Chapter # 44
X-Rays
SOLVED EXAMPLES QUESTIONS
1. 2. 3. 4. 5. 6.
FOR
SHORT
ANSWER
When a Coolidge tube is operated for some time it becomes hot. Where does the heat from ? In a Coolidge tube, electrons strike the target and stop inside it. Does the target get more and more negatively charged as time passes ? Can X-rays be used for photoelectric effect ? Can X-rays be polarized ? X-rays and visible light travel at the same speed in vacuum. Do they travel at the same speed in glass ? Characteristic X-rays may be used to identify the element from which they are coming. Can continuous Xrays be used for this purpose ?
7.
Is it possible that in a Coolidge tube characterstic L X-rays are emitted but not K X-rays ?
8. 9. 10.
Can L X-ray of one material have shorter wavelength than K X-ray of another ? Can a hydrogen atom emit characteristic X-ray ? Whh is exposure to X-ray injurious to health but exposure to visible light is not, when both are electromagnetic waves ?
Objective - I 1.
X-ray beam can be deflected (A) by an electric field (B*) by a magnetic field (C) by an electric field as well as by a magnetic field (D) neither by an electric field nor by a magnetic field X-fdj.k iqt a fo{ksfir fd;k tk ldrk gS (A) fdlh fo|qr {ks=k }kjk (B*) fdlh pqEcdh; {ks=k }kjk (C) fdlh fo|qr {ks=k }kjk lkFk gh fdlh pqEcdh; {ks=k }kjk Hkh (D) u rks fdlh fo|qr {ks=k }kjk vkSj u gh fdlh pqEcdh; {ks=k
}kjk
2.
Consider a photon of continuous X-ray coming from a Coolidge tube. Its energy comes from (A*) the kinetic energy of the strilking electron (B) the kinetic energy of the free electrons of the traged (C) the kinetic energy of the ions of the traget (D) an atomic transition in the target dwyht uyh ls vk jgh lrr~ X-fdj.kksa ds QksVkWu ij fopkj dhft;sA bldh ÅtkZ vkrh gS (A*) VDdj djus okys bysDVªkWu dh xfrt ÅtkZ ls (B) y{; ds eqDr bysDVªkWuksa dh xfrt ÅtkZ ls (C) y{; ds vk;uksa dh xfrt ÅtkZ ls (D) y{; esa ijekf.od LFkkukarj.k ls
3.
The energy of a photon of characteristic X-ray from a Coolidge tube comes from (A) the kinetic energy of the strilking electron (B) the kinetic energy of the free electron of the target (C) the kinetic energy of the ions of the target (D*) an atomic transition in the target dwyht uyh ls vkus okyh vfHkyk{kf.kd X-fdj.kksa ds QksVkWu dh ÅtkZ vkrh gS (A) VDdj djus okys bysDVªkWu dh xfrt ÅtkZ ls (B) y{; ds eqDr bysDVªkWuksa dh xfrt ÅtkZ ls (C) y{; ds vk;uksa dh xfrt ÅtkZ ls (D*) y{; esa ijekf.od LFkkukarj.k ls
manishkumarphysics.in
Page # 1
Chapter # 44 X-Rays 4. If the potential difference applied to the tube is doubled and the separation between the filament and the targer is also doubled, the cutoff wavelength (A) will remain unchanged (B) will be doubled (C) will be halved (D) will become four the original
;fn uyh ij vkjksfir oksYVrk nqxuh dj nh tk;s ,oa rUrq o y{; ds e/; dk vUrjky Hkh nqxuk dj fn;k tk;s rks] U;wure dkV rjaxnS/;Z (cutoff wavelength) (A) vifjofrZr jgsxh (B) nqxuh gks tk;sxh (C) vk/kh gks tk;sxh (D) ewy dh pkj xquh gks tk;sxh 5.
If the current in the circuit for heating the filament is increased, the cutoff wavelength (A) will increase (B) will decrease (C*) will remain unchanged (D) will change ;fn rUrq dks xeZ djus okys ifjiFk esa /kkjk c<+k nh tk;s rks dkV rjaxnS/;Z (A) c<+ tk;sxh (B) de gks tk;sxh (C*) vifjofrZr jgsxhA (D) ifjofrZr gks tk;sxhA
6.
