Scilab Textbook Companion for Nuclear Physics by D. C. Tayal1 Created by Arjun Singh Nuclear and Particle Physics Physics Shri Mata Vaishno Devi University College Teacher Mr. Pankaj Biswas Cross-Checked by Chaya Ravindra May 6, 2014
1 Funded by a grant from the National Mission on Education through ICT, http://spoken-tutorial.org/NMEICT-Intro. This Textbook Companion and Scilab codes written in it can be downloaded from the ”Textbook Companion Project” section at the website http://scilab.in
Book Description Title: Nuclear Physics Author: D. C. Tayal Publisher: Himalaya Publishing House, Mumbai Edition: 5 Year: 2011 ISBN: 978-93-5024-743-3
1
Scilab numbering policy used in this document and the relation to the above book. Exa Example (Solved example) Eqn Equation (Particular equation of the above book) AP Appendix to Example(Scilab Code that is an Appednix to a particular
Example of the above book) For example, Exa 3.51 means solved example 3.51 of this book. Sec 2.3 means a scilab code whose theory is explained in Section 2.3 of the book.
2
Contents List of Scilab Codes
5
1 General Properties of Atomic Nucleus
11
2 Radioactivity and Isotopes
24
3 Interactions of Nuclear Radiations with Matt er
37
4 Detection and Meas urement of Nuclear Radiations
43
5 Alpha Particles
55
6 Beta Decay
61
7 Gamma Radiation
74
8 Beta Decay
81
9 Nuclear Models
87
10 Nuclear Reactions
98
11 Particle Accelerators
110
12 Neutrons
121
13 Nuclear Fission and Fusion
130
15 Nuclear Fission Reactors
138
3
16 Chemical and Biological Effects of Radiation
146
18 Elementary Particles
148
4
List of Scilab Codes Exa 1.1 Exa 1.2 Exa 1.3 Exa 1.4 Exa 1.5 Exa 1.6 Exa 1.7 Exa 1.8 Exa 1.9 Exa 1.10 Exa 1.11 Exa 1.12 Exa 1.13 Exa 1.14 Exa 1.17 Exa 1.21 Exa 2.1 Exa 2.2 Exa 2.3 Exa 2.4 Exa 2.5 Exa 2.6 Exa 2.7
Distance of closest approach . . . . . . . . . . . . . . . 11 Nuclear Spin . . . . . . . . . . . . . . . . . . . . . . . 12 Kinetic energy and Coulomb energy for an electron confined within the nucleus . . . . . . . . . . . . . . . . . 13 Scattering of electron from target nucleus . . . . . . . 13 Positron emission from Cl33 decays . . . . . . . . . . . 14 Charge accelerated in mass spectrometer . . . . . . . . 15 Ionized atoms in Bainbridge mass spectrograph . . . . 15 Calculating the mass of hydrogen . . . . . . . . . . . . 16 Silver ions in Smith mass spectrometer . . . . . . . . . 16 Calculation of energy released during nuclear fusion reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Binding energies calculation . . . . . . . . . . . . . . . 18 Calculation of energy released during nuclear fusion reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 To find the stable Isobar . . . . . . . . . . . . . . . . . 20 To calculate the pairing energy term . . . . . . . . . . 21 Relative error in the el ectric potential at the first Boh r radius . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Spherical symmetry of Gadolinium nucleus . . . . . . 22 Weight of one Curie and one Rutherford of RaB . . . 24 Induced radioactivity of sodium by neutron bombardment . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Activity of K40 in man of weight 100 Kg . . . . . . . 25 Age of an ancient wooden boat . . . . . . . . . . . . . 26 Activity of the U234 . . . . . . . . . . . . . . . . . . 26 Number of alpha decays in Th232 . . . . . . . . . . . 27 Maximum possible age of the earth crust . . . . . . . . 28 5
Exa 2.8 Exa 2.9 Exa 2.10 Exa 2.13 Exa 2.14 Exa 2.15
Number of radon half lives . . . . . . . . . . . . . . . 28 Weight and initial acivity of Po210 . . . . . . . . . . . 29 Radioactive disintegration of Bi . . . . . . . . . . . . . 29 Half life of Pu239 . . . . . . . . . . . . . . . . . . . . . 30 Disintegration rate of Au199 . . . . . . . . . . . . . . 31 Activity of Na24 . . . . . . . . . . . . . . . . . . . . . 32
Exa 2.16 Exa 2.18
Radiation dose absorbed in 24 h r by the tissue in REP 33 Activity and the maximum amount of Au198 produced in the foil of Au197 . . . . . . . . . . . . . . . . . . . 34 Pu238 as power source in space flights . . . . . . . . . 34 Series radioactive decay of parent isotope . . . . . . . 35 Alpha particle impinging on an aluminium foil . . . . 37 Thickness of beta absorption . . . . . . . . . . . . . . 38 Beta particles passing through lead . . . . . . . . . . . 38 Thickness of gamma absorption . . . . . . . . . . . . . 39 The energy of recoil electrons . . . . . . . . . . . . . . 39 Average energy of the positron . . . . . . . . . . . . . 40 To calculate the refractive index of the material . . . . 41 Minimum kinetic energy of the electron to emit Cerenkov radiation . . . . . . . . . . . . . . . . . . . . . . . . . 41 Resultant pulse height recorded in the fission chamber 43
Exa 2.19 Exa 2.20 Exa 3.1 Exa 3.4 Exa 3.7 Exa 3.8 Exa 3.9 Exa 3.10 Exa 3.11 Exa 3.12 Exa 4.1 Exa 4.3 4.2 Exa Exa 4.4 Exa 4.5 Exa 4.6 Exa 4.7 Exa 4.8 Exa 4.9 Exa 4.10 Exa 4.11 Exa 4.12 Exa 4.13 Exa 4.14 Exa 4.15 Exa 4.16
Energy of of the the voltage alpha particles 43 Height pulse . .. .. .. .. .. .. .. .. .. .. .. .. .. .. 44 Radial field and life time of Geiger Muller Counter . . 45 Avalanche voltage in Geiger Muller tube . . . . . . . 46 Voltage fluctuation in GM tube . . . . . . . . . . . . . 46 Time measurement of counts in GM counter . . . . . . 47 Capacitance of the silicon detector . . . . . . . . . . . 47 Statistical error on the measured ratio . . . . . . . . . 48 Charge collected at the anode of ph oto multiplier tube 49 Charge collected at the anode of ph oto multiplier tube 49 Measurement of the number of counts and determining standard deviation . . . . . . . . . . . . . . . . . . . . 50 Beta particle incident on the scintillator . . . . . . . . 51 Time of flight of proton in scintillation counter . . . . 51 Fractional error in rest mass of the particle with a Cerenkov Detector . . . . . . . . . . . . . . . . . . . . . . . . . 52 Charged particles passing through the Cerenkov detector 53 6
Exa 5.1 Exa 5.2 Exa 5.3 Exa 5.4 Exa 5.6 Exa 5.8 Exa 5.9 Exa 5.10 Exa 6.1 Exa 6.2 Exa 6.3 Exa 6.4 Exa 6.5 Exa 6.6 Exa 6.7 Exa 6.9 Exa 6.10 Exa 6.11 Exa 6.12 Exa 6.13 Exa 6.14 Exa 6.15 Exa 7.1 Exa 7.2 Exa 7.3 Exa 7.4 Exa 7.5 Exa 7.8 Exa 7.9
Disintegration energy of alpha particle . . . . . . . . 55 Calculation of the barrier height . . . . . . . . . . . . 56 Speed and BR value of alpha particles . . . . . . . . . 57 Transmission probability for an alpha particle through a potential barrier . . . . . . . . . . . . . . . . . . . . 57 Difference in life times of Polonium isotopes . . . . . . 58 Half life of plutonium . . . . . . . . . . . . . . . . . . 59 Slope of alpha decay energy versus atomic number . . 59 Degree of hindrance for alpha particle from U238 . . . 60 Disintegration of the beta particles by Bi210 . . . . . . 61 Beta particle placed in the magnetic field . . . . . . . 62 K conversion . . . . . . . . . . . . . . . . . . . . . . . 62 Average energy carried away by neutrino during beta decay process . . . . . . . . . . . . . . . . . . . . . . . 63 Maximum energy available to the electrons in the beta decay of Na24 . . . . . . . . . . . . . . . . . . . . . . 64 Linear momenta of particles during beta decay process 64 Energies during disintergation of Bi210 . . . . . . . . . 65 The unstable nucleus in the nuclide pair . . . . . . . . 66 Half life of tritium . . . . . . . . . . . . . . . . . . . . 67 Degree of forbiddenness of transition . . . . . . . . . . 68 Coupling constant beta transitons . . and . . . .ratio . . . .of. .coupl . . . .ing . . .strengths . .. 69for Relative capture rate in holmium for 3p to 3s suble vels 70 Tritium isotope undergoing beta decay . . . . . . . . . 70 Fermi and Gamow Teller selection rule for allowed beta transitions . . . . . . . . . . . . . . . . . . . . . . . . 71 Bragg reflection for first order in a ben t crystal spectrometer . . . . . . . . . . . . . . . . . . . . . . . . . 74 Energy of the gamma rays from magnetic spectrograph data . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Attenuation of be am of X ra ys in p assing through human tissue . . . . . . . . . . . . . . . . . . . . . . . . 75 Partial half life for gamma emission of Hg195 isomer . 76 Estimating the gamma width from Weisskopf model . 76 K electronic states in indium . . . . . . . . . . . . . . 77 Radioactive lifetime of the lowest energy electric dipole transition for F17 . . . . . . . . . . . . . . . . . . . . 78 7
Exa 7.10 Exa 7.13 Exa 7.14 Exa 8.3 Exa 8.5 Exa 6.8 Exa 8.8 Exa 8.9 Exa 8.10 Exa 9.1 Exa 9.3 Exa 9.4 Exa 9.5 Exa 9.7 Exa 9.9 Exa 9.11
Electric and magnetic multipolarities of gamma rays from transition between Pb levels . . . . . . . . . . . . 79 Relative source absorber velocity required to obtain resonance absorption . . . . . . . . . . . . . . . . . . . . 79 Estimating the frequency shift of a photon . . . . . . . 80 Neutron and proton interacting within the deuteron . 81 Total cross section for np scattering at neutron energy 82 Beta decayed particle emission of Li8 . . . . . . . . . . 82 Possible angular momentum states for the deuterons in an LS coupling scheme . . . . . . . . . . . . . . . . . . 83 States of a two neutron system with given total angular momentum . . . . . . . . . . . . . . . . . . . . . . . . 84 Kinetic energy of the two interacting nucleons in different frames . . . . . . . . . . . . . . . . . . . . . . . . 85 Estimating the Fermi energies for neutrons and protons 87 General propeties of a neutron star . . . . . . . . . . . 88 Stability of the isobar using the liquid drop model . . 88 Energy difference between neutron shells . . . . . . . . 89 Angular frequency of the nuclei . . . . . . . . . . . . . 90 Angular momenta and parities . . . . . . . . . . . . . 91 Quadrupole and mag netic moment of ground state of
