Atomic Physics Modified-RP

ATOMIC PHYSICS

1MIT- MANIPAL

Early models of atom and Bohrs model

Quantum model of hydrogen atom

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

2MIT- MANIPAL BE-PHYSICS-ATOMIC PHYSICS-2010-11

1. Mention the postulates of Bohrs model of H-atom.[2]

2. Based on the Bohrs model for H-atom, obtain the expression for (a) the total energy of the H-atom (b) radii of the electron orbits. [5]

3. 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]

4. 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 to be ANSWERED today

BOHRS MODEL OF THE HYDROGEN ATOM

Bohrs postulates

1. The electron moves in circular orbits around the proton under the influence of the electric force of attraction as shown in the figure

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v+e

mee

r

Fr

2. Only certain electron obrits 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.


When the atom makes a transition from higher energy state (Ei) to lower energy state (Ef) [ie, the electron makes a transition from a

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v+e

mee

r

Fr

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 electrons orbital motion.


In his semiclassical model of the H-atom Bohr postulated that-

The angular momentum of the electron in any stable orbit is quantized-

mev r = n n = 1, 2, 3, . . . me = mass of the electron

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v+e

mee

r

Fr

v = speed of the electron in the orbitr = radius of the electrons orbit

h

=

2

hh


Electric potential energy of the H-atom is

ke= Coulomb constant


r

ekU

2e= v+e

mee

r

Fr

r

ek

2

vmUKE

2e

2e =+=The total energy

of the H-atom is

Newtons 2ND lawr

vmF

r

ek 2e2

2e ==

r2

ek

2

vmK

2e

2e ==


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r2

ekE

r

ek

r2

ekUKE

2e

2e

2e

=

=+=

The total energy of the H-atom is

From Newtons 2ND law equation and orbit quantization equation rm

ek

rm

nv

e

2e

22e

222 ==

h

2ee

22

nekm

nr

h=

Thus the electron orbit radii are quantized rn= n2 ao

n = 1, 2, 3, . . .

pm9.52ekm

a2

ee

2

o ==h

Bohr radius

+e

e4ao

ao

9ao


Energy quantization

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Substitute rn= n2 ao in

the total energy equation

==

2o

2e

2e

nn

1

a2

ek

r2

ekE

...,3,2,1n,n

eV606.13E

2n==

E1= 13.606 eV 21

nn

EE =


Ionization energy = minimum energy required to ionize the atom in its ground state

= 13.6 eV for H-atomFrom the equation Ei Ef = h fFrequency of the photon emitted during transition of the atom from state i to state f is

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=

=

2i

2fo

2efi

n

1

n

1

ha2

ek

h

EEf

Use c = f

=

2i2f

Hn

1

n

1R

1

==

2i2fo

2e

n

1

n

1

cha2

ek

c

f1

cha2

ekR

o

2e

H =

RH = 1.097 x 107 /m


Extension of Bohrs theory to other one-electron atoms - Nuclear charge = + Z e

radius

Energy

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( )Z

anr o2n =

...,3,2,1nn

Z

a2

ekE

2

2

o

2e

n =

=

Limitations of Bohrs 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.


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Bohrs correspondence principle:Quantum physics agrees with classical physics when the difference between quantized levels becomes vanishingly small.

SJ-Example-42.1 Spectral lines from the star -Puppis:Some mysterious lines observed in 1896 in the emission spectrum of the star -Puppis fit the empirical equation

=

2i

2

f

H

2

n

1

2

n

1R

1

Show that these lines can be explained by the Bohrs theory as originating from He+.


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SJ-Example-42.2 Electronic transition in hydrogen:

(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 radioastronomers 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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SJ-Problem-42.7 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.(Ans: r=0.212nm, 9.95 x 10-25 N-s, L=2.11x10-34 kgm2/s, K=3.4eV, U=-6.8eV, E=K+U=-3.4eV)

SJ-Problem-42.9 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.Ans: 410nm, 3.03eV, 7.32x1014 Hz


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SJ-Problem-42.13 (a) Construct an energy-level diagram for the He+ ion (Z = 2). What is the ionization energy for He+ ?

