chapter 5 x-rays (pp 126-147)

CHAPTER 05 X-RAYS 126

CHAPTER 5 X-RAYS

5-1 X-RAY SPECTRUM Example 5-1 An X-ray tube operates at 30 kV. Calculate the maximum speed of electrons striking the target. Solution Now eVvm 0

202

1=

0

02 2m

eVv =

)10109.9()10602.1)(1030(22

31

193

0

0

==

m

eVv

smv /10027.1 8= Example 5-2 Find the minimum wavelength and corresponding frequency of X-rays produced by an X-ray tube operated upon 10 6 volts. K.U. B.Sc. 2003 Solution The minimum wavelength

min is given by

)10602.1)(10()10998.2)(10626.6(

195

834

0min

==

eVch

011 124.010240.1 A==

The corresponding frequency is

Hzc 19118

min

10418.210240.110998.2

=

==


Example 5-3 Electrons are accelerated in television tubes through potential difference of about 10 kV. Find the highest frequency of the electromagnetic waves emitted when these electrons strike the screen of the tube? What kind of waves are these? Solution The minimum wavelength

min is given by

eVhc

0min =

heVc 0

minmax ==

Hz1834193

max 1042.210626.6)10602.1)(1010(

=

=

These waves are X-rays. Example 5-4 Calculate the wavelength

min for the continuous spectrum of X-ray emitted when electrons of energy 35000 eV fall on a target. P.U. B.Sc. 2000 Solution The minimum wavelength

min is given by

eVhc

0min =

)10602.135000()10998.2)(10626.6(

19

834

min

=

m1110543.3 = 3543.0=

Example 5-5 If V is the potential difference through which the electron is accelerated in an X-ray tube, the show that the minimum wavelength of emitted X-rays is given by

)(12400)(

0

minvoltsV

A =


Solution The minimum wavelength

min is given by

min = )10608.1(

)10998.2)(10626.6(19

834

=

VVehc

VVmetresin

106

min101240010240.1)(

=

=

)(12400)(

0

minvoltsV

A =

Example 5-6 An X-ray tube operates at 50 kV and current through it is 20 mA. Only one percent of the total energy supplied is emitted as X-rays. Calculate (i) the rate at which heat must be removed from target in order to keep it at steady temperature. (ii) the velocity of electrons when they hit the target. (iii) the lower wavelength limit of the X-rays emitted. B.U. B.Sc. (Hons.) 1991A Solution (i) P = V0I = )1020)(1050( 33 = 1000 W or J s -1 Rate of dissipation of heat energy

= 19901000

10099

10099

== JsP

(ii) The kinetic energy is given by eVvm 0

202

1=

smm

eVv /10326.1

10109.9)10602.1)(1050(22 8

31

193

0

0 =

==

(iii) The lower wavelength limit of emitted X-rays is given by )10602.1)(1050(

)10998.2)(10626.6(193

834

0min

==

eVch

= m1110480.2 or 0.248


Example 5-7 X-rays are produced in an X-ray tube by a target potential of 50 kV. If an electron makes three collisions in the target before coming to rest and loses one-half of its remaining energy on each of the first two collisions, determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms. Solution Energy of the incident electron

E = V0 e = )10602.1)(1050( 193 = J151001.8 Energy of the electron after first collision E 1 = (E/2) = J1510005.4 Energy of the electron after second collision E 2 = (E2/2) = J15100025.2 The wavelengths of the resulting photons are given by

mEhc 11

)15

834

11 10960.410005.4(

)10998.2)(10626.6(

=

==

mEhc 11

15

834

22 1092.9)100025.2(

)10998.2)(10626.6(

=

==

Example 5-8 Determine Plancks constant from the fact that the minimum X-ray wavelength produced by 40.0 keV electron is 31.1 pm. Solution The minimum wavelength