nuclidesenergy . . . . .of. iron . . . .nucleus . . . . . .. .. .. .. .. . . .. . . .. . .... . . 92 95 Kinetic Electric quadrupole moment of scandium . . . . . . . 96 Energy of lowest lying tungsten states . . . . . . . . . 96 Q value for the formation of P30 in the gr ound state . 98 Q value of the re action and atomic mass of th e residual nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . 98 Exa 10.3 Kinetic energy of the neut rons emitted at given angle to the incident beam . . . . . . . . . . . . . . . . . . . 99 Exa 10.4 Estimating the temperature of nuclear fusion reaction 100 Exa 10.5 Excitation energy of the compound nucleus . . . . . . 101 Exa 10.6 Excitation energy and parity for compound nucleus . . 102 Exa 10.7 Cross section for neutron induced fission . . . . . . . . 102 Exa 10.8 Irradiance of neutron beam with the thin sheet of Co59 103 Exa 10.9 Bombardment of protons on Fe54 target . . . . . . . . 104 Exa 10.10 Fractional attenuation of neutron beam on passing through nickel sheet . . . . . . . . . . . . . . . . . . . . . . . . 105 Exa 9.12 Exa 9.14 Exa 9.16 Exa 10.1 Exa 10.2
8
Exa 10.11 Scattering contribution to the resonan ce . . . . . . . . 106 Exa 10.12 Estimating the relative probabilities interactions in the indium . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Exa 10.13 Peak cross section during neutron capture . . . . . . . 107 Exa 10.14 Angle at which differential cross section is maximumat a givem l value . . . . . . . . . . . . . . . . . . . . . . 107 Exa 10.15 Estimating the angular momentum transfer . . . . . . 108 Exa 11.1 Optimum number of stages and ripple voltage in Cockcroft Walton accelerator . . . . . . . . . . . . . . . . . 110 Exa 11.2 Charging current and potential of an electrostatic generator . . . . . . . . . ...... ...... ...... 111 Exa 11.3 Linear proton accelerator . . . . . . . . . . . . . . . . 111 Exa 11.5 Energy and the freq uency of deuterons accelerated in cyclotron . . . . . . . . . . . . . . . . . . . . . . . . . 112 Exa 11.6 Protons extracted from a cyclotron . . . . . . . . . . . 113 Exa 11.7 Energy of the electrons in a betatron . . . . . . . . . . 114 Exa 11.8 Electrons accelerated into betatron . . . . . . . . . . . 115 Exa 11.9 Deuterons accelerated in synchrocyclotron . . . . . . . 116 Exa 11.10 Electrons accelerated in electron synchrotron . . . . . 116 Exa 11.11 Kinetic energy of the accelerated nitrogen ion . . . . . 117 Exa 11.12 Maximum magnetic flux densit y and frequency of proton in cosmotron proton synchrotron . . . . beam . . . . . .. . . 118 Energy of the single proton in the colliding 119 Energy of the electron during boson product ion . . . . 120 Maximum activity induced in 100 mg of C u foil . . . . 121 Energy loss during neutron scattering . . . . . . . . . 122 Energy loss of neutron during collision with carbon . . 123 Number of collisions for neutron loss . . . . . . . . . . 123 Average distance travelled by a neutron . . . . . . . . 124 Neutron flux through water tank . . . . . . . . . . . . 125 Diffusion length and ne utron flux for thermal neutrons 126 Diffusion length for thermal neutrons in gra phite . . . 127 Neutron age and slowing down leng th of neutrons in graphite and beryllium . . . . . . . . . . . . . . . . . . 128 Exa 12.10 Energy of the neutrons reflected from the crystal . . . 129 Exa 13.1 Fission rate and energy released during fission of U235 130 Exa 13.2 Number of free neutrons in the reactor . . . . . . . . . 131 Exa 13.3 Number of neutrons released per absorption . . . . . . 131 Exa 11.13 Exa 11.14 Exa 12.1 Exa 12.2 Exa 12.3 Exa 12.4 Exa 12.5 Exa 12.6 Exa 12.7 Exa 12.8 Exa 12.9
9
Exa 13.4 Exa 13.5 Exa 13.6 Exa 13.7 Exa 13.8 Exa 13.9
Excitation energy for uranium isotopes . . . . . . . . . 132 Total energy released in fusion reaction . . . . . . . . 133 Maximum temperature attained by thermonuclear device 134 Energy radiated and the temperature of the sun . . . 135 Estimating the Q value for symmetric fission of a nucleus 136 Estimating the asymmetric binding energy term . . . . 136
Exa 15.1 Exa 15.2 Exa 15.3
Exa 15.7 Exa 15.8 Exa 16.1 Exa 16.2 Exa 16.4 Exa 18.1 Exa 18.3
Estimation of the leakage factor for thermal reactor . . 138 Neutron multiplication factor of uranium reactor . . . 139 Multiplication factor for uranium graphite modera ted assembly . . . . . . . . . . . . . . . . . . . . . . . . . 140 Ratio of number of uranium atoms to graphite atoms . 141 Multiplication factor for LOPO nuclear reactor . . . . 142 Control poison required to maintain the crit icality of U235 . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Dimensions of a reactor . . . . . . . . . . . . . . . . . 143 Critical volume of the sp herical reactor . . . . . . . . . 144 Radiation dosimetry . . . . . . . . . . . . . . . . . . . 146 Conversion of becquerel into curie . . . . . . . . . . . 146 Amount of liver dose for a liver scan . . . . . . . . . . 147 Root mean square radius of charge distribution . . . . 148 Isospin of the strange particles . . . . . . . . . . . . . 148
Exa 18.4 Exa 18.9 Exa 18.10 Exa 18.13 Exa 18.15 Exa 18.16 Exa 18.18
Allowed forbidden laws 151 Decay ofand sigma particlereactions . . . . . . under . . . .conservation . . . . . . . 157 Estimation of the mean life of tau plus . . . . . . . . . 158 Possible electric charge for a baryon and a meson . . . 158 Branching ratio for resonant decay . . . . . . . . . . . 159 Ratio of cross section for reactions . . . . . . . . . . . 160 Root mean square rad ius of charge distribution . . . . 160
Exa 15.4 Exa 15.5 Exa 15.6
10
Chapter 1 General Properties of Atomic Nucleus
Scilab code Exa 1.1 Distance of closest approach
1 2 3 4 5
// Scil ab c o d e Exa1 .1 :
: Page 51 (2 01 1)
clc ; clear ; Z = 79; // At om ic num be r of Go ld z =1; // At omi c nu mb er of Hyd rog en e = 1.6 02 18e-01 9; // Charge of an electron
,
coulomb // Coulomb constant , ne wt on me tr e squar e per co ul om b squar e 7 E = 2*1 .60 218e-0 13 ; // En er gy of the pr ot on , joule 8 b = Z* z*e^ 2*K /E; // Di st an ce of closest approa ch , metre 9 printf ( ” \ nDi sta nce of clo se st ap pr oa ch : %7.5 e me tr e ” , b); 6 K = 9e+ 09;
10 11 // Res ult 12 / / Di st an ce of
closest
a p p r o ac h : 5. 69 57 5e
11
−014 meter
Scilab code Exa 1.2 Nuclear Spin
1 2 3 4 5 6 7 8
// Scil ab c o d e Exa1 .2 :
: Page 51 (2 01 1)
clc ; clear ; A = 14; // Number of protons Z =7; // Number of neutrons N = A-Z ; // Number of ele ctr on s i = modulo ((N+A),2); // Rem ain der
// Check for if
i == 0
ev en and odd number of pa rti cl es // Fo r ev en number of
!!!!!
then
particles printf ( ” \ nPartic les ha ve integral 9 sp in ” 10 s = 1; // Nu cl ea r spin 11 end 12 if i == 1 then // Fo r odd number of
);
particles 13 14
printf ( ” \nPar tic les s = 1/2;
ha ve half
integra
l sp in ”
15 end 16 if s == 1 then 17 printf ( ” \ nM eas ur ed value agree wi th the assumption” ) ; 18 end 19 if s == 1/2 then 20 printf ( ” \ nMeasured val ue disagree wi th th e assumption” ); 21 end 22 23 // Res ult 24 / / Partic les h ave half integra l sp in 25 // Meas ured valu e disagree wi th th e as su mp ti on
12
);
Scilab code Exa 1.3 Kinetic energy and Coulomb energy for an electron
confined within the nucleus 1 // Scil ab c o d e Exa1 .3 2 clc ; clear ; 3 p = 62;
:
: Page 52 (2 01 1) // Momentum of the
MeV/ c
ele ctr on ,
// Coulomb cons tant // Energy of th e electron
4 K = 9e+ 09; 5 E = 0. 51 1;
,
MeV 6 e = 1.6 02 18e-01 9; 7 Z = 23; 8 R = 0. 5* 10 ^-14;
// Cha rge o f an e lect ron , C // Ato mic nu mb er // Di am et er of th e nuc le us ,
meter 9 T = sqrt (p^2+E^2)-E;
// Ki ne ti c en er gy o f th e
el ec tr on ,MeV // Co ul om b ener gy , MeV 11 printf ( ” \ nKin etic en er gy of th e electro n : %5.2 f M eV \ nCoulomb energy per el ec tr on : %5.3 f MeV” ,T,E_c 10 E_c = -Z*K*e^2 /(R*1.60 218e-013 );
); 12 13 // Res ult 14 // Kin et ic en er gy of t he elect ron 15 // C oulomb en er gy pe r electro n :
: 61 .4 9 M eV −6.63 3 MeV
Scilab code Exa 1.4 Scattering of electron from target nucleus
1 // Scil ab c o d e Exa1 .4 : 2 clc ; clear ; 3 K = 500 *1. 602 18e-01 3;
th e electron
: Page 52 (2 01 1) // Kine tic
en er gy of
, joule // Pla nc k ’ s constant
4 h = 6. 62 62e-0 34 ;
joule sec 5 C = 3e+ 08;
// Vel oci ty of ligh t , 13
,
me tr e pe r sec // Momentum of the se c pe r m et e r 7 la mb da = h/ p; // de Broglie wavel ength , met re 8 A = 30 *%p i/18 0; // An gl e ( in radian ) 6 p = K/C;
ele ctr on , joule
9 r = la mb da/(A* 10 ^-15 );
nucleus , fem tome tre 10 printf ( ” \ n R a d i u s of th e target
// R a di u s of th e target nuc leu s : %4.2 f fm ”
r); 11 12 // Res ult 13 / / R a d i u s of t he tar get
nu cl eu s : 4. 74 f m
Scilab code Exa 1.5 Positron emission from Cl33 decays
1 2 3 4
// Scil ab c o d e Exa1 .5 : clc ; clear ; e = 1.6 02 18e-01 9; A = 33;
: Page 52 (2 01 1) // Charg e of an elec tron , C // At om ic mas s of Chlorine ,
amu // Coulomb constant , ne wt on me tr e sqaur e per co ul om b squar e E = 6.1 *1. 602 18e-01 3; // Co ul om b energy , joule R_0 = 3/5 *K/E*e^2* (A)^(2/3); / / Di st an ce of closest approa ch , metre R = R_ 0*A^ (1 /3 ); // R a di u s of th e nucleu s , me tr e printf ( ” \ nR ad iu s of th e nucl eus : %4.2 e me tr e” , R);
5 K = 9e+ 09; 6 7 8
9 10 11 // Res ult 12 / / R a d i u s of t he nu cl eu s : 4. 68 05e
