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 + 2n0 + 2p1] = {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
10
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 / 1460 = 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 01e 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 1018 4.9 3 = 6.47 10 years. 0.693 0.693 9 = 4.5 10 years. b) Half life of uranium = 4.9 1018 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. tt 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 2 103 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 4.4710 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 et
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 0.6938 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 0.69370 –t
23
23
23
= 6 10 e 14.324 = 6 10 0.868 = 5.209 10 . 23 5.209 1023 0.693 0.010510 dis/hour. 14.324 3600 –6 23 17 = 2.9 10 10 dis/sec = 2.9 10 dis/sec. 1ci 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 –0.028450 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 et
–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 1012 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 1019 )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