min is given by

minmax

hcE =

or )10998.2()101.31)(10602.11040(

8

12193minmax

==

c

Eh

Q JeV 1910602.11 = = sJ .10647.6 34


Example 5-9 What is the minimum potential difference across an X-ray tube that will produce X-rays with a wavelength of 0.126 nm? Solution The minimum wavelength

min is given by

eVhc

0min =

or )10602.1)(10126.0()10998.2)(10626.6(

199

834

min0

==

e

hcV

kVvoltsV 841.910814.9 30 == Example 5-10 What voltage must be applied to an X-ray tube for it to emit X-rays with a minimum wavelength of 30 pm? Solution The minimum wavelength

min is given by

eVhc

0min =

or )10602.1)(1030()10998.2)(10626.6(

1912

834

min0

==

e

hcV

kVvoltsV 3.411013.4 40 == Example 5-11 What is the voltage of an X-ray tube that produces X-rays with wavelength down to 0.01 nm but no shorter? Solution The minimum wavelength

min is given by

eVhc

0min =

)10602.1)(1001.0()10998.2)(10626.6(

199

834

min0

==

e

hcV

kVvoltsV 1241024.1 50 ==


Example 5-12 Duane and Hunt observed that an X-ray tube operated with 30,000 V emits a continuous X-ray spectrum with a shorter wavelength = 0.414 . (a ) Calculate (h/e) and (b) Calculate h using the accepted value of e Solution (a)

eVhc

0min =

CsJc

Ve

h /min0 =

(b) 15810

min0 10143.410998.2

)10414)(.30000(

=

==

c

Ve

h

eh )10143.4( 15=

343415 10637.6)10602.1)(10143.4( ==h J s Example 5-13 Find the minimum potential difference that must be applied to an X-ray tube to produce X-rays with a wavelength equal to Compton wavelength of the electron. Solution The Compton wavelength for the electron is defined as

cm

hc

0

=

Now eV

hcc

0min ==

Equating the right hand sides of above equations

eVhc

cm

h00

=

19

281920

0 10602.1)10998.2)(10109.9(

==

e

cmV

51011.5 = volts or 5.11 kV


5-2 MOSELEYS LAW Example 5-14 The wavelength of K line of an element is 1.790 . Can you identify the target element? Solution For K line we have

2)1(431

= ZR or 2)1(

34

= ZR

RZ

321+=

where 1710097373.1 = mR is the Rydberg constant. Hence

27)10097373.1)(10790.1(3

21710

=

+=

Z

The target element is cobalt. Example 5-15 The wavelength of the K line for a certain element is 0.3368 nm. What is the element? Solution For K line we have

2)1(431

= ZR or 2)1(

34

= ZR

RZ

321+=

20)10097.1)(103368.0(3

2179

=

+=

Z

The target element is Calcium. Example 5-16 Calculate the wavelength of K line for molybdenum target ( Z = 42). Solution Now 2)1(

431

= ZR


or 272 )142)(10097373.1(34

)1(34

=

=

ZR

= m111023.7

= 0.723 Example 5-17 Calculate the energy, in eV, of K of Sodium (Z = 11). Solution The energy of K line is given by

2)1(4

3== ZhcRhcE

k in Joules

2)1(4

3= Z

e

hcRE in eV

)10602.1(4)111)(10097.1)(10998.2)(10626.6(3

19

27834

=E

eV1020=

Example 5-18 The K line of Thallium (Tm-69) has a wavelength of 0.246 . Compare the energy of this K photon with the rest mass energy of an electron. Solution The energy of the given photon is calculated as

JchE 1510834

10075.810246.0

)10998.2)(10626.6(

=

==

keVeVeVE 41.5010041.510602.110075.8 4

19

15

==

=

The rest mass energy of electron is given by keVMeVcmE 511511.0200 ===

Hence

1.0099.0511

41.500

=

EE

01.0 EE =


Example 5-19 Calculate the ratio of the wavelengths of K line for Niobium (Z = 41) to that for Gallium (Z = 31). Solution For K line we have

