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Modern Physics/Sol. Target IIT-JEE/AIIMS 2018

Page # 1

CONTANT PAGE NO.

EXERCISE - 1 : Single Correct 2

EXERCISE - 2 : Multiple Correct 17

EXERCISE - 3 : NEET/AIIMS Special 21

EXERCISE - 4 : Miscellaneous 22

MODERN PHYSICSSOLUTION

Physics for IIT-JEE/AIIMS

By Shiv R. Goel (B.Tech., IIT-Delhi)

Physics Secrets by Shiv R. Goel(meant only for Waves classroom students)


Page # 2

EXERCISE #1 Only One Option is Correct

[SINGLE CORRECT CHOICE TYPE]

1 Pr = Pb

r

r hct

n =

b

b hct

n

b

r

b

r

nn

> 1

2/hc

Pe = i

e)12400(10 3–

(5000)(e) = 0.16 × 10–6

=50

1241016.0 3– = 0.4 × 10–3 = 0.04%

3 K =hc

–

K' = 4/3hc

– =

hc

34

– =hc

34

– 34

+ 3

= 34

–hc + 3

= 3

4K + 3

4 Vs = e–h

V's = e–)2(h

= e)–h(2

= 2VS + e

5 KEmax = eV 4 eV V= 4

6 10.4 + 1.7 = E = 12.1

7 One photon corresponds to one electron

8 KEmax = h – KEmax = h(2) – = 2 (h – ) +

/hcIA

dtdv

=h

IA dt

'dv = )2(h

A)I2( = dt

dv

is = is


Page # 3

9 Vs = e–h

= constant

10 i =

dtdv

e =

AR4P

2 ×h1

× × e

i 2R1

11 = h f0 hf1 = + KEmax KEmax = h1 – = h (1 – 0)

12

–5–2

KEKE

2

1

22

21

mv21

mv21

=41

2

1

vv

= 21

13 V = e

hc

=e

hc –

e

V2 – V1 = ehc

12

1–1

14 5 = + 2 = 3eV6 = f + KEmax KEmax = 3eV eVs = 3eV Vs = – 3V

15 (KEmax)emitted =

–hc = 200

1240 – 4.5 = 1.7 eV

(KEmax)reached = 1.7 + 2 = 3.7 eV

16 KEmax = hv –

dtdn

=

h

IA

17 hf – = eV (f0)y > (f0)x

18 Photo electric current depends only upon total no. of photons incident. It is independent the area over


Page # 4

which the photons are incident changing the fecal length results in changes in are only

19 i = ehIA

i' =

)2(hIA

e i' < i

20 i = ehIA

Increase in either of intensity or area will result in inerease in photo current

21 F = cIA2

F' =ccosIA2 2

= F cos2

22 = ph

= mK2h

= mqV2h

p

e

=e

e

mm

e > p

23 KE = 100 + 50 = 150 ev

= )KE(m2h

= 19–31–

34–

106.1150101.92106.6

= 1Aº

24 p =h

= 12–10h

= 1012 h

25 K1 =1

hc

– =22

hc

– K2 =2

hc –

K1 = 2K2

– = 2K 2 – 2

26 = d

= d

)ev(m2h

if &


Page # 5

27 mvr =2

h &

rmv2

=04

1 2r

q)q3(

mv2r = )q3(4

1 2

0

v

2

h =

0

2

4q3 v =

0

2

h2q3

28 n = 2r

nr n

J n J

29 For the anode voltange to be retarding and greater than Vs, more likely it is to make the photo currentzero.

30 mvr =2

nh =

2h3

n = 3

KE = – TE = 13.6 × 2

2

nz

= 13.6 × 2

2

)3()1(

= 1.51 ev

31

22

21 n

1–n11

A

1 = R

22A

1–411

A = R16

B

1 = R

22 31–

21

B = R536

= R2.7

C

1 = R

22 41–

21

C = R316

= R3.5

D

1 = R

22 41–

31

D = R7144

= R6.20

32 En = 21

mv2 ....(i)


Page # 6

mvr =2

nh....(ii)

equation )ii()i(

2

= nhE2 n

f =2 = nh

E2 n

33 E = 20

nE

E = 2

18–

)3(1018.2

= 2.42×10–19 J

34 E4n – E2n = E02

22 z)n4(

1–)n2(

1

E2n – En = E02

22 z)n2(

1–n1

nn2

n2n4

E–EE–E

=4nz 00

35 21–n C = 10 n = 6

36 r = r0 Zn 2

rn+1 – rn = rn–1 (n + 1)2 – n2 = (n + 1)2

(n + 2n + 1) – n2 = (n2 – 2n + 1) 4n = n2 n = 4

37 62

)1–n(n 22 n2 = 4

n1 = 2n1 = 1 More energy photon is not possiblen1 = 3 Less energy photon is not possible

