Page 1

THE NUCLEUS CHAPTER - 46 1.

M = Amp, f = M/V, mp = 1.007276 u 1/3 –15 1/3 –27 R = R0A = 1.1  10 A , u = 1.6605402  10 kg =

A  1.007276  1.6605402  10 27

4 / 3  3.14  R3 14 ‘f’ in CGS = Specific gravity = 3  10 . 2.

3.

4.

5.

6.

7.

8. 9.

= 0.300159  10

18

= 3  10

17

3

kg/m .

M M 4  1030 1 1 V    1013   1014 17 v f 0.6 6 2.4  10 3 V = 4/3 R . 1 1 3 1 3 3   1014 = 4/3   R  R =    1014 6 6 4  1 100 3  R =   1012  8 4 4  R = ½  10  3.17 = 1.585  10 m = 15 km. Let the mass of ‘’ particle be xu. ‘’ particle contains 2 protons and 2 neutrons. 2  Binding energy = (2  1.007825 u  1  1.00866 u – xu)C = 28.2 MeV (given).  x = 4.0016 u. 7 7 Li + p  l +  + E ; Li = 7.016u 4  = He = 4.0026u ; p = 1.007276 u 7 E = Li + P – 2 = (7.016 + 1.007276)u – (2  4.0026)u = 0.018076 u.  0.018076  931 = 16.828 = 16.83 MeV. 2 B = (Zmp + Nmn – M)C Z = 79 ; N = 118 ; mp = 1.007276u ; M = 196.96 u ; mn = 1.008665u 2 B = [(79  1.007276 + 118  1.008665)u – Mu]c = 198.597274  931 – 196.96  931 = 1524.302094 so, Binding Energy per nucleon = 1524.3 / 197 = 7.737. 238 4 234 a) U 2He + Th E = [Mu – (NHC + MTh)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev. 238 234 b) E = U – [Th + 2n0 + 2p1] = {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u = 0.024712u = 23.0068 = 23.007 MeV. 223 209 14 Ra = 223.018 u ; Pb = 208.981 u ; C = 14.003 u. 223 209 14 Ra  Pb + C 223 209 14 m = mass Ra – mass ( Pb + C)  = 223.018 – (208.981 + 14.003) = 0.034. Energy = M  u = 0.034  931 = 31.65 Me. 1 EZ.N.  EZ–1, N + P1  EZ.N.  EZ–1, N + 1H [As hydrogen has no neutrons but protons only] 2 E = (MZ–1, N + NH – MZ,N)c  f=

E2N = EZ,N–1 + 10 n . 2

2

Energy released = (Initial Mass of nucleus – Final mass of nucleus)c = (MZ.N–1 + M0 – MZN)c . 10. P32  S32 +

0v

0

 10

Energy of antineutrino and -particle = (31.974 – 31.972)u = 0.002 u = 0.002  931 = 1.862 MeV = 1.86. – 11. In  P + e We know : Half life = 0.6931 /  (Where  = decay constant). –4 Or  = 0.6931 / 1460 = 8.25  10 S [As half life = 14 min = 14  60 sec]. 2 2 Energy = [Mn – (MP + Me)]u = [(Mnu – Mpu) – Mpu]c = [0.00189u – 511 KeV/c ] 2 2 2 = [1293159 ev/c – 511000 ev/c ]c = 782159 eV = 782 Kev. 46.1


The Nucleus 12.

13.

226 58 Ra

 24   222 26 Rn

19 8 O

0 0  19 9 F  n  0 v

13 25 Al

25  12 MG  01e  00 v

64

Cu  64Ni  e  v

Emission of nutrino is along with a positron emission. a) Energy of positron = 0.650 MeV. Energy of Nutrino = 0.650 – KE of given position = 0.650 – 0.150 = 0.5 MeV = 500 Kev. b) Momentum of Nutrino = 14. a)

500  1.6  10 19 3  108

19 K

40

 20 Ca40  1e0  0 v 0

19 K

40

 18 Ar 40  1e0  0 v 0

19 K

40

 1e0  18 Ar 40

19 K

40

 20 Ca40  1e0  0 v 0 .

–22

 103 J = 2.67  10

kg m/s.

2

b) Q = [Mass of reactants – Mass of products]c 2 = [39.964u – 39.9626u] = [39.964u – 39.9626]uc = (39.964 – 39.9626) 931 Mev = 1.3034 Mev. 19 K

40

 18 Ar 40  1e0  0 v 0 2

Q = (39.9640 – 39.9624)uc = 1.4890 = 1.49 Mev. 19 K

40

0

 1e  18 Ar

40 2

Qvalue = (39.964 – 39.9624)uc . 15.