Moseley’s law of characteristic X-rays is = a(Z-b). in this, (A*) both a and b are independent of the material (B) a is independent but b depends on the material (C) b is independent but a depends on the material (D) both a and b depend on the material vfHkyk{kf.kd X-fdj.kksa ds fy;s ekSty + s dk fu;e = a(Z-b) gSA blesa (A*) a o b nksuksa gh inkFkZ ij fuHkZj ugha djrs gSaA (B) a vfuHkZj gS fdUrq b inkFkZ ij fuHkZj djrk gSA (C) b vfuHkZj gS fdUrq a inkFkZ ij fuHkZj djrk gSA (D) a o b nksuksa gh inkFkZ ij fuHkZj djrs gSaA
7.
Frequencies of K X-rays of deifferent materials are measured. Which one of the graphs in fig. may represent the relation between the frequency and the atomic number Z. fofHkUu inkFkks± dh K X-fdj.kksa dh vko`fÙk;k¡ ekih x;hA fuEu fp=k esa dkSulk xzkQ vko`fÙk rFkk ijek.kq Øekad z esa lac/a k iznf'kZr djrk gS -
(D*) 8.
The X-ray beam coming from an X-ray tube (A) is monochromic (B) has all wavelengths smaller than a certain maxiumu wavelength (C*) has all wavelength greater than a certain minimum wavelength (D) has all wavelength lying between a minimum and a maximum wavelength X-fdj.k xyh ls vkus okyh X-fdj.k iqt a (A) ,d o.khZ gksrh gSA (B) esa ,d vf/kdre rjaxnS/;Z ls de okyh leLr rjaxnS/;Z mifLFkr gksrh gSA (C*) esa ,d U;wure rjaxnS/;Z ls vf/kdokyh leLr rjaxnS/;Z mifLFkr gksrh gSA (D) ,d U;wure rFkk ,d vf/kdre rjaxnS/;Z ds chp okyh leLr rjaxnS/;Z mifLFkr gksrh
gSA
9.
One of the following wavelength is absent and the rest are presnet in the X-rays coming from a Cooling tube, Which one is the absent wavelength ? fuEu rjaxnS/;ks± esa ls ,d rjaxnS/;Z dwyht uyh ls vkus okyh X-fdj.kksa esa vuqifLFkr gS rFkk 'ks"k mifLFkr gSA vuqifLFkr rjaxnS/ ;Z dkSulh gS (A*) 25 pm (B) 50 pm (C) 75 pm (D) 100 pm
10.
Fig,. shows the intensity-wavelength relations of X-rays coming from two different Cooling tubes. The solid curve represents the relation for the tube A in which the potential difference between the target and the filament is VA and the atomic number of the target material is ZA. These quantities are VB and ZB for the other tube. Then, fp=k esa nks fHkUu&fHkUu dwyht ufy;ksa esa vkus okyh X-fdj.kksa ds fy;s rhozrk&rjaxnS/;Z lac/a k iznf'kZr fd;s x;s gSAa Bksl oØ uyh A ds fy;s gS] ftlesa y{; o rarq ds e/; foHkokarj VA ,oa y{; ds inkFkZ dk ijek.kq Øekad ZA gSA nwljh uyh ds fy;s manishkumarphysics.in
Page # 2
Chapter # 44
X-Rays
;g jkf'k;k¡ VB rFkk ZB gS] rks -
(A) VA > VB, ZA > ZB (C) VA < VB, ZA > ZB
(B*) VA > VB, ZA < ZB (D) VA < VB, ZA < ZB
11.
50% of the X-rays coming from a Coolidge tube is able to pass through a 0.1 mm thick aluminum foil. If the potential difference between the target and th filament is increased, the fraction of the X-ray passing through the same foil will be dwyht uyh esa vkus okyh X-fdj.kksa dk 50% 0.1 feeh eksVh ,Y;wfefu;e dh iRuh ls xqtj ldrk gSA ;fn rarq ,oa y{;ds e/; foHkokarj c<+k fn;k tk;s rks iUuh ls xqtjus okyh X-fdj.kksa dk Hkkx gks tk;sxk (A) 0% (B) < 50% (C) 50% (D*) > 50%
12.