38 ac = rv2

=

r

mvm1 2

= 2

2

rkze

m1

= 420

22

Arz

mkze


Page # 7

ac 4n1

Lc

mc

)a()a(

=4

m

L

nn

=

4

32

= 8116

39r

mv2

= 2

2

rkze

...(i)

mvr =2

nh...(ii)

equation )ii()i(

=nhr

)2)(kze( 2 = 2f

f =nh

kze2

2onrz

f 3n1

2

1

ff

=3

1

2

nn

= 27

1

1

2

nn

= 31

41 E = E0 2

2

nz

E = (13.6) 2

2

)2()1(

= 3.4

42 r = 22

22

mze4hn

e

muon

rr

=muon

e

mm

=e

e

m207m

rmuon = 207Å529.0

= 2.56×10–3 Å

43 2n C = 10 n = 5

12400 = 13.6

22 51–

11

= 6.13242512400

= 950Å = 95 nm


Page # 8

44 5 lines means (1 2, 1 3, 1 4, 1 5, 1 6) No. of lines = 6C2 = 15

45hc

= 13.6 × (1)2

22 51–

11

= 13.6 × 2524

eV

p =h

= 8

19–

3510

106.125246.13

= (1.67 × 10–27) v

v = 4.2 m/s

46 Largest wavelength of Balmer series [n = 3 n = 2] electron must be excited to n = 3 from n = 1 Eincident electron = En = 3 – Ex = 1 = 12.1 eV

471

n

= 3n1

log n

1

nlog

log

1

n = – 3log n

481

= Rz2

22

21 n

1–n1

2z1

49 r n2 1

2

nn

= 2

T n3 1

2

TT

= 18

50 n+1C2 = 2n)1n(

= 1 + 2 + 3 + ....... + n

51 PE = 2T E + u0 = 2 (–13.6 × 2

2

nz

) + u0

PEn = 2 = 2

)2(

)1(6.13–2

+ u0 = 0 u0 = 6.8 eV

TEn = 0 = –13.6 + u0 = 13.6 + 6.8 = – 6.8


Page # 9

52 2E

> 10.2 (for inelastic collision)

E > 20.4 eV

53 KE = – 2PE = – TE = 13.6× 2

2

nz

n KE PE TE

54 f 3n1

, r n2 , L n

frL n0

55 r = 0.529 ×z

n 2

= 0.529 × )2()2( 2

= 1.058

56 n = 2, z = 1

L =2

nh =

2h2

=h

57

K >

K Here Q > P

58 x Ray 1 Å to 100Å

59 Greater the frequency greater energy per photon

601

= R (57 – 1)2

22 21–

11

1

= R (29 – 1)2

22 21–

11

=2

2556

= 4

61 E3 – E1 = (E3 – E2) + (E2 – E1)

h

K = h

K + h

L

62 E =hc

= 021.012400

59 keV


Page # 10

63 After removeal of one electron it will follow Bohr's Model,

E = E1 + (13.6) 2

2

1)2(

= 24.6 + 54.4 = 79 eV..

64 c = evhc

'c = )v2(ehc

= 2c

K will remain contant as that is a material for

K – 'c = K – 2

c = cK –21

+2K

65 a.m.u is defined for carbon 12 only

66 A = 4R2 = 4(R0 A1/3)2 = 4 R0

2 A2/3

67 Q = BEA + BEB – BEx= (7.4) + (8.2) (110) + (8.2) (90) – (200)= (0.8) (200) = 160 MeV

68 C13 C12 + n0

Q = 13CBE – 12C

BE

= (7.5) (13) – (7.68) (12)= 97.5 – 92.16 = 5.34 MeV

69 2x Y Q = BEY – 2BEX = E2 – 2E1

70

dtdn

(200 × 106 × 1.6 × 10–19) = 1 × 1000

dtdn

=6.1200

1010 133

= 3.125 × 1013

71 Let E be energy liberated per deuteron

1040 E = Pt t =P

E1040

E is of the order of MeV

72 3B A


Page # 11

e = EA – 3EB 3Eb = Ea – e

73 Binding energy per nucleon

He 428

= 7 MeV

Li 752

= 7.4 MeV

C 1290

= 7.5 MeV Most stable

N 1498

= 7 MeV

74 When two small nuclear combine to form a larger nuclei, the reaction is said to be fusion.

75 Maximum loss of kinetic energy happens particles are same moving comes to rest.

76 1H2 p + n

Q = 1876 – 939 – 940 = –3Mev

77 x –1e0 + y

Ejection of electron does not changes the mas number

78 88Ra226 82Pb206 + 5 2H4 + 4 –1e

0

79 r–r rays has the highest penetrating power then b rays & a-rays has the least.

