Chapter 4 Atomic Structure (Pp 96-125)

5/28/2018 Chapter 4 Atomic Structure (Pp 96-125)

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CHAPTER 4

ATOMIC STRUCTURE

BOHRS THEORY OF HYDROGEN ATOMExample 4-1

Determine the radius of the first orbit of the hydrogen atom.SolutionThe radius of electron in nth orbit of hydrogen atom is given by

kme

hnrn 22

22

4=

)10988.8()10602.1)(10109.9(4)10626.6()1(

9219192

2342

1

=

r

53.010287.5 111 = mr

Example 4-2

Find the value of quantum number n for a hydrogen atomthat has an orbital radius of 847 pm.SolutionThe radius of in nth orbit of hydrogen atom is given by

211103.5 nrn

= in metres

4103.5

10847

103.5 11

12

11 =

=

=

nrn the desired quantum

number.Example 4-3

Determine the speed of electron in 10th orbit of hydrogen

atom. B.U. B.Sc. (Hons.) 1984ASolutionThe velocity of electron in nthorbit of hydrogen atom is givenby

nh

ekvn

22=


2/30

smv /10187.2)10626.6)(10(

)10602.1)(10988.8(2 534

2199

10 =

=

the desired speed.

Example 4-4

Find the de Broglie wavelength of an electron in the n = 3

orbit of hydrogen atom. In what region of the spectrumwould a photon of the same wavelength be classified?SolutionThe velocity of electron in nthorbit of hydrogen atom is given

byhn

ekvn

22= or

h

ekv

32 2

3

= for n = 3.

The de Broglie wavelength is defined as

2

2

3 2

3

ekm

h

vm

h

p

h

===

219931

234

)10602.1(10988.8)(10109.9(2

)10626.6(3

= 1010977.9 10 = m

The photon of this wavelength belongs to X-ray region.Example 4-5

Calculate the frequency of electron in the first Bohr orbit ofhydrogen atom. B.U. B.Sc. 2004ASolutionThe velocity of electron in terms of radius is given by

rm

nhv

2=

The orbital frequency is given byf = (speed of electron) (circumference of orbit)

22421

2 mrnh

rrm

nhf

==

Butkme

hnr

22

22

4= therefore


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=

=

33

2422

22

22

2

444 hn

kme

hn

kme

m

nhf

3343

29419312

)10626.6()1()10988.8()10602.1)(10109.9(4

=

f

1510577.6 = Hz

Example 4-6

On the average, an atom will exist in an excited state for810 second, before it makes a downward transition and

emits a photon. Assuming that the electron in hydrogen is inthe state n = 2, how many revolutions about the nucleus aremade before the electron jumps to the ground state?SolutionThe time period T of electron in nthorbit is given by

242

33

4

1

kme

hn

f

T

==

29419312

343

)10988.8()10602.1)(10109.9(4)10626.6()2(

=

T

sT 1510216.1 = Number of revolutions completed by electron is given by

615

8

10224.810216.1

10=

=

periodTime

stateexcitedinTimeRevolutions

Example 4-7

Show that the energy of electron in nthfor hydrogen atom is

given by 26.13

nEn = eV.SolutionThe energy of electron in nth orbit is given by

22

2422hn

kmeEn

= in joules


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22

2322

hn

kmeEn

= in eV

2342

29319312

)10626.6()10988.8()10602.1)(10109.9(2

=

nEn

2

6.13n

En = in eV

Example 4-8

Calculate the binding energy of the hydrogen atom in thefirst excited state.SolutionThe energy of electron in nth orbit is given by

2

6.13n

En = in eV

For first excited state we substitute n = 2, therefore

4.3)2(6.1322 ==E eV

Hence the binding energy of the hydrogen atom in the firstexcited state is 3.4 eV.Example 4-9

Calculate the energy of the electron for first two excitedstates of hydrogen atom. What will be the wavelength ofemitted (or absorbed) radiation if an electron makes atransition between these two states?Solution(a) The energy of electron in nth orbit is given by

