23-Modern Physics
TOPIC
MODERN PHYSICS
23
SECTION - I : STRAIGHT OBJECTIVE TYPE
[k.M-I: lh/ks oLrqfu"B izdkj
23.1 In an -decay the Kinetic energy of particle is 48 MeV and Q-value of the reaction is 50 MeV. The mass number of the mother nucleus is:- (Assume that daughter nucleus is in ground state)
,d–{k;esa&d.k dhxfrtÅtkZ48mevgSrFkkvfHkfØ;kdkQeku50Mev gSAiSr`dukfHkddhnzO;ekula[;k gS ¼ ekuk iq=kh ukfHkd ewy voLFkk esa gSA½
(A) 96 (B*) 100 (C) 104 (D) none of these
buesalsdksbZugh
my A 4
A 4
Sol. We have K =
my m .Q K = A
.Q 48 = A
.50 A = 100
my A 4
A 4
gy% ge tkurs gS K = my m .Q K = A
.Q 48 = A
.50 A = 100
23.2 A sample of radioactive material decays simultaneously by two processes A and B with half lives
1 and
2
1 hr respectively. For first half hour it decays with the process A, next one hr with the process B and
4
for further half an hour with both A and B. If originally there were N0 nuclei, find the number of nuclei after 2 hr of such decay.
,djsfM;ks,fDVoinkFkZdkuewuk,dlkFknksçfØ;kvksaArFk B 1 1
ftudhv)Zvk;qØe'k%2 rFk 4 ?kaVsgS]}kjk{kf;r
gksrkgSAigysvk/ks?k.Vsesa;gçfØ;kA}kjk{kf;rgksrkgS]vxys,d?k.Vs;gçfØ;kBrFk mldsvkxsvk/k ?k.Vsnksuksa çfØ;kvksaArFk B}kjk{kf;rgksrkgSA;fnçkjEHkesaN0ukfHkdgksrks2?k.Vsdsblçdkjds{k;dsi'pkr~ukfHkdksadh la[;k Kkr djks A
N0
(A*) (2)8
N0
(B) (2)4
N0
(C) (2)6
N0
(D) (2)5
Sol. After first half hrs N = N 1
0 2
1 1 1
1 4
1 5
for t = 2 to t = 1 2 N = N0 2
2
= N
0 2
1
1 1 1
for t = 1 2 to t = 2 hrs. [for both A and B
t1/ 2 = 1/ 2 + 1/ 4
= 2 + 4 = 6
t1/2 = 1/6 hrs.]
1 5
1 3
1 8
N = N0
= N
2
2 0 2
gy% igysvk/ks?k.VscknN=N 1
0 2
1 1 1
1 4
1 5
t = 2 lst = 1 2 ds fy;s N = N0 2
2
= N
0 2
t = 1 1
2
ls t = 2 hrs. ds fy;s [A vkSj B ds fy;s
1
t1/ 2
= 1
1/ 2
+ 1
1/ 4
= 2 + 4 = 6
1 5
1 3
1 8
t1/2 = 1/6 hrs.]
N = N0
= N
2
2 0 2
23.3 In which of the following process the number of protons in the nucleus increases .
(A) decay (B*) decay (C) + decay (D) k capture
uhpsfn;sx;sfdlçØeesaukfHkdesaçksVkWudhla[;kesao`f)gksrhgS&
(A) {k; ds fy, (B*) {k; ds fy, (C) + {k; ds fy, (D) k izxzg.k (capture) ds fy,
Sol. For - decay :
x Ay x 2By 4 +
For – – decay : For + – decay :
x Ay x 1By x Ay x 1By
+ 10
+ 10
For k - capture : there will be no change in the number of protons. Hence, only case in which no of protons increases is - decay Hence (B).
Sol. {k;dsfy, :
x Ay x 2By 4 +
{k;dsfy, :
+ {k;dsfy, :
x Ay x 1By x Ay x 1By
+ 10
+ 10
k-izxzg.k:ukfHkdesaçksVkWudhla[;kesaifjorZuughagksrkgSA dsoy-{k;esaçksVkWudhla[;kesao`f)gksrhgSA
vr% (B).
23.4 The angular momentum of an electron in first orbit of Li++ ion is :
Li++ vk;u dh çFke d{kk esa bysDVªkWu dk dks.kh; laosx gksrk gS :
3 h
(A) 2
9 h
(B) 2
h h
h
(C*) 2
h
(D) 6
Sol. Angular momentum (mvr) = n. 2
= 2
(n = 1)
h h
dks.kh; laosx(mvr) =n. 2 = 2
(n = 1)
23.5 A radioactive nucleus ' X' decays to a stable nucleus ' Y '. Then the graph of rate of formation of ' Y' against time ' t ' will be :
,djsfM;ks,fDVoukfHkd'X'LFk ;hukfHkd'Y'esaifjofrZrgksrkgSrksfuEuesalsdkSulkxkzQ'Y'dsmRiUugksusdhnj rFkk le; t dks çnf'kZr djrk gSA :
(A)
(B)
(C*)
(D)
Sol. N = N0e–t Ny = N0(1 – e–t)
Rate of formation of mRiUu gksus dh nj R = At t = 0 , R = N
dN = + N e–t X
dt
Y
t = , R = 0
23.6 A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass number 80 and 120. If binding energy per nucleon for parent atom is 6.5 M eV and for daughter nuclei is 7 MeV and 8 MeV respectively, then the energy released in the decay will be:
,dHk jhukfHkdftldknzO;ekuØekad200gSnksVqdM+sftudsnzO;ekuØekad80rFk 120gS]esafoHkDrgksrkgSAiSr`d ijek.kqdhçfrU;wfDyvkWuca/kuÅtkZ6.5MeVgSrFk iw=khukfHkdksadsfy,çfrU;wfDyvkWuca/kuÅtkZØe'k%7MeVrFk 8 MeV gks rks {k; ds le; eqDr ÅtkZ gksxh &
(A) 200 MeV (B) 220 MeV (C*) 220 MeV (D) 180 MeV
Sol. Energy released eqDr ÅtkZ= (B.E. of product – BE of reactant) (mRikn dh c0Å0 – vfHkdkjd dh c0Å0)
= (80 × 7 + 120 × 8 – 200 × 6.5) = 220 MeV.
23.7 If first excitation potential of a hydrogen like atom is V electron volt, then the ionization energy of this atom will be:
;fnH–tSlsijek.kqdsfy,çFkemÙkstufoHkoVbysDVªkWuoksYVgSrksblijek.kqdhvk;u ÅtkZfdruhgksxh&
(A) V electron volt (B) 3 V electron volt
4
4 V
(C*) 3
electron volt (D) cannot be calculated by given information.
(A) V bysDVªkWu oksYV (B) 3 V bysDVªkWu oksYV
4
4 V
(C*) 3
ugha
bysDVªkWu oksYV (D) nh xbZ tkudkjh ls x.kuk lEHko
1 1 3
Sol. First excitation energy çFke mÙkstu ÅtkZ = RhC 12 22 = RhC 4
3 RhC = V e.v.
4
RhC = 4V
3
e.v.
23.8 All electrons ejected from a surface by incident light of wavelength 200 nm can be stopped before travelling 1 m in the direction of uniform electric field of 4 N/C. The work function of the surface is:
fdlhlrgijvkifr çdk'kftldkrjaxnS/;Z200nmgS]lsfudyusokyslHkhbysDVªkWuksadks,dfo|qr{ks=kftldhrhozrk
4N/CgS]dhfn'k esa1m nwjhr;djuslsiwoZfojkeesayk;ktkrkgSrkslrgdkdk;ZQyugksxk:
(A) 4 eV (B) 6.2 eV (C) 2 eV (D*) 2.2 eV
Sol. The electron ejected with maximum speed vmax are stopped by electric field E = 4N/C after travelling a distance d = 1m
1 mV 2 = eE d = 4eV
2 max
1240
The energy of incident photon = 200
From equation of photo electric effect
= 6.2 eV
1 mv 2 = h –
2 max 0
= 6.2 – 4 = 2.2 eV.
gy ;fnbysDVªkWudsmRltZudkvf/kdreosxvmaxgSrksmlsd=1mpyus ij E=4N/ClsjksdktkrkgSA
1 mV 2 = eE d = 4eV
2 max
1240
vkifrr QksVkWu dh ÅtkZ = 200
çdk'kfo|qrçHk olehdj.kls
=6.2eV
1 mv 2 = h –
2 max 0
= 6.2 – 4 = 2.2 eV.