14
−015 metre
,
Scilab code Exa 1.6 Charge accelerated in mass spectrometer
1 2 3 4 5 6 7 8
// Scil ab c o d e Exa1 .6: clc ; clear ; V = 100 0; R = 18 .2e- 02 ; B = 10 00e- 04 ; e = 1.6 02 18e-01 9; n =1; v = 2*V/ (R* B);
: Page 53 (2 01 1)
// Pote nti al differ ence , volts // Rad iu s of th e orb it , m e t r e // M ag ne t ic field , tesla // Charg e of an elec tron , C // Number of the ion // Spe ed of th e io n , me tr e
pe r sec 9 M = 2*n *e*V /v^2 ; // Mass of th e io n , Kg 10 printf ( ” \ nS pe ed of the ion : %6.4 e m /s \ nMass of the io n : %4.2 f u ” , v, M/1. 67e-0 27 ); 11 12 // Res ult 13 // Sp ee d of the ion : 1.0989 e+ 05 m/s 14 / / M ass of t h e io n : 15 .8 9 u
Scilab code Exa 1.7 Ionized atoms in Bainbridge mass spectrograph
1 2 3 4
// Scilab
co de Ex a1 .7 : : Page 53 (2 011 )
5 6 7 8 9
B = 0.0 8; // M ag ne t ic field , tesla e = 1.6 02 18e-01 9; // Charg e of an elec tron , C n =1; // Number of the ion R_ 20 = M*v/ (B*n* e) // Ra di us of th e neon −20, me tr e R_ 22 = 22 /2 0*R_2 0; // Ra di us of th e neon −22, me tr e
clc ; clear ; M = 20* 1.6 605 4e-027 ; // v = 10^ 5; // Spe ed of th e io n , me tr e pe r
sec
15
10 printf ( ” \ nR ad iu s of the neon −20 : %5.3 f me tr e \ nR ad iu s of the neon −22 : %5.3 f me tr e” , R_2 0, R_ 22 ); 11 12 // Res ult 13 // Ra di us of th e neon −20 : 0.2 59 m e t r e 14 // Ra di us of th e neon
−22 : 0.2 85 m e t r e
Scilab code Exa 1.8 Calculating the mass of hydrogen
1 // Scil ab c o d e Exa1 .8 2 clc ; clear ; 3 a = 17 .7 8e-0 3;
diffe rence
: Page 53 (2 01 1) // First
// Se co nd doublet
mass
// Th ir d doublet
mass
, u
5 c = 87 .3 3e-0 3;
diffe rence
do ubl et mass
, u
4 b = 72 .9 7e-0 3;
diffe rence
:
, u // Mass of th e
6 M_H = 1+1/ 32*(4 *a+5*b-2*c);
hydrogen ,amu
7 printf ( ” \ nMas s of the hydr ogen : %8.6 f amu” 8 9 // Res ult 10 // Mass of the hy dr og en : 1.00816 6 amu
,M_H);
Scilab code Exa 1.9 Silver ions in Smith mass spectrometer
1 // Scilab co de Ex a1 .9 2 clc ; clear ; 3 e = 1.6 02 18e-01 9;
: : Page 54 (2 011 ) // Char ge of an
electron ,C 4 B = 0.6 5;
// Ma gn et ic fiel d ,
tesla 16
// Distan ce be tw ee n
5 d_S 1_S 2 = 27 .9 4e-02 ;
s l i t S1 an d S2 , met re // R a d i u s of orbit of ente rin g s l i t S2 , me tr e 7 d_S 4_S 5 = 26 .24 8e-02 ; // Distan ce be tw ee n s l i t S4 an d S5 , met re 6 R_1 = d_ S1_ S2/2 ;
ions
8 R_2 =
ions
d_ S4 _S 5/2 ;
leavin g s l i t S4 , me tr e
//R a d i u s of orbit
of
// Mass of an io n
9 M = 106 .9* 1.6 605 4e-02 7;
(Ag+)Kg, 10 T_1 = B^2*e^2 *R_1^2/(2 *M*1.6021 8e-019);
// Kine tic
en er gy of th e ion
enter ing
s l i t S2 ,e V
11 T_2 = B^2*e^2 *R_2^2/(2 *M*1.6021 8e-019);
// Kine tic
en er gy of th e ion leaving s l i t S4 ,e V en er gy of th e ion enter ing s l i t S2 : %d eV \ nKi net ic en er gy of th e io n leaving slit S4 : %d e V ” ,T_1,T_2)
12 printf ( ” \ nKi ne tic
13 14 // Res ult 15 // Kin et ic en er gy of t he io n ent eri ng
s l i t S2 : 3 7 2 1
eV 16 // Kin eV et ic en er gy of t he io n leav ing
s l i t S4 : 3 2 8 4
Scilab code Exa 1.10 Calculation of energy released during nuclear fusion
reaction 1 2 3 4 5 6 7
// Scil ab c o d e E x1.10
:
: Page 5 5 (2 01 1)
clc ; clear ; M_ Li = 7.0 116 004 ; // M ass of M_ Be = 7.0 169 29; // M ass of m_ e = 0. 51 1; // Mass of if (M_L i -M_Be)*931 .48 < 2*m_ e printf ( ” \ nThe Li −7 is n o t
8 else
17
lith ium nu cl eu s , u beryll ium nuc le us , u an electron , MeV then
a be ta em it te r ”
);
9 printf ( ” \ nThe 10 end 11 if (M_B e -M_Li)*931 12 printf ( ” \ nThe 13 else 14 printf ( ” \ nThe 15 16 17 18 19
Li −7 is a be t a em itt er ” .48 > 2*m_ e
);
then
Be −7 is a be t a em itt er ”
);
Be −7 is n o t a be ta em it te r ”
);
end
// Res ult // The Li −7 is n o t a b e t a em it te r // The Be −7 is a b e t a em it te r
Scilab code Exa 1.11 Binding energies calculation
1 2 3 4 5
// Scil ab c o d e E x1.11
:
clc ; clear ; M_n = 1. 00 86 65 ; M_p = 1.0 078 25; N_N i = 36 ;
: Page 5 5 (2 01 1) // Mass of neu tro n , amu // Mass of pro ton , amu // Number of neutron
in Ni −64 // At om ic nu mbe r of Ni
6 Z_N i = 28 ;
−64 7 N_C u = 35 ;
//
Number of neu tro n
//
Atom ic number of
in Cu −64 8 Z_C u = 29 ;
Cu−64 9 10 11 12
// // // //
A = 64; M_ Ni = 63. 927 958 ; M_ Cu = 63. 929 759 ; m_ e = 0. 51 1;
Ma ss nu mb er , amu Mass of Ni −64 Mass of Cu −64 Mass of an
ele ctr on , MeV 13 d_M_Ni
= N_N i*M_n+Z_Ni*M
_p-M_Ni;
// Ma ss
_p-M_Cu;
// Ma ss
defe ct , amu 14 d_M_Cu
= N_C u*M_n+Z_Cu*M
defe ct , amu 18
// Bind ing
15 B_E_N i = d_M _Ni*931. 49;
en er gy of Ni −64 , MeV // Bind ing
16 B_E_C u = d_M _Cu*931. 49;
en er gy of Cu −64 , MeV //
17 Av_ B_E _Ni = B_ E_ Ni/A;
A v e rag e
bi nd ing en er gy of Ni −64 , MeV 18 Av_ B_E _Cu = B_ E_ Cu/A;
bin ding en er gy 19 printf ( ” \ nB ind in g nBi nd in g ene rgy bi nd ing en er gy bin ding en er gy
// A v e rag e of Cu −64 , MeV ene rgy of Ni −64 : %7.3 f MeV \ of CU −64 : %7. 3 f MeV \ nAverage of Ni −64 : %5. 3 f MeV \ nAverage of Cu −64 : %5. 3 f MeV ” , B_ E_ Ni,
B_ E_C u , Av_ B_E _Ni , Av_B _E_C u); 20 if (M_ Cu - M_ Ni)* 93 1. 48 > 2*m_ e then 21 printf ( ” \ nN i −64 is n o t a be ta em it te r bu t C u is a be ta emit ter ” ) ; 22 end 23 24 // Res ult 25 // Bind in g en er gy of Ni −64 : 561.765 MeV 26 // Bi nd in g en er gy of C U −64 : 559.305 MeV
−64
27 // // Av Ni −64 −64 :: 8.778 MeV 28 Av era era ge ge bind bind ing ing en en er er gy gy of of Cu 8.739 MeV 29 // Ni −64 is n o t a b et a em itt er b ut C u −64 is a b e t a
emitter
Scilab code Exa 1.12 Calculation of energy released during nuclear fusion
reaction 1 // Scil ab c o d e Exa1.12 : 2 clc ; clear ; 3 M_n = 1. 00 86 65 *9 31 .4 9;
: Page 55 (2 01 1) // Mass of
neut ron , MeV // Mass of pr ot on
4 M_p = 1.0078 25*93 1.49;
, MeV 19
// Mass of He
5 M_ He = 2*M _p+2 *M_ n -2 8;
−4
nuc leu s , MeV // Mass of H
6 M_ H = M_ p+M_ n -2. 2;
−2 nucleus
, MeV // En ergy releas
7 d_E = 2*M _H-M_ He;
dur ing
fusion
ed
reacti on , MeV
8 printf ( ” \ nEnergy relea sed du ri ng fusio n reac tion : %4. 1 f MeV ” ,d_E); 9 10 // Res ult 11 // En erg y released du ri ng fusion reaction : 23. 6 M eV
Scilab code Exa 1.13 To find the stable Isobar
1 2 3 4 5 6 7 8 9 10 11 12 13 14
// Scil ab c o d e Ex1.13 : : P.N o.5 5 (2 01 1) // W e ha ve to det ermi ne for mass num bers 80 and 97. clc ; clear ; A = [80, 97]; // Mat rix of Mass nu mb er s Ele me nt = [ ”Br” , ”Mo” ]; // Ma tr ix of ele men ts M_ n = 93 9. 6; // M as s of neut ron , MeV M_ H = 93 8. 8; // M as s of prot on , MeV a_v = 14 .0 ; // V ol um e energy , MeV a_s = 13 .0 ; // Surface energ y , MeV a_ c = 0. 58 3; // Cou lo mb energy , MeV a_a = 19 .3 ; // A sy mm et ry energy , MeV a_p = 33 .5 ; // Pairing ene rgy , MeV for i = 1:1:2 Z = poly (0 , ’ Z ’ ) ; // Dec lare th e pol yno mia l
variable 15 M_A Z = M_ n*(A(i)-Z)+M_ H*Z-a_v*A(i)+a _s*A(i)^(2 /3)+ a_c*Z*(Z-1)*A(i)^(-1/3)+a_a*(A(i)-2*Z)^2/A(i)+a_p *A(i)^(-3/4); // Mass of the nuclide , MeV/cˆ2 16 Z = roots ( derivat (M_AZ)); 17 printf ( ” \ nFor A = %d, th e most stable isobar is %s(
%d,%d)” , A(i ), El em en t(i) , Z, A(i )); 20
18 19 20 21 22
end
// Res ult // For A = 80 , th e most stable // For A = 97 , th e most stable
isobar isobar
is Br(35 ,80 ) is Mo (42 ,97 )
Scilab code Exa 1.14 To calculate the pairing energy term
1 2 3 4
// Scilab
co de Exa1 .14
clc ; clear ; A = 50; M_ Sc = 49. 951 730 ;
:
: P. no . 56( 20 11)
// Ma ss nu mb er // Mass of sc an di um , a tom ic
mas s unit // Mass of tit ani um ,
5 M_ Ti = 49. 944 786 ;
at omi c mass unit // Mass of va na di um , atomic
6 M_V = 49. 947 167 ;
mas s unit // Mas s of ch ro mi um ,
7 M_ Cr = 49. 946 055 ;
at omi c mass unit 8 M_ Mn = 49. 954 215 ;
// Mas s of ma ng an es e ,
at omi c mass unit
9 a_ p = (M_ Mn-M_Cr+M_
V -M_Ti)/( 8*A^(-3/4 ))*931.5;
//
Pairing en er gy temr , mega electr on volts 10 printf ( ” \ nPairing energy te rm : %5.2 f MeV” , a_ p); 11 12 // Res ult 13 // Pairing
en er gy term : 23.08 M eV
Scilab code Exa 1.17 Relative error in the electric potential at the first
Bohr radius 1 // Scil ab c o d e E x1.17 2 clc ; clear ;
:
: Page 5 7 (2 01 1)
21
// Fo r simpli city as sume m in o r axis len gth to be un it y , uni t a = 10 /1 00 +b; // Ma jor axis le ngt h , unit A = 125; // M as s n um be r o f me dium nu cleus r = 0. 53e- 01 0; // Bohr ’ s radius , m ep s = (a-b )/(0 .5 *a+b); // Defor mati on para mete r
3 b =1; 4 5 6 7
8 R = 1. 2e-01 5*A^( 1/ 3); // Ra di us of th e nu cl eu s , m 9 Q = 1. 22 /1 5*R^ 2 // El ec tr ic Qu ad rup ole m oment ,
me tr e square // Re la ti ve err or in t h e potential 11 printf ( ” \nThe re la ti ve erro r in the e l e c t r i c poten tial at th e f i r s t Bohr radius : %e” , 10 V_r el_ err = Q/r^ 2;
V_rel_err); 12 13 // Res ult 14 // The rel ati ve
error in th e e l e c t r i c potential th e f i r s t Bohr radius : 1.04 2364e −09
at
Scilab code Exa 1.21 Spherical symmetry of Gadolinium nucleus
1 2 3 4 5
// Scil ab c o d e Exa1.21
:
: Page
− 58(2011)
clc ; clear ; Q = 130; // Quadr upol e m oment , square femt o met re A = 155; // Mass number of gadolinium R_0 = 1. 4*A^( 1/ 3) // Di st anc e of clo sest a p p ro ac h ,
fm 6 Z = 64; // Ato mic nu mb er 7 delR0 = 5*Q/( 6*Z*R_0^2 )*100;
// Change in t he val ue of R 0 , per ce nt 8 printf ( ” \ nChange in th e va lue of frac tio nal c h an ge in R 0 is on ly % 4.2 f per ce nt \ nThu s , we can assumed tha t Ga do li ni um nucl eus is sph erica l . ” , delR0); 9
22
10 // Res ult 11 // C hange in t he va lu e of
fraction al c h a n g e in R 0 is on ly 2. 99 pe rc en t 12 // Thus , we ca n as su me d that Ga do li ni um nucleus is spherical .