50% of the X-rays coming from a Coolidge tube is able to pass through a 0.1 mm tick aluminum foil. The potential difference between the target and the filament is increased. The thickness of aluminimum foil, which will allow 50% of the X-ray to pass through, will be dwyht uyh ls vkus okyh X-fdj.kksa dk 50% 0.1 feeh eksVh ,Y;wfefu;e dh iUuh ls xqtj ldrk gSA y{; rFkk rarq ds e/ ; foHkokarj c<+k fn;k tkrk gSA ,Y;wfefu;e iUuh dh og eksVkbZ ftlls 50%, X-fdj.ksa gh xqtj ldsxha] gksxh (A) zero (B) < 0.1 mm (C) 0.1 mm (D*) 0.1 mm
13.
X-rays from a Coolidge tube is incident on a thin aluminium foil. The intensity of the X-ray transmitted by the foil is found to be Io. The heating current is increased so as to increase the temperature of the filament. The intensity of the X-ray transmitted by the foil will be dwyht uyh ls vkus okyh X-fdj.ksa ,d iryh ,Y;wfefu;e dh iUuh ij vkifrr gSA iUuh ls xqtjus okyh X-fdj.ksa dh rhozrk Io ik;h tkrh gSA xeZ djus okyh /kkjk esa o`f) dj nh tkrh gS] ftlls rarq dk rki c<+ tkrk gSA iUuh ls xqtjus okyh X-fdj.kksa dh rhozrk gks tk;sxh (A) zero (B) < Io (C) Io (D*) > Io
14.
Visible light passing through a circular hole forms a diffraction disc of radius 0.1 mm on a screen. If X-rays is passed through the same set-up, the radius of the diffraction disc will be o`Ùkkdkj fNnz ls xqtjus okyk n`'; izdk'k] insZ ij 0.1 feeh f=kT;k dh foorZu pdrh cukrk gSA ;fn blh O;oLFkk ls X-fdj.ksa xqtkjh tk;s rks foorZu pdrh dh f=kT;k gks tk;sxh (A) zero (B*) < 0.1 mm (C) 0.1 mm (D) > 0.1 mm
Objective - II
1.
For harder X-rays, (A) the wavelength is higher (C*) the frequency is higher dBksj X-fdj.kksa ds fy;s (A) rjaxnS/;Z vf/kd gksrh gSA (C*) vko`fÙk mPp gksrh gSA
(B) the intensity is higher (D*) the photon energy is higher (B) rhozrk vf/kd gksrh gSA (D*) QksVkWu dh ÅtkZ mPp
gksrh gSA
2.
Cutoff wavelength of X-rays coming from a Coolidge tube depends on the (A) target material (B*) accelerating voltage (C) separation between the target and the filament (D) tepmerature of the filament dwyht uyh ls vkus okyh X-fdj.kksa dh dkV rjaxnS/;Z ¼U;wure rjaxnS/;Z½ fuHkZj djrh gS (A) y{; ds inkFkZ ij (B*) Rojd oksYVrk ij (C) Rojd oksYVrk ij (D) QksVkWu dh ÅtkZ mPp gksrh gSA
3.
Mark the correct options. (A) An atom with a vacancy has smaller energy than a neutral atom (B*) K X-ray is emitted when a hole makes a jump from the K shell to some other shell (C*) The wavelength of K X-ray is smaller than the wavelength of L X-ray of the same material (D) The wavelength of K X-ray is smaller than the wavelength of K X-ray of the same material manishkumarphysics.in
Page # 3
Chapter # 44
X-Rays
lgh fodYiksa dks fpfUgr dhft;s (A) ,d [kkyh LFkku okys ijek.kq dh ÅtkZ mnklhu ijek.kq ls de gksrh gSA (B*) K X-fdj.k rc mRiUu gksrh gS] tc ,d gksy K-d{kd ls fdlh vU; d{kd esa dwnrk gSA (C*) K X-fdj.kksa dh rjaxnS/;Z blh inkFkZ dh L X-fdj.kksa ls de gksrh gSA (D) K X-fdj.kksa dh rjaxnS/;Z blh inkFkZ dh K X-fdj.kksa ls de gksrh gSA 4.
For a given material, the energy and wavelength of characteristic X-ray satisfy fdlh fn;s x;s inkFkZ ds fy;s] vfHkyk{kf.kd X-fdj.kksa dh ÅtkZ ,oa rjaxnS/;Z larq"V djrs gSa (A) E(K) > E(K) > E(K) (B) E(M) > E(L) > E(K) (C*) (K) > (K) > (K) (D*) (M) > (L) > (K)
5.