Ans:

THE QUANTUM MODEL OF THE HYDROGEN ATOM

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The potential energy function for the H-atom is

r

ek)r(U

2e=

ke = 8.99 x 109 N.m2/C2

r = radial distance of electron from proton [H-nucleus]

The time-independent Schrodinger equation in 3-dimensional space is

Since U has spherical symmetry, it is easier to solve the Schrodinger equation in spherical polar coordinates (r, , ):

where is the angle between z-axis and

222 zyxr ++=rr

P

y

x

z

rr


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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 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. The quantum numbers are integers and correspond to the three independent degrees of freedom.

rr

P

y

x

z

rr


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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

o

2

en

n

1

a2

ekE

= ...,3,2,1n,

n

eV606.132

==

which is in agreement with Bohr theory. The polar function f() is associated with the orbital quantum

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 quantum

numbers: n can range from 1 to . l can range from 0 to n1 ; [n allowed values]. ml can range from l to +l ; [(2l+1) allowed

values].


All states having the same principal quantum no are said to form a shell. All states having the same values of n and llll are said

to form a subshell

n = 1 K shell llll = 0 s subshelln = 2 L shell llll = 1 p subshelln = 3 M shell llll = 2 d subshelln = 4 N shell llll = 3 f subshelln = 5 O shell llll = 4 g subshelln = 6 P shell llll = 5 h subshell. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


SJ-Example-42.3 The n = 2 level of hydrogen:For a H-atom, determine the number of allowedstates corresponding to the principal quantumnumber n = 2, and calculate the energies of thesestates.


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SJ-Problem-42.16A 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 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

2

2

2

1

2

en

n2

qqkE

h

=

21

21

mm

mm

+=

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.

For hydrogen

For positronium

21

Ans: Wavelength doubles. ie., 1.31 m, b) energy becomes 4 times. So , wavelength=656.3/4= 164nm

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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 energyand 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 theuncertainty in its position is approximately equal to theradius r of its orbit. The electrons average vectormomentum is zero, but its averaged squared momentumis equal to the squared uncertainty in its momentum, asgiven by the uncertainty principle.


e

2

m2

pK =

r

ekU

2e=

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SJ-Problem-42.17 continued

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 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 the

total 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.


e

2

m2

pK =

A

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1. Give a brief account of quantum model of H-atom. [2]

2. 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]

3. 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 to be Answered

o

3o

s1

ar

ea

1)r(

=

o

o

2

3

o

s2

ar

ea

r2

a

1

24

1)r(

=

Sketch schematically the plot of the radial probability

density vs. radial distance for H-atom in 1s-state

and 2s-state. [2]

Give the physical interpretation of the following:

(a) Orbital quantum number l [1]

(b) Orbital magnetic quantum number ml [4]

(c) Spin magnetic quantum number ms [3]

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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, l = 0):

ao = Bohr radius.

|1s|2 is the probability density for H-atom in 1s-state.

THE WAVE FUNCTIONS FOR HYDROGEN

o

3o

s1

ar

ea

1)r(

=

o

3o

2

s1

ar2

ea

1

=

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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

a

r

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 dr

P(r) = 4r2 ||2

Radial probability density for H-atom in its ground state:

29

Plot of the probability of finding the electron as a 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

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SJ-Example-42.4 The ground state of H-atom:

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.


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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.


32

The next simplest wave function for the H-atom is the

2s-state wave function (n = 2, l = 0):

o

o

2

3

o

s2

ar

ea

r2

a

1

24

1)r(

=

2s is spherically symmetric (depends only on r).

E2 = E1/4 = 3.401 eV

(1ST excited state).

rMOST PROBABLE = 5ao

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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

s1

ar

ea

1)r(

=

=

+

E

r

ek

rr

2

rm2

2e

2

22h

Question to be answered

Give the physical interpretation of the

following:

(a) Orbital quantum number l [1]

(b) Orbital magnetic quantum number ml [4]

(c) Spin magnetic quantum number ms [3]

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35

The orbital quantum number llll

According to quantum mechanics, an atom in astate whose principal quantum number n cantake on the following discrete values of themagnitude of the orbital angular momentum:

PHYSICAL INTERPRETATION OF THE QUANTUM NUMBERS

1n,...,2,1,0)1(L =+= lhll

SJ-Example-42.6 Calculating L for a p-state:Calculate the magnitude of the orbital angularmomentum of an electron in a p-state ofhydrogen.