80 Q = KE + KED

Q =m2

p2

+D

2

m2p

=m2

p2

Dmm1 = KE

Dmm1

50 = 48

Dmm1

Dmm

= 241

mD = 24 m = 96 amu

mp mD + m = 100 amu

81 92U238 82Pb206 + 8 2He4 + 6 –1e

0

82 When product has higher Binding energy than parent then it releases energy


Page # 12

83B

A

A

B

BB

AA

B

A

NN

TT

NN

AA

=

hr2hr4

2N2

N

0

0

= 1:2

84 N = N0 e–t/

at t = N = N0 (e–1)

0N

N = e

1

85 Any nucleus can decay at any time.

862

1

A

A

NN

= t–0

t10–0

0

0

eNeN

= t9– 0e

Now according to question e1e t9– 0

t =09

1

87 0.9 N0 = N0 e–t

t = ln

910

After time 2t N = N0 e

–(2t) = N0 (2e–t) = N0 (0.9)2 = 0.81 N0 N0 – N = 0.19N0

88 2

1

t–0

t–0

eN2eN

=2

1

AA

)t–t( 12e =2

1

AA2

t2 – t1 =

)A/A2ln( 21 = 2ln)A/A2ln( 21 T

t1 – t2 = 2ln)A2/Aln( 12 T

89 R2 = N02t–e

R1 = N01t–e


Page # 13

)t–t(–

1

2 12eRR = )t–t( 21e

90B

A

)dt/dN(–)dt/dN(–

= 1 =BB

AA

NN

1 =

21

B

A

NN

NA = 2NB

91 Probability of survival of a nucleus upto time 't' =0N

N = e–t

Number of nuclei disintegrating in 5T1/2 to 6 T1/2

2/1T5N – N6 T1/2

= 32N0 – 64

N0 = 64N0

32/N64/N

0

0 = 2

1

92 p = N)N–N( 0 = (1– e–t) = 1 –

105

ln–2

e

= 1 – 41

= 43

93 9A

3A

33A

3AA 0hour100hour10hour1

0

94 N0 – N = N0= (1– e–t)

=

2310023.6

2235

5243651620

2ln–e–1

= 3.2 × 1015

95 x yt = 0 16% = N0 0t = t 2% = N 14%

N = 8N0 8

1 = n2

1 n = 3

t = 3T1/2 = 135 yrs.


Page # 14

961

2

AA

= 1

2

t–0

t–0

eAeA

)t–t(

1

2 21eAA

A2 = A1 ePower –

Tt–t 12

97 N0 – N = (1 – e–t)

f1 = 1 –1–

e = 1 – e1

f2 = 21

f1 > f2

98 dtdN

= 10 – 21

N

10

0 N–20dN

= t

0

dt21

– ln

2010

= 21

t t = 2 ln2

= (0.693) 2 = 1.386 sec

99 R1 = N01T–e

R2 = N02T–e

2

1

RR

= )T–T( 12e

Number of nuclei disntegrated in time T1 to time T2

= N0 1Te1 – N0 2Te1

= N0 12 T–T– e–e = N0

0

1

0

2

NR–

NR

=

12 R–R

= 2lnT)R–R( 12

100 If it is end product, it means it is a stable nuclei, means its disintegration constant is zero.


Page # 15

101 N1t = N1

t– 2e

–dt

dNmix = –

dtdN

dtdN t21 = )N( t22t11

= 1 N1t2 1e + 2 N2

t2 2e

102 dtdN

= – 1 N – 2 N = – (1 + 2) N = – eff N

eff = 1 + 2

103dt

dNy = Nx = N0 e–t

104 If is the element that is disitegrating and rate of decay is independent of chemical surrounding conditions.

105 1000 = 60401

N N = 24 × 105 Ans.

106

t

N

O

24×105

20×105

107 t =2000

103200 3 = 1600 sec. = 2T1/2

N = 4N0 =

4108

= 25 × 106

108 V = RKQ

1 = 3

19109

106.1t106.11025.6109

= 18 sec Ans.

109 For reaction to be feasible, m > 0

(1) 0)m2MM(m eFeCo


Page # 16

183712935399.56936296.56

= (8.97 × 104 – 1.09 × 10–3) = – ve not possible

(2) m = 0)MM( FeCo Possible


Page # 17

EXERCISE #2 One or More Than One Option is Correct

1. c = Ehc

= 2000012400

= 0.62 Å = 62 pm

< c > 62 pm

2. h – = eVsVs is independent of intensity

3.1

= RZ2

22

21 n

1–n1

x =1–

4/336/5

=1–

275

= 527

p =h

y = x1

=1–

527

= 275

E =hc

z = y =1–

527

= x1

= 275

4. KE = – TE = 3.4 eV.

= ph

= mK2h

= 19–31–

34–

106.14.3101.92106.6

= 6.6×10

5. 122.4 = 13.6× 2

2

)1(Z

Z = 3

E1 2 = 13.6 × (3)2

22 21–

11

= 13.6 × 9 × 43

= 91.8 eV..

