Chapter 1 Origin of Quantum Theory (Pp 1-42)

5/28/2018 Chapter 1 Origin of Quantum Theory (Pp 1-42)

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CHAPTER 1

ORIGIN OF QUANTUM THEORY

1-1 BLACK BODY RADIATIONProblem 1-1

A cavity whose walls are held at K1900 has a small hole,mm00.1 in diameter, drilled in its wall. At what rate does

energy escape through this hole from the cavity interior?Solution

The rate of energy escape or radiated power is given by Stefan-Boltzmann Law

4TAP = where 428 /1067.5 KmW= . As for cavity resonator = 1,therefore

4238 )1900(])105.0([)1067.5)(1( = P

mWWP 580580.0 ==

Problem 1-2An object is at a temperature of C0400 . At whattemperature would it radiate energy twice as fast?Solution

Now12 2PP =

41

42 2 TATA =

41

42 2 TT =

CKTT

04/1

1

4/1

2 527800)273400()2()2( ==+==

the desired temperature.


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Problem 1-3In 1983 the Infrared Astronomical Satellite (IRAS) detecteda cloud of solid particles surrounding the star Vega,radiating maximally at a wavelength of m32 . What isthe temperature of this cloud of particles? Assume anemissivity of unity.

SolutionAccording to Wiens displacement lawKmTm =

310898.2

KTm

6.901032

10898.210898.26

33

=

=

=

Problem 1-4The sun approximates a black body at K5800 . Find thewavelength at which the sun emits the most energy.Solution

According to Wiens displacement law

KmTm = 310898.2

580010898.210898.2 33

=

=T

m

nmormm 5001057

= Problem 1-5At what wavelength does earth, approximated as a K286 blackbody, radiates energy?Solution

According to Wiens displacement law

KmTm = 3

10898.2

28610898.210898.2 33

=

=T

m

mormm 101015

=


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Problem 1- 6Estimate the peak wavelength of light issuing from the pupilof human eye (which approximates a black body) assumingnormal body temperature.Solution

Here we take the normal body temperature as

KCFT 3103798

00===

According to Wiens displacement law310898.2 =Tm

31010898.210898.2 33

=

=T

m

mm610348.9 =

Problem 1-7The shortest visible wavelength is about nm400 . What isthe temperature of an ideal radiator whose spectralemittance peaks at this wavelength?

SolutionAccording to Wiens displacement law

KmTm = 310898.2

KTm

724510400

10898.210898.29

33

=

=

=

Problem 1-8(a)Show that a human body of area 28.1 m , emissivity

0.1= and temperature C034 will emit radiation atthe rate of W910 .

(b)

Why, then, do people not glow in the dark?Solution(a)The rate of energy escape or power is given by Stefan-

Boltzmann Law4

TAP = WP 907)27334)(8.1)(1067.5)(1( 48 =+=

(b) Here we calculate m corresponding to body temperature


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i.e. 340 C.

)27334(10898.210898.2 33

+

=

=

Tm

mormm 440.910440.96

= As the wavelength of above radiation is in the infrared regionwhich is not visible to humans, therefore the human body doesnot glow in the dark.Problem 1-9An ideal radiator radiates with total intensity 2/94.6 mMW .At what wavelength does the spectral emittance )(I occurs?Solution

According to Stefan- Boltzmann law4)( TTI =

Now 262 /1094.6/94.6)( mWmMWTI == 428

/10670.5 KmW =

Therefore486 )10670.5(1094.6 T=

KT 3326= Now one can use Wiens displacement law

KmTm = 310898.2

332610898.210898.2 33

=

=T

m

871310713.8 7 ormm

=

Problem 1-10

An ideal radiator has a spectral radiancy at nm400 that is3.5 times its spectral radiancy at nm200 . What is itstemperature?Solution

The spectral radiancy of a radiator is given by Plancksradiation law


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1)/exp(1

)(5

=

Tb

aR

where a = 2 c 2h and b = h c / k are constants. According togiven conditions

)(5.3)( 21 RR =

1)/exp(

15.3

1)/exp(

1

252151

=

Tb

a

Tb

a

1)/exp(11

5.31)/exp(

11

2521

51

= TbTb

1)/exp(1

5.31)/exp(

1

2

5

2

1

1

=

TbTb

But 21 2 = , therefore

1)/exp(12

5.31)2/exp(

1

2

5

2

2

2

=

TbTb

1)/exp(112

1)2/exp(1

22 =

TbTb

]1)2/[exp(1121)/exp( 22 = TbTb

0111)2/exp(112)/exp( 22 =+ TbTb The above equation is quadratic in )2/exp( 2Tb with thefollowing factors.