2

6.13

n

En = in eV

The energy of the first two excited states can be calculated bysubstituting n = 2 and n = 3 respectively. Hence

eVEn 4.3)2(6.13

,222

===

eVEn 51.1)3(6.13

,323

===


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(b)Now

hc

hEE == 23

)10602.1)}(4.3(51.1{

)10998.2)(10626.6(19

834

23

=

=

EE

hc

m710561.6 = or 6561

Example 4-10

Calculate the first, second and third excitation energies forhydrogen atom.SolutionThe energy of electron in nth orbit is given by

2

6.13n

En = in eV

Now ,1=n eVE 6.13)1/()6.13( 21 ==

,2=n eVE 4.3)2/()6.13( 22 ==

,3=n eVE 51.1)3/()6.13( 23 ==

,4=n eVE 85.0)4/()6.13( 24 == Let 1, 2 and 3 be the first, second and third excitationenergies of the hydrogen atom, then

1= eVEE 2.10)6.13(4.312 == 2= eVEE 09.12)6.13(51.113 == 3= eVEE 75.12)6.13(85.014 ==

Example 4-11

Calculate the value of Rydberg constant for hydrogen atom.

SolutionThe Rydberg constant is defined by

3

4222ch

mekR

=

3348

41931292

)10626.6)(10998.2()10602.1)(10109.9()10988.8(2

=

R

1710097.1 = mR


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Example 4-12

The wavelength of the first line of Lyman series of hydrogenatom is 1216 . Calculate the wavelength of third line.

B.U. B.Sc. 1990ASolutionThe wavelength of different members of the Lyman series ofhydrogen atom is given by

=

22

1

1

11

nR

n , n = 2,3,4,.

For the first member n = 2 , therefore

43

21

11

12161

22

RR =

=

)1216)3(4

=R =9121

= -1

For the third member n = 4 and formula for Lyman seriesbecomes

48645

9121

1615

1615

41

111 22

3

===

= RR

8.9725

48643 ==

Example 4-13

The wavelength of the first line of Lyman series of hydrogenatom is 1216 . Calculate the wavelength of 5thline.

B.U. B.Sc. 1992ASolutionThe wavelength of different members of the Lyman series of

hydrogen atom is given by

=

22

1

1

11

nR

n , n = 2,3,4,.

For the first member n = 2 , therefore

43

21

11

12161

22

RR =

= or

)1216)3(4

=R 9121

= -1


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For the fifth member n = 6 and formula for Lyman seriesbecomes

3283235

9121

3635

3635

61

111

223

===

=

RR

93835

328323 ==

Example 4-14The wavelength of the first member of Balmer series of thehydrogen atom is 6563 . Calculate the wavelength of thefirst member of Lyman series in the same spectrum.

B.U. B.Sc. 2009ASolutionThe formula for Balmer series of hydrogen atom is

=

22

1

2

11

nR

, n = 3,4,5, . . .

For first member of this series we substitute n = 3

365

31

211 22

1

RR =

=

R536

1 = (1)

The Lyman series is given by

=

22

1

1

11

nR

, n = 2,3,4, . . . .

For first member of this series we substitute n = 2

43

2

1

1

1122

1

RR =

=

R34

1 = (2)

Divide Eq(2) by Eq(1)

275

365

34

1

1=

=

R

R


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37.121527

)6563(5275 1

1 ===

Example 4-15

Calculate the longest and shortest wavelength i.e. serieslimit of the Lyman series.Solution

The Lyman series is given by

=

22

1111

nR

, n= 2, 3, 4 . . .

For longest wavelength, substitute n = 2

43

2

1

1

1122

RR =

=

R34

=

1215105.121)10097.1(3

4 97

==

= m

For series limit, substitute =n

RRn

R =

=

=

222 )(

11

1

1

11

9121091210097.1

11 107 ==

==

m

R

Example 4-16

Calculate the longest wavelength in Balmer series. Also findthe value of the shortest wavelength. B.U. B.Sc. 1988SSolution

The formula for Balmer series of hydrogen atom is

=

22

1

2

11

nR

, n = 3,4,5,.