23
Chapter 2 Radioactivity and Isotopes
Scilab code Exa 2.1 Weight of one Curie and one Rutherford of RaB
1 2 3 4 5 6 7
// Scilab
co de Exa2 .1 : : Page
−88 (2011)
clc ; clear ; T = 26 .8 *6 0; // Ha lf l i f e of th e su bs ta nc e , s C = 3. 7e+0 10 ; // O ne cur ie , disintegration pe r se c N = 6.0 221 37e+02 6; // Avo gad ro nu mb er , per km ol m = 214; // Mole cular wei ght of RaB, kg/k mol R = 1e+ 00 6; / / One Ru th er fo rd , dis int egr ati on pe r
se c . 8 W_ C = C*T* m/(N* 0.6 93);
// Weight of on e Cur ie of RaB
, Kg // Weight of on e Ruth erf ord of RaB, Kg 10 printf ( ” \ nW ei gh t of on e Curie of RaB : %5.3 e Kg \ nW ei gh t of on e Rutherfo rd of RaB : %5.3 e Kg” , W_C 9 W_ R = R*T* m/(N* 0.6 93);
, W_ R); 11 12 // Res ult 13 // We ight of on e Cu ri e of R aB : 3.051e −011 Kg 14 // Weight of on e Rut he rfo rd of R aB : 8.245e −016 Kg
24
Scilab code Exa 2.2
Induced radioactivity of sodium by neutron bom-
bardment 1 2 3 4 5 6 7
// Scilab
co de Ex a2 .2 : : Page 88 (2 011 )
clc ; clear ; T_h = 14 .8 ; // Hal f l i f e of Na −24, hou rs Q = 1e+ 00 8; / / Pr od uc ti on rate of Na −24, pe r sec L = 0. 69 3/T _h; // Decay cons tant , per sec t = 2 ; // Time af te r the bom ba rdm ent , hours A = Q/3 .7e+01 0*1 000 ; // The m aximum ac ti vi ty of Na
−24, mCi // The ti me ne ed ed to pr odu ce d 90 the maximum act ivi ty , ho ur 9 N = 0. 9*Q*3 60 0/L*% e^(-L *t); // Nu mber of at om s of Na −24 l e f t two hours af te r bombardment was stopp ed 10 printf ( ” \nThe m aximum ac ti vi ty of Na −24 = % 3. 1 f mCi \ nThe ti me ne ed ed to pr od uc ed 90 perc ent of th e maximum ac ti vi ty = %4.1 f hrs \nNumber of at om s of 8 T = -1* log (0.1)/L;
% of
Na−24 l e f t tw o hours a ft er bo mb ar dm en t wa s stopped = % 4.2 e ” , A, T, N); 11 12 // Res ult 13 // The m aximum ac ti vi ty of Na −24 = 2.7 mCi 14 // T he ti me ne ed ed to pr od uc ed 90 perc ent of th e
maximum ac ti vi ty = 49.2
hrs
15 // Number of at om s of Na −24 l e f t two hou rs af ter
bo mb ar dm en t wa s stopped
= 6.3 0 e+0 12
Scilab code Exa 2.3 Activity of K40 in man of weight 100 Kg
1 // Scilab co de Exa2 .3 : : Page 89 (20 11) 2 clc ; clear ;
25
// Ha lf l i f e of th e
3 T = 1.3 1e +09*3 65*24* 60*60;
substance , sec 4 N = 6.0 221 37e+02 6; // Avoga dro nu mb er . 5 m = 0.3 5* 0.0 12* 10^-2; // Mass of K −40, Kg. 6 A = m*N*0 .69 3/(T*4 0); / / Ac ti vi ty of K −40 ,
dis int egr ati ons / sec . 7 printf ( ” \nThe activity
of K −40 = %5.3 e di si nt eg ra ti on s / sec = % 5.3 f mi cr o −cur ie ” , A, A
/3.7e+004); 8 9 // Res ult 10 / / T he activity
/ sec = 0.287
of K −40 = 1.06 1 e+ 004 dis int egr ati ons mi cr o −curie
Scilab code Exa 2.4 Age of an ancient wooden boat
1 2 3 4
// Scilab
co de Ex a2 .4 : : Page 89 (2 011 )
clc ; clear ; T = 556 8; // Ha lf l i f e of th e C lamb da = 0. 69 3/T; / / Disintegration
−14, yea rs co ns ta nt , ye ars
ˆ −1. 5 N_0 = 15 .6 /lamb da;
// Activity
of fresh
car bo n , dpm
. gm 6 N = 3. 9/la mb da;
// Activity
of an ancient
wooden
boat ,dpm.gm. 7 8 9 10 11
t = 1/ (lam bd a)* log (N_0/N); / / Age of th e bo at , year s printf ( ” \nThe ag e of th e bo at : %5.3 e year s ” , t);
// Res ult // The ag e of th e bo at : 1.114 e+ 004 year s
Scilab code Exa 2.5
Activity of the U234 26
1 2 3 4 5 6 7 8 9 10
// Scilab
co de Ex a2 .5 : : Page 90 (2 011 )
clc ; clear ; m_0 = 3e- 06 ; // I n i t i a l mas s of the U −234, Kg A = 6.0 221 37e+02 6; // Avagad ro ’ s nu mb er , atoms N_0 = m_ 0*A/2 34 ; // I n i t i a l nu mb er of atoms T = 2. 50e+ 05 ; / / Ha lf life , ye ar s lamb da = 0. 69 3/T; // Disintegration co nst ant t = 15 00 00 ; / / Disintegration time , yea rs m = m_ 0*%e^ (-lam bd a*t); // Mass aft er ti me t ,K g acti vity = m*lambd a /(365 *24*6 0*60)*A/23 4; //
Act ivi ty of U −234 aft er ti me t , dp s of U −234 af te r %6d yrs = %5.3 e dis int egr ati ons / sec ” , t, ac ti vi ty) ;
11 printf ( ” \nThe activity 12 13 // Res ult 14 // T he activit
y of U −234 afte r 15 00 00 yrs = +005 dis int egr ati ons / sec
4.47 8 e
Scilab code Exa 2.6 Number of alpha decays in Th232
1 2 3 4 5 6
// Scilab
co de Ex a2 .6 : : Page 90 (2 011 )
clc ; clear ; A = 6.0 221 37e+02 3; // Avagad ro N_ 0 = A/2 32 ; // I n i t i a l nu mb er t = 3. 15 0e+0 7; // De cay ti me , la mbd a = 1. 58e-0 18 ; // Disinte
’ s nu mb er , atoms of ato ms sec gration co nst ant , sec
ˆ−1 7 N = la mb da*t*N _0;
// Number of alp ha dec ays in Th
−232 8 printf ( ” \nThe number of alp ha dec ays in Th −232 = %5 .2 e ” , N); 9 10 // Res ult 11 // The nu mber of alp ha dec ays in Th −232 = 1. 29 e+0 11
27
Scilab code Exa 2.7 Maximum possible age of the earth crust
1 2 3 4 5
// Scilab
co de Ex a2 .7 : : Page 90 (2 011 )
clc ; clear ; T_ 23 8 = 4. 5e+09 ; // Ha lf l i f e of U −238, years T_ 23 5 = 7. 13e+0 8; // Ha lf l i f e of U −238, year s lambda _238 = 0.69 3/T_238; // Disintegration co nst ant
o f U−238, year sˆ −1 // Disintegration co nst ant o f U−235, year sˆ −1 7 N = 13 7. 8; // Ab un da nc es of U −238/U −235 8 t = log (N)/(lambd a_235 - lamb da_2 38); // Age of th e ear th ’ s crust , years 9 printf ( ” \nThe m aximum po ssi ble ag e of the earth crus t = % 5.3 e years ” , t); 6 lambda _235 = 0.69 3/T_235;
10 11 // Res ult 12 // T he m aximum pos sibl e ag e of th e eart h crust =
6.02 2 e+ 009 year s
Scilab code Exa 2.8 Number of radon half lives
1 2 3 4
// Scilab
co de Ex a2 .8 : : Page 91 (2 011 )
clc ; clear ; N = 10; // Number of at om s l e f t unde cay ed in Rn −222 n = log (10)/ log (2); // N umber of half li ves in R a
−222 5 printf ( ” \nThe n umber of half li ve s in rad on −222 = %5 .3 f ” , n); 6 7 // Res ult 8 // T he n umber of half liv es in radon −222 = 3.322
28
Scilab code Exa 2.9 Weight and initial acivity of Po210
12 3 4 5 6 7 8 9 10
// ab c;o d e Exa2 .9 : : Page 91 (2 01 1) clcScil ; clear M_ Po = 209 .98 29; // Mass of Po lo ni um , g M_ Pb = 205 .97 45; / / Mass of le ad , g A = 6.2 21 37e+02 3; // Avo gad ro ’ s nu mb er M_H e = 4.0 02 6; / / Mass of al ph a particle , g C = 3e+ 08; / / Ve loc ity of ligh t , m /s T = 138 *24 *36 00; / / H a l f life , se c P = 250; // Po wer pro duc ed , jo ul e / sec Q = [M_ Po-M_P b -M_He]*93 1.25 ; // disintegration ener gy , MeV
11 lamb da = 0. 69 3/T;
/ / Disintegration
co ns ta nt , pe r
year 12 N = P/(lamb da*Q*1.602
18e-013);
// Number of at om s ,
atom // Number of at om s presen t atom 14 W = N_ 0/A* 21 0; // We ig ht of Po −210 after o ne y e ar , g 15 A_0 = N_ 0*lambda/(3 .7e+010 ); // I n i t i a l ac ti vi ty , curie 16 printf ( ” \nThe weigh t of Po −210 aft er one ye ar = %5.2 f g \nThe i n i t i a l ac ti vi ty of the material = % 4.2 e curies ” , W, A_0); 13 N_0 = N*% e^(1. 833 );
ini tia lly ,
17 18 // Res ult 19 // T he wei ght of Po −210 after o ne y ea r = 10. 49 g 20 // The i n i t i a l ac ti vi ty of the material = 4.88 e+ 004
curies
Scilab code Exa 2.10 Radioactive disintegration of Bi
29
1 // Scilab co de Ex a2 .10 : : Page 91 (2 011 ) 2 clc ; clear ; 3 lambda _t = 0.693/ (60.5* 60); // Tot al de ca y con stan t ,
pe r sec 4 lambd a_a = 0.3 4*lambda _t; // Decay cons tant
for
al pha d ec ay , pe r sec 5 lambd a_b = 0.6 6*lambda _t; // Decay cons tant
for be ta d ec ay , pe r sec 6 printf ( ” \nThe de ca y cons tant for tota l emissi on = %4 .2 e / se c ” , lam bda _t); 7 printf ( ” \nThe de ca y cons tant for bet a deca y lam bd a b = %4. 2 e / se c ” , lam bda _b); 8 printf ( ” \nThe de ca y cons tant for alph a de cay lambda a = %4. 2 e / sec ” , lam bda _a); 9 10 // Res ult 11 // T he de c ay con st ant
/ se c 12 // The de ca y cons tant e −004 / se c 13 // The de ca y cons tant
for
total
emi ssi on = 1 .91e
for
bet a dec ay la mb da b = 1.26
for
alph a de cay la mb da a =
−004
6.49e −005 / sec
Scilab code Exa 2.13 Half life of Pu239
1 2 3 4 5 6 7
// Scil ab c o d e Exa2.13
:
: Page 93 (2 01 1)
clc ; clear ; M_ A = 4; // M ass of al ph a parti cle , a mu M_ U = 23 5; //M ass of U −235, amu M_ P = 23 9; // Mass of P −239, amu Amou nt = 12 0. 1; // qu an ti ty o f P −239, g E_ A = 5. 14 4; // Energy of emittin g alp ha
part icles ,
Mev
8 E_R = (2 *M_A)/(2*M_
// The rec oi l ene rg y
U)*E_A;
o f U−235, Mev 30
// The e ne rg y released pe r ion , Mev 10 P = 0. 23 1; // Evap ora tion rat e , wa tt 11 D = P/(E* 1.6 021 8e-01 3); // Disintegration ra te , pe r sec 12 A = 6.0 221 37e+02 3; // Ava gadr o ’ s nu mb er , atom s 9 E = E_R + E_A;
disin tegrat
13 N = Am ou nt/M_ P*A;
// Number of nuclei
o f P−239
in 120. 1g
14 T = 0. 693 /(D*3. 15e+0 7)*N;
// Half
years 15 printf ( ” \nThe hal f l i f e of Pu
l i f e of Pu 239 ,
−239 = % 3.2 e yea rs ” , T )
; 16 17 // Res ult 18 // The half
−239 = 2.4 2 e+ 004 year s
l i f e of Pu
Scilab code Exa 2.14 Disintegration rate of Au199
1 2 3 4 5
// Scilab
co de Ex a2 .14
: : Page 93 (2 011 )
clc ; clear ; T_ h_ 1 = 2.7 *24 *36 00; // Hal f l i f e of Au −198, sec T_h_2 = 3.15* 24*36 00; // Hal f l i f e of Au −199, sec S_1 = 99 e-02 8; // Crossection for f i r s t react ion , Sq