2)1(431

= ZR

2)1(34

=

ZR or 2)1(

1

Z

If 1 and 2 represent the wavelength of K lines for niobium and gallium respectively, then

169

141131

)1()1(

21

22

2

1=

=

=

ZZ

or 9:16

Example 5-20 Calculate the wavelengths of K line of Manganese (Z = 25) if K line of Titanium (Z = 22) is 275 pm. Solution If 1 and 2 represent the wavelength of K lines for manganese and titanium respectively, then

21

22

2

1

)1()1(

=

ZZ

pmZZ 211)275(

125122

11 2

2

2

1

21 =

=

=

Example 5-21 X-rays from a cobalt target (Z = 27) show a strong K line at 1.798 for cobalt and a weak line at 1.435 due to some impurity. What is the impurity element? Solution According to Moseleys law

2)1(431

= ZR


==

RZ

34)1( 2 Constant

211

222 )1()1( = ZZ

1

2112

2)1()1(

=

ZZ

30)127(435.1798.11)1(1 1

2

12 =+=+= ZZ

The impurity element is zinc.


5-3 ABSORPTION OF X-RAYS Example 5-22 The mass absorption coefficient of silver is 38 cm2 /g for X-rays of wavelength 0.4 . Determine the atomic absorption coefficient of silver for this measurement. (The average atomic mass of silver is 107.868 g / mol) Solution The atomic absorption coefficient is given by

230 10022.6

)868.107)(38(

==

NM

ma

= 211081.6 cm2 / atom or 251081.6 m2 / atom

Example 5-23 The mass absorption coefficient of X-rays of wavelength = 0.7 is 5 cm2 /g. The density of aluminum is 2.7 g /cm 3. What thickness in centimeters of aluminum is needed to reduce the intensity of the X-rays passing through it to one-half its initial value? Solution Now )exp(0 xII =

)exp(2 00 xI

I =

2)exp( =x )2(nx l=

cmnnn

xm

051.0)5)(7.2()2()2()2(

====

lll

Example 5-24 The absorption coefficient of low energy rays in lead is 1.50 cm-1. What thickness of lead is required to reduce the intensity of the rays (a) to half the original intensity? (b) to 0.01 of the original intensity? Solution According to exponential decay law for intensity of gamma rays


)exp(0 xII = )/()exp( 0 IIx = )/( 0 IInx l=

)/( 0 IInx l=

(a) I = 0.5 I0 , cmIInx 462.050.1)5.0/( 0

==

l

(b) I = 0.01 I0 , cmIInx 070.350.1)01.0/( 0

==

l


5-4 X-RAY DIFFRACTION Example 5-25 If d = 2.52 for a crystal grating and the wavelength of X-ray is 1.1 , find the directions of strong diffracted beams. P.U. B.Sc. 1980 Solution From Braggs law, we have

md =sin2

mm

dm )21825.0()1052.2(2

)101.1(2

sin 1010

=

==

Now 1=m , 21825.0sin = , 01 6.12= 2=m , 4365.0sin = , 01 9.25= 3=m , 65475.0sin = , 01 9.40= 4=m , 873.0sin = , 01 8.60= 5=m , 09125.1sin = As the value of sin can not

exceed unity, therefore only fourth- order diffracted beams exist. The angles of the directions of strong diffracted beams are 06.12 , 09.25 , 09.40 and 08.60 . Example 5-26 A beam of X-rays of wavelength 29.3 pm is incident on a Calcite )( 3CaCO crystal of lattice spacing 0.313 nm. Find the smallest angle between the crystal planes and the beam that will result in constructive reflection of the X-rays. Solution According to Braggs law md =sin2