Substitute n = 3

365

31

211

22

RR =

=


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656310563.6)10097.1(5

36536 7

7 ==

==

m

R

Substitute =n in above formula

41

2

1122

RR =

=

364610646.310097.1

44 77 ====

mR The desired longest and shortest wavelengths are 6563 and3636 respectively.Example 4-17

Calculate the longest and shortest wavelength i.e. serieslimit of the Paschen series. . B.U. B.Sc. 1991ASolutionThe formula for Paschen series of hydrogen atom is

=

22

1

3

11

nR

, n =4,5,6, . . .

Substitute n = 4

1447

41

311

22

RR =

=

)10097.1(7144

7144

7

==R

18750187510875.1 6 === nmm Substitute =n in above formula

91

311

22

RR =

=

710097.199

==R

82044.82010204.8 7 === nmm The desired longest and shortest wavelengths are 1875 nm and820.4 nm respectively.


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Example 4-18

Determine the largest wavelength when excited electronmake transition to n = 4level. K.U. B.Sc. 2001SolutionThe wavelength of the emitted photon will; be the longest whenelectron jumps from n = 5 to n = 4. Hence

400951411 22 RR =

=

)10097.1(9400

9400

7

==R

nmm 405110051.4 6 == Note that this is the longest wavelength in Brackett series ofHydrogen atom.Example 4-19

Compute the first three wavelengths for the Paschen seriesof hydrogen. In what region of the spectrum do the lines of

Paschen series lie?SolutionThe formula for Paschen series of hydrogen atom is

=

22

1

3

11

nR

, n =4,5,6, . . .

Substitute n = 4 ,1447

41

311

221

RR =

=

R7144

1 = )10097.1(7144

7

=

nmm 187510875.1 61

==


51

311

222

RR =

=

nmm

R

128210282.1

)10097.1(16225

16225

62

72

==

==


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1311

223

RR =

=

nmmR

109410094.110097.1

1212 673 ==

==

The wavelengths of desired lines are 1875 nm, 1282 nm and1094 nm. These lines lie in the infrared region of the spectrum.

Example 4-20Find the wavelength of the spectral line corresponding tothe transition in hydrogen atom from n = 6 to n = 3 statewhere 710097.1 =R m-1. B.U. B.Sc. 1991SSolutionThe desired line is the third line in Paschen series. The Paschenseries is given by

=

22

1

3

11

nR

, n =4,5,6, . . .

Substitute n = 6

1261

311

22RR =

=

710097.11212

==R

109399.1093100939.1 6 === nmm

Example 4-21

In the hydrogen atom, an electron experiences a transitionfrom a state whose binding energy is 0.54 eV to anotherstate whose excitation energy is 12.2 eV. (a) What are the

quantum numbers for these states? (b) Compute thewavelength of the emitted photon. (c) To what series doesthis line belong? K.U. B.Sc. 2008SolutionThe binding energy of the electron is given by

2

6.13n

En = in eV


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(a)The above equation can be rewritten as

nEn

6.13=

Now 54.0=nE eV, therefore 554.06.13==n .

The binding energy 0.54 eV corresponds to the state n = 5. Notethat the binding energy of the state n = 1 is 13.6 eV. As theexcitation energy is given as 10.2 eV, therefore

eVEn 4.32.106.13 ==

Hence 24.36.13==n

i.e. the given excitation energy corresponds to state n = 2.

(b)

hceVEE === 86.254.04.352

eV

hc

86.2

=

m7

19

834

10336.4)10602.1)(86.2(

)10998.2)(10626.6(

=

=

4336= the wavelength of the emitted photon.(c) As the wavelength of the emitted photon is in the visibleregion, therefore it belongs to Balmer series of the hydrogenatom.

Example 4-22

A collection of hydrogen atoms in the ground state isilluminated with ultraviolet light of wavelength

59.0 nm.