125 = 122.4 + –eKE

–eKE = 2.6 eV

6. Some of the wavelengths corresponding to which excitation can take place will be absent in out goingradiation. Some of the radiations will be in infrared and some will be in visible as well.

7. Because hydrogen atom are in ground state therefore all absorption liner will be abserved in lyman seriesonly


Page # 18

8. Potential energy of ground state is taken as zero it means potential energy crrosponding to all other stateswill be positive [as w = (–) ive in going from ground state to any other state]

E = 21

mv2 – + 27.2 = – 21

rkZe2

+ 27.2

= – 13.6 2

2

n)1(

+ 27.2

E = 13.6

2n1–2 E as n (B) is wrong

E = 13.6

2n1–2 E0 = 13.6 (for n = 1)

9. rD = 0.529 ×2n

× µme = 0.529 ×

zn 2

×

D

De

mmm

rD = 0.529 ×z

n 2

×

D

e

mm1

rH = 0.529 ×z

n2

×

H

e

mm1 rH > rD

10. Loss of energy in in elastic collision = 2E

(when m1 = m2)

2Emin = 10.2 eV [First excitation energy]

Emin = 20.4 eV

11. E = 2.55 eV (This is the difference correspoinding to n = 4 and n = 2 for H atom)It mean s hydrogen atom must have been excited to n = 4. This means that second photon will becorrespoinding n = 2 to n = 1 which means a photon of energy 10.2 eV eV. Total energy = 10.2 + 2.55= 12.75 eVMinimum KE of colliding neutron = 2 (12.75) = 25.5 eV

12. Minimum wavelength will be corresponding to cut-off wavelength continuous spectrum will becorresponding to continuous x-ray & same prominent wavelengths will be corresponding to charactenisticx - Ray.

13.min

hc = eV

V = 2–1030.6612420

= 18.75 KV


Page # 19

14. Continuous x-Ray will be starting from some lower wavelength means more no. of photons will beconitted which increases the intensity

15. T µ n3

2

1

TT

=3

2

1

nn

= 8

2

1

nn

= 2

16. ZXA Z–2YA–4 + 2He4

ZXA Z–1Y1A + +1e

0

ZXA Z+1Y''A + –1e0

ZXA ZYA +

17.

BE/A

56A

Binding energy per nucler increases upto 56 and then decreases as shown in figure. Initially neutronproton ratio is constant = 1

18. M1 < 10 (mp + mn)M2 < 20 (mp + mN) M2 < 2M1 Binding energy per nucleon will be larger for higher mass number (It increases upto 56 and then decreases),thus mass defect will be larger for M2 compared to M1

19. In a mean life time, nearly 63% nuclei decays

4N0 = N0 e–(0.173)t

t = 173.04ln

= 173.0)693.0(2 8 years

20. A stable nucleus forms when its rest mass is less then rest masses of its nucleous therefore some energyget liberated as result of loss of mass. Fission means fragementation of a heavy nucleus.


Page # 20

21. F = – drd

negative slope in AB F = Positive RepulsionPositive slope of BC F = Negative Attrection

22. 7 N14 + 3Li7 + 2He4 + 4 1p' + 2 –1 e

0

7 N14 + 3 Li7 + 4 1 p

1 + 4 0 n1

23. Excess nucleous both proton & Netron should be ejected or particleExcess proton proton should be ejected (+ emission)Excess neutrous0 n

1 + ¾ –1e0 + 1 p

1 (– emission)

24. Q = KE + KE + KE

25. n p+ e– + (– emission) for face nucleus > 0p n0 + e+ + (+ emission) for face proton < 0

26. particle is positively charged which results in negative charge being acqured by daughter particlewhereas – decay means emission of negative charge (e–) which means positive charge being acquiredby the daughter


Page # 21

EXERCISE #3 NEET/AIIMS SPECIAL

1 Stopping potential is defined only when there is photo electric effect taking place. For frequency lessthan 0, no photoelectric effect takes place.

2 Different photo electrons have different kinetic energy depending upon how much energy is lost by theelectron before coming out of the surface of metal.

3 An electron interacts with one photon only. If the energy of photon is greater than work function, thenonly enission take place.

4 = ph

= mk2h

= meV2h

As me < mp e > p

5 Their linear momentum may be in different direction so their linear momentum magnitude will be same forsure but linear momentum may be different.

6 Characteristic X - ray wavelength depends only upon nature of the target metal only.

7 Any nuclei can disintegrate at any time half life means half of the nuclei present decays when the size ofsample is large.