23.9 An electron of mass ' m ', when accelerated through a potential V has deBroglie wavelength . The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be:
,dbysDVªkWuftldknzO;eku 'm'gSdksVfoHkolsRofjrfd;ktkrkgS;fnbysDVªkWudkMhczksxyhrjaxnS/;ZgSrks çksVkWulslEc}MhczksxyhrjaxnS/;ZfdruhgksxkA;fnçksVkWudknzO;ekuMrFk çksVkWudkslekufoHko VlsRofjrdjs:
(A)
(B*)
(C) M
m
(D) m
M
h
Ans. =
e
P
K = qV is same for both proton and electron. K=qV izksVkWuobysDVªkudsfy,lekugSA
23.10 Two hydrogen atoms are in excited state with electrons residing in n = 2. First one is moving towards left and emits a photon of energy E1 towards right. Second one is moving towards right with same speed and emits a photon of energy E2 towards right. Taking recoil of nucleus into account during emission process nksgkbMªkstuijek.kqmÙksftrvoLFk esagSftlesabysDVªkWun=2esagSA,dcka;hrjQpyrkgSrFk ,dE1ÅtkZdkQksVkWu nka;hrjQmRlftZrdjrkgSAnwljknka;hvksjlekupkylspyrkgSrFk nka;hvksjE2ÅtkZdkQksVkumRlftZrdjrk gSA ukfHkdksa ds çfr{kIr ds nkSjku mRltZu çØe esa
(A) E1 > E2 (B*) E1 < E2 (C) E1 = E2
(D)informationinsufficientlwpuk vi;kZIr gSA
Sol.
In the first case K.E. of H-atom increases due to recoil whereas in the second case K.E. decreases due to recoil but E1 + KE1 = E2 + KE2.
E > E
fLFkfrI dsçFkeckjgkbMªkstuijek.kqdhxfrtÅtkZçfrf{kIrlsc<+rhgSAfLFkfr2 ds Hijek.kqdhxfrtÅtkZijUrq nwljh ckj ?kVrh gS] ysfdu E1 +KE1 =E2 +KE2
E > E
23.11 In a hydrogen atom following the Bohr’s postulates the product of linear momentum and angular momentum
is proportional to (n)x where ‘n’ is the orbit number. Then ‘x’ is :
cksjdhvoLFk dsvuqlkjHijek.kqesajs[kh;laosxrFk dks.kh;laosxdkxq.kuQy(n)xdslekuqikrhgS tgk¡‘n’d{k Øekad gS rks ‘x’ gksxk :
(A*) 0 (B) 2 (C) –2 (D) 1
Sol. Linear momentum mv 1
n
angular momentum mvr n
product of linear momentum and angular momentum n0
gy js[kh;laosx mv 1
dks.kh;laosx mvr n
dks.kh;ojs[kh;laosxdkxq.kuQy n0
23.12 The voltage applied to an X-ray tube is 18 kV. The maximum mass of photon emitted by the X-ray tube will be: X-rayuyhesa18kVoksYVstyxkusijmRlftZrQksVkWudkvf/kdrenzO;ekugksxk:
(A) 2 × 10–13 kg (B) 3.2 × 10–36 kg (C*) 3.2 × 10–32 kg (D) 9.1 × 10–31 kg
Sol. Energy of photon is given by mc2 now the maximum energy of photon is equal to the maximum energy of electron = eV QksVkWudhÅtkZE=mc2=QksVkWudhvf/kdreÅtkZrksQksVkWudhvf/kdreÅtkZbysDVªkWudhvf/kdreÅtkZdscjkcj
=eVgSA
eV
hence vr% , mc2 = ev m = 2
1.6 1019 18 103
= 8 2
= 3.2 × 10–32 kg
c (3 10 )
4 1
23.13 The wavelengths of K x-rays of two metals ‘A’ and ‘B’ are 1875 R
and
675 R
respectively, where ‘R’
is Rydberg's constant. The number of elements lying between ‘A’ and ‘B’ according to their atomic numbers is
4
nks/k rq,sa'A"rFk 'B'dsfy,x-fdj.k sadhK rjaxnS/;ZØe'k%1875 R rFk
1
675 R gStgka'R'–fjMoxZfu;rkadgSA
rks 'A' rFkk 'B' ds e/; ijek.kq Øekad ds vuqlkj fdrus rRo gSA
(A) 3 (B) 6 (C) 5 (D*) 4
1 1 1
Sol. Using = R(z – 1)2 n 2 n 2
2 1
For particle ; n = 2, n = 1
For metal A ;
1875R
= RZ
12 3
z = 26
4
For metal B; 675R = RZ
4
12 3
1
z = 31
4 2
Therefore, 4 elements lie between A and B, i.e. with Z= 27, 28 ,29 , 30
1 1 1
gy. = R(z – 1)2 n 2 n 2 dk iz;ksx djus ij
2 1
d.k ds fy;s; n1 = 2, n2 = 1
A /kkrqds fy;s; 1875R = RZ
12 3
z = 26
4 4 1
B /kkrq ds fy;s 675R = RZ
12 3
z = 31
4 2
vr% A rFkk B ds e/; pkj rRo gSa vFkkZr Z= 27, 28 ,29 , 30
23.14 One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is :
Li2+dhmRltZdLisDVªedh,djs[k dkrjaxnS/;ZgkbMªkstuijek.kqesackejJs.khdhf}rh;js[k dsrjaxnS/;Zds cjkcj gS rks ml js[kk ds fy, bysDVªkWu dh d{kk esa ifjorZu gksxk :
(A) n = 4 n = 2 (B) n = 8 n = 2 (C) n = 8 n = 4 (D*) n = 12 n = 6
Sol. For 2nd line of Balmer seires in hydrogen specturm
gkbMªkstuLizsDVªeesackejJs.khdhf}rh;js[k dsfy;s
1 R(1) 1 1 3
22 42
= 16 R
1 2 1
1 3
For Li2+ : (Li2+ ds fy;s ) R(3)
62 122 16 R
which is satisfied by only (D).
tksdsoy(D)dkslUrq"VdjrhgSA
23.15 If the short wavelength limit of the continuous spectrum coming out of a coolidge tube is 10 Å, then the debroglie wavelength of the electrons reaching the target metal in the coolidge tube is approximately : dqyhtufydklsfudyusokyhrjaxksesalw{ererjaxnS/;Zdkeku10Å,gSrksdqyhtufydkesafLFkry{;/kkrq rd igq¡pus okys bysDVªkWu ds fy, Mh czksxyh rjaxnS/;Z dk eku fdruk gksxk :
(A*) 0.3 Å (B) 3 Å (C) 30 Å (D) 10 Å
Sol. We have : K.E. =
P2
2me
hc
= min . P =
h
Also, de broglie = p =
hmin
2meC
for min = 10Å : de broglie 0.3Å
P2
gy% ge tkurs gS: K.E. = 2me
h
hc
= min . P =
= =
Mh czksxyh p
min = 10Å : Mh czksxyh 0.3Å
23.16 The photon radiated from hydrogen corresponding to 2nd line of Lyman series is absorbed by a hydrogen like atom ‘X’ in 2nd excited state. As a result the hydrogen like atom ‘X’ makes a transition to nth orbit. Then,
Hijek.kqdhykbeuJs.khdhf}rh;js[k esafofdjrQksVkWudksgkbMªkstutSlsijek.kq'X'f}rh;mÙksftrvoLFk esagS]}kjkvo'k sf"krfd;ktkrkgSftldsifj.k eLo:i'X'nth ohad{k esalaØe.kdjrkgSA
(A) X = He+, n = 4 (B) X = Li++, n = 6 (C) X = He+, n = 6 (D*) X = Li++, n = 9
Sol. (D) Energy of nth sate in Hydrogen is same as energy of 3nth state in Li++.