Ans: l=1 for p state, L=1.49 x 10-34 J.s

36MIT- MANIPAL

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


.Br

r

r

Br

.Brr

= -U

Lrr

=

em2

e

Lr

Lr

Br

37

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 llll = 2.

Lr

Br

THE ALLOWED

VALUES OF LZ

LIES ON THE SURFACE OF A CONE AND PRECESSES ABOUT THE DIRECTION OF

Lr

Br

is quantized 0 )1(

mLcos Z

+==

llr

l

L

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The Zeeman effect:splitting of energy levels and hence spectral lines in magnetic field

ENERGY

n=1, llll=0

n=2, llll=1

hfohfo

h(fof)

h(fo+f)

mllll=0

mllll=0mllll=1

mllll=+1NO MAG-FIELD MAG-FIELD PRESENT

fo fo (fo+f)(fof)

SPECTRUM WITHOUT

MAG-FIELD

SPECTRUM WITH MAG-FIELD PRESENT

39MIT- MANIPAL


SJ-Example-42.7 Space quantization for H-atom:Consider the H-atom in the llll = 3 state. Calculate the

magnitude of the allowed values of LZ, and the corresponding angles that makes with the z-axis. For an arbitrary value of llll, how many values of mllllare allowed.

,Lr

||Lr

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The spin magnetic quantum number msThe quantum numbers n, llll, mllll 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:


,Sr

41MIT- MANIPAL

Spin is an intrinsic property of a particle, like mass and charge. The spin angular momentum magnitude Sfor the electron is expressed in terms of a single quantum number (spin quantum number), s = :


( ) hh2

31ss =+=S

Sr

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 = /2Sr

42MIT- MANIPAL

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 the spin magnetic moment:

Bohr magneton


SSPIN

rr

=

em

eSrSPIN

r

em2

e h=ZSPIN,

T/J10x27.9m2

e 24

e

==h

B



SJ-Example-42.8 Putting some spin on H-atom:For a H-atom, determine the quantum numbers associated with the possible states that correspond to the principal quantum number n = 2.



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.

MIT-MANIPALBE-PHYSICS-QUANTUM MECHANICS-2010-2011

45



SJ-Problem-42.31 The -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 -mesons. List the possible sets of quantum numbers for -mesons in the 3d-subshell.

Ans: The 3d subshell has l=2, n=3; ml = -2,-1,0,1,2; s=1 and ms= -1,0,1

Questions to be answered

Explain the continuous x-ray spectrum with a schematic plot of the

spectrum. [2]

Obtain an expression for the cutoff wavelength in the continuous x-

ray spectrum. [4]

Explain the characteristic x-ray spectrum with a schematic plot of the

spectrum. [2]

Explain the origin of characteristic x-ray spectrum with a sketch of x-

ray energy level diagram. [3]

Write Moseleys relation for the frequency of characteristic x-rays.

sketch schematically the Moseleys plot of characteristic x-rays. [2]

Obtain Moseleys relation for characteristic x-ray frequency from Bohr

theory. [4]

MIT-MANIPALBE-PHYSICS-QUANTUM MECHANICS-2010-2011

47

48MIT- MANIPAL

THE X-RAY SPECTRUM OF ATOMS

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:

The x-rays are emitted by

atoms in a target when the

atoms are bombarded with

high energy electrons.

The x-ray spectrum 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 kVMIN = 35.5 pm



Consider an electron accelerated through a potential 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

MIN

MAX

chhfVe

==

Ve

chMIN

=

MIN depends only on V



The peaks in the x-ray spectrum is the 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 Enf (| Enf | K, nf = integer),

the vacancy in the inner shell is filled up by an electron from

the outer shell (energy = Eni, ni = integer).

The characteristic x-ray photon emitted has the energy:

nfni EEch

hf =

=

X-RAY ENERGY LEVEL DIAGRAM

FOR MOLYBDENUM EK= 17.4 keV

K= 71 pm



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.

HRK-Sample Problem 48-1: Calculate the cutoffwavelength for the continuous spectrum of x-raysemitted when 35-keV electrons fall on amolybdenum target.Ans: 35.5pm


HRK-Exercise 48.1: Show that the short-wavelength cutoff in the continuous x-ray spectrum is given by


V

pm1240MIN

=

where V is the applied potential difference in kilovolts.

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) MIN (b) K (c) K 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.