8 Proton Repulsion makes it easy to remove proton than neutron.

9 Positive charge of-particle & proton makes then get repeted by the nucless

10 Remaining energy is relased an energy of nitrogen and energy of v .


Page # 22

EXERCISE #4 Miscellaneous

1 No. of photons per unit time = )/hc(P

No. of photoelectrons ejected per unit time = 91051

)/hc(P

Photo current = 919–

3–

1051

106.15400

1240010663

× 1.6 ×10–19

i = 5.76 × 10–11 A

2 e3300

12400

= eV0 +

e2200

12400

= e (2 V0) +

V0 = 660012400

(3 – 2) = 1.87 V

3 (a) Stopping potential is independent of intersity of light it is just a function of the wavelength of photons. V0 = 0.6

e 2r1

1

2

ii

= 22

21

rr

= 2

2

)6.0()2.0(

= 231

= 91

i2 = 9i1 = 9

18 = 2mA

4 (a)

2R4p

× r2 ×/hc

1 = dt

dn

dtdn

= 2

2

R4pr

× )/hc(1 = 2

211–

)1.0(4)105(10

× 19–106.1124099

16

5

(b) dtdn

= dtdn

× 2r1

= 165

×

100

1 ×

14.31

× (5×10–11)2 =80

1020

5 hs = = 2eV

h = 15

19–

1049.0106.12

= 6.53×10–34


Page # 23

6 As we know T =m2

p2

=2h

m21

T =m2

h2

T 1

A

B

TT

=2

B

A

=2

A

A

2

= 41

TB = 4TA & TB = TA – 1.5

4TA = TTA – 1.5 TA = 3

45.1 = 2eV

TB = 4TA = 0.5 eV

(a) EA = A + TA A = EA – TA = 4.25 – 2 = 2.25 eV B = EB – TB = 4.7 – 0.5 = 4.2 eV

7 p =h

= Ft = eEt

= eEth

dtd

= 2eEth–

8 = Diameter (D) = ph

K.E. =m2

p2

= m21·

Dh 2

J

= w eVemD2

h2

2

9 E = 13.6

61–

112 = 13.6 × 36

35 = 13.22 ev

mH vH = pphoton = CE

VH = 27–8

19–

106.1103106.122.13

= 4.41 m/s


Page # 24

10 13.6× 2

2

)1()2(

=hc

= E

13.6 × 4 =

1240

= 22.8 nm

11 Paschen series1

= 1.097 × 107

2n

191

106

91

– 2n1

97.101

2n1

97.10997.1

19710979

n2

n 1971097

7

n = 3 4 , 3 5, 3 6, 3 7 4 lines

123

hc =

21

hchc

[E3 = E1 + E2]

3

1 =

21

11

13 (i) 47.2 = 13.6 × Z2

22 31–

21

47.2 = 13.6 (Z2)

365

Z = 5

(ii) E = (13.6 Z2)

22 41–

31

= 13.6× 25 × 1697

= 16.5 eV

(iii) E = (13.6) Z2

221–

11

= 13.6 × 25 = 340 eV l = nm65.3340

1240

(iv) KE = 13.6 × 2

2

)1(Z

= 340 eV

PE = – 2KE = – 680 eV

L =2

nh =

2

h

(v) r = 0.529 ×Zn 2

= 0.529 × )5()1( 2

Å = 1.06×10–11 m


Page # 25

14 41

42

TT

= 16 T2 = 2T1

1122 TT 4500TT

2

112

Å

45124

450012400hc

2

eV

KEmax = 13.6

22 3

121

= 1.9 eV

=2

hc – KEmax = 45

124 – 1.9 = 0.88 eV

15 (EH)n = (EH)1

13.6× 2

2

n)3(

= 13.6 × 2

2

11

n = 3

r =

zn0529.0

2

Å

H

Li

rr

=2

H

Li

nn

×

Li

H

zz

=

31

13 2

= 13

= 3

16 K n = 2 to n = 1K n = 3 to n = 1L n = 3 to n = 2 E3 – E1 = (E3 – E2) + (E2 – E1)

KE =

LE +KE

hKf = h

Lf + hkf

hKf = h

Lf + hKf

Kf =Lf +

Kf

17 E = E1 + E2

20,000 = 3.112400

+

12400

1.61 = 3.11

+1


Page # 26

1.61 =

3.13.12

2

1.61 2 + 0.093 – 1.3 = 0 = 0.87 Å

1 = 87 pm2 = 217 pm

18 –

L

hc = –

K

hc +

K

hc

L

1 =

K

1–

K

1 = 2

1–11

= 21

L = 2Å

E =

L

hc = 2

12420 = 6210 eV

19 K =hc

– = 4001242

–1.9 = 1.2 eV

EHe = – 13.6 × 2

2

nZ

= – 13.6 × 2

2

)5()2(

= – 2.176 eV

Ephoton = K – HeE

= 1.2 + 2.176 = 3.376 eV (during combination)