0}12/{exp(}111)2/{exp( 22 = TbTb 0111)2/(exp( 2 =Tb 111)2/exp( 2 =Tb

)111(2 2

nT

b l=

)111ln(2)111ln(2 22 khcb

T ==

KT 7651)111ln()10200)(1038.1(2

)103)(1063.6(923

834

=

=


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1-2 THE QUANTIZATION OF ENERGYProblem 1-11According to Plancks theory, what is the minimum non-zero energy of a molecule with vibration frequency

14104.3 Hz?

SolutionAccording to Plancks theory the energy is given byhnE=

JE 191434 10253.2)104.3)(10626.6)(1( ==

eVE 406.1= Q JeV 1910602.11 =

Problem 1-12Find the energy of a nm700 photon.Solution

The energy of a photon is given by

JchE 199

834

10838.210700

)10998.2)(10626.6(

=

==

eVeV

E 772.110602.1

10838.219

19

=

=

Problem 1-13A microwave oven uses electromagnetic radiation at

GHz4.2 . What is the energy of each microwave oven?Solution

The energy of the photon is given by

JhE 6934 10590.1)104.2)(10626.6(

===

eVeVeVE 925.910925.910602.110590.1 6

19

6

==

=


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Problem 1-14An atom absorbs a photon having a wavelength of nm375 and immediately emits another photon of wavelength

nm580 . What was the net energy absorbed by the atom inthis process? P.U. B.Sc. 2002Solution

Net energy absorbed by the atom.= (Energy of the absorbed photon) (Energy of the emittedphoton)

=

=

)(chchch

=)10580)(10375(

10)375580)(10998.2)(10626.6(99

9834

J1910872.1 = = 168.110602.110872.1

19

19

=

eV

Q JeV 1910602.11 = Problem 1-15Calculate the critical potential in volts corresponding toexcitation of mercury line 2536 . B.U. B.Sc. 2007ASolution

Now Jch

E10

834

102536)10998.2)(10626.6(

==

eVeVE 890.4)10602.1)(102536()10998.2)(10626.6(

1910

834

=

=

Hence the desired value of critical potential is 4.89 volts.

Problem 1-16Silicon becomes better electrical conductor whenilluminated by photons with energies of eV14.1 or greater.What is the corresponding wavelength? K.U. B.Sc. 2002


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Solution


chE=

)10602.114.1()10998.2)(10626.6(

19

834

==

E

ch

Q JeV 1910602.11 = mm 0878.1100878.1 6 ==

Problem 1-17Find energy in electron-volts of

(a) a 0.1 MHz radio photon,(b) a 14100.5 Hz optical photon and(c) a 18100.3 Hz X-ray photon.

Solution

(a) hE= JE 28634 10626.6)101)(10626.6( ==

eVeVE 919

28

10136.410602.110626.6

=

=

(b) hE= JE 191434 10313.3)105)(10626.6( ==

eVeVE 068.210602.110313.3

19

19

=

=

(c) hE= JE 151834 109878.1)103)(10626.6( ==

keVeVeVE 41.1210241.110602.1

109878.1 419

15

==

=

Q JeV 1910602.11 =

Problem 1-18What will be the frequency of the photon whose energy is

eV626.6 ?


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Solution

Now hE=

Hzh

E 1534

19

10602.110626.6

)10602.1)(626.6(=

==

Problem 1-19

What will be the wavelength and frequency of a keV100 photon?Solution

Now

ch

hE == therefore

)10602.1)(10100(

)10998.2)(10626.6(193

834

==

E

ch

011 1240.010240.1 Am ==

And

)10626.6(

)10602.1)(10100(34

193

==

h

E

Hz1910418.2 =

Problem 1-20Find the wavelength and frequency of a MeV100 photon.Solution

The energy of the photon is given by

chhE ==

Now JeVMeVE )10602.1)(101(10100100 1986 ===

Q JeV 1910602.11 = JE 1110602.1 =

Therefore

mE

ch 1411

834

10240.110602.1

)10998.2)(10626.6(

=

==

Hzh

E 2234

11

10418.210626.610602.1

=

==


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Problem 1-21The light from a high way sodium lamp has a wavelength

nm589 . What is the energy, in eV, of a photon from such alamp? P.U. B.Sc. 2007Solution


chhE ==

JE9

834

10589)10998.2)(10626.6(

=

Q JeV 1910602.11 =

eVeVE 21)10602.1)(10589()10998.2)(10626.6(

199

834

=

=

Problem 1-22Under favourable circumstances the human eye can detect

J18101 of electromagnetic energy. How many nm600

photons does this represent?Solution

The energy of 600 nm photon is given by

Jch

E 199

834

1 10311.310600)10998.2)(10626.6(

=

==

Let n be the number of photons in JE 18101 = , thenEEn =1

photonsE

En 3

10311.3101

19

18

1

==

Problem 1-23A 1.00 kWradio transmitter operates at a frequency of 880kHz.How many photons per second does it emit?Solution

The energy of each photon is given byJhE 28334 10831.5)10880)(10626.6( ===


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and WkWP 31000.100.1 == . Hence the desired rate ofemission of photons from the transmitter will be

sphotonsE

P/10715.1

10831.51000.1 30

28

3

=

=


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1-3 THE PHOTOELECTRIC EFFECTProblem 1-24Find the work function for silver surface for which thethreshold frequency of the incident light is Hz15101.1 .

P.U. B.Sc. 2000

SolutionThe work function 0 is defined as

0 =191534

0 10289.7)101.1)(10626.6(

==h J

eVeV 55.410602.110289.7

19

19

0 =

=

Q JeV 1910602.11 = Problem 1-25The stopping potential for electron ejected from a Zincsurface is 2.42 eV for the 1849 ultraviolet mercury line.What is the stopping potential for the 2537 mercury line?