HRK-Exercise 48.12: The binding energies of K-shelland L-shell electrons in copper are 8.979 keV and 0.951keV, respectively. If a K x-ray from copper is incidenton a sodium chloride crystal and gives a first-orderBragg reflection at 15.9 when reflected from thealternating planes of the sodium atoms, what is thespacing between these planes ?

HRK-Exercise 48.9: X-rays are produced in an x-raytube by a target potential of 50.0 keV. If an electronmakes three collisions in the target before coming torest and loses one-half of its remaining kinetic energyon each of the first two collisions, determine thewavelengths of the resulting photons. Neglect the recoilof the heavy target atoms.

54MIT- MANIPAL

X-RAYS AND THE NUMBERING OF THE ELEMENTS

Moseleys 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 arrangedaccording to their atomicnumbers in the periodic table



Bohr theory and the Moseley plot: Bohrs formula forthe frequency of radiation corresponding to a transitionin a one-electron atom between any two atomic levels iand f differing in energy by E is

=

=

2

i

2

f

32

o

42

n

1

n

1

h8

eZm

h

Ef

In a many-electron atom, for a K transition, theeffective nuclear charge felt by an L-electron can bethought of as equal to +(Zb)e instead of +Ze, where bis the screening constant due to the screening effect by

the only K-electron.

MOSELEY PLOT OF THE K X-RAYS

56


HRK-Sample problem 48-2: Calculate the value of the constant C in the Moseleys relation for x-ray frequency and compare it with the measured slope of the straight line in Moseley plot.

( )bZh32

em3fand

2

1

32

o

4

=

Frequency of the K x-ray is

( )

=2232

o

42

2

1

1

1

h8

ebZmf

( ) 1bcesin1ZCfor =

Bohr theory and the Moseley plot:



HRK-Sample Problem 48-3: A cobalt target is bombarded

with electrons, and the wavelengths of its characteristicx-ray spectrum are measured. A second, faintercharacteristic spectrum is also found, due to an impurityin the target. The wavelengths of the K lines are178.9 pm (cobalt) and 143.5 pm (impurity). What is theimpurity ?

58MIT- MANIPAL

1. 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]

2. Explain the principle of a laser. [5]

3. Give a brief account of a He-Ne laser. [4]

QUESTIONS


LASERS AND LASER LIGHT

Characteristics of laser light: Laser light is highly monochromatic, highly coherent, highly directional and can be sharply focused.

Interaction of radiation with matter

Absorption: Absorption of a photon of frequency ftakes 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



Spontaneous Emission: The average life time of the atomic system in the excited state is of the order of 108 s. After the life time of the atomic system in the excited state, it comes back to the state of lower energy on its own accord by emitting a photon of energy hf = E2 E1

In an ordinary light source the radiation of light from different atoms is not coherent. The radiations are emitted in different directions in random manner. Such type of emission of radiation is called spontaneous emission.



Stimulated Emission: When a photon (stimulatingphoton) of suitable frequency interacts with an excitedatomic system, it comes down to ground state beforeits life time. Such an emission of radiation is calledstimulated emission.In stimulated emission, both the stimulating photon

and the stimulated photon are of same frequency,same phase and are in same state of polarization,they are emitted in the same direction.

In other words, these two photons are coherent.Thus amplified radiation is got by stimulated emission

Population inversion: Boltzmann statistics gives the population of atoms in various energy states at temperature T.

k = Boltzmann constant. n(E1) = density of atoms with energy E1 , n(E2) = density of atoms with energy E2 . n(E2) < n(E1) if E2 > E1 (Figure a).



( )( )

=

Tk

EEexp

En

En 12

1

2

Metastable state: A metastable state is an excited energy state of an atomic systemfrom which spontaneous transitions to lower states is forbidden (not

allowed by quantum mechanical selection rules).

The average life time of the atomic system in the metastable

state is of the order of 103 s which is much longer than that in an

ordinary excited state.

Stimulated transitions from the metastable state are allowed. An

excited atomic system goes to metastable state (usually a lower

energy state) due to transfer of its extra energy by collision with

another atomic system.

Thus it is possible to have population inversion of atomic systems

in a metastable state relative to a lower energy state.



64MIT- MANIPAL


Principle of a Laser: The main parts of a laser are lasing medium,

resonant cavity and pumping system.

In a laser the medium chosen to amplify light is called lasing

medium (active medium).

