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


3/128

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


4/128

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


6/128


Ernest Rutherford

1911


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8/128

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


11/128

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


12/128

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


16/128


17/128

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


21/128

(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


22/128

[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+.


23/128

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

[ ]


31/128

[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]


33/128

[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


34/128

Solution a:

Solution c:

[42 3 P 13] ( ) C t t l l di f


35/128

[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 + ?


36/128

(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


38/128

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


39/128

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


40/128

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


41/128

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


42/128

[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


43/128

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


44/128

[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


45/128

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.


46/128

THE WAVE FUNCTIONS FOR HYDROGEN


47/128

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


48/128

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


49/128

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


50/128

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 (=


51/128

[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


52/128

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


53/128

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 ]


54/128

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


55/128

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


56/128

[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

)(


57/128

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


58/128

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


59/128

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


60/128

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:


61/128

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


62/128

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


63/128

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 =


64/128

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


65/128

[ ] 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


66/128

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


97/128

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
http://en.wikipedia.org/wiki/Image:Helium_neon_laser_spectrum.png


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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 !


118/128

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

r

ea

r

31 1)(

o

oo

sa

r

ear

ar

21

241)(

2

3

2

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]

Documents

Atomic Physics Official Slide