The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation, (A) the intensity increases (B) the minimum wavelength increases (C*) the intensity remains unchanged (D*) the minimum wavelength decreases fdlh X-fdj.k ufydk ij vkjksfir foHkokarj c<+k fn;k x;k gSA ifj.kkeLo:i mRlftZr fofdj.k esa (A) rhozrk c<+ tk;sxhA (B) U;wure rjaxnS/;Z c<+ tk;sxh (C*) rhozrk vifjofrZr jgsxhA (D*) U;wure rjaxnS/;Z de gks tk;sxhA
6.
When an electron strikes the target in Coolidge tube, its entire kinetic energy (A) is conberted into a photon (B*) may be converted into a photon (C) is converted into heat (D*) may be converted into heat
dwyht uyh esa tc bysDVªkWu y{; ls Vdjkrk gS rks bldh xfrt&ÅtkZ (A) QksVkWu esa ifjofrZr gks tkrh gSA (B*) QksVkWu esa ifjofrZr gks ldrh gSA (C) Å"ek esa ifjofrZr gks tkrh gSA (D*) Å"ek esa ifjofrZr gks ldrh gSA 7.
8.
X-ray incident on a material (A*) exerts a force on it (B*) transfers energy to it (C*) transfers momentum to it (D*) transfers impluse to it tc X-fdj.ksa fdlh inkFkZ ij vkifrr gksrh gS rks (A*) bl ij cy yxkrh gSA (B*) bldks ÅtkZ LFkkukarfjr djrh (C*) bldks laosx LFkkukarfjr djrh gSA (D*) bldks vkosx iznku djrh gSA
gSA
Consider a photon of continuous X-ray and a photon of characteristic X-rays of the same wavelength. Which of the following is/are different for the two photons ? (A) frequency (B) energy (C) penetrating power (D*) method of creation tc X-fdj.k ds QksVkWu rFkk leku rjaxnS/;Z okyh vfHkyk{kf.kd X-fdj.k ds QksVkWu ij fopkj dhft;sA bu nksuksa QksVkWuksa ds fy;s fuEu esa ls dkSulk@dkSuls fHkUu gS (A) vko`fÙk (B) ÅtkZ (C) Hksnu {kerk (D*) mRiUu gksus dh izfØ;k
WORKED OUT EXAMPLES 1. Sol.
Find the maximum frequency of the X-rays emitted by an X-ray tube operating at 30- kV. For maximum frequency, the total kinetic energy (eV) should be converted into an X-ray photon. Thus, hv = eV or,
v=
= = 2. Sol.
e V h e 30 10 3 V
4.14 10 15 eV s
30 × 1018 Hz = 7.2 × 1018 Hz. 4.14
An X-ray tube operate at 20 kV. A particular electron loses 5% of its kinetic energy of to emit an X-ray photon at the first collision. Find the wavelength corresponding to this photon. Kinetic energy acquired by the by the electron is K = eV = 20 × 103 eV. The energy of the photon = 0.05 × 20 = 103 eV = 103 eV. manishkumarphysics.in
Page # 4
Chapter # 44
X-Rays hv 10 3 eV
Thus,
=
= 3.
Sol.
( 4.14 10 15 eV s) (3 10 8 m / s) 10 3 eV
1242 eV nm 103 eV
1.24 nm
An X-ray tube is operated at 20 kV and the current through the tube is 0.5 mA. Find (a) the number of electrons hitting this target per second, (b) the energy falling on the target per second as the kinetic energy of the electrons and (c) the cut off wavelength of the X-rays emitted. (a) i = ne = 0.5 × 10–3 A
0.5 10 3 A or,
n=
1.6 10
10
C
3.1 1015 / s
(b) The kinetic energy of an electron reaching the target is K = eV. The energy falling on the target per second = n eV – iV = (0.5 × 10 3 A) × (20 × 10 3 V) = 10 J/s (c)
hc eV min
or,
min
=
4.
Sol.
hc eV
1242 eV e(20 10 3 V )
0.062 nm
Find the constants a and b in Moseley’s equation
v a( Z b) from the following data.