SolutionThe work function of Zinc is calculated from the relation

00

=hc

eV or eVch

00 =

)10602.142.2(101849

)10998.2)(10626.6( 1910

834

0

=

Q JeV 1910602.11 = J

190 10867.6

=

Let V1be the stopping potential for 2537 mercury line, then

01

1

= cheV or eechV 01

=

19

19

1910

834

1 10602.1

10867.6

)10602.1)(102537(

)10998.2)(10626.6(

=V

voltV 60.01 =


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Problem 1-26A metal surface has a photoelectric cutoff wavelength of325.6 nm. The radiation from a 5.0 mWHelium-Cadmiumlaser ( = 3250 ) ejects electrons from a Cesium surfacethat has a stopping potential 1.91 V.(a) What is the work function in electron volts for Cesium?(b) What will be the stopping potential when the incidentradiation is 10.0 mW?Solution

(a) For calculation of work function of Cesium we use the

relation 00

=ch

eV

eVch

00 =

in Joules

00 Ve

ch=

in eV

eV91.191.1)10602.1)(103250( )10998.2)(10626.6( 1910834

0 ==

(b) As stopping potential is independent of the intensity ofincident radiation, therefore eV91.10 = .Problem 1-27Find the maximum kinetic energy in eV of photoelectrons ifthe work function of the material is 2.33 eV and thefrequency of the radiation is Hz151019.3 .

B.U. B.Sc. 2004SSolution

The energy of the incident radiation is given byhE= JE 181534 10114.2)1019.3)(10626.6( ==

eVeVE 20.1310602.110114.2

19

10

=

=

Q JeV 1910602.11 =


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The maximum kinetic energy of photoelectrons is given byEinsteins photoelectric equation.

87.1033.220.13.).( 0max === hEK eVProblem 1-28The minimum frequency of photoelectric emission in copperis 15101.1 Hz. Find the maximum energy of the

photoelectrons (in electronvolts) when light of frequency15105.1 Hz is directed on a copper surface.Solution

According to Einsteins photoelectric equation

JinhhhvmEK m )(21

).( 002

0max ===

eVine

hEK )(.).( 0max =

eVEK19

151534

max 10602.1

)101.1105.1(10626.6(.).(

=

eVEK 65.1.).( max = Problem 1-29Light of frequency 14105.7 Hz falling on the surface of ametal emits electrons of maximum kinetic energy

19106.1 J. Calculate the maximum frequency of lightwhich can just emit electrons from the surface?

F.P.S.C. 1979Solution


)(2

1.).( 00

20max === hhhvmEK m

0max.).( =

h

EK

h

EK max0

.).(=

Hz1434

1914

0 1009.510626.6106.1

105.7 =

=


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Problem 1-30If the photoelectric threshold wavelength of sodium is

nm542 , calculate the maximum velocity of photoelectronsejected by photons of wavelength nm400 .

B.U. B.Sc. 2004SSolution

Now

00

202

1

hchc

hhvm m ==

00

0

00

2 )(2112

m

ch

m

chvm

=

=

00

0 )(2

m

chvm

=

)10542)(10400)(10109.9(}10)400542){(10998.2)(10626.6(2

9931

9834

=mv

smvm /10345.55

= Problem 1-31If the work function for zinc is 4.3 eV, what is the maximumkinetic energy of electrons out of a clean surface by 2537 ultraviolet line of mercury? B.U. B.Sc. 2007ASolution

The energy of the incident photon is given by

Jch

E10

834

102537

)10998.2)(10626.6(

==

eVeVE 888.4)10602.1)(102537( )10998.2)(10626.6( 1910834

=

=

The maximum kinetic energy of the electrons is given byeVeVEEK 6.0588.03.4888.4.).( 0max ===


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Problem 1-32When a copper surface is illuminated by radiation ofwavelength 2537 from a mercury arc, the value ofstopping potential is found to be 0.24 eV. Calculate(a) the threshold wavelength of copper and(b) the work done by the electrons in escaping through

the surface of copper. B.U. B.Sc. 1986ASolution

(a)According to Einsteins photoelectric equation0

02

021

hchchhvm m ==

=

00

11

hceV

0

0 11

=ch

eV

hc

eVhc

hc

eV 00

0

11 ==

eVhc

hc

00

=

)]102537)(10602.1)(24.0()10998.2)(10626.6[(

)]102537)(10998.2)(10626.6[(1019834

108340

=

2668102668 100 ==

m (b) The work done by the electrons in escaping through thecopper surface is defined as work function and given by

J

ch 1910

834

00 10446.7)102668(

)10998.2)(10626.6(

=

==

eV19

19

0 10602.110446.7

= Q JeV 1910602.11 =

= 4.648 eV


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Problem 1-33Nickel has a work function of 5.0 eV.(a) What is the maximum kinetic energy of photoelectronsknocked out of a Nickel surface by a 1.0 mW ultravioletsource of 2000 ?(b) What is the maximum kinetic energy of photoelectronsknocked out by a 15 Wargon laser source at a wavelengthof 4658 ?Solution

The maximum kinetic energy of photoelectrons is given by

0max.).(

=ch

EK

(a)Now Jch10

834

102000)10998.2)(10626.6(

=

eVeVch

20.6)10602.1)(102000()10998.2)(10626.6(

1910

834

=

=

eVEK 20.00.520.6.).( max ==

(b)Now Jch10

834

104658)10998.2)(10626.6(

=

eVeVch

66.2)10602.1)(104658()10998.2)(10626.6(

1910

834

=

=

As energy of the incident photon is less than the work functionof Nickel, therefore no photoelectrons will be emitted.Problem 1-34The emission of photoelectrons due to incident photons onthe sodium surface stopped when a stopping potential of

4.92 eV is applied. Find the wavelength of the incidentphotons (work function of sodium is 2.28 eV.)