6 3 Li  n 8 3 Li

 73Li ; 73 Li  r  83Li

 84Be  e  v 

8 4Be

 42He  24He +

16. C  B +  + v mass of C = 11.014u ; mass of B = 11.0093u Energy liberated = (11.014 – 11.0093)u = 29.5127 Mev. For maximum K.E. of the positron energy of v may be assumed as 0.  Maximum K.E. of the positron is 29.5127 Mev. 17. Mass 238Th = 228.028726 u ; 224Ra = 224.020196 u ;  = 24He  4.00260u 238

224

Th  Ra* +  224 Ra*  Ra + v(217 Kev) 224 Now, Mass of Ra* = 224.020196  931 + 0.217 Mev = 208563.0195 Mev. 226Th 224 – E( Ra* + ) KE of  = E = 228.028726  931 – [208563.0195 + 4.00260  931] = 5.30383 Mev= 5.304 Mev. 12 12 + 18. N  C* + e + v 12 12 C*  C + v(4.43 Mev) 12 12 + Net reaction : N  C + e + v + v(4.43 Mev) + 12 12 Energy of (e + v) = N – (c + v) = 12.018613u – (12)u – 4.43 = 0.018613 u – 4.43 = 17.328 – 4.43 = 12.89 Mev. Maximum energy of electron (assuming 0 energy for v) = 12.89 Mev. 19. a) t1/2 = 0.693 /  [  Decay constant]  t1/2 = 3820 sec = 64 min. b) Average life = t1/2 / 0.693 = 92 min. –t c) 0.75 = 1 e  In 0.75 = – t  t = In 0.75 / –0.00018 = 1598.23 sec. 20. a) 198 grams of Ag contains  N0 atoms. 224

1 g of Ag contains  N0/198  1 g =

6  1023  1 10 6 atoms 198 46.2


The Nucleus Activity = N =

0.963 0.693  6  1017 N = disintegrations/day. t1/ 2 198  2.7

0.693  6  1017 0.693  6  1017 disintegration/sec = curie = 0.244 Curie. 198  2.7  3600  24 198  2.7  36  24  3.7  1010 A0 0.244  = 0.0405 = 0.040 Curie. b) A = 7 2t1/ 2 2 2.7 21. t1/2 = 8.0 days ; A0 = 20  Cl a) t = 4.0 days ;  = 0.693/8 =

= 20  10  e( 0.693 / 8)4 = 1.41  10 Ci = 14  Ci 0.693 –6 b)  = = 1.0026  10 . 8  24  3600 –18 –1 22.  = 4.9  10 s 1 1 1 238 a) Avg. life of U =    10 18 sec.  4.9  1018 4.9 3 = 6.47  10 years. 0.693 0.693 9  = 4.5  10 years. b) Half life of uranium =  4.9  1018 A A c) A = t / t0  0  2t / t1/ 2 = 22 = 4. A 2 1/ 2 23. A = 200, A0 = 500, t = 50 min –t –50 60 A = A0 e or 200 = 500  e    –4   = 3.05  10 s. 0.693 0.693 = 2272.13 sec = 38 min. b) t1/2 =   0.000305 5 24. A0 = 4  10 disintegration / sec 6 A = 1  10 dis/sec ; t = 20 hours. A A A = t / t0  2t / t1/ 2  0  2t / t1/ 2  4 A' 2 1/ 2 1/2  t / t1/ 2 = 2  t = t/2 = 20 hours / 2 = 10 hours. 

–t

A = A0e

A0 t / t1/ 2

–6

–5

4  106

6 3 = 0.00390625  10 = 3.9  10 dintegrations/sec. 2100 / 10 2 25. t1/2 = 1602 Y ; Ra = 226 g/mole ; Cl = 35.5 g/mole. 1 mole RaCl2 = 226 + 71 = 297 g 297g = 1 mole of Ra.

A =

 A  

1 0.1 6.023  1023 22 = 0.02027  10  0.1 mole of Ra = 297 297 –11  = 0.693 / t1/2 = 1.371  10 . –11 20 9 9 Activity = N = 1.371  10  2.027  10 = 2.779  10 = 2.8  10 disintegrations/second. 26. t1/2 = 10 hours, A0 = 1 ci 0.1 g =

0.693 –t

Activity after 9 hours = A0 e = 1 e 10 th No. of atoms left after 9 hour, A9 = N9  N9 =

9

= 0.5359 = 0.536 Ci.