E54 = 13.6 × (2)2

51–

41

= 13.6 × 4 × 25169

= 1.224 eV

E53 = 13.6 × (2)2

22 51–

31

= 13.6 × 4 × 25916

= 3.87 eV

E43 = 13.6 5 (2)2

321–

31

= 13.6 × 4 × 1697

= 2.64 eV

20 EL – EK =hc

= 2–103.2112400

= 58.2 keV

EL = – 11.3 keV

EK = EL – 58.2 = – 11.3 –58.2

EK = – 69.5 keV

eV = EK V = 69.5 kV


Page # 27

21 (i) KE = – TK = 3.4 eV

(ii) = ph

= mK2h

= 19–31–

34–

106.14.3101.92106.6

= 6.63 Å

22 min = Ehc

= 2000012400

= 0.62 Å

23 1H2 + 1H

2 2He4

E = 4 (BE/A) – 4 (BE/A)H = 4 (7 – 1.1) = 23.6 MeV

24 N = N 0 e–t

N0 – N = N0 (1 – e–t) 1 × 105 = N0 [1 – e–(36)]& 1.11 × 105 = N0 [1 – e–(108)]

1.11 =

36–

108–

e–1e–1

[Assume e–36 = x]

1.11 =x–1x–1 3

1.11 = )x–1()xx1)(x–1( 2

x2 + x – 0.11 = 0

x = )1(244.011–

= 0.1

e–36 = 0.1 e–36 = 0.1 36 = ln 10

36 )T(2ln

2/1 = ln 10

T1/2 = 36 10ln2ln

= 36 10log2log

= 10.8 sec

25 19K40 18Ar40 + 1e

0 +

N–NN

0 = 7

1

N = 8N0


Page # 28

N0 e–t = 8

N0

t = ln8

2/1T2ln

t = 3 ln 2

t = 3T1/2 = 4.2 × 109 years

26 R Unstable Nuclide 2/1T Stable

dtdN

= R – N

At equilibrium dtdN

= 0 R = N N =R

/R8.0

0 NRdN

= t

0

dt

–l

ln

R

R8.0R= t

ln 0.2 = – t = –2/1T2ln t

– ln 5 = t2ln–

t = 2ln5ln

27

Asun N

Mm

4Q

= Pt

1107.1 33

× 6.023×10234

)106.1()1026( 19–6 = (3.9 × 1026) t

t = 17109.34

023.66.1267.1

= 27.3 × 1017 secs

= 38

× 1018 sec


Page # 29

28 –eemm c2 =

hc

×2

2 × 0.5 = 21024.1 12–

= 2.48 pm

29 Q = BEHe – 4 (BE/A)Deutron BEHe = 23.6 + 4 (1.1) = 28 MeV

30 + + +

Q = k + E

From conservation of momentum we get

|p|

= |p| r

E = pc = p c

Q = k + (

mk2 ) × c [m = rest mas of +]

2)kQ( = 2mc2

k [mc2 = 100 MeV]

Q2 +2k – 2Q

k = 2(mc2) k [Q = 50 MeV]

2k – 300 k + 502 = 0

k =

2504300300 22 =

2100)223(

= 9 MeV ]

31 Nuclear Fusion Reaction

Q = [2 )m( 21 H

– )m( 42 He ]C2

= [2 (2.0141) – 4.0024](931MeV)Q = 24.02 MeV

32 (a) dtdN

= – N


Page # 30

N

N0N–

dN =

t

0

dt

–1

ln

0N–N–

= t

0N–

N–

= e–t

N =

(1– e–t) + N0 e–t

(b) NT/2 =0N2

(1 – e–t) + N0 e–t

= 2N0

2/12/1

TT

2ln–

e–1 + N0

2/12/1

TT

2ln–

e

= 2N0

211 + N0

21

2/1TN = 2

N3 0

N = 2N0 [1 – e–()] + N0 e–()

N = 2 N0

33 E54 = 13.6

22 5

141

(3)2 = 13.6 × 9 × 25169

= 2.754 eV

E43 = 13.6

22 4

131

(3)2 = 13.6 × 9 × 1697

= 5.95 eV

E43 = + e(Vs)43 5.95 = + 3.95 = 2eVE54 = + e(Vs)54 (Vs)54 = 2.754 – 2 = 0.754 eV Ans.

34 E1 = 414412420

= 3eV > capable of ejecting photoelectron

E2 = 497212420

= 2.5 eV > capable of ejecting photoelectron


Page # 31

E3 = 621612420

= 2 eV < non-capable of ejecting photoelectron

3I

A × t ×1E

1 +

A

3I

t ×2E

1 = n

n = 3IAt

21 E1

E1

=

3210106.3 43

5.21

31

× 19106.11

n = 2.4 × 1012 × 1511

× 6.11

= 1.1 × 1012 Ans.

35

)r(R4P

dtdN 2

2e ×

/hc1

×

= 2

3

)8.0(4102.3

× (8 × 10–3)2 × 19106.151

× 610

1

= 6192

63

1010510642106410

= 105 e/sec Ans.

(b) Ein =hc

= 5eV = 512400

= 2480Å

KEmax = Ein – = 2eV

e– = p

h = km2

h

e = 1931

34

106.12101.92

106.6

= 25

34

1063.7106.6

= 8.65Å

18.286e

Ans.