This medium has atomic systems (active centers), with special

system of energy levels suitable for laser action (see figure). This

medium may be a gas, or a liquid, or a crystal or a semiconductor.

The atomic systems in this may have energy levels including a

ground state (E1), an excited state (E3) and a metastable state (E2).

65MIT- MANIPAL


The atoms in the state E3 may come down

to state E1 by spontaneous emission or

they may come down to metastable state

(E2) by collision. The atoms in the state

E2 come down to state E1 by stimulated

emission.

In ruby laser the lasing medium is a ruby rod. Ruby is Al2O3doped with Cr2O3.

Cr3+ ions are the active centres, which have approximately similar

energy level structure shown above.

The resonant cavity is a pair of parallel mirrors to 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 the pumping is done by

xenon flash lamp.

66MIT- MANIPAL


These radiations may be reflected due to mirror action of the end

faces (see figure).

When population inversion takes place at E2, a stray photon of

right energy stimulates chain reaction, accumulates more photons, all

coherent.

The reflecting ends turn the coherent beam back into active

region so that the regenerative process continues and part of the

light beam comes out from the partial mirror as a laser pulse.

The out put is an intense beam of coherent light.

The ruby laser gives red light

67


He-Ne Laser has a glass discharge tube filled with He (80%)

and Ne (20%) at low pressure. He-gas is the pumping medium and Ne-gas is the lasing medium. The simplified energy level diagram (see figure) shows 4 levels: Eo, E1, E2 and E3. Electrons and ions in the electrical gas discharge occasionally collide with He-atoms, raising them to level E3 (a metastable state).

68MIT- MANIPAL


During collisions

between He- and Ne-

atoms,

the excitation energy of He-atom is transferred to Ne-atom (level E2).

Thus, population inversion occurs between levels E2 and E1. This

population inversion is maintained because (1) the metastability of

level E3 ensures a ready supply of Ne-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 mirror M1 is fully

reflective and the mirror M2 is partially reflective to allow the laser

beam to come out. The Brewsters windows W & W are at polarizing

angles to the mirrors, to make the laser light linearly polarized.



HRK-Sample Problem 48-7: A three level laser of the type shown in figure below, emits laser light at a wavelength 550 nm, near the centre of the visible band. If the optical mechanism is shut off, what will be the ratio of the population of the upper level E2 to that of the lower level E1 at 300 K ? At what temperature for the condition of (a) would the ratio of populations be half ?



HRK-Sample Problem 48-8: A pulsed ruby laser has a ruby rod (Al2O3 doped with Cr2O3) as an active medium, which is 6 cm long and 1 cm in diameter. There is one aluminium ion (active centre, with energy levels of the type shown in the figure) for every 3500 chromium ions. The ruby laser light has a wavelength of 694.4 nm. Suppose that all the chromium ions are in metastable state (E2) and none are in ground state (E1). How much energy is there in a single laser pulse if all these ions come down to ground state in

a single stimulated emission chain reaction episode ? Density of Al2O3 is 3700 kg/m

3. Molar mass of Al2O3 is 0.102 kg/mol.



HRK-Exercise 48.28: A ruby laser emits light ata wavelength of 694.4 nm. If a laser pulse isemitted for 12.0 ps and the energy release perpulse is 150 mJ, (a) what is the length of thepulse, and (b) how many photons are there ineach pulse ?



HRK-Exercise 48.29: It is entirely possible thattechniques for modulating the frequency or amplitude ofa laser beam will be developed so that such a beamcan serve as a carrier for television signals, much asmicrowave beams do now. Assume also that lasersystems will be available whose wavelengths can beprecisely tuned to anywhere in the visible range (400 nmto 700 nm). If a television channel occupies abandwidth of 10 MHz, how many channels could beaccommodated with this laser technology ?



HRK-Exercise 48.30: A He-Ne laser emits light at a wavelength of 632.8 nm and has an output power of 2.3 mW. How many photons are emitted each minute by this laser when operating ?

HRK-Exercise 48.33: An atom has two energylevels with a transition wavelength of 582 nm. At 300K, 4.0 x 1020 atoms are in the lower state. (a) Howmany occupy the upper state under conditions ofthermal equilibrium ? (b) Suppose, instead, that 7.0 x1020 atoms are pumped into upper state, with 4.0 x 1020

atoms in the lower state. How much energy could bereleased in a single laser pulse ?

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