.m 6 S_2 = 2. 6e-02 4;
/ / Cros sect ion
for
se co nd rea cti on ,
Sq .m 7 I = 1e+ 01 8;
// Intensity
of radi atio n , pe r Sq .m pe r
sec 8 L_1 = I*S _1 ; // Decay constant of Au −197, pe r sec 9 L_2 = 0.6 93/T_h_ 1+I*S_2; // Decay constant of Au
−198, pe r sec 10 L_3 = 0. 69 3/T_h _2;
// Decay constant
of Au −199, per
sec 11 N_ 0 = 6.0 221 37e+02 3; 12 N_1 = N_ 0/19 7;
// Avoga dro nu mb er
// I n i t i a l nu mb er of at om s of Au 31
−197
13 14 15 16 17 18
t = 30 *3 60 0; // Gi ven tim e , sec p = [ exp (-L_1*t)]/[(L_2-L_1)*(L_3-L_1)]; q = [ exp (-L_2*t)]/[(L_1-L_2)*(L_3-L_2)]; r = [ exp (-L_3*t)]/[(L_1-L_3)*(L_2-L_3)]; N3 = N_ 1*L_ 1*L_ 2*[p+ q+r] ; N_1 99 = N3;
19 L = L_ 3*N_ 19 9;
sec
// Disint
egrat ion
rat e of A u −199, per
20 printf ( ” \nThe disintegration rat e of A u −199 = %3. 1 e ” , L); 21 22 // Res ult 23 // T he disintegration rat e of A u −199 = 1.9 e+ 012 (
Wrong ans we r in the textbook
)
Scilab code Exa 2.15 Activity of Na24
1 2 3 4 5
// Scilab
co de Ex a2 .15
: : Page 94 (2 011 )
clc ; clear ; Y = 11 0e-0 3; // Yi el d of N a −24, mCi/hr T = 14. 8; // Hal f l i f e of Na −24, hou rs t = 8 ; // Time aft er whic h act ivi ty to be compute ,
hours 6 lamb da = 0. 69 3/T; // Disintegration
co ns ta nt , ho ur s
ˆ−1 7 A = 1. 44 *Y*T ; // M aximum ac ti vi ty of Na −24, Ci 8 A_C = A*[ 1-%e^(-lambd a*t)]; // Ac ti vi ty after a
con tin uou s bo mb ar dm en t , Ci da*t)); // Ac ti vi ty after 8 hour s , Ci 10 printf ( ” \nThe m aximum ac ti vi ty of Na −24 = % 5.3 f Ci \ nThe ac ti vi ty af ter a continuo us bombardment = %6 .4 f Ci \nThe activ ity after 8ho ur s = %7.5 f Ci” ,A , 9 Activit y = A_ C*(%e^(-lamb
A_ C , Acti vity); 11
32
12 // Res ult 13 // The m aximum ac ti vi ty of Na −24 = 2.344 Ci 14 // The act ivi ty aft er a con tinu ous bombardment =
0.73 24 Ci 15 // The activity
after
8ho ur s = 0. 50 36 0 Ci
Scilab code Exa 2.16 Radiation dose absorbed in 24 hr by the tissue in
REP 1 // Scilab co de Ex a2 .16 : : Page 94 (2 011 ) 2 clc ; clear ; 3 A_0 = 3. 7e+07 ; // I n i t i a l ac ti vi ty ,
disintegr
ations p e r se c // Ha lf l i f e of I −13 0, hou rs // t im e for do se a bs or be d calculation , sec E = 0.2 9*1 .6e-0 6; // Av er ag e en er gy of be ta ray s , ergs m =2; // M ass of iodine thy roi d tissue , g m lambd a = 0.69 3/(T*360 0); // Disintegration con stan t , secˆ −1 N_0 = A_ 0/lam bd a; // I n i t i a l nu mb er of ato ms N = N_ 0*[ 1-%e^ (-lam bd a*t)] ; // Number of average atoms disi nteg rated E_ A = N* E; // En ergy of be ta ray s em it te d , ergs E_ G = E_ A/(2*9 7.0 003 5); // Energy of be ta ra ys emit ted per gram of tissu e , REP printf ( ” \nThe en er gy of be ta ray s em itt ed pe r gram of ti ss ue = %6.1 f REP” , E_ G);
4 T = 12. 6; 5 t = 24 *3 60 0; 6 7 8 9 10 11 12 13
14 15 // Res ult 16 // T he en er gy of be ta ray s em itt ed pe r gram of
ti ss ue = 4245.0
REP
33
Scilab code Exa 2.18 Activity and the maximum amount of Au198 pro-
duced in the foil of Au197 1 2 3 4 5 6
// Scilab
co de Ex a2 .18
: : Page 95 (2 011 )
clc ; clear ; N_ 0 = 6.0 221 37e+02 3; // Av ag ad ro nu mbe r d = 0.0 2; / / Th ic kn es s of t he foil , cm R = 19. 3; // Dens ity of Au, g/cc N_1 = d*R/1 97 *N_0; // I n i t i a l nu mb er of Au
−197 pe r un it ar ea of foil ,c mˆ −2 T_ H = 2.7 *24 *36 00; // Hal f l i f e of Au −198, se c L = log (2)/T_H; / / Decay cons tant for Au −198, sec ˆ −1 I = 10 ^1 2; // Int ens ity of neu tro n beam , neutr ons/c m ˆ2/ se c S = 97 .8e- 02 4; / / Cr os s section for rea cti on ,c mˆ −2 t = 5*6 0; // Reactio n ti me , s A = S*I*N _1* (1-% e^(-L *t)) ; // Acti vity of A u −198,cm ˆ−2secˆ −1 nuclei
7 8 9 10 11 12
// The maximum amo unt of Au −198 produced , cmˆ −2 14 printf ( ” \nThe activ ity of A u −198 = % 5. 3 e per Sq .c m pe r sec \ nThe maximum amoun t o f Au −198 produ ced = %4. 2 e per Sq .c m” , A, N_ 2); 13 N_2 = S*I*N _1/L ;
15 16 // Res ult 17 // T he activity
of A u −198 = 1.02 8 e+ 008 per Sq .c m per sec 18 // The maximum amo unt o f Au −198 produced = 3.8 8 e+0 16 per Sq . cm
Scilab code Exa 2.19 Pu238 as power source in space flights
34
1 2 3 4 5 6
// Scilab
co de Ex a2 .19
: : Page 95 (2 011 )
clc ; clear ; N_ 0 = 6.0 221 37e+02 3; // Av ag ad ro nu mbe r T_P = 90*36 5*24*3 600; // Hal f l i f e of Pu −238,s L_P = 0. 69 3/T _P ; // Decay constant of Pu −238,sˆ −1 E = 5.5; // Energy of al ph a partic le , MeV
7 P =E*L _P*N _0;
/ / Power releas of Pu −238,MeV/s
ed by th e gm mole cul e
// Time in wh ic h po we r redu ces to 1/8 ti me of it s i n i t i a l val ue 9 printf ( ” \nThe power releas ed by th e gm mole cul e of Pu−238 = %4. 2 e MeV/ s \nThe tim e in wh ic h po we r redu ces to 1/ 8 ti me of it s i n i t i a l val ue = %d yrs ” ,P,t) 8 t = log (8)/(L_P*365*24*3600);
10 11 // Res ult 12 // The power releas
ed by th e g m mole cul e of Pu −238 = 8 .0 9 e+014 MeV/ s 13 // T he ti me in whic h power redu ces to 1/ 8 ti me of it s i n i t i a l value = 27 0 yrs
Scilab code Exa 2.20 Series radioactive decay of parent isotope
1 2 3 4 5 6 7
// Scilab
co de Ex a2 .20
: : Page 96 (2 011 )
clc ; clear ; N_ 1 = 10 ^2 0; // N umber of nuclei of pa re nt isotopes T_P = 10 ^4 ; // Hal f l i f e of par ent nu cle us , year s T_ D = 20; // Half l i f e of dau ght er nuc leus , years T = 10^ 4; // Gi ve n tim e , year s L_P = 0. 69 3/T _P ; // Decay cons tant of par en t
year sˆ −1 / / Decay cons tant of da ugh te r nucleus , year sˆ −1 9 t_ 0 = log (0.03)/(L_P-L_D); // Re qu ir ed ti me for nucleus ,
8 L_D = 0. 69 3/T _D ;
de ca y of dau ght er nuc leus , years 35
(-L_P*T)-%e^ (-L_D*T))*N_ 1; // Number of nucle i of da ug ht e r isot ope 11 printf ( ” \nThe required ti me for de ca y of da ug ht er nucleus = %d yr \nThe nu mber of nuc lei of daug hte r isot ope = %1.0 e ” , t_0, N) ; 10 N = L_ P/L_D*(%e^
12 13 // Res ult 14 // T he requi red
t im e for de c ay of da ug ht er nuc leu s = 10 1 yr 15 // The nu mber of nuclei of dau gh te r isotope = 1 e+ 017
36
Chapter 3 Interactions of Nuclear Radiations with Matter
Scilab code Exa 3.1 Alpha particle impinging on an aluminium foil
1 2 3 4 5 6
// Scilab
co de Exa3 .1 : : Page
−123 (2011)
clc ; clear ; E =9; // En ergy of th e al ph a partic le , MeV S = 170 0; // St op pi ng po wer of Al D = 270 0; // De nsi ty of Al , Kg pe r cubi c me tr e R_ai r = 0.00 318*E^(3/ 2); // Range of an alph a
par ticl e in air , m e t r e R_air/ S; / / Range of an al ph a particle in Al , me tr e 8 T = D* 1/S ; / / Th ic kn es s in Al of 1m air , Kg pe r square met re 9 printf ( ” \nThe ra ng e of an al ph a pa rti cle = %4.2 e metre \nThe thickne ss in Al of 1 m a ir = %4.2 f Kg pe r squ are me tre” , R_Al, T) ; 7 R_Al =
10 11 // Res ult 12 // The ra ng e of an al ph a particle 13 // The thic knes s in Al of 1 m air
square
met re 37
= 5.0 5e −05 metre = 1.5 9 Kg pe r
Scilab code Exa 3.4 Thickness of beta absorption
1 // Scilab co de Exa3 .4 : : Page −124 (2011) 2 clc ; clear ; 3 E_ ma x = 1. 17 ; // M aximum ene rgy of the bet a par ticl e
, mega electr
on volts
4 D = 2.7; // Den sity of Al , gram per cubic me tr e 5 u_m = 22 /E_m ax; / / Mass abso rpt ion coef fici ent ,
centi metre
squ are pe r gram // Ha lf va lu e thic kne ss for bet a absorpt ion , cm 7 printf ( ” \nThe Ha lf va lue thickness for be ta absor ption = % 5.3 f cm” , x_ h); 6 x_ h =
log (2)/(u_m*D);
8 9 // Res ult 10 / / T he Ha lf va lue
thickness
for
be ta abs orpt ion =
0.014 cm
Scilab code Exa 3.7 Beta particles passing through lead
1 2 3 4 5 6
// Scilab
co de Exa3 .7 : : Page 125 (20 11)
clc ; clear ; Z = 82; // Ato mic nu mb er E = 1 ; / / Energy of th e be ta paricle , MeV I_ l = 80 0; // Ionisation loss , MeV R = Z* E/I _l; // R a t i o of rad iat ion loss
ion isat ion
to
loss
// E nerg y of th e be ta pa rtic le w hen ra di at io n ra di at io n loss is e q u al t o ioni sati on lo ss , MeV
7 E_1 = I_ l/Z ;
8
38
9 printf ( ” \nThe rati o of rad iat ion
loss t o ionisati on lo ss = % 5.3 e \ nThe e ne rg y of th e be t a particle = %4. 2 f MeV ” , R, E_1);
10 11 // Res ult 12 / / T he rat io of ra di at io n loss
t o ioni sat ion
1.025e −01
13 // T he en er gy of th e be ta part icle
loss
=
= 9.76 M eV
Scilab code Exa 3.8 Thickness of gamma absorption
1 2 3 4
// Scilab
co de Exa3 .8 : : Page 12 5( 201 1)
clc ; clear ; x = 0.2 5; // Thi ckn ess of A l , me tre U_ l = 1/x* log (50); // Li ne ar abs or pt ion
coefficient 5 d = 270 0;
// dens ity
of th e Al , Kg pe r
cubi c centim etre 6 x_ h =
log (2)/U_l;
// Ha lf va lu e thic kne ss of A l ,
metre
U_l/ d; // Mass abso rpt ion coef fici ent , squ are me tr e pe r K g 8 printf ( ” \nThe half val ue thickness of Al = %6.4 f Kg pe r cubi c me tr e \nThe mass a bs o rp ti on c o e f f i c i e n t = %7.5 f squar e me tr e per Kg ” ,x _h, U_ m); 7 U_m =
9 10 // Res ult 11 // T he half