Find the kinetic energy of the emitted electron.SolutionThe energy of the incident photon is given by

Jch

E 189

834

10367.3100.59

)10998.2)(10626.6(

=

==


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eVeVE 0.2110602.110367.3

19

18

=

=

The energy needed to separate the electron from hydrogen atomis given by

eVE 6.131 =

The kinetic energy of the emitted electron will beeVEEK 4.76.130.211 ===


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4-2 HYDROGEN-LIKE ATOMSExample 4-23

How much energy is required to remove the electron from asingly ionized helium atom (He+1) in its ground state?Solution

The energy of electron in nth

orbit for hydrogen-like atom isgiven by

eVinn

Z

hn

ZkmeEn 2

2

22

2242 )6.13(2==

Now n = 1 and Z = 2, therefore the desired energy is

eVE 4.54)1(

)2)(6.13(2

2

1 ==

Example 4-24

Apply the Bohrs theory to He+and calculate for n = 1(a) the radius, (b) the frequency of revolution, (c) the linear

speed of the electron, (d) the total energy of the electron, (d)the angular momentum and (e) the ratio v/c, and (f) decidewhether the classical treatment can be used or not.

Solution(a) The radius of nthorbit of electron in hydrogen-like atom isgiven by

Z

n

eZmk

hnrn

211

22

22 )10287.5(4

==

Substitute n = 1, Z = 2 in above equation

mr11

211

1 10644.2

2

)1)(10287.5(

=

=

(b) The orbital frequency of the electron in hydrogen-like atomis given by

33

42224

hn

eZkmfn

=

)10626.6()1()10602.1()2)(10988.8)(10109.9(4

343

41929312

1

=

f


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Hz1610631.2 = the desired frequency.(c) The velocity of electron in nthorbit of hydrogen-like atom isgiven by

nh

eZkvn

22=

)10626.6)(1(

)10602.1)(2)(10988.8(234

2199

1

=

v sm/10375.4 61 =

(d) The energy of electron in nth orbit for hydrogen-like atom isgiven by

eVinn

Z

hn

ZkmeEn 2

2

22

2242 )6.13(2==

eVE 4.54)1(

)2)(6.13(2

2

1 ==

(e) The angular momentum of electron in first orbit is given by

)10644.2)(10375.4)(10109.9( 11631111

== rvmL 123410054.1 = smkg

(f) 015.010998.2

10375.48

6

=

=

c

v As the ratio (v/c) is small i.e. v

is very small as compared with c, therefore classical treatment isapplicable.Example 4-25

Determine the radius of the second Bohr orbit for doublyionized Lithium.Solution

The radius of nthorbit for Hydrogen-like atom is given by

Z

rnrn

12

=

where 53.01 =r is the radius of the first Bohr orbit ofHydrogen atom. Hence

706.03

)53.0()2( 22 ==r


16/30

Example 4-26

How much energy is required to remove the electron from adoubly ionized lithium atom (Li+2) in its ground state?SolutionThe energy of electron in nth orbit for hydrogen-like atom isgiven by

eVinn

Zhn

ZkmeEn 2

2

22

2242

)6.13(2 ==

Now n = 1 and Z = 3, therefore the desired energy is

eVE 4.122)1(

)3)(6.13(2

2

1 ==

Example 4-27

(a) Compute the first and second excitation potentials forsingly ionized helium atom. (b) What wavelengths areemitted when the He+ returns to the ground state from theseexcited states?

SolutionThe energy of electron in nth orbit for hydrogen-like atom isgiven by

eVinn

Z

hn

ZkmeEn 2

2

22

2242 )6.13(2==

eVnn

22

2 4.54)2)(6.13(== for He+

Now n = 1 , eVE 4.54)1(

4.5421

==

n = 2 , eVE 6.13)2(

4.5422 ==

n = 3 , eVE 04.6)3(

4.5423

==

Let 1 and2be the first and second excitation energies of theHe+atom, then


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1= eVEE 8.40)4.54(6.1312 == 2= eVEE 46.48)4.54(04.613 ==

The desired excitation potentials are 40.8 V and 48.46 V.

(b) Now 11

hc= or

)10602.1)(8.40()10998.2)(10626.6(

19

834

11

==

hc

30410039.3 81 = m

22

hc= or

)10602.1)(46.48()10998.2)(10626.6(

19

834

212

==

hc

25610559.2 81 = m

Example 4-28

Determine the wavelength of the first two lines of singlyionized helium that corresponds to the first two lines of theBalmer series.SolutionThe formula for Balmer series of hydrogen-like atom is

=

222 1

2

11

nZR

, n = 3,4,5,..

For singly ionized helium Z = 2 and above equation becomes

=

22

1

2

14

1

nR

, n = 3,4,5,..