P.U. B.Sc. 2001Solution


o

cheV

=0


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cheV =+ 00

00

+=

eV

ch

)10602.128.2()10602.1)(92.4(

)10998.2)(10626.6(1919

834

+

=

nmmm 172101721072.1 97 === Problem 1-35Electromagnetic radiation of wavelength 4000 falls on ametal surface having a work function of 2.1 eV. What is theminimum kinetic energy of the emitted electrons?

B.U. B.Sc. (Hons.) 1990ASolution

The energy of the incident radiation is

E =

hc

h = in J =e

hc

in eV

= 1.3)10602.1)(104000()10998.2)(10626.6(

1910

834

=

eV

Work function = 1.200 == h eVThe maximum kinetic energy of the emitted electrons will be

eVhhEK MAX 0.11.21.3.).( 0 ===

Problem 1-36(a) If the work function for a metal is 1.85 eV, what wouldbe the stopping potential for light having a wavelength of410 nm? P.U. B.Sc. 2008

(b)What should be the maximum speed of the emittedphotoelectrons at the metals surface?

Solution

(a) According to Einsteins photoelectric equation

00

=ch

eV


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ee

chV 00

=

)10602.1(

)10602.185.1(

)10410)(10602.1(

)10998.2)(10626.6(19

19

919

834

=

174.10 =V V

(b) ( ) =max..EK 21 eVvm m 020 =

0

02 2

m

eVvm =

0

02

m

eVvm = = 31

19

10109.9()10602.1)(174.1(2

= 51043.6 1ms

Problem 1-37A maximum wavelength of 5450 is required to ejectphotoelectrons from Sodium metal.(a) Determine the maximum velocity of electrons ejected by

light of wavelength 2000 ?(b) What is the stopping potential for photoelectrons ejectedfrom Sodium by light of wavelength 2000 ?Solution

(a) The maximum kinetic energy of the electron is given by

=

chchvm m

202

1

=

=

00

2 )(2112

m

ch

m

chvm

=0

)(2 mchvm

)105450)(102000)(10109.9(

}10)20005450){(10998.2)(10626.6(2101031

10834

=mv

smvm /10175.16

= (b) For calculation of stopping potential we use the relation


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200 2

1mvmeV =

)10602.1(2

)10175.1)(10109.9(

2 19

263420

0

==

e

vmV m

voltsV 93.30 =

Problem 1-38A metal surface has a photoelectric cutoff wavelength of325.6 nm. It is illuminated with light of wavelength 259.8nm. What is the stopping potential?Solution


=

chcheV0

)()(11

0

=

=

e

ch

e

chV

)108.259)(106.325)(10602.1(]10)8.2596.325)[(10998.2)(10626.6( 9919

9834

0

=V

voltsV 645.90 = Problem 1-39You wish to pick a substance for a photocell operable withvisible light. Which of the following will do (work functionin parentheses): tantalum (4.2 eV), tungsten (4.5 eV),aluminum (4.2 eV), barium (2.5 eV), lithium (2.3 eV) andcesium (1.9 eV)?Solution

The violet light in the visible region will have the mostenergetic photons possessing a wavelength of about 400 nm.The energy corresponding to this wavelength is

199

834

10966.4)10400(

)10998.2)(10626.6(

=

==

hcE J

= 3.1 eV Q JeV 1910602.11 =


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Hence barium, lithium and cesium can be used because theirwork function is less than 3.1 eV.

Problem 1-40Incident photons strike a sodium surface having workfunction of 2.28 eV causing photoelectric emission. Thephotoelectric current is stopped by applying 4.92 V asstopping potential. Calculate the wavelength of the incidentphoton. K.U. B.Sc. 2001Solution


00

=hc

eV or

hc

eV =+ 00

)10602.128.2()10602.1)(92.4(

)10998.2)(10626.6(1919

834

00

+

=

+=

eV

hc

m710722.1 = or 1722

Problem 1-41In the photoelectric effect, the target has a stoppingpotential of 4.0 volts when the incident photon has amomentum of smkg /1050.3 27 . What is the thresholdfrequency?Solution


00 hheV =

eVhh 00 =

h

eV00 =

Nowc

h

c

hhp

===

)/( or

h

pc= , therefore above

equation becomes

h

eVpc

h

eV

h

pc 000

==


22/42

34

19278

0 10626.6

)10602.1)(0.4()1050.3)(10998.2(

=

Hz14

0 10165.6 = which is the desired value of threshold frequency.