)10313.0)(2()103.29)(1(

2sin 9

12

==

dm

04681.0sin = or 00 68.295042 or=


Example 5-27 Monochromatic high-energy X-rays are incident on a crystal. If first-order reflection is observed at Bragg angle 3.400, at what angle would second-order reflection be expected? Solution According to Braggs law

md =sin2 For first and second order reflections it becomes =1sin2d (1) 2sin2 2 =d or =2sind (2) Dividing Eq.(2) by Eq.(1) 1

sin2sin

1

2=

or 11861.040.3sin2sin2sin 012 ===

002 81.634846 or=

Example 5-28 At what angle an X-ray beam of wavelength 0.110 nm must be diffracted if the interplanar spacingd of NaCl is 0.252 nm? K.U. B.Sc. 2001 Solution According to Braggs law md =sin2

21825.0)252.0)(2()110.0)(1(

2sin ===

dm

01 6.12)04681.0(sin == Example 5-29 X-rays of wavelength 1 are reflected strongly at glancing angle of 4100 in the second order. What is the interplanar spacing of the reflecting surface? Solution According to Braggs law md =sin2


mmd 100

10

1072.5)410sin(2)101)(2(

sin2

=

==

Example 5-30 The smallest angle of Bragg scattering in potassium chloride (KCl) is 28.40 for 30 nm X-rays. Find the distance between atomic planes in potassium chloride. Solution According to Braggs law md =sin2

nmmd 7.31)4.28sin(2

)30)(1(sin2 0

===

Example 5-31 X-rays of wavelength 0.122 nm are found to reflect in the second order from the face of lithium fluoride crystal at a Bragg angle of 28.10. Calculate the distance between adjacent crystal planes. Solution According to Braggs law md =sin2

nmmd 259.0)1.28sin(2

)122.0)(2(sin2 0

===

Example 5-32 How far apart are the planes of NaCl crystal for radiation of wavelength 1.54 making glancing angle 150 54 in first order? K.U. B.Sc. 2003 Solution According to Braggs law md =sin2

81.2)4515sin(2)54.1)(1(

sin2 0=

==

md


Example 5-33 A beam of X-rays of = 0.842 is incident on a crystal at a glancing angle of 5380 when the first order Braggs reflection occurs. Calculate the glancing angle of third order reflection. (K.U. B.Sc. 2001) Solution First the value of d is calculated for given data. According to Braggs law

md =sin2 821.2)538sin(2

)842.0)(1(sin2 0

=

==

md

Now

44771.0)821.2)(2()842.0)(3(

2sin ===

dm

0553260 = the desired glancing angle of third order reflection.

Example 5-34 Using Braggs X-ray spectrometer, the glancing angle for the first order spectrum was observed to be 8o. Calculate the wavelength of X-rays if d = 2.82 . B.U. B.Sc. (Hons.) 1993A Solution According to Braggs law sin2dm =

7849.01

8sin)82.2(2sin2 0===

m

d

Example 5-35 A monochromatic beam of X-rays produces a first order Bragg maximum when reflected off the face of a NaCl crystal with glancing angle = 200 . The spacing of the relevant plane is d = 0.28 nm. What is the minimum possible voltage of the tube that produced the X-rays?


Solution According to Braggs law sin2dm =

mm

d 1009 10915.11

20sin)1028.0(2sin2

=

==

Now min = Ve

hc

voltse

hcV 31019834

min

1048.6)10915.1)(10602.1()10998.2)(10626.96

=

==

Example 5-36 A first order Bragg reflection occurs when a monochromatic X-ray incident at an angle of 300 is reflected from rock salt (NaCl) crystal. Determine the wavelength of the incident X-rays (d = 2.82 ). Solution According to Braggs law sin2dm =

m

d sin2=

82.21082.21

30sin)1082.2(2 10010==

=

m


5-5 PAIR PRODUCTION Example 5-37 A positron collides head on with an electron and both are annihilated. Each particle had a kinetic energy of 1.00 MeV. Find the wavelength of the resulting photons. Solution The energy of the resulting photon is given by KcmE += 202