3 1 transition in H would give same energy as the 3×3 1× 3
transition in Li++.
gy% (D) H ijek.kq dh noh d{kk dh ÅtkZ Li++ dh 3noh voLFkk dh ÅtkZ ds cjkcj gksrh gSA
H ijek.kq esa 3 1 voLFkk esa ifjorZu leku ÅtkZ nsrk gS tks fd
Li++ esa 3×3 1× 3
ifjorZu ds fy,A
23.17 In a photoelectric experiment, with light of wavelength , the fastest electron has speed v. If the exciting
3
wavelength is changed to 4 , the speed of the fastest emitted electron will become
çdk'kfo|qrçHk oesaç;qDrçdk'kdkrjaxnS/;ZgSrFk bysDVªkWudhvf/kdrepkyv gSA;fnmÙksturjaxnS/;Z dks 3 fd;ktkrkgSrksmRlftZrbysDVªkWudhvf/kdrepkygksxhA
(A) v (B) v
(C) less than v (D*) greater than v
(A) v (B) v
(C) v
ls de gSA (D*) v
ls vf/kd gSA
1 2 hc
Sol. (D)
mv
2
1 mv´2
2
hc
= (3 / 4)
– =
4hc
3
Clearly Li"Vr%
v' v
23.18 An element X decays, first by positron emission and then two -particles are emitted in successive radioactive decay. If the product nuclei has a mass number 229 and atomic number 89, the mass number and atomic number of element X are
,drRoX{kf;rgksrkgSigysikWftVªkWumRlftZrgksrkgSrFk fQjnks-d.kmÙkjksÙkjjsfM;ks,fDVo{k;esamRlftZr gksrsgSA;fnmRiknukfHkddknzO;ekuØekad229gSrFk ijek.kqØekad89gSrksrRoXdknzO;ekuØekadoijek.kq Øekadgksxk
(A) 237, 93 (B*) 237, 94 (C) 221, 84 (D) 237, 92
Sol. (B)
Given (fn;k gS) A – 8 = 224
& Z – 5 = 89 A = 237, Z = 94.
23.19 1.5 MW of 400 nm light is directed at a photoelectric cell. If 0.10% of the incident photons produce
photoelectrons, the current in the cell is
1.5MW rFkk400nm çdk'k ,d çdk'k fo|qr lsy dh rjQ vkifrr gSA ;fn 0.10%vkifrr QksVkWu bysDVªkWu mRiUu djrs gS rks lsy esa fo|qr /kkjk gksxh &
(A) 0.36 µA (B*) 0.48 µA (C) 0.42 mA (D) 0.32 mA
Power (watt) Emission %
Sol. (B) no. of photoelectron emitted per second, n =
Energy of 1 photon(inJ)100
1.5 103 W (103 ) 0.1
= 1240(nm)(eV) e 100 =
400(nm)
0.48 106
e
. (energy of 1 photon =
1240 nm / eV 400 nm e
Joule)
Photo current = ne = 0.48 A
' kfDr¼okWmVR½ltZu%
gy (B) çfrlSd.M mRlftZrQksVks bysDVªkWu,n= 1 QksVdkhWÅutkZ¼twy½100
1.5 103 W (103 ) 0.1
= 1240(nm)(eV) e 100 =
400(nm)
0.48 106 .
e
QksVks/kkjk =ne = 0.48A
23.20 The element which has a k x-rays line of wavelength 1.8 Å is
fdl rRodsfy, k x-fdj.kksadhrjaxnS/;Z1.8ÅgS
(R = 1.1 × 107 m–1, b = 1 and vkSj
5 / 33
= 0.39)
(A*) Co, Z = 27 (B) Iron, Z = 26 (C) Mn, z = 25 (D) Ni, z = 28
Sol. (A) for K
1 3R (Z 1)2 transition is from n =2 to n =1
(Z – 1) = =
78
= = 3
= 26. Z = 27
23.21 When an electron accelerated by potential difference U is bombarded on a specific metal, the emitted X-ray spectrum obtained is shown in adjoining graph. If the potential difference is reduced to U/3, the correct spectrum is
UfoHkokUrjlsRofjrbysDVªkWu]tcfo'ks"k/k rqijvkifr gksrks]mRlftZrX-fdj.kdsLisDVªedkxzkQfuEugksrk gSA ;fn foHkokUrj ?kVdj U/3 gks tk;s rks lgh LisDVªe gksxkA
m m X-ray wavelength
(A)
(A)
(B*)
(B)
(C)
(C)
(D)
(D)
Sol. m will increase to 3 m due to decrease in the energy of bombarding electrons. Hence no characteristic x-rays will be visible, only continous X-ray will be produced.
vkifrrbysDVªkWudhÅtkZdegksusdsdkj.km ,3m rdc<+tkrkgSAblfy,x-fdj.kdhdksbZHkhvfHkykf{k.kd n`'; ugha gSA blfy, dsoy lrr~ X-fdj.k mRiUu gksxhA
23.22 In the hydrogen atom, an electron makes a transition from n = 2 to n = 1. The magnetic field produced by the circulating electron at the nucleus
(A) decreases 16 times (B) increases 4 times
(C) decreases 4 times (D*) increases 32 times
,dgkbMªkstuijek.kqesabysDVªkWun=2lsn=1esavkrkgSrksukfHkd dspkjksvksjpDdjyxkusokysbysDVªkWu ds }kjk mRié pqEcdh; {ks=k dk eku A
(A) 16 xquk de (B) 4 xquk vf/kd
(C) 4 xquk de (D*) 32 xquk vf/kd
Sol. B = B =
B
0
2r
0 e 2rT
1
n5
and rFkk = e
[r n2, T n3]
23.23 90% of a radioactive sample is left undecayed after time t has elapsed. What percentage of the initial sample will decay in a total time 2t. tle;dsckn90%jsfM;ks,sfDVoinkFkZ'ks"kjgrkgSAdqy2tle;esaçkjfEHkdinkFkZdkfdrukçfr'kr{kf;rgksxkA
(A) 20% (B*) 19% (C) 40% (D) 38%
Sol. 90% of sample left undecayed after time t t le;i'pkr~90%uewuk'ks"kjgrkgSA
9
10 N0 = N0 e–t
= 1 ln 10
(1)
t 9
After time 2t, 2t le; ckn
1
10
t ln 9 2t
N = N0 e–(2t) = N0 e
10 2
2
ln
9
9
N = N0 e
= N0
(2)
10
19% of initial value will decay in time 2t. 2t le; ckn çkjfEHkd dk 19% de gks tk,xkA
23.24 A radioactive element X converts into another stable element Y. Half life of X is 2 hrs. Initially only X is present. After time t, the ratio of atoms of X and Y is found to be 1 : 4, then t in hours is :
(A) 2 (B) 4 (C*) between 4 and 6 (D) 6
,djsfM;ks,fDVorRo XLFk ;hukfHkdYesaifjofrZrgksrkgSAXdhv)Zvk;q2?kaVsgSAçkjEHkesadsoyXmifLFkr gS ijUrq t le; i'pkr~ X o Y dk vuqikr 1 : 4 gS rks t dk eku ?kaVks esa gksxk :
(A) 2 (B) 4 (C*) 4 o 6 ds e/; (D) 6
Sol. Let N2 be the number of atoms of X at time t = 0. Then at t = 4 hrs (two half lives)
ekukN2t=0,Xds ijek.kq dh la[;k gSA blfy, t=4hrs (nks v)Z vk;q)
N = N0 rFkk N
= 3N0
x 4 y 4
N / N = 1/3
x y
rFkk at t = 6 hrs (three half lives)
t = 6 hrs ij (rhuv)Z vk;q)
N = N0 andrFk N
= 7N0
x 8 y 8
Nx 1
or Ny = 7
1 1 1
The given ratio 4 lies between 3 and 7 .