The desired wavelengths are calculated by substituting n = 3and n = 4 respectively in above relation. Now

95

31

21

41

221

RR =

=

164110641.1)10097.1(5

959 7

71 ==

==

mR

43

4

1

2

14

122

2

RR =

=

121510215.1)10097.1(3

434 7

72 ==

==

mR


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Example 4-29

Calculate the wavelengths of first two lines of the Balmerseries of triply ionized Beryllium (Z = 4).SolutionThe formula for Balmer series of Hydrogen-like atom is

=

222 1

2

11

n

ZR

, n = 3,4,5,..

For present case Z = 4 , therefore

= 22

2 121

161

nZ

, n = 3,4,5,..

Now n = 3,9

2031

21

161

221

RR =

=

nmmR

0.411010.4)10097.1(20

920

9 871 ==

==

n = 4, RR 34

1

2

116

1

222

=

=

nmmR

4.301004.3)10097.1(3

13

1 872 ==

==

Hence the wavelengths of the first two lines of the Balmerseries of triply ionized Beryllium are 41.0 nm and 30.4 nmrespectively.Example 4-30

An atom of tungsten has all of its electrons removed exceptone. (a) Calculate the ground-state energy for this oneremaining electron. (b) Calculate the energy and wavelength

of the radiation emitted when this electron makes adownward transition from n = 2to n = 1. (c) In what portionof the electromagnetic spectrum is this photon?Solution(a)The energy of electron in nth orbit for hydrogen-like atom isgiven by


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eVinn

Z

hn

ZkmeEn 2

2

22

2242 )6.13(2==

For tungsten Z = 74, therefore

keVn

eVnn

En 22

4

2

2 5.741044736.7)74)(6.13(

==

keVE 5.741 = the desired ground state energy.(b) When electron makes a downward transition from n = 2 ton = 1, the energy E of the emitted photon is given by

keVkeVkeVEEE 9.55875.5511

21

5.742212 =

==

The wavelength of the emitted photon is calculated from therelation

)10602.1)(10875.55()10998.2)(10626.6(

193

834

==

E

hc

222.010219.211

=

m (c) The emitted photon belongs to X-rays.


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4-3 THE FRANK-HERTZ EXPERIMENTExample 4-31

In Franck-Hertz experiment the ionization potential ofsodium is 5.13 V. Calculate the velocity of the electron.Solution

The kinetic energy of the electron is given byVevm =

2

21

1631

19

10343.1)10109.9(

)10602.1)(13.5(22

=

== sm

m

Vev

Example 4-32

What will be the minimum energy of electron to excite2536 spectral line of mercury in a Franck-Hertzexperiment?

SolutionThe minimum kinetic energy of electron to excite a spectral lineof wavelength is given by

10

934

min 102536

)10998.2)(10626.6(.).(

==

hcEK

eVorJ 890.410833.7 19=


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4-4 ANGULAR MOMENTUMExample 4-33

Calculate the value of Bohr magneton.SolutionThe Bohr magneton is defined as

ee

Bm

eorm

eh24

h

=

12431

3419

10273.9)10109.9(4

)10626.6)(10602.1(

=

= TJB

The accepted value of Bohr magneton is 12410274.9 TJ or

.10778.5 15 TeV Example 4-34

Calculate the magnitude of the orbital angular momentumof an electron in a state with 4=l .

SolutionThe magnitude of orbital angular momentum is given by

sJL3434 10718.4)10055.1()14(4)1( =+=+= hll

Example 4-35

Write down the quantum numbers for all the hydrogenatom states belonging to subshells for which 4=n and 3=l .SolutionNow 4=n and 3=l . As

ll

= ..,.........2,1,0m therefore the desired magnetic quantum numbers are

3,2,1,0,1,2,3 =lm

The spin quantum number is given by21

=Sm .


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Example 4-36

A hydrogen state is known to have the quantumnumber 3=l . What are possible n,

lm and Sm .

SolutionThe principal quantum number n must be greater than l which is 3 for present case. Hence n = 4 and the magnetic

quantum number lm value are as under3,2,1,0,1,2,3 =

lm

The spin quantum number is given by21

=Sm .

