Problem book for the course Nuclear Physics. Educational edition

6.15. Based on the laws of conservation of charge, lepton and baryon numbers, determine whether the following reactions are possible:

90 Th emits an -particle and goes to the ground (unexcited) state of 228

88 Ra. Determine the kinetic energy of an -particle and nuclei. What portion of the total

energy is spent on the kinetic energy of the daughter nucleus? What is the recoil velocity of the daughter nucleus?

6.17. Resting nuclei of 192

84 Po decay (from the ground state) with the emission of three groups of particles: with energy1 = 6416 keV (0.7 %), 2 = 6611 keV (1.4 %), 3 = 7167 keV (97.3 %). Find the energy of -decays of these nuclei

and the energy of - quanta emitted by the daughter nuclei.

directly in the ground state. Find the amount of heat that 5 mg of the preparation will release in a time equal to 3 1/2.

6.19.Derive the formulae (6.7).

6.20.Determine, if these processes are possible: −-decay,

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6.21. A nucleus of 66 −-decay, the daughter 29Cu at rest has a

nucleus is in the ground state. Determine the maximum kinetic energy of an electron and its velocity. What are the kinetic energy of the daughter nucleus and its speed?

6.22. A nucleus of 66 −-decay, the daughter 29Cu at rest has a

nucleus is in the ground state. Determine kinetic energies of an electron and an antineutrino if the daughter nucleus has

nucleus is in the ground state. Determine the maximum kinetic energy of a positron and its velocity. What are the kinetic energy of the daughter nucleus and its speed in this case?

nucleus is in the ground state. Determine kinetic energies of a positron and a neutrino if the daughter nucleus has no recoil.

6.25. A part weighing 1 kg contains 0.005 mass percent of the stable silver isotope 109Ag. The part is placed in a neutron

flux of 2 · 1013 n/(cm2·s). Find the activity of the radioactive isotope of silver 110Ag 5 min after the start of irradiation, if

the cross-section of the radiation capture of the silver isotope 109Ag is 91 barn, and the half-life of the isotope 110Ag is 25 s.

6.26. A steel sample weighing 2 kg contains 0.5 % of manganese 55Mn, it was placed in a nuclear reactor with a

thermal flux of 5 · 1013 n/(cm2·s). As a result of radiation capture of neutrons a radioactive isotope of manganese 56Mn is

formed. The sample was removed after 10 days of irradiation. What number of -quanta will it emit per second after one

hour of the exposure? The neutron capture cross-section for 55Mn is 13.4 barn. The −-decay of 56Mn is accompanied by

the emission of the following -quanta: 0.85 MeV (in 99 % of decays), 1.81 MeV (26 %), 2.11 MeV (15 %).

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Answers

1.2.

939.6 MeV, 938.3 MeV, 0.511 MeV.

1.5.

3.70 GeV, 2.76 GeV, 3.58 GeV/ .

1.6.

= 2.97 · 108 m/s.

1.7.

+ = 4.1 MeV, = 29.6 MeV.

1.8.

− = + = 66.98 MeV.

1.9.

− = 246.3 MeV, ˜ = 246.8 MeV.

1.10.

− = 108.54 MeV, 0 = 110.58 MeV.

1.11.

60 .

1.12.

= 11 0.

1.13.

~ = 197.33 MeV · fm ≈ 200 MeV · fm.

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1.14.

= 1.59 · 10−5 s, = 3.45 · 10−8 s, = 3.33 · 10−8 s.

1.15.

¬ = 0.2 fm.

1.16.

1969 fm, 141 fm, 0.2 fm.

1.17.

20 GeV, 19 GeV.

1.18.

rel is the electron velocity by relativistic formulae, cl – by non-relativistic formulae:

a)rel = 4.16 · 107 m/s, cl = 4.19 · 107 m/s, 0.7 %,

b)rel = 1.24 · 108 m/s, cl = 1.33 · 108 m/s, 7.2 %,

c)rel = 2.22 · 108 m/s, cl = 2.97 · 108 m/s, 33.5 %,

d)rel = 2.97 · 108 m/s, cl = 10.3 · 108 m/s (> ), 246 %.

1.19.

rel is the proton velocity by relativistic formulae, cl – by non-relativistic formulae:

a)rel = 4.21 · 107 m/s, cl = 4.24 · 107 m/s,

cl/ rel = 1.007,

b)rel = 1.25 · 108 m/s, cl = 1.34 · 108 m/s,

cl/ rel = 1.07,

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c)rel = 2.60 · 108 m/s, cl = 4.24 · 108 m/s (> ),

cl/ rel = 1.6,

d)rel = 2.99 · 108 m/s, cl = 13.4 · 108 m/s (> ),

cl/ rel = 4.5.

1.20.

rel – by relativistic formulae, cl – by non-relativistic formulae:

a)rel = 4.69 · 10−2 MeV, cl = 4.69 · 10−2 MeV,rel/ cl = 1.00007,

b)rel = 4.73 MeV, cl = 4.69 MeV, rel/ cl = 1.008,

c)rel = 145 MeV, cl = 117 MeV, rel/ cl = 1.24,

d)rel = 1214 MeV, cl = 380 MeV, rel/ cl = 3.2.

1.21.

68.6 m/s, 80 m/s.

1.22.

= 1.32 MeV, = 0.9 MeV.

1.23.

√

The dimensionality method: = / ; the exact answer:

√

= 2 / .

1.24.

= 1/( 2).

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1.25.

√

= / 3, matches the exact formula.

1.26.

12 = ( 1 − 2)/(1 − 1 2/ 2) = /3.

1.27.

thres = ( 2 − 21 − 22) 2/2 2.

2.1.

≈ 1.45 · 1014 g/cm3.

2.2.

≈ 7 · 1018 C/cm3.

2.3.

≈ 9 · 1037 1/cm3.

2.4.

≈ 2 fm.

2.5.

a.e.m. 2 = 121 126 C 2 = 931.5 MeV.

2.6.

= ( − , − ) + ( , ) − ( , ), = − .

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2.7.

W is the binding energy in MeV, is the binding energy per

2.12.

14.6 MeV.

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2.13.

-0.09 MeV.

2.14.

7.27 MeV.

2.15.

6.5 MeV, 7.3 MeV.

2.16.

= 6.5 fm.

2.17.

sin( min) = 2 ¬ · 0.61/ , = 1,2,3, = 3.3 − 4.1 fm.

2.18.

1.9 fm.

2.19.

1.1 = 0.42 MeV. 2. 2 = 5.49 MeV. 3. 3 = 12.86 MeV.

tot = 24.68 MeV.

3.1.

opt ≈ /(0.015 2/3 + 1.973).

3.2.

+, −, −, +.

3.3.

10, 10, 50, 50, 52, 53, 54, 55.

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3.4.

45 46 47 48 49 50 51 52 53 54 55 Z 45 46 47 48 49 50 51 52 53 54 55 Z

3.5.

≈ 194 MeV.

3.6.

≈ 17.

3.7.

≈ 49.

3.8.

3.9.

32 2

= 5 · 4 0 , J.

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3.10.

3.15.

Ellipsoid flattened along the axis of symmetry , ellipsoid stretched along the axis of symmetry . -0.063 b and 0.92 b.

3.16.

0.09 and -0.15.

3.17.

( )max = (~ )/ 0 · (9 /8)1/3 = 214 MeV.

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