vuqikr 1 , 1 1
4 3 rFkk 7 ds e/;vkrk gSA
Therefore, t lies between 4 hrs and 6 hrs.
blfy, le;t,4?k.Vs ls 6?k.Vsds e/; gksxkA
23.25 An electron in a hydrogen atom makes a transition from first excited state to ground state. The equivalent current due to circulating electron
gkbMªkstu ijek.kq esa ,d bysDVªkWu çFke mÙksftr voLFkk ls ewy voLFkk esa vkrk gSA pDdj yxkus okys
(A) increases 2 times (B) increases 4 times
(C*) increases 8 times (D) remains the same (A)nksxqukc<+tk;sxhA (B) pkjxqukc<+tk;sxhA (C*) vkBxqukc<+tk;sxhA (D) vifjofrZrjgsxhA
Sol. i = q T
Now T2 r3 n6 i 1
n3
T n3
i H3
1 2
i2 1
(1)3
(2)3
= i2
= 8i1
SECTION - II : MULTIPLE CORRECT ANSWER TYPE
[k.M- II: cgqy lgh fodYi izdkj
23.26 When a hydrogen atom is excited from ground state to first excited state then
tcgkbMªkstuijek.kqewyÅtkZLrjlsizFkemÙksftrÅtkZLrjes amÙksftrgksrkgSrks&
(A) its kinetic energy increases by 10.2 eV. (B*) its kinetic energy decreases by 10 .2 eV. (C*) its potential energy increases by 20.4 eV.
(D*) its angular momentum increases by 1.05 × 10–34 J-s. (A)bldhxfrtÅtkZ10.2bysDVªkWuoksYVlsc<+tk;sxhA (B*)bldhxfrtÅtkZ10.2bysDVªkWuoksYVls ?kVtk;sxhA (C*)bldhfLFkfrtÅtkZ20.4bysDVªkWuoksYVlsc<+tk;saxhA
(D*) bldk dks.kh; laosx 1.05 × 10–34 twy × lS0 ls c<+ tk;sxkA
Sol. ground state n = 1
ewy voLFkkn=1
first excited state n = 2
çFke mÙkstu voLFkkn=2
KE =
1
40
e2
(z = 1)
2r
KE =
14.4 1010
eV
2r
vc Now r = 0.53 n2 Aº (z = 1)
14.4 1010
(KE)1 =
2 0.53 1010
eV = 13.58 eV
14.4 10 10
(KE)2 = 2 0.53 1010 4 ev = 3.39 ev
KE decreases by = 10.2 ev
xfrt ÅtkZ10.2ev ls ?kV tkrh gSA
vc 1 e2
14.4 1010
Now PE = 40 r = r ev
– 14.4 10 10
(PE) =
0.53 10 –10
14.4 1010
ev = –27.1 ev
(PE)2 = 0.53 1010 4
= –6.79ev
PE increases by = 20.4 ev
fLFkfrt ÅtkZ =20.4ev lsc<+tkrhgSA
Now Angular momentum ; vr%dks.kh;laosx
nh
L = mvr =
h
L –L =
2
6.6 1034
=
= 1.05 × 10–34 J–sec.
2 1 2
6.28
23.27 In an xray tube the voltage applied is 20KV. The energy required to remove an electron from L shell is 19.9 KeV. In the xrays emitted by the tube
,dX–fdj.kufydkesavkjksfir oksYVrk20KVgSALdks'kls,dbysDVªksudksgVkus dsfy;s vko';dÅtkZ19.9 KeVgSAufydk }kjkmRlZftrX–fdj.k saesa]
(A*) minimum wavelength will be 62.1 pm
(B*) energy of the characterstic xrays will be equal to or less than 19.9 KeV (C*) L xray may be emitted
(D) L xray will have energy 19.9 KeV
(A) U;wurerjaxnZ/;Z62.1Pmgksxh
(B) vfHkyk{kf.kd X–fdj.k sadhÅtkZ19.9KeVdscjkcj;kdegksxhA
(C) L X–fdj.kmRlftZrgksldrh gSA N
(D) Lx–fdj.k dh ÅtkZ 19.9KeVgksxh A m
Sol.
12400
= v0 Å
-19.9 keV L
K
12400
20,000 = .62 Å
62·1pm
Any transition to L will have energy less than or equal to 19·9 kev so B.
Sol.
12400
= v0 Å
N
m
-19.9 keV L
12400
20,000 = .62 Å
62·1pm K
L rd fdlh Hkh laØe.k es ÅtkZ 19·9kevls de ;k cjkcj gksxhA vr%B.
Ke 2
23.28 Suppose the potential energy between electron and proton at a distance r is given by –
3r 3
of Bohr’s theory to hydrogen atom in this case shows that
Ke2
. Application
ekukbysDVªkWuçksVkWudschpfLFkfrtÅtkZ– 3r 3 gStgk¡rnksuksadschpdhnwjhgSAblfLFkfresacksgjdsfl}kUrdk
vuqç;ksx djus ij
(A*) energy in the nth orbit is proportional to n6 nohad{k esaÅtkZn6 dslekuqikrhgksrhgSA
(B*) energy is proportional to m–3 (m : mass of electron)
ÅtkZm–3 dslekuqikrhgS(m :bysDVªkWudknzO;eku)
(C) energy of the nth orbit is proportional to n–2 nohad{k dhÅtkZn–2 dslekuqikrhgSA
(D) energy is proportional to m3 (m = mass of electron)
ÅtkZm3 dslekuqikrhgksrhgSA(m :bysDVªkWudknzO;eku)
dU
Sol. |F| = dr
Ke2
= r 4
(1)
Ke2
r 4
mv2
=
r
nh
(2)
and vkSj mvr = 2
By (2) and o (3) ls,
Ke2 42 m m
(3)
r = h2
n2 = K1 n2
(4)
Total energy =
1 (potential energy)
2
dqy ÅtkZ= 1 (fLFkfrt ÅtkZ)
Ke 2
= 6r 3
Ke2
= K m 3
6
Ke2n6
= 6K 3m3
n2
Total energy n6
dqy ÅtkZ n6 Total energy m–3 dqy ÅtkZ m–3
(A) and (B) are correct.
(A) o(B) lgh gSA
23.29 Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of 𝑙n(An/Al) against 𝑙n(n)
gkbMªkstuijek.kqesanohad{k dk{ks=kQyAn gSrks𝑙n(An/Al)rFk 𝑙n(n)dse/;xzkQesa
(A*) will pass through origin ewy fcUnq ls xqtjsxkA
(B*) will be a straight line with slope 4 ljy js[kk gksxk ftldh ço.krk 4 gksxhA
(C) will be a monotonically increasing nonlinear curve
lr ~:ilsc<+rkgqvkvjs[kh;xzkQA
(D) will be a circle ,do`Ùkgksxk
Sol. rn = n2r1
4ln(n)
A
r 2
𝑙n n
A1
n
r 2
= 𝑙n n4 = 4 𝑙n (n)
1
ln (An/A1)
SECTION - III : ASSERTION AND REASON TYPE
[k.M- III: dFku vkSj dkj.k izdkj
23.30 STATEMENT-1 : Though light of a single frequency (monochromatic light) is incident on a metal, the energies of emitted photoelectrons are different.
STATEMENT-2 : The energy of electrons just after they absorb photons incident on metal surface may be lost in collision with other atoms in the metal before the electron is ejected out of the metal. oDrO;-1: ;+|fiçdk'kdh,dyvko`fÙk(,do.khZçdk'k) dks/kkfRodlrgijvkifrrgksusdsmijkUrHkhmRlftZr QksVksbysDVªkWuksadhÅtkZ,savyx&vyxgksrhgSA
oDrO;-2: tcQksVkWu/kfRodlrgijfxjrsgSrksbysDVªkWuksa}kjkmudsvo'ks"k.kdsmijkUrbysDVªkWuksads/krqlsckgj mRltZugksuslsigysbysDVªkWuksadhÅtkZ/krqesavU;ijek.kqvksadslkFkVDdjesau"VgksldrhgSA
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS, oDrO;-2 lR; gSA
Sol. (A) Energy of photoelectron emitted is different because after absorbing the photon electrons within metals collide with other atom before being ejected out of metal.