(c) This is because of positive charge acquired by the sphere due to the emission of photoelectron. Thussphere starts attracting the outgoing photoelectron and when this charge is sufficeint enough for the mostenergetic photoelectron can not leave the surface the sphere, then photo-electric emission stops.

maxKEer

kQ


Page # 32

(d)

Eer

tdt

dNk e

3

59

108t)10109(

(1.6 × 10–19) = 2 × 1.6 × 10–19

t = 2014

3

10161091016

= 91000

= 111 sec.

36m2

jpipV 21CM

CM/1P = m CM1 VV

= m 1V – CMVm

=

2jpipip 21

1 =2

jpip 21

2pp 2

221 = p1/CM

1/RM =CM/1ph

= 22

21 pp

h2

p1 = 1

h p2 = 2

h

=

22

22

21

2 hh

h2

=2

22

1

212

37 2

= 2.5 – 2 = 0.5Å

dmin = 2

= 0.5 Å

p =h

= 10–

34–

10106.6

= 6.6 × 10–24 kg m/s

KE =e

2

m2p

= )101.9(2)106.6(

31

224

= 2.39 × 10–17 J 151 eV


Page # 33

38 Ephoton = 13.6 × Z2

2

221 n

1n1

= 13.6 (2)2

22 2

111

= 13.6 × 3 = 40.8 eV

KE = Ephoton – EH = 40.8 – 13.6 = 27.2 eV

v =m

)KE(2 = 31

19

101.9106.12.272

= 3.1 × 106 m/s

40 wavelength of 1215Å corresponds to an energy of 10.2 eV, i.e. energy of 2®1 transition inhydrogen atom. Therefore,h – = 13.6 eV ...(i)

6h5

– = 10.2 eV ...(ii)

solving the above two equation we get = 6.8 eVand = 5 × 1015 Hz

41 68 = (13.6) z2 =

22 3

121

68 = 13.6 (z2)

365

z = 6 Ans.

KE1 = –TE1 = 13.6 × 2

2

nz

= 13.6 × 2

2

)1()6(

= 489.6 eV

1

hc = TE – TE1 = 0 – TE1

1 = 6.48912400

= 25.28Å Ans.

42 eVmin = 3 × 10–15 J

Vmin = 19

15

106.1103

= 18750V.. Ans.

(ii) E = EL – EK = –3 × 10–16 + 3 × 10–15 = 27 × 10–16 = 2.7 × 10–15 J Ans.

(iii)hc

= E = 15

834

107.2)103)(106.6(

= 7.3 × 10–11 = 0.73Å

(iv) E = (E – EL) + KE


Page # 34

KE = E + EL – EA = 2.7 × 10–15 – 3 × 10–17 = 2.67 × 10–15 J

43Hen

p0

N

He

p2

p1

p0 represents momentum of neutron before collisionp1 represents momentum of neutron after collisionp2 represents momentum of helium atom after collisionfrom momentum conservationp0 = p2 cos ...(i)p1 = p2 sin ...(ii)From (i) and (ii)p0

2 + p12 = p2

2 ...(iii)Let E be the energy of transition, then using energy equation

m2p2

0 =m2

p21 + )m4(2

p22 + E ...(iv)

Divide (iii) by 2m

m2p2

0 +m2

p21 =

m2p2

2

65 eV + E1 = E2 × 4 ...(v)From (iv)65eV = E1 + E2 + E ...(vi)(v) – (vi) × 4 gives65 – 4 × 65 + E1 = – 4E1 – 4 E 5E1 = 195 – 4E

E1 = 39 – 54

E

Now possible value of E are 40.8eVm, 48.36eV, 51eV etccorreponding values of E1 are 6.36, 0.312 eV, only (other are –ve)Allowed values of E2 are 17.84 eV, 16.32eV(ii) De-excitation may take place from 3 1, 2 1 and 3 2correponding frequencies are 11.6 × 1015 Hz, 9.845×1015Hz and 18.23 × 1014 Hz


Page # 35

44 * The energy of photon in the beam vary between 2001240

to 1001240

, i.e. 6.2 to 12.4 eV

* This beam can get electrons in H atom upto 2nd excited state i.e. n = 3 (for which excitation energy is12.1 eV). As a result, the emission spectrum will have 3 distinct wavelengths, with energy 10.2, 12.1 and1.9 eV.* For Beam A , Emax = 12.4 eV

KEmax = Emax – = 5 = 7.4 eV* For Beam B , Emax = 12.1 eV

KEmax = Emax – = 12.1 –7.4 4.7 eV

45 10.2 + 17 = 13.6 Z2

22 n1–

21

27.2 = 13.6 Z2

2n1–

41

2 = Z2

2n1–

41

...(i)

4.25 + 5.95 = 13.6 Z2

22 n1–

31

10.2 = 13.6 Z2

2n1–

91

2

2

n1–

91Z

43

...(ii)