va lue thickness of Al = 0. 044 3 K g per cubic me tr e 12 // The ma ss abs orp tio n c o e f f i c i e n t = 0.005 80 square me tr e per Kg
Scilab code Exa 3.9 The energy of recoil electrons
39
1 // Scilab co de Exa3 .9 2 clc ; clear ; 3 E_ g = 2.1 9*1 .6e-01 3;
: : Page
− 125(2011)
// En er gy of the gamma ray s ,
joule 4 m_ e = 9. 109 39e-0 31 ; / / Mass of th e electron 5 C = 3e+ 08; / / Ve loc ity of ligh t , m /s 6 E_m ax = [E_g/(1 +(m_e*C^2 )/(2*E_g))]/(
, Kg
1.6e-013 );
En er gy of the compton re co il electron , MeV 7 printf ( ” \nThe en er gy of th e compton rec oil electrons = %5 . 3 f MeV” , E_ ma x); 8 9 // Res ult 10 // T he en er gy of th e compton rec oi l electrons
//
=
1. 96 1 MeV
Scilab code Exa 3.10 Average energy of the positron
1 2 3 4
// Scilab
co de Exa3 .10
clc ; clear ; m_e = 9. 1e-31 ; e = 1. 6e-1 9;
: : Page
− 125(2011)
// M ass of th e posit ron , K g // Charge of th e posit ron ,
coulomb 5 c = 3e+ 08;
// Vel oci ty of th e ligh t ,
me tr e pe r sec // Ab so lu te permittivity of free sp ac e , pe r N pe r metre −square per co ul om b square 7 h = 6. 6e-3 4; // P lan ck ’ s con sta nt , joule sec 8 E = e^2*m_ e*c/(eps*h*1 .6e-13); // Ave rag e en er gy of th e pos itr on , mega electron volts 9 printf ( ” \nThe ave rag e en er gy of th e positron = %6.4 fZ MeV” , E); 6 eps = 8. 85e-1 2;
10 11 // Res ult 12 // The ave rag e en er gy of th e positron
40
= 0.0075Z
MeV
Scilab code Exa 3.11 To calculate the refractive index of the material
12 // − 125(2011) clcScilab ; clearco; de Exa3 .11 : : Page 3 P =1; // M omentum o f the proton , G eV/c 4 M_0 = 0. 94 ; // Res t ma ss of the proto n , G eV/c
−
square 5 G = sqrt ((P/M_0)^2+1) 6 V = sqrt (1-1/G^2);
// Lo re nt z factor // Minimum vel oc ity
of th e m/s 7 u = 1/V; // R efr acti ve i nd e x of t he ga s 8 printf ( ” \nThe re fra ct iv e ind ex of th e ga s = % 4.2 f ” electron ,
,
u); 9 u = 1.6; // Refract ive in de x 10 the ta = round ( acos (1/(u*V))*180/3.14);
// An gl e at whi ch ce re nko v radia tin is em it te d , degr ee 11 printf ( ” \nThe angl e at whi ch Cer enk ov radiatio n is emitted = % d degree ” ,theta) 12 13 // Res ult 14 / / The refractive in d e x of t h e g a s = 1 .3 7 15 // T he angl e at whi ch Cer en ko v radiati on is
em itt ed
= 3 1 deg ree
Scilab code Exa 3.12
Minimum kinetic energy of the electron to emit Cerenkov
radiation 1 2 3 4
// Scilab
co de Exa3 .12
: : Page
− 126(2011)
clc ; clear ; n = 1+ 1. 35e-0 4; // Refractive ind ex of th e medium V_ mi n = 1/n; // Minimum vel oc ity of the electron , m /
s 41
// It to ta ke th e pr od uc t 6 G_ mi n = 1/ sqrt (p); // Lo r e n t z 7 m_ e = 9. 109 39e-0 31 ; / / Mass of 8 C = 3e+ 08; // Ve lo ci ty of li ght
fact or th e electron , metre per
9 T_m in = [(G_ mi n -1)*m_ e*C^2]/(1.
602e-013);
5 p = (1 +V_m in)* (1-V_ mi n);
is n ot hi n g bu t jus t
, Kg se c // Minimum
kinet ic e n e r g y re qu ir ed b y a n elect ro t o e m i t ce re nk ov radi atio n , mega electron volts 10 printf ( ” \nThe m inimum ki ne ti c ene rgy required to elec tro n to em it cer enk ov radi atio n = %5.2 f MeV” T_min); 11 12 // Res ult 13 // T he m inimum kinetic
en er gy requ ired to to em it ce re nko v radiati on = 30.64 M eV
42
electron
,
Chapter 4 Detection and Measurement of Nuclear Radiations
Scilab code Exa 4.1 Resultant pulse height recorded in the fission cham-
ber 1 2 3 4 5
// Scilab
co de Ex a4 .1 : : Page 1 7 8 (2 011 )
clc ; clear ; N = 20 0e+006 /35 ; // Tot al n umber of ion −pairs e = 1.6 02 18e-01 9; // Ch ar ge of an i on , cou lo mb Q = N*e; // Tot al char ge pr od uc ed in the ch amber ,
coulomb 6 C = 25e-0 12 ; // C ap ac it y of th e collector , fa ra d 7 V = Q/C; // R es ult ant pu lse h ei gh t , volt 8 printf ( ” \nThe resultant puls e hei ght re cor ded in th e f i s s i o n cham ber = %4. 2 e vo lt ” , V); 9 10 // Res ult 11 // T he result ant pu lse he ig ht re co rd ed in th e
f i s s i o n cham ber = 3. 66 e
−002 volt
43
Scilab code Exa 4.2 Energy of the alpha particles
1 2 3 4
// Scilab
co de Ex a4 .2 : : Page 1 7 8 (2 011 )
clc ; clear ; V = 0. 8/ 4; // Pu ls e hei gh t , volt e = 1.6 02 18e-01 9; // Ch ar ge of an i
5 C = 0. 5e-0 12 ;
farad
6 Q = V*C; 7 N = Q/e; 8 E_ 1 = 35;
on , cou lo mb
// C ap ac i ty o f th e collector
,
// Total charge pr od uc ed , co ul om b // Number of ion pairs // E n ergy of on e ion pa ir , electron
volt // E nerg y of th e alp ha particles , mega electron volt 10 printf ( ” \nThe en er gy of th e al ph a par tic les = %4.3 f MeV”, E); 9 E = N*E _1/ 10 ^6 ;
11 12 // Res ult 13 // The en er gy of th e al ph a par tic les
= 21. 845 M eV ( The an sw er is wrong in the tex tboo k )
Scilab code Exa 4.3 Height of the voltage pulse
1 // Scilab co de Ex a4 .3 : : Page 1 7 8 (2 011 ) 2 clc ; clear ; 3 E = 10e+ 06 ; // En ergy pr od uc ed b y the ion 4 5 6 7 8
pairs , ele ctr on volt s N = E/ 35; // Number of ion pair pr od uc ed m = 10^ 3; // Multi plicat ion factor N_ t = N* m; // To ta l number of ion pairs produced e = 1.6 02 18e-01 9; // Cha rg e of an io n , cou lo mb Q = N_t*e ; // To ta l ch ar ge fl ow in th e counter , co ul om b
9 t = 10^-3 ;
// Pul se tim e , sec 44
// Resistanc e , ohm // Cu rr en t passes th ro ug h th e res isto r , ampere 12 V = I*R; // H e ig h t o f t he vol tag e p ul se , volt 13 printf ( ” \ nT ot al number of ion pai rs pr od uc ed : %5.3 e \ nT ot al ch ar ge flo w in th e co unt er : %5.3 e 10 R = 10^ 4; 11 I = Q/t;
coulomb \ nH ei gh t of th e volt age vo lt ” , N_t, Q, V);
puls e : %5.3 e
14 15 // Res ult 16 // Tot al number of ion pair s pr od uc ed : 2.857 e+ 008 17 / / T ot al ch ar ge fl ow in t he co un te r : 4. 57 8e −011
coulomb 18 // He i gh t of t he vol tag e pu lse
: 4. 57 8e
−004 volt
Scilab code Exa 4.4 Radial field and life time of Geiger Muller Counter
1 2 3 4 5 6 7 8
// Scilab
co de Ex a4 .4 : : Page 1 7 8 (2 011 )
clc ; clear ; V = 100 0; // Ope ra ti ng voltage of Co un ter , volt x = 1e-0 04 ; // Time ta ke n , sec b =2; // Ra di us of the ca th od e , cm a = 0.0 1; // Di am et er of the wi re , cm E_r = V/ (x * log (b/a)); // Radial e l e c t r i c fi el d , V/m C = 1e+ 00 9; // To ta l co unt s in th e GM
counter 9 T = C/(50 *60 *60 *20 00);
// Life
of the G .M. Co un te r ,
year 10 printf ( ” \nThe r a d i a l e l e c t r i c
f i e l d : %4. 2 eV/m \nThe , E_r, T) ; l i f e of the G. M. Co unt er : %5.3 f years ”
11 12 // Res ult 13 // The r a d i a l e l e c t r i c f i e l d : 1. 89 e+0 06V/m 14 // The l i f e of th e G. M. Co un ter : 2.77 8 years
45
Scilab code Exa 4.5 Avalanche voltage in Geiger Muller tube
1 2 3 4
// Scilab
co de Ex a4 .5 : : Page 1 7 8 (2 011 )
clc ; clear ; I = 15. 7; b = 0. 02 5;
// Ionisation potenti al of ar g o n , e V // Ra di us of the ca th od e ,
metre 5 a = 0. 00 6e-0 2; // Ra di us of th e wi re , m et re 6 L = 7. 8e-0 6; // Mean fr ee pa th , me tr e 7 V = round (I*a* log (b/a)/L); // Av al an ch e voltage
in G. M. tu be , volt 8 printf ( ” \nThe avala nche vo lt ” , V); 9 10 // Res ult 11 // The ava lan che voltage
voltage
in G. M. tu be = %d
in G .M. tu be = 72 9 volt
Scilab code Exa 4.6 Voltage fluctuation in GM tube
1 2 3 4 5
// Scil ab c o d e Exa4 .6 :
: Page 17 9 (2 01 1)
clc ; clear ; C_r = 0. 1e-02 ; // Co un ti ng rate of GM tu be S =3; // Sl op e of t he cu rv e V = C_ r*100 *10 0/S; // Vo lt ag e fluctuat ion ,
volt 6 printf ( ” \nThe voltage flu ctu at ion GM tub e = %4.2 f vo lt ” , V); 7 8 // Res ult 9 // The volt age fluctuation GM tu be = 3.3 3 volt
46
Scilab code Exa 4.7 Time measurement of counts in GM counter
1 2 3 4 5 6
// Scilab clc ; R_ t = R_ B = V_S = R_ S =
co de Exa4 .7 : : Page
clear ; 10 0; 25; 0. 03 ; R_t-R_ B;
−179 (2011)
// Ac tu al c ou nt ra te , p er sec // Backward co un t rate , per sec // Coeff icient of var iat ion // Sour ce counti ng rate , per
sec // Time co un t , sec // Time measure ment
7 T_t = (R _t+ sqrt (R_t*R_B))/(V_S^2*R_S^2);
me as ur em en t for
actual
8 T_ B = T_t* sqrt (R_B/R_t);
for
ba ck war d co un t , sec
9 printf ( ” \ nTime me as ur em en t fo r actu al cou nt : %5.3 f
s e c \ nTime me asu re me nt fo r bac kw ar d count se c ” , T_t, T_ B);
: %4.1 f
10 11 // Res ult 12 // 13 //
Time me a suremen t for Time me as urement for
actu al co un t : 29. 630 backward co un t : 14.8
sec sec
Scilab code Exa 4.8 Capacitance of the silicon detector
1 // Scilab co de Exa4 .8 2 clc ; clear ; 3 A = 1.5e- 4;
square
: : Page
−179 (2011)
// Area of capacito
r pla tes ,
met re
// Dielectric co ns ta nt // Electrical perm itt ivi ty of the medium , per ne wt on −metre −square co ul omb square
4 K = 12; 5 D = K*8. 854 2e-01 2;
47
// Width of depletion
6 x = 50e- 06 ;
laye r ,
metre // Cap ac it an ce of
7 C = A*D/ x*10 ^1 2;
detector
th e si li co n
, pF // En er gy pr od uc ed by the ion
8 E = 4. 5e+0 6;
pairs , eV 9 10 11 12
N = E/ 3. 5; e = 1.6 02 18e-01 9; Q = N*e; V = Q/ C*1 0^ 12 ;