Element
Z
Wavelength of Ka X-ray
M0 C0
42 27
71 pm 178.5 pm
Moseley’s equation is
v a( Z b) Thus,
c a( Z1 b) 1
....(i)
and
c a( Z 2 b) 2
....(ii)
From (i) and (ii)
1 1 a ( Z1 Z 2 ) c 2 1
or,
=
a=
(3 10 8 m / s)1/ 2 42 27
1 c 1 ( Z1 Z 2 ) 1 2
1 1 12 1/ 2 12 1/ 2 (178.5 10 m) (71 10 m) = 5.0 × 107 (Hz)1/2
Dividing (i) by (ii),
manishkumarphysics.in
Page # 5
Chapter # 44
X-Rays
2 Z b 1 1 Z2 b or, or,
178 .5 42 b 71 27 b b = 1.37
EXERCISE Iykad fu;rkad h = 4.14 × 10–15 eV - ls-] izdk'k dh pky C = 3 × 108 eh@ls 1. Ans: 2.
Ans:. 3. Ans. 4.
Ans: 5.
Find the energy the frequency and the momentum of an x-ray photon of wavelength 0.10 nm. 0.10 nm rjaxnS/;Z okys X-fdj.k QksVkWu dh ÅtkZ] vko`fÙk rFkk laox s Kkr dhft;sA 12.4 ke V, 3 × 10 18 Hz 6.62 × 10 – 24 kg m/s Iron emits k x-ray of energy 6.4 ke V and calcium emits k X-ray of energy 3.69 ke V Calculate the times taken by an iron Ka photon and a calcium Ka photon to cross through a distance of 3km yksgk] 6.4 ke V ÅtkZ okyh K X-fdj.ksa mRlftZr djrk gS rFkk dSfY';e 3.69 keV ÅtkZ okyh K X- fdj.kksa dk mRltZu djrk gSA yksgs ds K QksVkWu rFkk dSfY'k;e ds K QksVkWu dks 3 fdeh nwjh r; djus esa yxs le;ksa dh x.kuk dhft;sA Find the cutoff wavelength for the continuous X-rays coming from an X -rays tube operating at 30 kV Find the cutoff wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV. 30 kV ij dk;Z'khy ,d X-fdj.k uyh ls vkus okyh lrr X-fdj.kksadh dkV rjaxnS/;Z Kkr dhft;sA 41.4 pm What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule? fdlh X-fdj.k uyh ij fdruk foHkokarj vkjksfir fd;k tk;s fd blls 0.10 nm rjaxnS/;Z ls de rjaxnS/;Z okyh X-fdj.ksa izkIr ugaha gks\ bu X-fdj.kksa ds fdlh QksVkWu dh vf/kdre ÅtkZ dk eku twy esa fdruk gksxk\ 12.4 kV, 2.0 × 10 – 15 The X-ray coming from a collide tube has a cutoff wavelength of 70 pm Find the kinetic energy of the electrons hitting the target ,d dwyht uyh ls vkus okyh X-fdj.kksa dh dkV rjaxnS/;Z 80 pm gSA y{; ls Vdjkus okys bysDVªkWuksa dh xfrt ÅtkZ Kkr
dhft;sA Ans:
15.5 ke V
6.
If the operating potential in an X -ray tube is increased by 1% by what percentage does the cutoff wavelength decrease? ;fn ,d X-fdj.k uyh dh izpkyu oksYVrk 1% c<+k nh tk;s rks dkVk rjaxnS/;Z fdrus izfr'kr de gks tk;sxh\ approximately
Ans: 7.
The distance between the cathode (filament ) and the target in an x-ray tube is 1.5 m If the curoff wavelength is 30 pm, find the electric field between the cathode and the target ,d X-fdj.k uyh ds y{; rFkk dSFkksM ¼rarq½ ds e/; nwjh 1.5 eh gSA ;fn dkV rjaxnS/;Z 30 pm gS] dSFkksM rFkk y{; ds
chp esa fo|qr {ks=k Kkr dhft;sA Ans:
27.7 kV/m
8.
The short wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube in increased to 1.5 times the original value .What was the original value of the operating voltage ? tc X-fdj.k uyh esa izpkyu oksYVrk ewy dh 1.5 xquh dj nh tkrh gS rks U;wure rjaxnS/;Z lhek 26 pm ls foLFkkfir gks
tkrh gSA izpkyu oksYVrk dk ewy eku fdruk gS\ Ans:
15.9 kV
9.