Example 4-37

If an electron in a hydrogen atom is in a state with 5=l ,what is the smallest possible angle between L and ZL .Solution

The minimum angle occurs when ll =m and is given by

[ ] )1()1(cos min

+=

+==

ll

l

hll

hl

LLZ

[ ] 91287.0)15(5

5cos min =

+=

245241.24 00min = or

Example 4-38

Calculate and tabulate for a hydrogen atom in a state 3=l ,

the allowed values of ZL , Z and. Find the magnitude of L and.Solution

The corresponding values of magnetic quantum numberl

m for

3=l are 3,2,1,0,1,2,3 =l

m .The desired values arecalculated and tabulated below. Here we have employed thefollowing values


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sJ3410055.1 =h and TJB /10274.924

=

lm

smkg

mLZ

/2h

l=

TJ

m BZ

/

l

=

+=

)1(cos 1

ll

lm

-3 3410165.3 2310782.2 00.150-2 3410110.2 2310855.1 03.125

-1 3410055.1

2410274.9

08.106 0 0 0 00.90 1 3410055.1 2410274.9 02.73 2 3410110.2 2310855.1 07.54 3 3410165.3 2310782.2 00.30 The magnitude of orbital angular momentum is given by

sJL 3434 10655.3)10055.1()13(3)1( =+=+= hll The magnitude of dipole moment is given by

TJB /10213.3))13(3)(10274.9()1(2324

=+=+=

ll


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4-5 THE ZEEMAN EFFECTExample 4-39

A hydrogen atom gas is placed in intense magnetic field of0.8 tesla. Compute the separation between two consecutivestates (Normal Zeeman Effect) when = 5000 .

SolutionThe wavelength separation is given by

hc

B

cm

eB B

e

22

4

== where

ee

Bm

eor

m

eh

24h

=

)10998.2)(10626.6(

)105000)(8.0)(10274.9(834

21024

=

0937.010337.9 12 orm=

Example 4-40

What will be the separation, in , due to normal Zeemansplitting of 4916 line in the mercury spectrum in amagnetic field of 0.5 tesla?SolutionThe wavelength separation is given by

)10998.2)(10626.6(

)104916)(5.0)(10274.9(834

10242

==

hc

BB

212 1064.51064.5 == m

Example 4-41

Calculate the wavelengths of the normal Zeeman Effecttriplet of a spectral line 5500 placed in a magnetic field of10 tesla.SolutionThe wavelength separation is given by

)10998.2)(10626.6(

)105500)(10)(10274.9(834

210242

==

hc

BB


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412.110412.1 10 == m The wavelengths of Zeeman triplet are

588.5498412.155001 === 55002 ==

412.5501412.155003 =+=+= Example 4-42

The Zeeman components of a 500 nm spectral line are0.0116 nm apart when the magnetic field is 1.00 tesla. Findthe ratio of e/mof the electron from this data.SolutionThe wavelength separation is given by

cm

eB

e

4

2

=

or2

)(4

B

c

m

e

e

=

29

98

)10500)(1( )100116.0)(10998.2(4

=

eme

kgCm

e

e

/10733.1 11=

Example 4-43

A spectral line having a wavelength of 5500 shows anormal Zeeman splitting of 2101.1 . What is themagnitude of the magnetic field causing the splitting?Solution

The wavelength separation is given by

hc

BB2

=

or21024

102834

2 )105500)(10274.9()}10)(101.1){(10998.2)(10626.6()(

=

=

B

hcB

teslaB 078.0=


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Example 4-44

Find the minimum magnetic field needed for the ZeemanEffect to be observed in a spectral line of 400 nmwavelengthwhen a spectrometer whose resolution is 0.010 nmis used.SolutionThe wavelength separation is given by

hcBB

2

=

or 2)(

B

hcB

=

2924

9834

)10400)(10274.9()}10010.0){(10998.2)(10626.6(

=B

teslaB 339.1=


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CONCEPTUAL QUESTIONS(1) What is meant by a stationary orbit in Bohrs theory ofHydrogen atom?Answer: - A stationary or allowed orbit is one in which electrondoes not radiate any energy i.e. the total energy of the electronremains constant.(2) What should be the angular momentum of the electronmoving in an allowed orbit?Answer: - The angular momentum of an electron should be anintegral multiple of )2/( h=h while moving in an allowed

orbit i.e. hnL = .(3) How many electronic orbits are there in a Hydrogenatom?Answer: - Theoretically there are an infinite number of orbits ina Hydrogen atom. As there is only one electron in the Hydrogenatom, therefore one of these orbits will be occupied at one time.