Hence statement 2 is correct explanation of statement 1.
Sol. (A) mRlftZrQksVksbysDVªkWuksadhÅtkZ,savyxgksxhD;ksafdQksVksuksadksvo'ksf"krdjusdscknbysDVªkWu/krqdsvUnjvU; ijek.kqvksalsVdjkrsgSaAfQj/kfRodlrglsckgjfudyrsgSaA
vr%oDrO; 2 lR;gS rFkkoDrO; 1dhlgh O;k[;kdjrk gSA
23.31 STATEMENT-1 : The de-Broglie wavelength of a molecule (in a sample of ideal gas) varies inversely as the square root of absolute temperature.
STATEMENT-2 : The rms velocity of a molecule (in a sample of ideal gas) depends on temperature.
oDrO;-1 : (vkn'kZ xSl ds ,d uewus esa) ,d v.kq dh Mh&czksxyh rjaxnS/;Z
ije rki ds oxZewy ds O;qRØekuqikrh gksrh gSA
oDrO;-2: (vkn'kZxSlds,duewusesa)v.kqdkoxZek/;ewyosxrkiijfuHkZjdjrkgSA
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA
(B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement- 1
oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA
(C) Statement-1 is True, Statement-2 is False
oDrO; -1 lR; gS, oDrO;-2 vlR; gSA
(D) Statement-1 is False, Statement-2 is True.
oDrO; -1 vlR; gS , oDrO;-2 lR; gSA
Sol. de-Broglie wavelength associated with gas molecules varies as .
xSldsv.kqvksadhMh&czksxyhrjaxnS/;ZblizdkjifjofrZrgksrhgS 1 .
T
23.32 STATEMENT-1 : Heavy nuclides tend to have more number of neutrons than protons.
STATEMENT-2 : As there is coulombic repulsion between protons, so in heavy nuclei, excess of neutrons are preferable.
oDrO;-1:HkjhukfHkdizksVkWuksadhvis{k U;wVªkWuvf/kdla[;kesaj[kusdhizo`fÙkj[krsgSaA
oDrO;-2:pqfdizksVkWuksadschpdwykWfEcdizfrd"kZ.kgSvr%HkjhukfHkdksaesa]U;wVªkWuksadhvf/kdrkdksojh;rkgksrhgSA
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True.
(A*) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA
(B) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k ugha gSA
(C) oDrO; -1lR; gS, oDrO;-2vlR; gSA
(D) oDrO; -1vlR; gS, oDrO;-2lR; gSA
Sol. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA
23.33 STATEMENT–1 : Z XA
undergoes 2 decays, 2 decays (negative ) and 2 decays. As a result the
daughter product is Z2 YA8 .
STATEMENT–2 : In decay the mass number decreases by 4 unit and atomic number decreases by 2 unit. In decay (negative ) the mass number remains unchanged and atomic number increases by 1 unit. In decay, mass number and atomic number remains unchanged.
oDrO;–1 : Z XA ,2,2 (_.kkRed )o 2{k; djrk gS ftlls ifj.kkeh iq=kh mRikn Z2 YA8 gSA oDrO;–2: {k;esanzO;ekula[;k4bdkbZlsrFk ijek.kqla[;k2bdkbZlsdegksrhgSA{k;esa(_.kkRed) nzO;ekula[;kvifjofrZrjgrhgSoijek.kqla[;k1bdkbZlsc<+rhgSA{k;esanzO;ekula[;krFk ijek.kqla[;k vifjofrZrjgrhgSA
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True.
(A*) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k gSA
(B) oDrO;-1lR; gS,oDrO;-2lR; gS;oDrO;-2oDrO;-1dk lgh Li"Vhdj.k ugha gSA
(C) oDrO; -1lR; gS, oDrO;-2vlR; gSA
(D) oDrO; -1vlR; gS, oDrO;-2lR; gSA
Sol. Statement-2 is true by definition and correctly explains the staement-1, namely, Z XA
undergoes 2 de-
cays, 2 decays (negative ) and 2 decays. As a result the daughter product is Z2 YA8 .
dFku-2ifjHkk"k dsdkj.klR;gSrFkkdFku-1dkslghizdkjlsifjHk f"krdjrkgSA Z XA esals2,2 _.k Red rFk 2{k;dsckn Z2 YA8 iq=khukfHkdcurkgSA
SECTION - IV : COMPREHENSION TYPE
[k.M- IV:cks/ku izdkj
COMPREHENSION # 1 (SAME QUESTION 21.48)
The radionuclide 56Mn is being produced in a cyclotron at a constant rate P by bombarding a maganese target with deutrons. 56Mn has a half life of 2.5 hours and the target contains large number of only the stable maganese isotope 55Mn. The reaction that produces 56Mn is :
55Mn + d 56Mn + p
After being bombarded for a long time, the activity of 56Mn becomes constant equal to
13.86 × 1010 s–1 . (Use 𝑙n2 = 0.693; Avogadro No = 6 × 1023; atomic weight 56Mn = 56 gm/mole) esXuhty{;ijM~;wVªkWuksadhckSNkj(bombarding)lslkbDyksVªkWu(cyclotron)esajsfM;ks/kehZukfHkd56Mndkfu;r nj PlsmRiknu gksrkgSA 56Mndk v/kZvk;q&dky (half lifetime) 2.5?k.VsrFkk y{;esadsoyesXuht leLFkkfud (Isotope) 55MnLFkk;hukfHkdgSrFkk bldhla[;k vf/kdgSA vfHkfØ;ktks 56MnmRiUudjrhgSA
55Mn + d 56Mn + p
yEcs le; rd ckSNkj djus ds ckn 56Mn dh lfØ;rk fu;e gks tkrh gSA ftldk eku 13.86 × 1010 s–1 gSA (𝑙n2= 0.693; vkoxzknksa la[;k = 6× 1023; ijek.kqHkkj 56Mn= 56xzke/eksy)
23.34 At what constant rate P, 56Mn nuclei are being produced in the cyclotron during the bombardment ?
lkbEyksVªksu(cyclotron)esa56MnukfHkd]ckSNkjdsnkSjkufdlfu;rnj]PlsmRiUugksrsgS&
(A) 2 × 1011 nuclei/s (B*) 13.86 × 1010 nuclei/s
(C) 9.6 × 1010 nuclei/s (D) 6.93 × 1010 nuclei/s
(A) 2 × 1011 ukfHkd/lS- (B*) 13.86 × 1010 ukfHkd/lS-
(C) 9.6 × 1010 ukfHkd/lS- (D) 6.93× 1010 ukfHkd/lS-
Sol. In equilibrium, rate of decay = rate of production
lkE;koLFkkij fo?kVudhnj= mRiUugksusdhnj
23.35 After the activity of 56Mn becomes constant, number of 56Mn nuclei present in the target, is equal to
56MndhlfØ;rkfu;rgkstkusdsckn]y{;esamifLFkr56MnukfHkdksadhla[;kgksxh&
(A) 5 × 1011 (B) 20 × 1011 (C) 1.2 × 1014 (D*) 1.8 × 1015
Sol. As Rate of decay = Rate of production
pwfd fo?kVudh nj = mRiUugksus dhnj
P = N N = P
= Pt1/ 2
𝑙n2
= 1.8 × 1015
23.36 After a long time bombardment, number of 56Mn nuclei present in the target depends upon
(a) the number of 56Mn nuclei present at the start of the process.
(b) half life of the 56Mn
(c) the constant rate of production P.