The electron beam in a colour TV is accelerated through 32 kv and then strikes the screen. What is the wavelength of the most energetic X-ray photon ? ,d jaxhu Vhoh dk bysDVªkWu iqt a 32 kv ls Rofjr gksus ds i'pkr~ insZ ls Vdjkrk gSA lokZf/kd ÅtkZ okys X-fdj.k QksVkWu
dh rjaxnS/;Z fdruh gS\ Ans:
38.8 pm
10.
When 40 kv is applied across an X-ray tube X-ray tube X-ray is obtained with a maximum frequency of 9.7 × 1018 Hz Calculate the value of planck constant from these data. manishkumarphysics.in
Page # 6
Chapter # 44
X-Rays
tc X-fdj.k uyh ij 40 kv vkjksfir fd;k tkrk gSA vf/kdre 9.7 × 1018 gVZt ~ vko`fÙk dh X-fdj.ksa izkIr gksrh gSA bu vkadM+kas dh lgk;rk ls Iykad fu;rkad ds eku dh x.kuk dhft;sA – 15
Ans :
4.12 × 10
eV – s
11.
An X-ray tube operates at 40 kv suppose the electrons converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides. ,d X-fdj.k ufydk 40 kv ij izpkfyr dh tkrh gSA ekuk fd izR;sd VDdj esa bysDVªkWu dh ÅtkZ dk 70% QksVkWu esa
ifjofrZr gks tkrk gSA ufydk ls mRlftZr gksus okyh U;wure rhu rjaxnS/;ks± ds eku Kkr dhft;sA ftl ijek.kq ls bysDVªkWu Vdjkrk gSA mldks nh x;h ÅtkZ ux.; eku yhft;sA Ans:
44.3 pm , 493 pm
12.
The wavelength of ka X-ray of tungsten is 21.3 pm It takes 11.3 ke to knock out a electrons from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of k X-ray ? VaxLVu dh K X-fdj.kksa dh rjaxnS/;Z 21.3 pm gSA VaxLVu ijek.kq dh L-d{kd ls bysDVªkWu dks ckgj fudkyus ds fy;s 11.3 keV dh vko';drk gksrh gSA VaxLVu y{; okyh X-fdj.k ufydk ij vkjksfir Rojd oksYVrk dk U;wure eku fdruk gks ftlls K X-fdj.kksa dk mRiknu laHko gks lds\ 69 .5 kV
Ans: 13.
Ans:
The K X-ray of argon has a wavelength of 0.36 nm . The minimum energy needed to ionize an argon atom is 16 eV Find the energy needed to knock out an electrons from the K shall of an argon vkxZu dh K X-fdj.kksa dh rjaxnS/;Z 0.36 nm gSA vkxZu ijek.kq dks vk;fur djus ds fy;s U;wure 16eV ÅtkZ dh vko';drk gksrh gSA vkxZu ijek.kq ds K d{kd ls bysDVªkWu ckgj fudkyus ds fy;s vko';d ÅtkZ Kkr dhft;sA 3.47 keV
14.
The K x-ray of aluminium (Z = 13) and zinc (Z = 30 ) have wavelengths 887 pm and 146 respectively use mostly’s law
,Y;wfefu;e Ans: 15.
v = a (Z – b) to find to the wavelength of the k X-ray of iron (Z = 26) (Z = 13) rFkk ftad (Z = 13) dh K X-fdj.kksa dh rjaxnS/;Z Øe'k% 887 pm o 146 pm gSA ekslys
v = a (Z – b) dk 198 pm
ds fu;e
mi;ksx yksgs (Z = 26 ) dh K X-fdj.ksa dh rjaxnS/;Z Kkr djus ds fy;s dhft;sA
A certain element emits ka X-ray of energy 3.69 ke V use the data from the previous problem to identify the element. ,d fof'k"V rRo 3.69 keV ÅtkZ dh K X-fdj.ksa mRlftZr djrk gSA fiNys iz'u esa fn;sx;s vkadM+kas dk mi;kxs djds bl
rRo dh igpku dhft;sA Ans:
calcium
16.