A 9 0.536  10  3.7  1010  3600 10 13 = 28.6176  10  3600 = 103.023  10 .   0.693 –t

Activity after 10 hours = A0 e = 1 e th No. of atoms left after 10 hour A10 = N10

0.693 9 10

= 0.5 Ci.

46.3


The Nucleus

A10 0.5  3.7  1010  3600 10 13  = 26.37  10  3600 = 96.103  10 .  0.693 /10 13 13 No.of disintegrations = (103.023 – 96.103)  10 = 6.92  10 . 27. t1/2 = 14.3 days ; t = 30 days = 1 month As, the selling rate is decided by the activity, hence A0 = 800 disintegration/sec. –t [ = 0.693/14.3] We know, A = A0e A = 800  0.233669 = 186.935 = 187 rupees. 28. According to the question, the emission rate of  rays will drop to half when the + decays to half of its original amount. And for this the sample would take 270 days.  The required time is 270 days. + + 29. a) P  n + e + v Hence it is a  decay. b) Let the total no. of atoms be 100 N0. Carbon Boron 10 N0 Initially 90 N0 Finally 10 N0 90 N0  N10 =

0.693 –t

t

Now, 10 N0 = 90 N0 e  1/9 = e 20.3 [because t1/2 = 20.3 min] 1 0.693 2.1972  20.3  In  = 64.36 = 64 min. tt 9 20.3 0.693 23 30. N = 4  10 ; t1/2 = 12.3 years. dN 0.693 0.693 a) Activity =  n  N  4  1023 dis/year. dt t1/ 2 12.3 14

= 7.146  10 dis/sec. dN 14 b)  7.146  10 dt 14 17 19 No.of decays in next 10 hours = 7.146  10  10  36.. = 257.256  10 = 2.57  10 . 0.693 –t

23

6.16

23

= 2.82  10 = No.of atoms remained c) N = N0 e = 4  10  e 20.3 23 23  No. of atoms disintegrated = (4 – 2.82)  10 = 1.18  10 . 2 31. Counts received per cm = 50000 Counts/sec. 16 N = N3o of active nucleic = 6  10 2 Total counts radiated from the source = Total surface area  50000 counts/cm 4 4 9 = 4  3.14  1  10  5  10 = 6.28  10 Counts = dN/dt 1 cm2 dN We know,  N dt

6.28  109

1m

–7 –7 –1 = 1.0467  10 = 1.05  10 s . 6  1016 32. Half life period can be a single for all the process. It is the time taken for 1/2 of the uranium to convert to lead.

Or  =

No. of atoms of U

238

=

6  1023  2  103 12 20 =  10 20 = 0.05042  10 238 238

6  1023  0.6  10 3 3.6   1020 206 206 3.6   12 20  Initially total no. of uranium atoms =    10 = 0.06789  235 206  No. of atoms in Pb =

0.693

0.693 9

–t

N = N0 e  N = N0 e t / t1/ 2  0.05042 = 0.06789 e 4.4710 0.693t  0.05042   log    0.06789  4.47  109 9

 t = 1.92  10 years. 46.4


The Nucleus 33. A0 = 15.3 ; A = 12.3 ; t1/2 = 5730 year =

0.6931 0.6931 1  yr T1/ 2 5730

Let the time passed be t, We know A = A 0 et 

0.6931  t  12.3 = 15.3  e. 5730

 t = 1804.3 years. 34. The activity when the bottle was manufactured = A0 Activity after 8 years = A 0 e

0.693 8 12.5

Let the time of the mountaineering = t years from the present A = A0e

0.693 t 12.5

; A = Activity of the bottle found on the mountain.

A = (Activity of the bottle manufactured 8 years before)  1.5%  A0e

0.693 12.5

= A0e

0.693 8 12.5

 0.015

0.693 0.6938 t  In[0.015] 12.5 12.5  0.05544 t = 0.44352 + 4.1997  t = 83.75 years. 9 –1 35. a) Here we should take R0 at time is t0 = 30  10 s 

 30  109  i) In(R0/R1) = In   30  109  = 0  

30 25

 30  109 ii) In(R0/R2) = In  9  16  10

  = 0.63 

Count rate R(109 s–1) 20

 30  109 iii) In(R0/R3) = In  9  8  10

  = 1.35 

10

 30  10 iv) In(R0/R4) = In  9  3.8  10

  = 2.06 

9

15 5 25

 30  109  v) In(R0/R5) = In   2  109  = 2.7   –1 b)  The decay constant  = 0.028 min c)  The half life period = t1/2. 0.693 0.693 = 25 min.   0.028 9 9 36. Given : Half life period t1/2 = 1.30  10 year , A = 160 count/s = 1.30  10  365  86400 0.693  A = N  160 = N t1/ 2 

t1/2 =

 N=

160  1.30  365  86400  109 18 = 9.5  10 0.693 23

 6.023  10

No. of present in 40 grams.