// Number of ion pairs // C ha rg e of ea ch io n , cou lo mb // Total cha rge , co ul om b // Pot en ti al ap pl ie d acr oss
th e cap aci tor , volt 13 printf ( ” \nThe capacita
nce of th e detector : %6.2 f pF \ nThe potenti al app lie d acr oss t he capa cit or : % 4 .2 e volt ” , C, V);
14 15 // Res ult 16 // T he ca pac ita nce of th e det ect or : 31 8. 75 p F 17 // T he potenti al app lie d acr oss th e capa cit or : 6. 46
e −004 volt
Scilab code Exa 4.9 Statistical error on the measured ratio
1 // Scilab co de Exa4 .9 : : Page −180 (2011) 2 clc ; clear ; 3 N_A = 10 00 ; // Number of co unt obse rved
for
radiation A 4 N_B = 20 00 ;
// Number of co unt obse rved
for
radiation B 5 r = N_ A/N_ B;
// Ra ti o o f c ou nt A to th e c ou nt
B 6 E_ r = sqrt (1/N_A+1/N_B); // S t a t i s t i c a l e rr o r 7 printf ( ” \nThe s t a t i s t i c a l er ro r of the measu red ra ti o = %4.2 f ” , E_ r*r); 8 9 // Res ult
48
10 // The s t a t i s t i c a l err or
0.02
of the me as ur ed ra ti o = (W rong an sw er in the text book )
Scilab code Exa 4.10 Charge collected at the anode of photo multiplier
tube 1 // Scilab co de Exa4 .10 : : Page 18 0 (20 11) 2 clc ; clear ; 3 E = 4e+ 00 6; // Energ y lost in th e
scintill
ator ,
eV // Number of
4 N_ pe = E/10^ 2*0 .5* 0.1 ;
photoelectrons
emi tte d // Gain of
5 G = 10^ 6;
photomultiplier
tu be // Charge of th e
6 e = 1. 6e-0 19 ;
electron
, C
// Charg e collec ted at th e a no de of ph o t o multipl ier tube , C 8 printf ( ” \nThe ch ar ge colle cted at th e a n od e of p h ot o 7 Q = N_ pe* G* e;
mult ipli er tu be : %6.4 e C ” , Q);
9 10 // Res ult 11 // T he ch ar ge colle
multiplier
cted at th e a n od e of p ho t o t u b e : 3. 20 00e −010 C
Scilab code Exa 4.11 Charge collected at the anode of photo multiplier
tube 1 // Scilab co de E xa11 : : Page 1 80 (20 11) 2 E = 4e+ 06; // En ergy lost in t h e scintill
, eV // Number of
3 N_ pe = E/10^ 2*0 .5* 0.1 ;
photoelectrons
emi tte d 49
ator
// Ga in // Charge of th e
4 G = 10^ 6; 5 e = 1. 6e-0 19 ;
electron
, C
// Charg e colle cted at th e a no d e of p h ot o multipl ier tub e , C 7 printf ( ” \ nCha rge collec ted at th e an o de of p ho t o 6 Q = N_ pe* G* e;
mul tipl ier
tu be : %6.4 e C ” ,Q);
multiplier
a t t he an od e of p h o t o t u b e : 3. 20 00e −010 C
8 // Res ult 9 / / Cha rg e collected
Scilab code Exa 4.12 Measurement of the number of counts and deter-
mining standard deviation 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
// Scilab co de Exa4 .12 // Def ini ng an ar ra y
: : Page 18 1 (20 11)
clc ; clear ; n = ce ll (1, 6); // Dec lare th e c e l l ma tr ix of 1X6 n(1, 1).en tr ies = 10 00 0; n(2, 1).en tr ies = 10 20 0; n(3, 1).en tr ies = 10 40 0; n(4, 1).en tr ies = 10 60 0; n(5, 1).en tr ies = 10 80 0; n(6, 1).en tr ies = 11 00 0; g = 0 ; // k =6; H =0; for i = 1:k ; g = g + n( i, 1).e nt ri es end ; N = g/k; // Mean of the co un t D = sqrt (N); for i = 1:k ; H = H+((n(i,1). entr ies -N)*(n(i ,1).entri es -N))
21 end ;
50
22 S_ D = round ( sqrt (H/(k-1))); 23 printf ( ” \ n St an da r d deviat ion of th e re adi ng : % d” , S_D); 24 delt a_N = sqrt (N); 25 if (S_ D > de lt a_ N) then 26 printf ( ” \ nThe f o i l ca nno t be considered uni for m 27 28 29 30 31 32 33
. . ! ” );
else
printf ( ” \ nThe f o i l ca n be cons ider ed unif orm . ” end
// Res ult // St a n d a r d dev iat ion of th e re ad in g : 3 74 // The f o i l can not be cons ider ed unifo rm . . !
Scilab code Exa 4.13 Beta particle incident on the scintillator
1 // Scilab co de Exa4 .13 2 clc ; clear ; 3 4 5 6 7 8
: : Page 18 1 (20 11)
V = 2e-0 3; // Vo lt age im pu ls e , volt C = 12 0e-0 12 ; // Cap ac it an ce of th e cap aci tor , F e = 1. 6e-0 19 ; // C h arge of th e elec tron , C n = C* V/( 15 *e); // No. of elect ons N = n^( 1/ 10 ); // No. of electrons in t he o ut p u t printf ( ” \nNo. of electrons in th e ou t p ut : %4. 2 f ( approx ) ” , N);
9 10 // Res ult 11 // N o. of electrons
in th e ou tp ut : 3.1 6 ( a p pr ox )
Scilab code Exa 4.14 Time of flight of proton in scintillation counter
1 // Scilab
co de Exa4 .14
: : Page 18 1 (20 11) 51
);
2 3 4 5 6 7 8 9 10 11
clc ; clear ; m_ p = 0. 93 8; // M ass of the pr oto n , G eV E = 1.4; // Tot al ene rgy of pr ot on , GeV ga ma = E/m _p; // Boo st para mete r bt a = sqrt (1-1/gama^2); // Relativ istic fac tor d = 10; // Distance be tw ee n two counters ,m C = 3e+ 08; // Vel oci ty of light ,m/s t_p = d/(b ta*C ); // Time of fl i gh t of pr ot on , sec T_ e = d/ C; // Time of fl i gh t of electro n , sec printf ( ” \ nTime of f l i gh t of pro ton : %4.2 f ns \ nTime of fli gh t of elect ron : %4. 2 f n s ” , t_ p/1e-00 9, T_e/1e-009);
12 13 // Res ult 14 // Time of fl i gh t of pr ot on : 44.9 0 ns 15 // Time of fl igh t of elect ron : 33 .3 3 n s
Scilab code Exa 4.15 Fractional error in rest mass of the particle with a
Cerenkov Detector 1 2 3 4 5 6 7
// Scilab
co de Exa4 .15
: : Page 18 2 (20 11)
clc ; clear ; p = 100; // Momentum of the pa rt ic le , GeV n = 1+ 1. 35e-0 4; // Ref ract ive in de x of th e g as m_ 0 = 1; // Mass , GeV per square c oul om b gam a = sqrt ((p^2+m_0^2)/m_0); // Boo st param ete r bt a = sqrt (1-1/gama^2); // Relativ istic
parameter 8 d_t het a = 1e-0 03 ;
// Err or in
th e emi ssi on an gl e ,
radian 9 the ta =
// Em is io n angl e of
acos (1/(n*bta));
ph ot on , radian 10 F_e rr = (p^2*n^2* 2*theta*10
Fra cti onal
^-3)/(2*m_0^2
);
error
11 printf ( ” \nThe fractional
err or in rest 52
mass of t h e
//
par ticl e = %4.2 f ” , F_ er r); 12 13 // Res ult 14 / / T he fractional
=
err or in rest
mass of t h e particle
0.13
Scilab code Exa 4.16 Charged particles passing through the Cerenkov de-
tector 1 2 3 4
// Scilab
co de Exa4 .16
: : Page 18 2 (20 11)
clc ; clear ; u = 1.4 9; // Refrac E = 20* 1.6 021 8e-019 ;
tive in de x // En ergy of th e
elec tron , joule 5 m_e = 9. 1e-03 1;
// Mass of th e elec tron ,
Kg // Ve loc ity
6 C = 3e-0 8;
of t he lig ht ,
m/ s 7 bt a = (1 + {1 /(E/ (m_ e*C^ 2)+ 1)} ^2 );
// Boo st
parameter 8 z =1; // 9 L_1 = 40 00e-0 10 ;
// I n i t i a l waveleng th ,
metre 10 L_2 = 70 00e-0 10 ;
// Final
wav ele ngt h ,
metre 11 N = 2*%p i*z^2/137*
(1/L_ 1 -1 /L_2)*(1-1/
(bta^2*u^2));
// Number of qua n t a of vi sib le light , qu an ta per centimetre 12 printf ( ” \nThe tot al number of qua nta s durin g emi ssi on of vis ib le ligh t = %d qu an t a/ cm” , round ( N/100)); 13 14 // Res ult 15 // T he tota l number of qu an ta s dur ing
vis ib le
ligh t = 2 70 q u a n t a/ cm 53
emissi on of
54
Chapter 5 Alpha Particles
Scilab code Exa 5.1 Disintegration energy of alpha particle
1 2 3 4 5
// Scilab
co de Ex a5 .1 : : Page 2 0 3 (2 011 )
clc ; clear ; E_ a = 8. 76 6; A = 212; M_ a = 4;
// En ergy of th e al ph a parti cle , MeV // At om ic mas s of Po −212, amu // Atomic mass of al ph a parti cle ,
amu 6 e = 1. 6e-0 19 ;
// Char ge of an electron
,
coulomb 7 Z = 82; 8 R_0 = 1. 4e-01 5;
// At om ic nu mb er of Po // Di st an ce of closest
−212
approa ch , metre 9 K = 8. 99e+ 09 ; 10 E = E_ a*A/ (A-M_ a);
// Coulom b const ant // Disinteg ration en er gy , m ega
ele ctr on vol ts /3)*1.6* 10^-13); // Barr ier he ig ht for an alp ha particle wi th in t he nuc leu s , MeV 12 printf ( ” \nDisi ntegr atio n ene rgy : %5.3 f MeV \ nBar rie r hei ght for a lp h a − p a rt ic l e : %5.2 f MeV” , E 11 B_H = 2*Z*e^2 *K/(R_0*A^(1
,B_H); 13
55
14 // Res ult 15 // Disintegration en er gy : 8.9 35 M eV 16 / / Bar rie r he igh t for a l p h a − pa rti cl e : 28.26
MeV
Scilab code Exa 5.2 Calculation of the barrier height
1 // Scilab co de Ex a5 .2 : : Page 2 0 3 (2 011 ) 2 // We h a ve to m ake calculat ion for al ph a part icle
and for
p r ot o n
3 clc ; clear ; 4 E_ a = 8. 76 6;
mega electron 5 A_ Bi = 20 9;
// E nerg y of th e al ph a particle , volts // Atomic mass of Bi −20 9, atom ic
ma ss unit // Atomic mass of al ph a parti cle , at omi c mass unit A_ p = 1; // At om ic ma ss of pro ton , atom ic ma ss unit e = 1. 6e-0 19 ; // Ch ar ge of an electro n , cou lo mb Z = 83; // At om ic num be r of bis mut h R_0 = 1. 4e-01 5; // Di st anc e of clos est a p p ro a c h , metre K = 8. 99e+ 09 ; // Coulomb cons tant
6 A_ a = 4; 7 8 9 10
11 12 B_H _a = 2*Z*e^2 *K/(R_0*1. 6e-013*(A_B i^(1/3)+A_ ^(1/3))); // Bar rie r heig ht for a n a lp h a
particle
, mega electron
a
volts
13 B_H _p = 1*Z*e^2 *K/(R_0*1. 6e-013*(A_B i^(1/3)+A_ p ^(1/3))); // Ba rrier height for pr ot on , m ega
ele ctr on volt s 14 printf ( ” \nBa rr ie r he ig ht for
t he al ph a particle = %5 . 2 f MeV \nBarrie r heigh t for th e pr ot on = %5.2 f MeV”, B_ H_ a ,B_H_ p);
15 16 // Res ult 17 // Bar rie r he igh t for
th e al ph a particle 56
= 22 .6 7 M eV
18 // Barrie
r hei ght
for
th e pr ot on = 12 .30 M eV
Scilab code Exa 5.3 Speed and BR value of alpha particles
1 // Scilab co de Ex a5 .3 : : Page 2 0 3 (2 011 ) 2 // We h a ve also calculate th e va lu e of ma gn et ic
f i e l d in a particula
r orbit .