The Kp X-ray from certain elements are given below draw a moseley type plot of radiation. Element. Ne P Ca Mn Zn Br Energy (keV) 0.858 2.14 4.02 6.51 9.57 13.3.
dqN fuf'pr rRoksa dh K X-fdj.ksa uhps nh x;h gSA K fofdj.k ds fy;s rRo Ne P Ca Mn Zn Br ÅtkZ (keV) 0.858 2.14 4.02 6.51 9.57 13.3. 17.
Ans : 18.
v
v versus Z for Kp
o Z ds chp ekstys tSlk xzkQ [khafp,A
Use moseley’s law with b = 1 to find the frequency of the K X-ray of La (Z = 57) if the frequency of the X-ray of Cu(Z=29 is known to be 1.88× 10 18 Hz ;fn Cu(Z=29) ds fy;s K X-fdj.kksa dh vko`fÙk Kkr gSA ftldk eku 1.88 × 1018 gVZ~t gSA ekslys ds fu;e dk mi;ksx djds (b = 1 ekudj½ La (Z = 57) dh K X-fdj.kksa dh vko`fÙk Kkr dhft;sA 7.52 × 10 18 The K and K x-rays of molyblenum have wavelengths 0.71 Å and 0.63 Å respectively. Find the wavelength of L x-ray of molybdenum. eksfyfcMsue~ dh K rFkk K x-fdj.kksa dh rjaxnS/;Z Øe'k% 0.71 Å rFkk 0.63 Å gSA eksfyfcMsue dh L x-fdj.kksa dh
rjaxnS/;Z Kkr dhft;sA Ans:
5.64 Å
19.
The wavelengths of K and L X-rays of material are 21.3pm and 141 pm respectively Find the wavelength of K X-ray of the material. manishkumarphysics.in
Page # 7
Chapter # 44
X-Rays
fdlh inkFkZ ds fy;s K rFkk L X-fdj.kksa dh rjaxnS/;Z Øe'k% 21.3pm rFkk 141 pm gSA bl inkFkZ ds fy;s K X-fdj.kksa dh rjaxnS/;Z Kkr dhft;sA Ans:
18.5 pm
20.
The energy of a silver atom with a vacancy in K shell is 25.31 ke V, in L shall 3.56 ke V, M shell is 0.530 ke V higher than the energy of the atom with no vacancy. Find the frequency of K k and L X-ray of silver fcuk fjDr LFkku okys ijek.kq dh rqyuk esa K-d{kd esa ,d [kkyh LFkku okys pkanh ds ijek.kq dh ÅtkZ 25.31 keV, L-d{kd esa fjDr 3.56 keV rFkk M-d{kd esa fjfDr okys dh 0.530 keV vf/kd gksrh gSA pkanh ds fy;s K K rFkk L X-fdj.kksa dh
vko`fÙk Kkr dhft;sA 18
Hz 5.98 × 10 18 Hz 7.32 × 10 17 Hz
Ans:
5.25 × 10
21.
Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray The energy levels of the tungsten atom with an electron knocked out are as follows . Cell containing vacancy K L M Energy in ke V 69.5 11.3 2.3 VaxLVu y{; okyh X-fdj.k uyh ij vkjksfir fd;k tk ldus okyk og vf/kdre foHkokarj Kkr dhft;sA fd blls vfHkyk{kf.kd K ;k L X-fdj.kksa dk mRiknu ugha gks ldsA ,d bysDVªkWu ckgj fudys gq, VaxLVu ijek.kq ds ÅtkZ Lrj fuEukuqlkj gS : fjDr LFkku okyk d{kd K L M keV esa ÅtkZ 69.5 11.3 2.3 k less than 11.3 kV
Ans: 22.
Ans: 23.
Ans: 24.