6.023  1023 = 40 g  1 = 18

40 6.023  1023

40  9.5  1018

–4 = 6.309  10 = 0.00063. 6.023  1023  The relative abundance at 40 k in natural potassium = (2  0.00063  100)% = 0.12%.

 9.5  10

present in =

50

75

100

Time t (Minute)

46.5


The Nucleus 7

-1/2

37. a) P + e  n + v neutrino [a  4.95  10 s

; b  1]

b)

f = a(z – b)

c /  = 4.95  10 (79 – 1) = 4.95  10  78  C/ = (4.95  78)  10

7

 =

3  108 14903.2  1014

7

= 2  10

–5

–6

2

–4

 10 = 2  10

14

m = 20 pm.

dN dN R R= dt dt Given after time t >> t1/2, the number of active nuclei will become constant. i.e. (dN/dt)present = R = (dN/dt)decay  R = (dN/dt)decay  R = N [where,  = Radioactive decay constant, N = constant number] Rt1/ 2 0.693  R= (N)  Rt1/2 = 0.693 N  N = . t1/ 2 0.693

38. Given : Half life period = t1/2, Rate of radio active decay =

39. Let N0 = No. of radioactive particle present at time t = 0 N = No. of radio active particle present at time t. –t  N = N0 e [ - Radioactive decay constant] –t –t  The no.of particles decay = N0 – N = N0 – N0e = N0 (1 – e ) We know, A0 = N0 ; R = N0 ; N0 = R/ From the above equation R –t (substituting the value of N0) N = N0 (1 – e ) = (1  e t )  23 40. n = 1 mole = 6  10 atoms, t1/2 = 14.3 days t = 70 hours, dN/dt in root after time t = N 0.69370 –t

23

23

23

= 6  10  e 14.324 = 6  10  0.868 = 5.209  10 . 23 5.209  1023  0.693  0.010510 dis/hour. 14.324 3600 –6 23 17 = 2.9  10  10 dis/sec = 2.9  10 dis/sec.  1ci  Fraction of activity transmitted =    100%  2.9  1017 

N = No e

 1 3.7  108  –11    2.9  1011  100  % = 1.275  10 %.   41. V = 125 cm3 = 0.125 L, P = 500 K pa = 5 atm. 8 T = 300 K, t1/2 = 12.3 years = 3.82  10 sec. Activity =   N 5  0.125 23 22 N = n  6.023  10 =  6.023  1023 = 1.5  10 atoms. 8.2  10 2  3  102 0.693 –8 –9 –1 = = 0.1814  10 = 1.81  10 s 3.82  108 –9 22 3 Activity = N = 1.81  10  1.5  10 = 2.7  10 disintegration/sec = 42.

2.7  1013 3.7  1010

Ci = 729 Ci.

212 83 Bi

208  81 Ti  24He( )

212 83 Bi

212 212  84 Bi  84 P0  e 

t1/2 = 1 h. Time elapsed = 1 hour 212 Present = 1 g at t = 0 Bi 212  at t = 1 Bi Present = 0.5 g Probability -decay and -decay are in ratio 7/13.  Tl remained = 0.175 g  P0 remained = 0.325 g 46.6


The Nucleus 108

110

8

43. Activities of sample containing Ag and Ag isotopes = 8.0  10 disintegration/sec. 8 a) Here we take A = 8  10 dis./sec  i) In (A1/ A 01 ) = In (11.794/8) = 0.389 12

ii) In (A2/ A 02 ) = In(9.1680/8) = 0.1362

10

iii) In (A3/ A 03 ) = In(7.4492/8) = –0.072

8

iv) In (A4/ A 04 ) = In(6.2684/8) = –0.244 v) In(5.4115/8) = –0.391 vi) In(3.0828/8) = –0.954 vii) In(1.8899/8) = –1.443 viii) In(1.167/8) = –1.93 ix) In(0.7212/8) = –2.406 b) The half life of 110 Ag from this part of the plot is 24.4 s. 110 c) Half life of Ag = 24.4 s.  decay constant  = 0.693/24.4 = 0.0284  t = 50 sec, –t 8 –0.028450 8 The activity A = A0e = 8  10  e = 1.93  10 d)