(A) All (a), (b) and (c) are correct (B) only (a) and (b) are correct
(C*) only (b) and (c) are correct (D) only (a) and (c) are correct vf/kdle;rdckSNkjdjusijy{;esamifLFkr56MnukfHkdksdhla[;kfuHkZjdjsxhA (a)izkjfEHkdfLFkfrijmifLFkr56MndsukfHkdksdhla[;kij
(b) 56Mndsv/kZ vk;qdkyij
(c) mRiUu gksus dh fu;r nj P ij
(A) (a), (b) rFkk (c) lHkh lgh gSA (B) dsoy (a) rFkk (b) lgh gSA
(C*) dsoy (b) rFkk (c) lgh gSA (D) dsoy (a) rFkk (c) lgh gSA
Sol. As N =
Pt1/ 2
𝑙n2
it is dependent on P and t
1/2
. Initial no. of 56Mn nuclei will make no difference as in equilibrium rate of
production equals rate of decay. Large initial no. will only make equilibrium come sooner.
gy% N =
Pt1/ 2
𝑙n2
;gPrFk t1/2 ijfuHkZjdjrkgSAD;ksfdlkE;oLFk esamRiknudhnjrFk {k;dhnjcjkcjgksusls MnukfHkd
56
dhizkjfEHkdla[;kdsdkj.kdksbZvarjughgksxkAvr%izkjfEHkdukfHkdksadhla[;kijfuHkZjughdjrhgSAvf/kd la[;kdsoylkE;fLFkfrdkstYnhyknsxhA
Comprehension # 2
Pertain to the statement and diagram below :
4
3
2
1
The figure given shows an energy level diagram for the hydrogen atom. Several transitions are marked as
, . The diagram is only indicative and not to scale.
mi;qZDrfp=kesa]gkbMªkstuijek.kqdsfy;s]ÅtkZLrjfp=kn'k Z;kgSAfofHkUulaØe.k , lsvafdrgaSAfp=k dsoylwpukFkZgSAfdlhiSekuslsughcuk;sx;sgSA
23.37 In which transition is a Balmer series photon absorbed ?
fdllaØe.kesackejJs.khQksVkWuvo'ksf"krgksaxs&
(A) (B) (C) V (D*) V
Sol. For Balmer series, n1 = 2, n2 = 3,4,........
(lower) (higher)
In transition (V), Photon of Balmer series is absorbed.
gy ckejJs.khdsfy,, n1 = 2, n2 = 3,4,........
(fuEu) (mPp)
laØe.k (V), esa okej Js.kh dk QksVkWu vo'kksf"kr gks tk,xkA
23.38 The wavelength of the radiation involved in transition is
laØe.k lslEcfU/krfofdj.kdhrjaxnS/;Zgksxh&
(A) 291 nm (B) 364 nm (C*) 487 nm (D) 652 nm
Sol. In transition laØe.k esa
E2 = – 3.4 eV, E4 = – 0.85 eV
E = 2.55 eV
E =
hc = hc
E
= 487 nm.
23.39 Which transition will occur when a hydrogen atom is irradiated with radiation of wavelength 103nm?
tc gkbMªkstu ijek.kq ,103 (nm) uSuksehVj ds rjaxnS/;Z ds fofdj.k fodhfjr djrk gS rks dkSulk laØe.k gksxk ?
(A) I (B) II (C) IV (D*) V
Sol. Wavelength of radiation = 103 nm = 1030 Å
fofdj.k dh rjaxnS/;Z=103nm =1030Å
E =
12400
1030 Å
~– 12.0 eV
So difference of energy should be 12.0 eV (approx)
vr%ÅtkZesavUrjyxHkx12.0eVgkstk,xkA
Hence vr% n1 = 1 andrFk n2 = 3
(–13.6)eV (–1.51)eV
Transition is V.
laØe.kVagSA
SECTION - V : MATRIX - MATCH TYPE
[k.M- V:eSfVªDl&lqesy izdkj
23.40 In the shown experimental setup to study photoelectric effect, two conducting electrodes arc enclosed in an evacuated glass-tube as shown. A parallel beam of monochromatic light, falls on photosensitive electrodes. The emf of battery shown is high enough such that all photoelectrons ejected from left electrode will reach the right electrode. Under initial conditions photoelectrons are emitted. As changes are made in each situation of column I; Match the statements in column I with results in column II. izdk'koS|qrizHk odsv/; udsfy,fn[k ;sx;siz;ksfxdO;oLFk esa]nkspkydbysDVªksMvkdZ(arc)fn[k ;hxbZ
,dfuokZfr dk¡pdhuyhesacUngSA,do.khZizdk'kdk,dlekUrjiqatizdk'klaosnhbysDVªkWMijfxjrkgSAfn[k bZ xbZcSVjhdkfook-cyi;kZIrmPpgSrkfd]ck;sbysDVªkWMlsfudyusokyslHkhQksVksbysDVªkWunka;sbysDVªksMij igq¡ptk;sAizkjfEHkdizfrcU/ksadsvUrxZrQksVksbysDVªkWumRlftZrgksrsgSaAdkWye-IdhizR;sdfLFkfresaifjorZufd;s tkrsgSAdkWye-IesadFkuksadksdkWye-IIesafn;sifj.k eksalslqesfyrdhft,A
Column I Column II
(A) If frequency of incident light is increased (p) magnitude of stopping potential will increase keeping number of photons per second constant
(B) If frequency of incident light is increased (q) current through circuit may stop and number of photons per second is decreased.
(C) If work function of photo sensitive (r) maximum kinetic energy of ejected electrode is increased photoelectrons will increase
(D) If number of photons per second of incident light
is increased keeping its frequency constant (s) saturation current will increase
(t) saturation current will decrease
dkWye-I dkWye-II
(A) ;fnvkifr izdk'kdhvko`fÙk]rhozrkdks (p) fujks/khfoHkodkifjek.kc<+sxkA dksfu;rj[krsgq;sc<+kbZtkrhgSA
(B) ;fnvkifr izdk'kdhvko`fÙkc<+kbZtkrh (q) ifjiFkesaizokfgr/k jk:dldrhgSA gS rFkk bldh rhozrk ?kVkbZ tkrh gSA
(C) ;fnizdk'klaosnhbysDVªksMdkdk;Z (r) fudyusokysQksVksbysDVªkWuksadhvf/kdre Qyuc<+k;ktkrkgSA xfrtÅtkZc<+sxhA
(D) ;fnvkifrrizdk'kdhrhozrk]bldh (s)lar`Ir/kjkc<+sxhA vko`fÙkfu;rj[krsgq;sc<+kbZtkrhgSA
Ans. (A) p,r (B) p,r, t (C) q (D) s
Sol. Consider two equation
1
(t) lar`Ir/kjk?kVsxhA
eVs = 2
2
max
= h –
.... (1)
no of photoelectrons ejected/sec. no. of photons/second (2)
(A) As frequency is increased keeping intensity constant.
|V | will increase, 1 m(v 2
) will increase and saturation current will remains same.
s 2 max
(B) As frequency is increased and intensity is decreased.
|V | will increase, 1 m(v 2
) will increase and saturation current will decrease.
s 2 max
(C) It work function is increased photo emission may stop.
(D) If intensity is increased and frequency is decreased. saturation current will increase.
gy% fuEunkslehdj.kyhft,
1
eVs = 2
2
max
= h –
.... (1)
izfrlsd.MfudyusokysQksVksbysDVªkWuksadhla[;k izfrlsd.MQksVksudhla[;k (2)
(A) tcrhozrkdksfu;rj[krsgq;svko`fÙkc<+kbZtkrhgS]
|V |c<+sxk, 1 m(v 2
) c<+sxhrFk lar`Ir/k jkoghjgsxhA
s 2 max
(B) tcvko`fÙkc<+kbZtkrhgSrFk rhozrk?kVkbZtkrhgS]
|v |c<+sxk, 1 m(v 2
) c<+sxhrFk lar`Ir/kjk?kVsxhA
s 2 max
(C) ;fndk;Z Qyuc<+k;k tkrkgS] rksQksVks mRltZu:d ldrkgSA (D);fnrhozrkc<+kbZtkrhgSrFk vko`fÙk?kVkbZtkrhgS]rkslar`Ir/k jkc<+sxhA
23.41 In column-I, consider each process just before and just after it occurs. Initial system is isolated from all other bodies. Consider all product particles (even those having rest mass zero) in the system. Match the system in column-I with the result they produce in column-II.
dkWye-IesaizR;sdizØe(process)?kfVrgksusdsBhdigysoBhdcknfopkjdhft,AizkjfEHkdfudk;lHkhvU;oLrqvksa
ls foyfxr gSA lHkh mRikn d.kksa (pkgs mldk fojke nzO;eku 'kwU; gh gks) dks fudk; esa ekusA dkWye-I esa fudk; dks dkWye-II esa muds laxr ifj.kke ls lqesfyr dfj,A
Column-I Column-II
(A) Spontaneous radioactive decay of an (p) Number of protons is increased uranium nucleus initially at rest
as given by reaction
238U 234Th 4He ...