The electric current in a X-ray tube (from the target to the filament ) operating at 40K V is MA. Assume that on an average 1% of the total kinetic energy of the electrons hitting the target ate converted into Xrays. (a) What is the total power emitted as X-rays and (b) how much heat is produced in the target every second ? 40 kV ij izpkfyr ,d X-fdj.k ufydk esa fo|qr /kkjk ¼y{; ls rarq dh vksj½ 10 mA gSA eku yhft;s fd y{; ls Vdjkus okys bysDVªkWuksa dh dqy xfrt ÅtkZ dk vkSlru 1% gh X-fdj.kksa esa ifjofrZr gksrk gSA (a) X- fdj.kksa ds :i esa mRlftZr dqy 'kfDr fdruh gS rFkk (b) y{; esa izfr lsd.M fdruh Å"ek mRiUu gks jgh gS\ (a) 4 W (b) 396 J Heat at the rate of 200 W is produced in an X-ray tube operating at 20 kV Find the current in the circuit Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays. 20 kV ij izpkfyr ,d X-fdj.k ufydk esa 200 okWV dh nj ls Å"ek mRiUu gksrh gSA ifjiFk esa /kkjk Kkr dhft;sA ;g eku yhft;s fd bysDVªkWu dh xfrt ÅtkZ dk vR;Yi Hkkx gh X-fdj.kksa esa ifjofrZr gksrk gSA 10 mA Continues X-ray are made to strike a tissue paper soaked with polluted water. The incoming X-ray excite the atoms of the sample by knocking out the electrons from the inner shells. Characteristic Xray are analysed and the intensity is plotted against the wavelength (figure 44 E1) Assuming that only K intensities are detected list the elements present in the sample from the plot use mostly’s equation. v = ( 25 ×1014 Hz ) (Z – 1)2 iznfw "kr ikuh dks vo'kksf"kr fd;s gq, fV'kw isij ij lrr~ X-fdj.ksa vkifrr dh tkrh gSA vkus okyh X-fdj.ksa vkarfjd d{kdksa ls bysDVªkWuksa dks ckgj fudkydj izfrn'kZ ds ijek.kqvksa dks mÙksftr djrh gSA ftlls vfHkyk{kf.kd X-fdj.ksa mRiUu gksrh gSA mRlftZr X-fdj.kksa dk fo'ys"k.k fd;k tkrk gS rFkk rjaxnS/;Z ds lkis{k rhozrk dk xzkQ vkjsf[kr fd;k tkrk gSA ;g ekurs gq, fd dsoy K rhozrk,¡ gh lalfw prdh tkrh gS] xzkQ dh lgk;rk ls izfrn'kZ esa mifLFkr rRoksa dh lwph cukb;sA ekstys dh lehdj.k dk mi;ksx dhft;s % v = ( 25 ×1014 Hz ) (Z – 1)2 k
manishkumarphysics.in
Page # 8
Chapter # 44 Ans : Zr , Zn, Cu, Fe 25.
Ans: 26.
X-Rays
A free atom of iron emits K x-ray of energy 6.4 ke V calculate the recoil kinetic energy of the atom mass of an iron atom = 9.3 × 10 26 kg. yksgs dk ,d eqDr ijek.kq 6.4 keV ÅtkZ dh K X-fdj.ksa mRlftZr djrk gSA ijek.kq dh izfr{ksi rst ÅtkZ dh x.kuk dhft;sa yksgs ds ,d ijek.kq dk nzO;eku = 9.3 × 10–26 kg A 3.9 × 10 – 4 eV The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength (1/ ) of the light falling on the cathode The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength (1/ ) of the X-ray emitted show that the slopes of the lines in the two cases are equal and find its value. ,d izdk'k fo|qr&izHkko esa fujks/kd foHko] dSFkksM ij vkifrr izdk'k dh rjaxnS/;Z ds O;qRØe (1/ ) ij jSf[kd :i ls fuHkZj djrk gSA X-fdj.k ufydk ij vkjksfir foHkokarj] mRlftZr X-fdj.kksa dh dkV rjaxnS/;Z ds O;qRØe (1/ ) ij jSf[kd :i ls
fuHkZj djrk gSA O;Dr dhft;s fd bu nksuksa fLFkfr;ksa esa js[kkvksa ds
he = 1.242 × 10 e
–6
V–m
Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through a young’s double slit and the interference pattern is observed on a photographic plate separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm? [M_Bank_Chp-44_Ex-_27] ekuk fd ;ax f}&fLyV iz;ksx esa 100 pm rjaxnS/;Z okyh ,d o.khZ X-fdj.ksa iqt a iz;D q r fd;k tkrk gS rFkk fLyVksa ls 40 lseh
nwj fLFkr QksVksxkz fQd IysV ij O;fDrdj.k izfr:i izfs {kr fd;k tkrk gSA fLyVksa ds e/; dh nwjh fdruh j[kh tk;s] ftlls fd insZ ij izkIr nks Øekxr mfPp"Bksa ds chp dh nwjh 0.1 mm gSA Ans:
4 × 10
–7
m
manishkumarphysics.in
Page # 9