MevE 022.21)511.0(2 =+= MeVcm 511.020 =Q JJE 13196 10239.3)10602.1)(10022.2( == The wavelength of the resulting photon is calculated as

chE =

13

834

10239.3)10998.2)(10626.6(

==

Ech

pmm 887.01087.8 13 == Example 5-38 What is the energy and wavelength of the photon that will just create an electron-positron pair? Solution The desired energy is given by

JJcmE 13283120 10637.1)10998.2)(10109.9(22 ===

eVeVE 61913

10022.110602.110637.1

=

=

The corresponding wavelength will be

mEch 12

13

834

10213.110637.1

)10998.2)(10626.6(

=

==

Example 5-39 Determine the total kinetic energy of the electron and positron formed by pair production from a -ray of wavelength 0.00247 .


Solution The total kinetic energy of electron-positron pair will be

202 cm

ch

283110

834

)10998.2)(10109.9(21000247.0

)10998.2)(10626.6(

=

J1310604.4 =

MeVeVeV 410410602.110604.4 6

19

13

=

=

Example 5-40 What is the energy and wavelength of the photon that will just create a proton-antiproton pair? Solution The desired energy will be equal to twice the rest mass energy of photon i.e.

JJcmE p1028272

0 10007.3)10998.2)(10673.1(22 ===

GeVeVeVE 877.110877.110602.110007.3 9

19

10

==

=

The corresponding wavelength will be

mEch 16

10

834

10606.610007.3

)10998.2)(10626.6(

=

==


5-6 ANNIHILATION OF MATTER Example 5-41 What is the momentum of the photons created in annihilation of a proton and antiproton, each with an original kinetic energy of 1.00 MeV? Solution The energy of each photon will be

MeVGeVEKcmhE p 00.1)2/877.1(..21 2

0 +=+==

MeVMeVMeVE 5.93900.110)2/877.1( 3 =+= The desired momentum of each photon will be

cMeVc

Ec

chhp /5.9391 ==

==


CONCEPTUAL QUESTIONS (1) What is the order of wavelength of X-rays? Answer: - m1010 (2) What are two general kinds of X-ray tubes? Answer: - (a) The Gas Tube and (b) The Collidge Tube. (3) What is the order of distance between two adjacent atoms of a typical solid? Answer: - m1010 (4) What is the largest Bragg angle through which a beam of X-rays can be diffracted by a crystal? Answer: - 090


ADDITIONAL PROBLEMS (1) Calculate the wavelength min for continuous

spectrum of X- rays produced due to falling of 30 keV electrons on a molybdenum target.

(P.U. B.Sc. 2003) (2) At what angles an X-ray beam of wavelength 0.110

nm must be diffracted if the interplanar spacing d of NaCl crystal is 0.252 nm? (K.U. B.Sc. 2001)

(3) The spacing between principal planes of Calcite is 3.04 . Calculate the wavelength of X-rays for which the first order reflection occurs at 14.7o.

{B.U. B.Sc.(Hons.) 1989A} (4) Calculate the wavelength of the K line of rhodium

(Z=45). (5) Calculate the wavelength of the K line of uranium

(Z=92). (6) The wavelength of the K line for a certain element is

0.0794 nm. What is the element? (7) Calculate the energy, in eV, of K line of Magnesium

(Z = 12). (8) What is the longest wavelength in the K series of X-

rays emitted by cobalt (Z = 27)? (9) A sample of unknown element is used as target for

the electron beam in an X-ray tube. It emits a series of characteristic spectral lines in which the two longest wavelengths are 0.228 and 0.192 . Identify the element.

Answers (1) m1010133.4 or 0.4133 (2) 12.60, 25.90. 40.90 and 60.80 (3) 1.543 (4)

m1110676.6

(5) m1110467.1 (6) Z = 40 , Zirconium (7) 1234 eV (8) 798.110798.1 10 m (9) Tungsten

chapter 5 x-rays (pp 126-147)

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