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1-4 THE COMPTON EFFECTProblem 1-42What is the frequency of an X-ray photon whosemomentum is 23101.1 kg m/s.Solution

The momentum of a photon is given by

c

h

c

hhp

===

)/(

Hzh

pc 1834

238

10977.4)10626.6(

)101.1)(10998.2(=

==

Problem 1-43Find the frequency, energy and momentum of a photonhaving wavelength of 41.6 pm. P.U. B.Sc. 2003Solution

The frequency of the given photon is calculated from therelation.c=

Hzc 18

12

8

10208.7106.41

10998.2=

==

The energy of the given photon is given by

Jch

E 1512

834

10775.4)106.41(

)10998.2)(10626.6(

=

==

eVeVE4

19

15

10981.210602.1

10775.4=

=

Q JeV 1910602.11 = = 29.806 keV

The momentum of the given photon is given by

2312

34

10593.1106.41

10626.6

=

==

hp kg m s -1 or N s

Note that


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ckeVc

E

c

hchp /81.29

1==

==

Problem 1-44Find (a) the frequency (b) the wavelength and (c) themomentum of a photon whose energy equals the rest mass

energy of the electron.SolutionThe rest mass energy of the electron is given by

JcmE1428192

0 10187.8)10998.2)(10109.9(

=== (a) The frequency of the given photon is calculated from therelation. hE=

Hzh

E 2034

14

10236.110626.6

10187.8=

==

(b) For wavelength we use the relation

hcE=

mE

hc 1214

834

10426.2)10187.8(

)10998.2)(10626.6(

=

==

(c) The momentum of the photon is defined as

12212

34

10731.210426.2

10626.6

=

== smkg

hp

Problem 1-45The quantity )/( cmh , which has the dimensions of length, iscalled the Compton wavelength. Determine the wavelengthfor (a) an electron (b) a proton. (c) Show that if a photon haswavelength equal to Compton wavelength of a particle, thephotons energy is equal to the rest mass energy of theparticle.


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Solution

(a) mcm

h

e

12931

34

10426.2)10998.2)(10109.9(

10626.6

=

=

(b) mcm

h

p

15927

34

10321.1)10998.2)(10673.1(

10626.6

=

=

(c) 2)/(cm

cmhchchE ===

Problem 1-46An X-ray photon whose initial frequency was 19105.1 Hzemerges from a collision with an electron with a frequencyof 19102.1 Hz. How much kinetic energy was imparted tothe electron?Solution

The kinetic energy of the electron will be equal to the difference

of energies of incident and scattered photons i.e.0

hh }10)2.15.1){(10626.6()( 19340 ==

h

J1510988.1 = or 12.4 keV Q JeV 1910602.11 =

Problem 1-47Incident photons of energy 10.39 keV are Comptonscattered and the scattered beam is observed at 450relativeto the incident beam.(a) What is the energy of the scattered photon at this angle?(b) How much kinetic energy, in eV, is given to the scatteredphoton? B.U. B.Sc. 2009ASolution

(a) The energy of the scattered photon is given by

)cos1)(/(1 20 +=

cmE

EE


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)45cos1)(411/39.10(139.10

0+

=E

keVMeVcm 511511.020 ==Q

keVE 33.10= (b)The kinetic energy of the recoiled electron is given by

keVEEKe 06.033.1039.10 === Problem 1-48Find the maximum wavelength shift for a Compton collisionbetween a photon and a free proton.Solution

The Compton shift in wavelength is given by

)cos1( =cm

h

p

It is clear that the above expression will have maximum value ifthe photon is scattered through 180 0. Hence

)10998.2)(10673.1(

)180cos1)(10626.6(827

034

=

fmm 642.210642.2 15 == Problem 1-49X-rays with = 10 nm are scattered through a carbontarget. The scattered radiation is viewed at 900 to theincident radiation. Calculate the Compton shift.

K.U. B.Sc. 2002Solution


)cos1(0 == cmh

m12

831

034

10426.2)10998.2)(10109.9(

)90cos1)(10626.6(

=

=


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Problem 1-50X-rays are scattered from a carbon target. The scatteredradiation is viewed at 900 to the incident beam. What isCompton shift? P.U. B.Sc. 2006Solution


)cos1(0

= cm

h

)10998.2)(10109.9()90cos1)(10626.6(

831

034

=

pmm 426.210426.2 12 ==

Problem 1-51X-rays with wavelength of 110 pm are scattered off freeelectrons at angle of 200. Find the change in .Solution

The Compton shift in wavelength is given by)cos1(

0

=cm

h

)10998.2)(10109.9(

)20cos1)(10626.6(831

034

=

pmm 146.01046.1 13 ==

Problem 1-52Photons of wavelength 2.17 pm are incident on freeelectrons.

(a) Find the wavelength of a photon that is scattered 350from the incident direction.