6 4 2

20 40 60 80 100 200 300

2 4

400

500

Time

6 4 2 O

20 40 60 80 108

e) The half life period of Ag from the graph is 144 s. 44. t1/2 = 24 h tt 24  6  t1/2 = 1 2  = 4.8 h. t1  t 2 24  6 A0 = 6 rci ; A = 3 rci A 6 rci t  A = t / t0  3 rci = t / 4.8h  = 2  t = 4.8 h. 24.8h 2 2 1/ 2 45. Q = qe  t / CR ; A = A0e

–t

Energy 1q2  e 2t / cR  Activity 2 CA 0 e t Since the term is independent of time, so their coefficients can be equated, 2t 2 1 2  = t or,  = or,  or, R = 2 (Proved) So,  CR CR CR C 46. R = 100  ; L = 100 mH After time t, i = i0 (1  e t / Lr )

i i0 (1  e  tR / L )  N N0 et

–t

N = N0 (e )

i/N is constant i.e. independent of time.

Coefficients of t are equal –R/L = –  R/L = 0.693/t1/2 –3 –4 = t1/2 = 0.693  10 = 6.93  10 sec. 235 23 47. 1 g of ‘I’ contain 0.007 g U So, 235 g contains 6.023  10 atoms. So, 0.7 g contains

6.023  1023  0.007 atom 235

1 atom given 200 Mev. So, 0.7 g contains

6.023  10 23  0.007  200  106  1.6  10 19 –8 J = 5.74 10 J. 235

48. Let n atoms disintegrate per second 6 –19 Total energy emitted/sec = (n  200  10  1.6  10 ) J = Power 6 300 MW = 300  10 Watt = Power 46.7


The Nucleus 6

6

300  10 = n  200  10  1.6  10 3 3  n=  1019 =  1019 2  1.6 3.2 6  10

23

–19

atoms are present in 238 grams

3 238  3  1019 –4 = 3.7  10 g = 3.7 mg.  1019 atoms are present in 3.2 6  1023  3.2 8 49. a) Energy radiated per fission = 2  10 ev 8 7 –12 Usable energy = 2  10  25/100 = 5  10 ev = 5  1.6  10 8 8 Total energy needed = 300  10 = 3  10 J/s

3  108

20 = 0.375  10 5  1.6  1012 20 24 No. of fission per day = 0.375  10  3600  24 = 3.24  10 fissions. 24 b) From ‘a’ No. of atoms disintegrated per day = 3.24  10 23 We have, 6.023  10 atoms for 235 g 235 for 3.24  1024 atom =  3.24  1024 g = 1264 g/day = 1.264 kg/day. 23 6.023  10

No. of fission per second =

50. a)

2 2 1 H  1H

 13H  11H

Q value = 2M(12 H) = [M(13 H)  M(13 H)] = [2  2.014102 – (3.016049 + 1.007825)]u = 4.0275 Mev = 4.05 Mev. b)

2 2 1 H  1H

 32H  n

Q value = 2[M(12 H)  M(32 He)  Mn ] = [2  2.014102 – (3.016049 + 1.008665)]u = 3.26 Mev = 3.25 Mev. c)

2 3 1 H  1H

 24H  n

Q value = [M(12 H)  M(13 He)  M( 24 He)  Mn ] = (2.014102 + 3.016049) – (4.002603 + 1.008665)]u = 17.58 Mev = 17.57 Mev.

Kq1q2 9  109  (2  1.6  1019 )2 = r r –23 1.5 KT = 1.5  1.38  10  T

51. PE =

Equating (1) and (2) 1.5  1.38  10

–23

…(1) …(2)

T=

9  109  10.24  10 38 2  10 15

9  109  10.24  10 38

9 10 = 22.26087  10 K = 2.23  10 K. 2  10 15  1.5  1.38  10 23 4 4 8  Be 52. H + H 2  4.0026 u M( H) 8  8.0053 u M( Be) 2 8 Q value = [2 M( H) – M( Be)] = (2  4.0026 – 8.0053) u = –0.0001 u = –0.0931 Mev = –93.1 Kev. 23 53. In 18 g of N0 of molecule = 6.023  10

 T=

6.023  1026 25 = 3.346  10 18 26  % of Deuterium = 3.346  10  99.985 25 Energy of Deuterium = 30.4486  10 = (4.028204 – 3.016044)  93 5 = 942.32 ev = 1507  10 J = 1507 mJ In 100 g of N0 of molecule =

 46.8

46.The Nucleus  

E = [M u – (N HC + M Th )]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev. b) E = U 238 – [Th 234 + 2n 0 + 2p 1 ] 0 0 0 1...