The remaining orbits will be vacant.(4) How is the spacing of adjacent electronic energy levels ofan atom affected by the value of n?Answer: - The spacing of adjacent electronic energy levels ofan atom decreases as n increases.(5) What is indicated by the existence of sharp lines in thespectrum of Hydrogen atom?Answer: - It indicates that the electron is moving in a givenorbit with certain frequency only.

ORIt indicates the presence of discrete energy levels in the

Hydrogen atom.(6) Which transition of the electron in Hydrogen atom willemit highest energy photon?Answer: - The transition of the electron from =n to

1=n will emit maximum energy photon. The energy of thisphoton is eVeVEE 6.13)6.13()0(1 == .


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(7) The total energy of the electron in Hydrogen atom isnegative. What is the physical significance of negative sign?Answer: - It reflects the fact that the electron is bound to thenucleus. An energy of 13.6 eV is needed to separate the electronfrom the nucleus of hydrogen atom.(8) Can a Hydrogen atom absorb a photon whose energyexceeds its biding energy?

Answer: - Yes. The excess energy will appear as kinetic energyof the detached electron of Hydrogen atom.(9) Which series in the emission spectrum of Hydrogen atomhas the highest frequencies? In what part of the spectrumare these lines?Answer: - Lyman series. The spectral lines of Lyman series arein the ultraviolet region.(10) Any series of atomic Hydrogen yet to be discoveredwill probably found to be in what region of the spectrum?Answer: - In infrared region of the spectrum.(11) If an electron moves to a larger orbit, does its total

energy increase or decrease? Does its kinetic energyincrease or decrease?Answer: - The total energy of an orbiting electron is given by

r

eZkE

2

2

=

As rincreasesEbecomes less negative and therefore increases.The orbital kinetic energy of the electron is given by

r

eZkK

2

2

=

It is clear that Kdecreases with increase in r.

(12) Under what conditions an electron moving in the orbitwill have de Broglie wavelength?Answer: -The electron orbit (i.e. the circumference of circularorbit) must contain an integral number of de Brogliewavelengths.


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(13) What are two essential features of the vector atommodel?Answer: - (a) Space quantization and (b) Hypothesis ofspinning electron.(14) What is minimum energy needed to create an Electron -positron pair?Answer: - The minimum energy needed to create an electron-

positron pair is given by MeVcm 022.12 20 = .(15) What is the wavelength of the photon corresponding tothe threshold energy needed to produce electron-positronpair i.e. 1.022 MeV?Answer: - 0.012


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ADDITIONAL PROBLEMS(1)Compute the first three wavelengths of the Bracket

series of hydrogen.(2)The binding energy of electron in a certain state of

hydrogen atom is 0.85 eV. What is the correspondingquantum number?

(3)A photon of energy 12.1 eVabsorbed by a hydrogenatom, originally in the ground state, raises the atomto an excited state. What is the quantum number ofthis state?

(4)Determine the speed of electron in the n = 8orbit ofhydrogen atom. Use this value to find thecorresponding de Broglie wavelength.

(5)The lifetime of an excited state is about 810 second.Compute how many revolutions an electron ofhydrogen atom in the excited state n = 4will make

before jumping in the ground state?(6)Which of the spectral lines of the Bracket series isclosest in wavelength to the first spectral line of thePaschen series of hydrogen atom? What is differencein their wavelengths?

(7)Find the radius of the smallest Bohr orbit in doublyionized lithium ( ++Li ). What will be the energy ofelectron in this orbit?

Answers

(1) mandmm 666 10166.210625.2,10051.4

(2) 4 (3) 3 (4) 1510734.2 sm , 2661 (5) 71004.1 Revolutions (6) Fifth line (n = 9)(7) 0.18 , -122 eV

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Chapter 4 Atomic Structure (Pp 96-125)