(b) Do the same if the scattering angle is 1150.Solution


)cos1( =cm

h

p


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(a))10998.2)(10109.9(

)35cos1)(10626.6(831

034

=

pmm 439.01039.4 13 ==

607.2439.017.2' =+=+= pm

(b)

)10998.2)(10109.9(

)115cos1)(10626.6(831

034

=

pmm 45.31045.3 12 ==

62.545.317.2' =+=+= pmProblem 1-53An X-ray photon of energy ''E , moving along the x-axis,undergoes Compton scattering with an electron whichinitially at rest. The photon is scattered at angle '' withthe x-axis. Show that the energy ''E of the scatteredphoton is given by

)2/(sin)/2(1 220 cmE

EE

+

=

Solution


)cos1(0

=cm

h

The energy of the scattered photon is given by

)cos1)(/( 0 +=

+=

=

cmh

chchchE

)cos1)(/(1)cos1)}(/()/{(1

)/(2

0

2

0

+

=

+

=

cmE

E

cmch

chE

)2/(sin)/2(1 220 cmEE

E+

= )2/(sin2)cos1( 2 =Q


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Problem 1-54A photon of 1 MeVcollides with a free electron and scattersthrough 900. What are the energy of the scattered photonand the kinetic energy of the recoiling electron?Solution

The energy of the scattered photon is given by

)2/(sin)/2(1 220 cmE

EE

+=

MeVE 338.0)45(sin}511.0/)1(2{1

102 =

+=

The kinetic energy of the recoiling electron is given byMeVEEK 662.0338.01 ===

Problem 1-55Gamma rays of energy 0.662 MeVare Compton scattered.(a) What is the energy of the scattered photon observed at a

scattering angle of 60

0

?(b) What is the kinetic energy of the scattered electrons?Solution

(a) The energy of the scattered photon is given by

)cos1()/(1 20 +=

cmE

EE

MeVE 402.0)60cos1}(511.0/)662.0{(1

662.00 =

+=

MeVcm 511.020 =Q (b) The kinetic energy of the recoiling electron is given by

MeVEEK 260.0402.0662.0 ===

Problem 1-56At what scattering angle will incident 100 keVX-rays leavea target with energy of 90 keV?


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Solution

The energy of the scattered photon in terms of energy of theincident photon and scattering angle is given by

)cos1)(/(1 20 +=

cmE

EE

Rearranging the above equation we get

201cos cm

EEEE

=

43222.0)90)(100(

)511)(90100(1cos =

=

keVkeVMeVcm 51110511.0511.0 320 ===Q 01 4.64)43222.0(cos ==

Problem 1-57If the Compton shift in an experiment is found to be0.0121 , then calculate the scattering angle.Solution


)cos1(0

=cm

h

)cos1()(0

=

h

cm

h

cm )(1cos 0

=

34

10831

10626.6)100121.0)(10998.2)(10109.9(

1cos

=

50130.0cos = or 0545590 = or 09.59Problem 1-58In a Compton experiment, the wavelength of the incident X-rays is m1110078.7 while the wavelength of the outgoingX-rays is m1110314.7 . At what angle the scatteredradiation was measured? B.U. B.Sc. 2007S


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Solution

Now )cos1(0

=cm

h

)cos1()(0

=

h

cm

h

cm )(1cos 0

=

34

10831

10626.6}10)078.7314.7){(10998.2)(10109.9(

1cos

=

02734.0cos = or

42884.88)02734.0(cos 001 == or Problem 1-59On scattering via the Compton Effect, a photon undergoes afractional wavelength change ]/)[( equal to 6 %. Ifthe incident photon has a wavelength of 0.020 nm, at whatangle is the detector to the incident beam?

Solution


)cos1(0

==cm

h

)cos1(0

=

=

cm

h

cos10 =

h

cm

=

hcm01cos

=

1006

10636.6

)10020.0)(10998.2)(10109.9(1cos

34

9819

50543.0cos = or 428359)50543.0(cos 01 ==


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Problem 1-60In Compton scattering, calculate the maximum kineticenergy given to the scattered electron for a given photonenergy.Solution

The kinetic energy given to the recoiled electron is given byEEEK =..

The above energy will have a maximum value if the energy ofthe scattered photon i.e. E is a minimum. Now

)cos1)(/( 0 +=

+=

=

cmh

chchchE

It is clear that E will be a minimum for 0180= i.e.

)/2(1 0min

cmh

chE

+=

)/1)(/(21)/(

)/2(1)/(

200 cmch

ch

cmh

ch

+=

+=

Ecm

Ecm

cmE

EE

2)/2(1 20

20

20

min+

=+

=

Hence

Ecm

EcmEEcm

Ecm

EcmEEK

2

2

2.).(

20

20

220

20

20

max+

+=

+=

20

2

max 2

2.).(

cmE

EEK

+=

Problem 1-61If the maximum kinetic energy given to the electrons in a

Compton scattering experiment is 10 keV, what is thewavelength of the incident X-rays?Solution

Now2

0

2

max 2

2.).(

cmE

EEK

+=


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51122

102

+=

E

E

22511020 EE =+ 05100202 2 = EE 02555102 = EE

Using quadratic formula for positive root we get

keVE 8.55)1(2

)2555)(1(4)10()10( 2=

+=

The wavelength of the incident X-rays is calculated as

chE= or

E

ch=

nmm 022.0102.2)10602.1108.55()10998.2)(10626.6( 11

193

834

==

=

Problem 1-62What percentage increase in wavelength leads to 75percent

loss of photon energy in a collision of photon with astationary electron? B.U. B.Sc. 2004ASolution

Let E and E be the energies of incident and scattered photonsrespectively, then fractional loss of photon energy is given by

+

=

=

=

=

=

= 1

)/()/(

11hc

hc

E

E

E

EEf

375.01

75.01

=

=

=

f

f

Hence a 300 percent increase in wavelength will lead to 75

percent loss of photon energy.Problem 1-63When photons of wavelength 0.024 are incident on atarget, scattered photons are detected at an angle of 60o.Calculate

(a) the wavelength of the scattered photon.(b) the angle at which the electron is scattered.