92 90 2
(B) Fusion reaction of two hydrogen nuclei (q) Momentum is conserved
as given by reaction
1H 1H 2H .....
1 1 1
(C) Fission of U235 nucleus initiated by a (r) Mass is converted to energy thermal neutron as given by reaction or vice versa
1n 235U 144Ba 89Kr 31 n ...
0 92 56 36 0
(D) – decay (negative beta decay) (s) Charge is conserved
(t) No. of protons is decreased
dkWye-I dkWye-II
(A) ,d;wjsfu;eukfHkdtksfdizkjEHkesafojke (p) izksVkWuksdhla[;kc<+rhgSA esaFk ]mldkfuEuvfHkfØ;k}kjkLor%
jsfM;ks,fDVo{k;gksrkgS238U 234Th 4He ...
92 90 2
(B) nksgkbMªkstuukfHkdksadhlay;uvfHkfØ;k (q) laosxlajf{krjgrkgSA
fuEu}kjknhtk;sxh 1H 1H 2H .....
1 1 1
(C) U235 ukfHkddkrkih;U;wVªkWu}kjkfo[k.Mu (r) nzO;ekuÅtkZesa:ikUrfjrgksrkgS izkjEHkfd;ktkrkgS ;kbldkO;qRØe(vice-versa)gksrkgS
1n 235U 144Ba 89Kr 31 n ...
0 92 56 36 0
(D) – {k;¼_.k RedchVk{k;½ (s) vkos'klajf{krjgrkgSA
(t) izksVksuksdhla[;k?kVrhgSA
Ans. (A) q,r,s, (B) q,r,s (C) q,r,s (D) p,q,r,s
Sol. (A) In the given spontaneous radioactive decay, the number of protons remain constant and all conservation principles are obeyed.
(B) In fusion reaction of two hydrogen nuclei a proton is decreased as positron shall be emitted in the reaction. All the three conservation principles are obeyed.
(C) In the given fission reaction the number of protons remain constant and all conservation principles are obeyed.
(D) In beta negative decay, a neutron transforms into a proton within the nucleus and the electron is ejected out.
(A) fn;sx;sLor%jsfM;ks,sfDVofo?kVuesa]izksVkWuksadhla[;kfu;rjgrhgSrFk lHkhlaj{k.kfu;eksadkikyugksrkgSA
(B) nksgkbMªkstuukfHkdksadhlay;uizfØ;kesa],dizksVkWu],diksftVªkWuds:iesa?kVtkrkgStksvfHkfØ;kesamRlftZr gksrkgSAlHkhrhulaj{k.kfu;eksadkikyugksrkgSA
(C) nhxbZfo[k.MuizfØ;kesa]izksVkWuksadhla[;kfu;rjgrhgSrFk lHkhlaj{k.kfu;eksadkikyugksrkgSA
(D) _.k RedchVkfo?kVuesa],dU;wVªkWuukfHkdesa,dizksVkWuds:iesa:ikUrfjrgkstkrkgSrFk ,dbysDVªkWufudy tkrkgSA
SECTION - VI : INTEGER TYPE
[k.M-VI:iw.kkZadizdkj
SHORT SUBJECTIVE
y?kqÙkjkRediz'u
23.42 Photons of energy 5 eV are incident on cathode as shown in the figure. Electrons reaching the anode have kinetic energies varying from 6eV to 8eV. Find the work function of the metal in ev. & state whether the current in the circuit is less than or equal to saturation current. 5eVÅtkZdkQksVkWufp=kesafn[k ,x;sdSFk sMijvkifr gksrkgSA,uksM+ijigq¡pusokysbysDVªkWudhxfrtÅtkZ6eV ls8eVdschpifjofrZrgksrhgSA/k rqdkdk;ZQyuev. esaKkrdjksrFk KkrdjksifjiFkesa/k jkdkekular`Ir/k jk dscjkcj ;kdegSA
Ans. 2
Sol. KEmax = (5 – ) eV
when these electrons are accelerated through 5V, they will reach the anode with energy = (5 – + 5)eV
10 – = 8
φ = 2eV Ans.
Current is less than saturation current Ans.
Because if slowest electron also reached the plate it would have 5eV energy at the anode, but there it is given that the minimum energy is 6eV.
gy KEmax = (5 – ) eV
tcbysDVªkWudks5VlsRofjrdjrsgSA rksbysDVªkWu,uksMijigq¡psxkrcmldhÅtkZ=(5–+5)eV
10 – = 8
φ = 2eV Ans.
/k jklar`Ir/k jklsdegSA
D;ksafdeUnrebysDVªkWuIysVij5eVÅtkZdslkFk,uksMijigq¡psxkijUrq;gk¡fn;kgSfdU;wureÅtkZ6eVgSA
23.43 Consider a nuclear reaction A + B C. A nucleus ‘A’ moving with kinetic energy of 5 MeV collides with a nucleus ‘B’ moving with kinetic energy of 3 MeV and form a nucleus ‘C’ in excited state. Find the kinetic energy of nucleus ‘C’ just after its formation is x/100 mev. and it is formed in a state with excitation energy 10 MeV. Take masses of nuclei of A, B and C as 25.0, 10.0, 34.995 amu respectively. 1 amu = 930 MeV/c2. ukfHkdh;vfHkfØ;k A+BCijfopkjfdft,AukfHkd‘A’ftldh xfrtÅtkZ5MeVgS]ukfHkd‘B’ftldhxfrtÅtkZ 3MeVgS]lsVdjkrkgSrFk ukfHkd‘C’mÙksftrvoLFk dkfuekZ.kdjrsgSAcu sdsrqjUrckn‘C’dhxfrtÅtkZKkr djks ;fn;g 10MeVmÙkstr ÅtkZ ds lkFk curk gSA ukfHkd A,B rFkkCdsnzO;eku Øe'k% 25.0,10.0,34.995amu
gSA
1 amu = 930 MeV/c2 gSA
Sol. Applying conservation of energy ÅtkZ laj{k.k ds vuqlkj
m c2 + K
+ m c2 + K
= m c2 + K
+ excitation energy mÙkstuÅtkZ
(mA
+ mB
– m ) c2 + K
+ KB
= KC
+ excitation energy mÙkstuÅtkZ
4.65 + 5 + 3 = Kc + 10
or Kc = 2.65 MeV Ans. 2.65 MeV
23.44 A sample has two isotopes A150 and B having masses 50 g and 30 g respectively. A is radioactive and B is stable. A decays to A by emitting particles. The half life of A is 2 hrs. Find the mass of the sample after 4 hour in gm.
,duewusesanksleLFk fudA150rFk BftudsnzO;ekuØe'k%50grFk 30ggSAA,djsfM;ks,fDVogSrFk B,dLFk ;h ukfHkdgSAA,d.kmRlftZrdjA'cukrkgSA;fnAdhv)Zvk;q2?kaVsgSrks4?kaVscknuewusdknzO;ekuxzkeesafdruk gksxk
Ans. 79 gm.
Sol. After 4 hrs. sample will contain 4?kaVscknuewusesa 30 gm B
12.5 gm A
146 37.5
150
gm A
Total mass = 30 + 12.5 + 146 × 37.5 = 79 gm.
150
146
dqy mRikn= 30 + 12.5 + 150 × 37.5 = 79 gm.