)ii()i( 3

8 =

2

2

n44–n

9–n

n92

2

9–n4–n

2

2

= 2732

n =5

180 = 6

z = 3

46 T–0235

T–0238

235

238

eNNeNN

1140

= T)–( 238235e

474 m 222 m

226 m

v1 v2

E = )m4(2p2

+ )m222(2p2

=

22211

)m4(2p2

= E4m

111113

4.75

111113

= Mev87.4


Page # 36

48 T1/2 of Rn = 3.8 days

NoT1/2

3.8 days 2N0 T1/2

7.6 days 4N0

4N0 × 4000

1 = 1.2 × 106

N0 × 160001

×2/1T2ln

= 1.2 × 106

N0 = 693.06060248.316000102.1 6

= 9.1 × 1015 atoms

Mass =A

0

NN

MRn = 23

15

10023.6101.9

× 222 = 3.3×10–6 gm

49 N = dtdN

2/1T2ln

ANMm

= dtdN

230105.2

T063.0 3–

2/1

× 6.023×1023 = 8.4

T1/2 = 5.4×1017 sec = 6.25×1012 days T1/2 = 1.7 × 1010 years

50tdt

dN

= 320 disintegration per minute

0tdtdN

= 12 × 50 = 600 disintegration per minute

0t

t

)dt/dn()dt/dn(

=

0NN

= 600320

= 158

e–t = 158

t =

815ln =

1630ln = ln 3 + ln 10 – 4 ln 2

57302ln

t = ln 3 + ln 10 – 4 ln 2

t =

2log

2log410log3log 5730 = 5196 years


Page # 37

51 I T

vI

OP

v0

vP

Using COM –x direction mIvI = mPvP cosy direction m0v0 = mPvP sinmI

2 VI2 + m0

2 v02 = mp

2 vp2

2 mI KI + 2 m0 K0 = 2 mp Kp K0 =0

ppII

mkmKm–

Q = Kp + k0 – KI

= Kp – KI +0

ppII

mkmKm–

= Q = Kp

0

p

mm

1 – KI

0

I

mm1 Hence Proved

52 mv

v1

–dm m+dmv + dv

v1 + v + dv = u mv = (m + dm) (v + dv) – v1 × dm mv = mv + mdv + (dm) dv + (dm) u – (dm) v – (dm) (dv) mdv = – udm

dv = – u mdm

m = m0 e–t dm = – m0 e–t

dv = (–v)

t–0

t–0

emem–

= u

v = u t

53 N

dtdN

= – N

N

0 N–dN

= t

0

dt 1– ln

N–

= t

N–

= e–t


Page # 38

N =l (1 – e–t)

dt

dNdecay = N = (1 – et)

decayN

0decaydN =

t

0

t– dt)e–1(

Ndecay = at

0

t–et

=

1–et

t–

Ndecay E0

10020

= mS T

T = ms

1–et5E t–

0

=ms

)e–1(–tE2.0 t–0

54 Q = [mcm – (mpu + m)]C2

= [248.07222 – ( 244.0641 + 4.002603] × 931 mevQ = 5.14 mev

P =

dtdN

(0.08 Qfission + 0.92 Q)

= N [0.08 Qfission + 0.92 Q]

mT1

= )10(10

1 2013 [(0.08) (200) + (0.92) (5.14)]

= 107 [16 + 4.72] = 2.072 × 108 Mev/sec = (2.072 × 108) (106) (1.6 × 10–19) J/sec = 3.3 × 10–5 W

= 33 µ W

55 N0 = 1 µ Ci = (1 × 10–6) (3.7 × 1010) = 3.7 × 104 dps

N = N0 e–t = (3.7 × 104)

)5(15

2ln–e = (3.7 × 104) × 2–1/3

=26.1107.3 4

= 2.96 × 104

Vcm1

)60/296(3 = 2.96 × 104

V = 601002961096.2 4

cm3 = 6000 cm3 = 6L

56 T1/2 = 10 sec [T1/2 is the time taken to fall the no. of undecayed nucleus to half their original value]


Page # 39

T1/2 = 1

= 2lnT 2/1 = 693.0

10 = 14.43 sec

N = N0 e–t

0.0625 N0 = N0

tT

2ln–2/1e

ln 16 =2/1T2ln

t = 4 ln 2

t = 4 T1/2 = 40 sec

57 R = R0 A1/3

d =AR

34

106.130

27

= 3

32

R34

102

R3 = 6.12

× 1059 × 1.13 × 10–45

R3 = 6.120

× 1.13 × 1014

R = (1.25)1/3 × 1.1 × 105 = 5.5 × 104 mL = const.I11 = I22

52

m R12 ×

1T2

= 52

m R22 ×

2T2

T2 = 21

22

RR

× T1 =

2

8

4

105.5105.5

× 108 = 1 sec.

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