B.U. B.Sc. (Hons.) 1991A


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Solution

(a) The wavelength of the scattered photon is given by

)cos1( +=cm

h

o

+=

)10998.2)(10108.9()60cos1)(10626.6(

)10024.0( 83134

10o

= m12106.3 = 0.036 (b) The scattering angle of the electron is given by

cossin

cossin

tan

=

=

= 866.060cos)024.0(036.0

60sin)024.0(=

o

o

oor 9.403335400 =

Problem 1-64

X-ray photons of wavelength 0.220 nm are Comptonscattered at 450. Calculate the energy of the scatteredphoton in eV.Solution


)cos1(0

=cm

h

)10998.2)(10109.9(

)45cos1)(10626.6(831

034

=

m131011.7 =

The wavelength of the scattered photon is)1011.7()10220.0( 139 +=+=

m101020711.2 =

The energy of the scattered photon will be

Jinch

E

= or eVine

chE

=


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keVeVE 62.51062.5)10602.1)(1020711.2(

)10998.2)(10626.6( 31910

834

==

=

Problem 1-65X-rays of wavelength 0.040 are scattered from a carbonblock. Determine

(a) the momentum of a photon scattered at angle of 300

and(b) the kinetic energy of the recoil electron.Solution

(a) The Compton shift in wavelength is given by

)cos1(0

=cm

h

m13

831

034

1025.3)10998.2)(10109.9(

)30cos1)(10626.6(

=

=

)1025.3()10040.0( 1310 +=+=

m1210325.4 =

The momentum of the scattered photon is given by

msJh

p /10532.110325.410626.6 22

12

34

=

==

(b) The kinetic energy of the recoil electron is given by

=

=

)(chchchK

)10325.4)(10040.0()1025.3)(10998.2)(10626.6(

1210

13834

=K

JK 1510732.3 =

keVeVeVK 3.231033.210602.110372.3 3

19

15==

=

Problem 1-66How many head-on Compton scattering events arenecessary to double the wavelength of a photon havinginitial wavelength 200 pm?


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Solution

The Compton shift in wavelength is independent of thewavelength of the incident photon and will have maximumvalue for head-on collision i.e. 1800. Now

)cos1(0

=cm

h

m12831

034

10853.4)10998.2)(10109.9()180cos1)(10626.6(

=

=

and mmpm 1012 10210200200 === . HenceNumber of Compton Scattering events necessary to double

= 4110853.4

10212

10

=

=

Problem 1-67An X-ray photon of frequency 1019Hz is scattered throughan angle of 450 with the stationary electron.

(a) What is its new frequency?(b) What is the kinetic energy of electron after collision?Solution

(a) The Compton shift in wavelength is given by

)cos1(0

==cm

h

or )cos1(0

= cm

hcc

)cos1(11

20

+= cm

h

1

20

)cos1(1

+=

cm

h

1

2831

034

19 )10998.2)(10109.9()45cos1)(10626.6(

1011

+

=

Hz1810768.9 =


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the desired frequency of the scattered photon.(b) The kinetic energy of electron after collision is given by

)( == hhhK

)}10768.9()101){(10626.6( 181934 = K

J1610537.1 = = 959 eV 1 keV

Q JeV 1910602.11 =

Problem 1-68A 300 keV photon undergoes a Compton Scattering. Thekinetic energy of recoil electron is 250 keV. Calculate thewavelength of the scattered photon.Solution

Energy of the scattered photonE= Energy of the incident photon Energy of recoil electron

= (300 250) = 50 keVJJ 15196 1001.8)10602.1)(1050( ==

Now

=hc

E

mE

hc 1115

834

10480.2)1001.8(

)10998.2)(10626.6(

=

=

=

= 24.8 pm


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CONCEPTUAL QUESTIONS(1) How does the total intensity of thermal radiation varywhen temperature of an object is doubled?Answer: - The total intensity of thermal radiation is given byStephens law

4

TI =

If the temperature of an object is doubled, then new intensitywill be 1624 = times the initial intensity.(2) If all objects radiate energy, then why we are unable tosee in the dark?Answer: - As the radiation emitted by the object is in theinfrared region which is not visible to eye, therefore we areunable to see the objects in the dark.(3) What is ultraviolet catastrophe of classical physics?Answer: - The discrepancy between theoretical (Rayleigh-Jeansformula) and experimental spectral energy density towards the

ultraviolet end of spectrum of black body is termed asultraviolet catastrophe.(4) How ultraviolet catastrophe was resolved by quantumaction?Answer: - It was resolved by Plancks law for black bodyradiation.(5) What is Plancks concept of energy quantization?Answer: - The energy of a photon is given by

hnE= ........3.2,1,0=n where h is Plancks constant and is the frequency of thephoton.