23.45 A radioactive source, in the form of a metallic sphere of radius 10–2 m emits –particles at the rate of 5 × 1010 particles per second. The source is electrically insulated. How long in -sec.will it take for its potential to be raised by 2 volt, assuming that 40% of the emitted –particles escape the source. 10–2 m dhf=kT;k dh/kkrq dsxksys dhvkd`fr okysjsfM;ks,fDVoL=kksr5×1010 d.k@lsd.Mdh njls -d.k mRlftZrdjrkgSAL=k srfo|qrjks/khcuk;kx;kgS;gekudjpyksafd40%mRlftZr–d.kL=kksrdksNksM+dj pystkrsgS]rks2 oksYVfoHkoizkIrdjusdsfy,fdrukle; -sec.esa½yxsxk\
Sol. Let t be the time required to raise to potential by 2V. Then number of –particles emitted in t second is 5 × 1010 t. Now the number of –particles escaping from sphere is 40% i.e., 2 × 1010 t. So, the charge developed.
Q = (2 × 1010 t) (1.6 × 10–19) Coulomb
= (3.2 × 10–9 t) Coulomb
But Q = (40 R)V
102 2
= 9 109
Coulomb
102 2 9 109
= 3.2 × 10–9 t
or t = 700 × 10–6 sec. t = 700 sec.
Sol. ekuk 2VfoHko izkIrdjus esa t le;yxsxk rct lsd.M esamRlftZr –d.kksa dhla[;k 5× 1010 tgSAvc 40% –
d.k sadhla[;ktksxksysdksNksM+djtkrsgS]2×1010 tgSA vr%mRiUuvkos'k
Q = (2 × 1010 t) (1.6 × 10–19) dwyke
= (3.2 × 10–9 t) dwyke ijUrq Q = (40 R)V
102 2
= 9 109
dwyke
102 2 9 109
= 3.2 × 10–9 t
;k t = 700 × 10–6 sec. t = 700 sec.
LONG SUBJECTIVE
fucU/k Rediz'u
23.46 (a) magnitude of Q value of the reaction is x/10 mev. then x is fuEuvfHkfØ;k dsfy, Qdkeku x/10mev.
gksrksxKkrdjks [MB_Q.4.4]
N14 +
O17 + p
The masses of N14, He4, H1, O17 are respectively N14,He4,H1,O17 ds nzO;eku Øe'k%
14.00307 u, 4.00260 u, 1.00783 u and 16.99913 u.
14.00307 u, 4.00260 u, 1.00783 u rFkk 16.99913 u
(b) the total kinetic energy of the products if the striking particle has the minimum kinetic energy required to initiate the reaction is x/100 mev. then x is.
;fnd.kdhU;wurexfrtÅtkZvfHkfØ;kdksçkjEHkdjusdsfy,vko';dÅtkZdscjkcjgSrksmRikndhdqyxfrt ÅtkZx/100mev. gksrksx KkrdjksA
Ans. Q value eku = – 1·20 Mev, K· ε · = 0.34 Mev
Sol. N14 + O17 + proton
Q value = (14.00307 + 4.00260 – 1.00783 – 16.99913) 931.5 = – 1.20 MeV
Let m and M be mass of particle and nitrogen nucleus respectively and let minimum KE of particle be 1
2
mu2.
N14 + O17 + proton
Q eku = (14.00307 + 4.00260 – 1.00783 – 16.99913) 931.5 = – 1.20 MeV
ekukmrFkkM,d.koukbVªkstuukfHkddsnzO;ekugSrFk d.kdhU;wurexfrtÅtkZ
From energy equation ÅtkZlehdj.kls
1 mu2 = |Q| + minimum KE of system
2
1 mu2 =|Q|+ fudk; dh U;wure xfrt ÅtkZ
1 mu2 gSA
1 mu 2
= |Q| + 2 (m + M) (m M)
1 M
mu2 m M = |Q|
2
1
m M
or ;k
mu2 = |Q|
2 M
1 mu 2 1
m
KE of products mRikn dh xfrt ÅtkZ =
(m + M)
(m M) =
mu2
m M
= |Q|
2 2
m M
4
= 1.2 × 14 = 0.34 MeV Ans.
23.47 A sample of hydrogen atom gas contains 100 atoms. All the atoms are excited to the same nth excited
4800
state. The total energy released by all the atoms is 49 Rch (where Rch = 13.6 eV), as they come to
the ground state through various types of transitions. Find
(i) maximum energy of the emitted photon is x/49 Rch. then x is
(ii) n
(iii) maximum total number of photons that can be emitted by this sample.
,d gkbMªkstu ijek.kq xSl ds 100 ijek.kq ds uewus esa lHkh ijek.kq noh
4800
mRrsftr voLFkk esa gSA lHkh ijek.kq }kjk dqy mRlftZr ÅtkZ 49 Rch
(tgk¡ Rch = 13.6 eV), tc os fofHkUu laØe.k }kjk ewy d{kk okyh fLFkfr esa vkrs gS rks crkvksa
(i) mRlftZrQksVksudhvf/kdreÅtkZ x/49Rchgksrksx gksxkA
(ii) n
(iii) QksVksudhvf/kdrela[;ktksbluewus}kjkmRlftZrfd;stkldrsgSA
4800 Rch
Sol. (i) maximum energy of emitted photon = 49 =
100
48
49 Rch
1 1
(ii) Energy released if electron jumps from level n to level 1 = Rch 12 n2
1 1 48
Rch 12 n2 = 49 Rch
n = 7 then n excited state = n´ – 1
n = 6
(iii) Each atom can emit a maximum of 6 photons
there are 100 atoms, maximum number of photons that can be emitted = 600.
48
[Ans. (i) 49 Rch, (ii) n = 6, (iii) 600].
4800 Rch
Sol. (i) mRlftZr QksVkWu dh vf/kdre ÅtkZ = 49 = 48 Rch
100 49
1 1
(ii) ;fn bysDVªkWu nLrj ls ,d Lrj ij dwnrk gS rc mRlftZr ÅtkZ = Rch 12 n2
1 1 48
Rch 12 n2 = 49 Rch
n = 7 rc mRlftZr n mRlftZr voLFkk = n´– 1
n = 6
(iii) izR;sd ijek.kq vf/kdre 6 QksVWku fudky ldrs gSA
;gk¡100ijek.kq gSAvr% QksVkWu dh vf/kdrela[;k tksfudy ldrsgS= 600.
48
[Ans. (i) 49 Rch, (ii) n = 6, (iii) 600].
23.48 A positronium consists of an electron and a positron revolving about their common centre of mass. if calculate d is x/100 A° then x is
,dikWthVªksfu;eesa,dbysDVªkWu,oa,dikWthVªkuviusmHk;fu"BnzO;ekudsUnzdspkjksavksjpDdjyxkjgsgSA lEcU/k LFkkfir ,oa x.kuk dhft,& [6]
(i) Separation between the electron and positron in their first excited state.is x/100 A° then x is
izFkemRrsftrvoLFk esabysDVªku,oaikWftVªkWudschpnwjhA
(ii) Kinetic energy of the electron in ground state.is y/10 eV then y is
ewyÅtkZLrjesabysDVªkudhxfrtÅtkZA
Ke2
Sol. (i) r 2
= m2 r
2
r r
.....(1)
nh
and rFkkm
2 2
× 2 = 2
...(2)
By (1) and (2) r =
n2 h2
2 2 =
n2 h2
2
2 × 2 = (0.529 Å) × 2 n2
2 mK e
4 mK e
(1) rFk (2)}kjk r =
n2 h2
2 2 =
n2 h2
2
2 × 2 = (0.529 Å) × 2 n2
2 mK e
For first excited state n = 2
izFkemRrsftrvoLFkkdsfy,n=2
r = 8 × 0.529 Å r = 4.232 Å
4 mK e
1 r 2
1 m.2r 2
Ke2
Ke2
13.6 eV
(ii) KE = 2 m 2
= 2 .
= = 4 (0.529Å 2) = 4
= 3.4 eV
Ans. (i) 423 (ii) 34
4 4r
Comments
Post a Comment