that lie deep within
multi-electron atoms, one needs to consider the x-ray
spectrum of atoms, shown in the figure below:
80MIT- MANIPAL
THE X-RAY SPECTRUM OF ATOMS
Consider an electron accelerated through a potential difference of ∆V (x-ray tube voltage) , hitting a target atom. The electron’s 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: Breaking radiation). ∆K can have any value from 0 to K. Thus the emitted x-rays can have any value for the wavelength above λMIN in the continuous x-ray spectrum. Thus
MINMAX
chhfVe
Vech
MIN
λMIN depends only on ∆V
Continuous spectrum
81MIT- MANIPAL BE-PHYSICS-ATOMIC PHYSICS-2010-11
THE X-RAY SPECTRUM OF ATOMS
The peaks in the x-ray spectrum have wavelengths 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 LEVEL DIAGRAM
FOR MOLYBDENUM EK= 17.4 keV
λK= 71 pm
Characteristic spectrum
82MIT- MANIPAL
THE X-RAY SPECTRUM OF ATOMS
A K x-ray results due to the transition of the electron from L-shell to K-shell. A K x-ray results due to the transition of the electron from 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-series of x-rays can be explained.
KKLL L
KE
OE
LE
NE
ME
0E n
K1n
L2n
M3n N4n O5n
I
min K K LLL
X-RAY ENERGY LEVEL
DIAGRAM FOR MOLYBDENUM
EK= 17.4 keV λK= 71 pm
HRK-Sample Problem 48-1: Calculate the cutoff wavelength
for the continuous spectrum of x-rays emitted when 35-keV
electrons fall on a molybdenum target.
Solution:
Vech
MIN
pmmxMIN
5.35 1055.3 11
HRK-Exercise 48.1: Show that the short-wavelength cutoff in the continuous x-ray spectrum is given by
where ΔV is the applied potential difference in kilovolts.
V
pm1240MIN
Solution: The highest energy x-ray photon will have an
energy equal to the bombarding electrons,
Vech
MIN
V
pm
1240
m10x68.4910x6.1x10x25
10x3x10x625.6
E
hc
ch
2
KeV50E
)electronincident(KeV50E
12
193
834
Photon1
1
1
Photon1
o
Solution
HRK-Exercise 48.9: X-rays are produced in an x-ray tube by a target potential of 50.0 keV. If an electron makes three collisions in the target before coming to rest and loses one-half of its remaining kinetic energy on each of the first two collisions, determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.
m10x375.9910x6.1x10x5.12
10x3x10x625.6
E
hc
chKev5.12E
KeV5.12collisionthirdthebeforeelectronofEnergy
12
193
834
Photon3
3
3
Photon3
m10x375.9910x6.1x10x5.12
10x3x10x625.6
E
hc
ch
2
KeV25E
KeV25collisionondsecthebeforeelectronofEnergy
12
193
834
Photon2
2
2
Photon2
HRK-Exercise 48.12: The binding energies of K-shell and L-shell electrons in copper are 8.979 keV and 0.951 keV, respectively. If a K x-ray from copper is incident on a sodium chloride crystal and gives a first-order Bragg reflection at 15.9 when reflected from the alternating planes of the sodium atoms, what is the spacing between these planes ?Solution:
nm154.0
10x6.1x10x)951.0979.8(
10x3x10x625.6
EE
hc
hchEE
K
193
834
12
K
K
K12
.pm282)9.15sin(2
m10154.0
sin2d
1n,orderfirstfor,nsind29
K
2nL
1nK
keVBE 951.02
keVBE 979.81
HRK-Exercise 48.5: Electrons bombard a molybdenum
target, producing both continuous and characteristic x-
rays. If the accelerating potential applied to the x-ray
tube is 50.0 kV, what values of (a) λMIN (b) λKβ (c) λK
result ? The energies 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.
pmpm
V
pmMIN 8.24
50
12401240
pm39.71
10x6.1x10x)6.220(
10x3x10x625.6
EE
hc
hchEE
K
193
834
12
K
K
K12
pm37.63
10x6.1x10x)4.020(
10x3x10x625.6
EE
hc
hchEE
K
193
834
12
K
K
K12
X-RAYS AND THE NUMBERING OF THE ELEMENTS
Moseley’s observation on the characteristic K x-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 1ZCf
C is a constant. Based on this observation, the elements are arranged according to their atomic numbers in the periodic table
Bohr theory and the Moseley plot: Bohr’s
formula for the frequency of radiation
corresponding to a transition in a one-electron
atom between any two atomic levels differing in
energy by ΔE is
2i
2f
32o
42
n
1
n
1
h8
eZm
h
Ef
In a many-electron atom, for a K transition, the
effective nuclear charge felt by an L-electron can
be thought of as equal to +(Z–b)e instead of +Ze,
where b is the screening constant due to the screening
effect of the of the only K-electron.
MOSELEY PLOT OF THE K X-RAYS
bZh32
em3fand
2
1
32o
4
Frequency of the K x-ray is
2232
o
42
2
1
1
1
h8
ebZmf
1bcesin1ZCfor
HRK-Sample problem 48-2: Calculate the value
of the constant C in the Moseley’s relation for x-
ray frequency and compare it with the measured
slope of the straight line in Moseley plot.
SOLUTION:
2
1
32
4
32
3
h
emc
o2/3
2
32
3
h
emc
o
2/17 Hz1095.4 xc
2/17 Hz1096.4 xcfromgraph
HRK-Sample Problem 48-3: A cobalt target is bombarded with electrons, and the wavelengths of its 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 K lines are 178.9 pm (cobalt) and 143.5 pm (impurity). What is the impurity ?
c
f
1
1
Co
X
X
Co
z
z
1 ZCf
1 coco
ZCc
1 X
X
ZCc
and
127
1
5.143
9.178
Xz
pm
pm
)( 0.30 ZincZ X
96MIT- MANIPAL BE-PHYSICS-ATOMIC PHYSICS-2010-11
ATOMIC PHYSICS
01. Mention the postulates of Bohr’s model of H-atom. [2]02. Based on the Bohr’s model for H-atom, obtain the expression 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
each level and the transition lines for the Lyman series, 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]
QUESTIONS
97MIT- MANIPAL BE-PHYSICS-ATOMIC PHYSICS-2010-11
ATOMIC PHYSICS
05. Give a brief account of quantum model of H-atom.
[2]06. The wave function for H-atom
in ground state is
Obtain an expression for the radial probability density of H-atom in ground state. Sketch schematically the plot of this vs. radial distance. [4]
07. The wave function for H-atom in 2s state is
Write the expression for the radial probability density of H-atom in 2s state. Sketch schematically the plot of this vs. radial distance. [2]
QUESTIONS
o
3o
s1ar
ea
1)r(
o
o
23
os2
ar
ear
2a1
24
1)r(