(6) UV light causes sunburn whereas visible light does not.Explain.Answer: -As the energy of ultraviolet (UV) photon is greaterthan that of a visible photon, therefore it is capable of causingsunburn.(7) Which photon is more energetic- violet or red?Answer: - Violet because the frequency of violet photon isgreater than the red photon.


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(8) Do we get quantization of energy in classical mechanics?Answer: - No. One can assign any value of energy from zero tomaximum to the classical system instead of certain discretevalues.(9) What is the rest mass of a photon?Answer: - The rest mass of a photon is zero.(10) If an X-ray photon is scattered by an electron, does its

wavelength change? If so, does it increase or decrease?Answer: - Yes. The wavelength of the scattered photon isgreater than that of the incident photon.(11) At what angle maximum shift in wavelength will beobserved in Compton Effect experiment?Answer:- 0180= (12) What is the physical significance of Compton Effect?Answer: - The Compton Effect confirms the particle nature ofelectromagnetic radiation.(13) Can we observe Compton Effect if electron is replacedby a proton?Answer: - No. A proton is much heavier than electron (1836times), therefore the Compton shift in wavelength will be toosmall to be observed experimentally.(14)Why is Compton Effect not observed with visible light?Answer: - The energy of the visible photon is far too sufficientto eject even the most loosely bound electron.(15) Why does not the photoelectric effect work for freeelectrons?Answer: - The rest mass of photon is zero. Since any materialparticle has nonzero rest mass, the photoelectric cannot takeplace with a free electron.

(16) Why the alkali metals i.e. Sodium, Potassium andCalcium etc. are most suitable for studying the photoelectricemission?Answer: - The alkali metals have a low value of work functionand show the photoelectric effect for visible light (= 4000 to 7500 ).


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(17) Is it possible to observe photoelectric emission at allfrequencies?Answer: - No. In order to observe photoelectric emission, thefrequency of the incident photon must be greater or equal to thethreshold frequency of the given material i.e. 0 .

(18) What will be the velocity of emitted photoelectrons ifthe frequency of incident photon is just equal to thresholdfrequency?Answer: - Zero.(19) Which parameter governs the kinetic energy of thephotoelectrons emitted from a material?Answer: - The kinetic energy of photoelectrons is governed bythe frequency of the incident radiation provided that it is greaterthan threshold frequency of the given material.(20) What will happen to the velocity of the photoelectrons ifthe wavelength of incident radiation is increased?Answer: - The velocity of the photoelectrons will decrease withincrease in wavelength of the incident photon.(21) Name two effects which support the photon theory oflight?Answer: - Compton Effect, Photoelectric effect, Pairproduction etc.


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ADDITIONAL PROBLEMS

(1) Taking the sun as a 5800 K blackbody how does itsultraviolet radiance at 200 nm compare with itsvisible radiance at its peak wavelength of 500 nm?

(2) The photoelectric threshold of tungsten is 2300 .Determine the energy of electrons ejected from thesurface by ultraviolet radiation of wavelength 1800. {B.U. B.Sc.(Hons.) 1983S}

(3) The photoelectric threshold of tungsten is 2300 .Determine the energy of electrons ejected from thesurface by ultraviolet radiation of wavelength 1500. Express your answer in eV. (B.U. B.Sc. 1990S)

(4) Calculate the value of Plancks constant when a lightof frequency 11.510 14 Hertz is incident upon amaterial of work function 4.410 14 Hertz and the

stopping potential is 3 volts. (B.U. B.Sc. 1992A)(5) If the photoelectric threshold wavelength of sodiumis 542 nm, calculate the maximum velocity of photoelectrons ejected by photons of wavelength 400 nm.

(B.U. B.Sc. 2004S)(6) Photoelectrons emitted from a photocell by light

with a wavelength of 2500 can be stopped byapplying a potential of 2 volts to the collector.Calculate the work function (in eV) for the surface.

{B.U. B.Sc.(Hons.) 1988S}(7) Monochromatic X-rays of wavelength = 0.124

are scattered from a carbon block. Determine thewavelength of the X-rays scattered through 180o.(B.U. B.Sc. 1986S)

(8) X-rays with = 1.0 are scattered from a carbonblock. The scattered radiation is viewed at 90oto theincident beam. What is the Compton shift inwavelength? (B.U. B.Sc. 1987A)


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(9) Prove that the wavelength difference in Compton

Effect is equal to 0.0243 if the scattering angle is90o. (B.U. B.Sc. 1989S)

(10) An X-ray photon is scattered from a stationaryelectron at an angle of 45oto the original direction ofpropagation. Calculate the Compton shift inwavelength and the energy loss by the photon.

{B.U. B.Sc.(Hons.) 1989A}

Answers

(1) nmnm RR 500200 057.0= (2) 2.39910-19J or 1.498 eV

(3) 2.875 eV (4) 3410769.6 J s (5) 510345.5 m s-1(6) 2.96 eV (7) 0.173 (8) 0.02426 (10) m1310109.7 = , MeVE 744.1=

Documents

Chapter 1 Origin of Quantum Theory (Pp 1-42)