Atomic Physics

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



[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

e

2

2

22 2

2

o

o

oaadr

d ar

e 115

By Substituting the above values

[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 ≠ 0 (ml is never greater than l means L never align parallel or antiparallel to magnetic field B.

)1(

mLcos Z

L

θ is quantized.

The Zeeman effect:

The phenomenon of splitting of energy levels and

hence spectral lines in magnetic field is known an

Zeeman effect.

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 (for H-atom) :

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 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.

Pauli’s Exclusion Principle

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:

77MIT- 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.

78

The x-ray spectrum has two parts: The X ray spectrum consists of a continuous spectrum having radiations of all possible wavelengths within a certain range and superimposed on it a sharp line spectrum known as characteristic spectrum of definite wavelengths which are characteristic of the material used as target.

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:

79MIT- MANIPAL


Sharply defined cutoff wavelength (λMIN) is a prominent feature of the continuous x-ray spectrum. 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

80MIT- MANIPAL BE-PHYSICS-ATOMIC PHYSICS-2010-11


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

81MIT- MANIPAL


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

KK12

.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

hchfEE

K

193

834

12

K

K

K12

pm37.63

10x6.1x10x)4.020(

10x3x10x625.6

EE

hc

hchfEE

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

32o

4

h32

em3C

2/3

o

2

h32

em3C

2/17 Hz10x95.4C

2/17 Hz10x96.4C

graphfrom

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


LASERS AND LASER LIGHT

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

Interaction of radiation with matter Absorption: Absorption of a photon of frequency f takes place when the energy difference E2 – E1 of the allowed energy states of the atomic system equals the energy hf of the photon. Then the photon disappears and the atomic system moves to upper energy state E2 (see figure).



Spontaneous Emission: The average life time of the atomic system in the excited state is of the order of 10–8 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 (see figure). 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 (stimulating photon) of suitable frequency interacts with an excited atomic system, it comes down to ground state before its life time. Such an emission of radiation is called stimulated 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(see figure).

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). This is the normal condition in which the population of the atoms in upper energy state is less than that in lower energy state. For the stimulated emission rate to exceed the absorption rate it is necessary to have higher population of upper energy state than that of lower energy state. This condition is 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

EEexp

En

En 12

1

2

Metastable state: A metastable state is an excited energy state of an atomic system from 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 10–3 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. 99MIT- MANIPAL BE-PHYSICS-ATOMIC PHYSICS-2010-11


100

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

LASERS AND LASER LIGHTPrinciple 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).

101



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 Al2O3 doped 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.

102


LASERS AND LASER LIGHTThese 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

103



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).

104



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 Brewster’s 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 emits light of wavelength 550 nm. (a) What is the ratio of population of the 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 E2 to that of E1 becomes half?

kT

EEexp

N

N.a

12

1

2

eV26.210x6.1

J10x616.3

J10x616.3 10x550

10x3x10x625.6hchfEE

19

19

19

9

834

12

923.86expN

N

1

2

38

1

2 10x77.1N

N

KT = 0.026eV

K=1.38 x 10-23 /1.6 x 10-19 = 8.625 x 10-5eV/K

2

1

N

N)b

1

2

2

1

kT

EEexp 12

)2ln(

kT

EE 12

K37800)2ln(k

EET 12

HRK-Exercise 48.28: A ruby laser emits light at wavelength 694.4nm. If a laser pulse is emitted for 12ps and 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?

m10x6.310x12x10x3

ctpulsetheofLength)a(3128

1710x25.5hc

En

cnhnhfE,pulseperofEnergy)b(

HRK-Exercise 48.29. Assume that lasers are available whose

wavelengths can be precisely "tuned" to anywhere in the visible

range - that is, in the range 400 nm < λ < 700 nm. If every

television channel occupies a bandwidth of 10MHz,how many

channels can be accommodated within this wavelength range?Solution:The lower frequency isf1=c/ λ1 =4.29 x 1014 HzThe higher frequency isf2=c/ λ2 =7.50 x 1014 HzThe number of signals that can be sent in this range is(f2-f1)/10 = 3.21 x 107

That's quite a number of television channels.

HRK-Exercise 48.30. A He-Ne laser emits light of 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?

nhfPtime/PowerE

hc

P

hf

Pnphotonsof.no

ondsecper10x3x10x626.6

nm8.632x10x3.2n

834

3

ondsecper10x325.7n 15

ondsecper10x325.7n 15

min/60x10x325.7n 15

min/10x4.4n 17

HRK-Exercise 48.33: A atom has two energy levels with a transition wavelength of 582 nm. At 300 K, 4 x 1020 atoms are there in the lower state. (a) How many occupy the upper state under conditions of thermal equilibrium? (b) Suppose, instead, that 7.0 x 1020 atoms are pumped into upper state, with 4.0 x 1020 remaining in the lower state. How much energy could be released in a single laser pulse?

kT

EEexp

N

N)a

12

1

2

eV13.2hc

hfEE 12

eV026.0kT,Also

kT

EEexpNN 12

12

92.81expx10x4N 202

162 106.6 xN

nhfE)b

J10x6.1x13.2x10x7E 1920 J240E

0N2 That's effectively none.

111


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

112


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

113


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(

114


ATOMIC PHYSICS

08. Sketch schematically the plot of the radial probability density 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 l [1](b) Orbital magnetic quantum number ml [4](c) Spin magnetic quantum number ms [3]

10. Explain the continuous x-ray spectrum with a schematic plot of the spectrum. [2]

11. Obtain an expression for the cutoff wavelength in the continuous x-ray spectrum.

[4]

QUESTIONS

115


ATOMIC PHYSICS

12. Explain the characteristic x-ray spectrum with a schematic plot of the spectrum. [2]

13. Explain the origin of characteristic x-ray spectrum with a sketch of x-ray energy level diagram. [3]

14. Write Moseley’s relation for the frequency of characteristic x-rays. Sketch schematically the

Moseley’s plot of characteristic x-rays.[2]

15. Obtain Moseley’s relation for characteristic x-ray frequency from Bohr theory.

[4]

16. Mention the characteristics of a laser beam. [2]

QUESTIONS

116


ATOMIC PHYSICS

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]

QUESTIONS

Atomic Physics

Documents

series n

n j2 n

series r h

series spectrum

value of n

e t f e

hydrogen atom

orbit r