3 clc ; clear ; 4 C = 3e+ 08; // Velo city of light 5 M_0 = 6.6 44e-027* (C)^2/(1.60 218e-013 );
, m/ S //
Re st mass of alp ha parti cle , MeV // Kine tic en er gy of al ph a part icle em it te d by P o −218 7 q = 2*1 .60 218e-0 19 ; // Cha rge of al ph a partic le , C 8 V = sqrt (C^2*T*(T+2*M_0)/(T+M_0)^2); // Velo city of al ph a particle , m e tr e pe r sec 6 T = 5. 99 8;
9 B_r = V*M_0*(1.6
0218e-01 3)/(C^2*q*
sqrt (1-V^2/C^2));
// ma gn et ic f i e l d in a par ticu lar orbit , Web per mt er e
10 printf ( ” \nThe velocity
of al ph a par ticl e : %5.3 e m/s \ nThe ma gn et ic f i e l d in a part icul ar orbit : %6.4 f Wb/m” , V , B_r);
11 12 // Res ult 13 // The velocity of al ph a par ticl e : 1.6 99 e+ 007 m/s 14 // The ma gn et ic f i e l d in a particul ar orbit : 0.3 528
Wb/m
Scilab code Exa 5.4
Transmission probability for an alpha particle through
a potential barrier 1 // Scilab
co de Exa5 .4 : : Page 2 04 (20 11) 57
2 clc ; clear ; 3 a = 10^- 14;
// Width of t he
potentia
l
barrie r , m // Energy of th e al ph a
4 E = 5*1 .60 218e-0 13 ;
partic
le , joule
5 V = 10* 1.6 021 8e-013 ;
joule
he ig ht ,
// Re st mass of th e al ph a
6 M_0 = 6. 64 4e-02 7;
partic
// Potent ial
le , joule // Reduc ed value
7 h_ re d = 1. 054 57e-0 34 ;
Pl an ck ’ s constant
of
, jo ul e sec
// Probabil ity of lea kage th ro ug h th ro ug h potential barrier 9 printf ( ” \nThe probability of lea kag e of al p ha − particle t hr o ug h potent ial barrier = %5. 3 e ” ,T); 8 T = 4 * exp (-2*a* sqrt (2*M_0*(V-E)/h_red^2));
10 11 // Res ult 12 // T he probabi
lity of le ak age of a l p h a − particle t h r ou gh potenti al barrier = 1.27 1e −008
Scilab code Exa 5.6 Difference in life times of Polonium isotopes
1 2 3 4 5 6 7 8
// Scilab
co de Exa5 .6 : : Page 2 04 (20 11)
clc ; clear ; Z_ D = 82; // A to mi c n umb er of P o E_Po 210 = 5. 3; // Al ph a −sou rce fo r Po2 10 , MeV E_Po 214 = 7. 7; // Al ph a −sou rce fo r Po2 14 , MeV log_l ambda_ Po210 = -1* 1.7 2*Z_D*E_Po 210^(-1/2 ); log_l ambda_ Po214 = -1* 1.7 2*Z_D*E_Po 214^(-1/2 ); delta_ OM_t = log_la mbda_P o214 - log_l ambda_ Po210;
// Difference in ord er of m a gn i t u d e of l i f e times of Po2 14 an d Po2 10 9 printf ( ” \nThe disintegration co ns ta nt increases by a fa cto r of some 10 ˆ% 2d” , delt a_OM _t); 58
10 11 // Res ult 12 / / T he disintegration
co ns ta nt increases
by a factor
of some 10ˆ 10
Scilab code Exa 5.8 Half life of plutonium
1 2 3 4 5
/ / Scilab
co de E xa5 .8 :
: Page 2 0 5 (2 011 )
clc ; clear ; N = 120.1 *6.0 23e+023/23 9; // Number of Pu nucl ei P_re l = 0.2 31 ; // Power released , wa tt E_rel = 5.323 *1.6 026e-13; // En er gy released ,
joule // Decay rate
6 decay_ rate = P_r el/E_re l;
of
Pu23 9 , per hou r 7 t_h al f = N* log (2)/(decay_rate*365*86400);
l i f e of Po239 , sec 8 printf ( ” \nThe hal f l i f e of Pu = % 4.2 e yr”
// Hal f , t_ ha lf);
9 10 // Res ult 11 // The hal f l i f e of Pu = 2.46 e+ 004 yr
Scilab code Exa 5.9 Slope of alpha decay energy versus atomic number
1 2 3 4 5 6 7 8
// Scilab
co de Exa5 .9 : : Page 20 5( 201 1)
clc ; clear ; a_ v = 14; a_ s = 13; a_c = 0. 60 ; a_ a = 19; A = 202; Z = 82;
// Volume energy c onstant , MeV // Surface ene rgy con stan t , MeV // Coulomb energy constant , MeV // As ym me tr ic energy constant , MeV // Ma ss nu mb er //
At om ic num be r 59
9 dE_by_dN = -8/ 9*a_s/A ^(4/3) -4/ 3*a_c*Z/A ^(4/3)*(1-4*Z /(3*A))-16*a_a*Z/A^2*(1-2*Z/A); // Slop e ,
mega electron
volts pe r nu cl eo n of al ph a de ca y en er gy versu s nu mb er = % 7. 5 f MeV/ nucl eon ” , dE_ by_ dN);
10 printf ( ” \nThe slope
atomic 11
12 // Res ult 13 // T he slope
number =
of al ph a de c ay en er gy ver sus −0.15007 MeV/nucleon
at om ic
Scilab code Exa 5.10 Degree of hindrance for alpha particle from U238
1 // Scilab co de Exa5 .10 2 clc ; clear ; 3 h_ kt = 1.0 54 57e-3 4;
co ns ta nt , joule 4 e = 1.6 02 18e-1 9;
: : Page 20 6 (20 11) // Re duc ed Plan ck ’ s
sec // Char ge of an electron
,
coulomb // Orb ita l angul ar momentum // Ab so lu te permittivity of free sp ac e , coulomb squ are pe r newton pe r me tr e square 7 Z_ D = 90; // At om ic number of daught er nucleus 8 m = 6. 64 4e-2 7; // Mass of al ph a particle , Kg 9 R = 8. 62 7e-1 5; // Ra di us of dau ght er nuc leu s , metre 5 l =2; 6 ep s_ 0 = 8.5 54 2e-12 ;
sqrt (%pi*eps_0/(Z_D* 10 T1_by _T0 = exp (2*l*(l+1)*h_kt/e* m*R))); // Hi nd ra nc e facto r 11 printf ( ” \nThe hi nd ra nc e factor for al p ha particle = %5 . 3 f ” ,T1_by_T0); 12 13 // Res ult 14 // T he hi nd ra nc e factor for al ph a particle = 1. 76 8
60
Chapter 6 Beta Decay
Scilab code Exa 6.1 Disintegration of the beta particles by Bi210
1 // Scilab co de Exa6 .1 : : Page − 2 40 (201 1) 2 clc ; clear ; 3 T = 5*2 4*6 0*6 0; // Hal f l i f e of th e sub st anc e ,
sec 4 N = 6.0 23e+026* 4e-06/210 ; 5 lamb da = 0. 69 3/T;
// Number of ato ms // Disintegration
co nst ant , pe r sec 6 K = la mb da* N;
// Rate of
disintegration , // Ene rg y of th e be ta
7 E = 0.3 4*1 .60 218e-0 13 ;
partic
le , joule
// Rate a t wh i ch en er gy is em it te d , wat t 9 printf ( ” \nThe rate at whic h en er gy is emi tt ed = %d watt” , P); 8 P = E*K;
10 11 // Res ult 12 // The rate
at whi ch en er gy is em it te d = 1 wa t t
61
Scilab code Exa 6.2 Beta particle placed in the magnetic field
1 // Scilab co de Exa6 .2 2 clc ; clear ; 3 M_ 0 = 9. 109 39e-0 31 ;
electron ,
: : Page
// Re st mass of th e
Kg
4 C = 2. 92e+ 08 ;
// Vel oci ty of th e ligh t ,
me tr e pe r sec
// Ene rg y of th e be ta
5 E = 1.7 1*1 .60 218e-0 13 ;
partic
−241 (2011)
le , joule // Charge of th e
6 e = 1.6 02 18e-01 9;
electron
, C // R a di u s of th e orb it ,
7 R = 0.1;
metre 8 B = M_ 0*C*(E/(M_
0*C^2)+1)*1/
// Magnetic of t he partic le ,
(R*e);
per pen dic ula r t o th e b eam we be r per squar e me tr e
field
9 10 printf ( ” \nThe mag net ic f i e l d perpendicu
beam of the
lar to the pa rt ic le = %5.3 f Wb/squa re −metre” , B )
; 11 12 // Res ult 13 // The mag ne tic
f i e l d perpendicu lar to the beam of th e pa rti cle = 0.075 W b/squ ar e −metre
Scilab code Exa 6.3 K conversion
1 // Scilab co de Exa6 .3 2 clc ; clear ; 3 m_ 0 = 9. 109 63e-0 31 ;
electron ,
−241 (2011)
// Re st mass of th e
Kg
4 e = 1.6 02 18e-01 9;
electron
: : Page
// Charge of th e
, C
5 c = 2. 99 79e+0 8;
// Vel oci ty of th e ligh t , 62
me tr e pe r sec 6 BR = 33 81e-0 06 ; // Fie ld −radius pr od uc t , tesla −m 7 E_ k = 37 .4 4; // B in di ng e ne rg y of k −electron 8 v = 1 / sqrt ((m_0/(BR*e))^2+1/c^2); // Vel oci ty of t he
conv erson
electron
, m/s
9 E = m_ 0*c^2* (1 / sqrt (1-v^2/c^2)-1)/(e*1e+003);
//
Energy of th e electron
, keV // Energy of th e conv ert ed gamma ray photon , KeV 11 printf ( ” \nThe en er gy of th e electro n = %6.2 f keV \ nThe energy of the converted gamma ray ph ot on = %6. 2 f keV” , E, E_C); 10 E_C = E+E _k ;
12 13 // Res ult 14 // The en er gy of th e electron = 624 .11 keV 15 // T he ene rgy of the conver ted gamma ra y ph ot on =
661.55
keV
Scilab code Exa 6.4 Average energy carried away by neutrino during beta
decay process 1 // Scilab co de Exa6 .4 2 clc ; clear ; 3 E = 18. 1;
parti cle ,
: : Page
−241 (2011)
// Ene rgy carried
by be ta
keV
// Av er ag e en er gy carrie d away by bet a par tic le , keV 5 E_r = E-E _a v; // The rest en er gy carried ou t by the neut rino , keV 4 E_a v = E/3 ;
6 7 printf ( ” \nThe rest
en er gy carried neutri no : %5.3 f KeV” , E_ r);
ou t by th e
8 9 // Res ult 10 // T he rest
e ne rg y carrie
d o u t by t he ne ut ri no : 63
12. 067 KeV
Scilab code Exa 6.5 Maximum energy available to the electrons in the
beta decay of Na24 1 2 3 4 5
// Scilab
co de Exa6 .5 : : Page
clc ; clear ; M_ Na = -8 42 0. 40 ; M_M g = -13 933 .56 7; E = (M_ Na-M_M g)/100 0;
− 242(2011)
// Mass of s od iu m 24 , keV // Mas s of ma gn es iu m 24 , k eV // E n erg y of th e elec tron ,
MeV 6 printf ( ” \nThe m aximum energy
av ai la bl e to the ele ctr on s in the bet a de ca y = %5.3 f MeV” , E);
7 8 // Res ult 9 // T he m aximum en er gy avai labl e to th e elect rons
the bet a de ca y = 5.513
in
MeV
Scilab code Exa 6.6 Linear momenta of particles during beta decay pro-
cess 1 // Scilab co de Exa6 .6 : : Page −242 (2011) 2 clc ; clear ; 3 c =1; // F or simplicity as sume sp ee d of
light
to
be unit y , m/s 4 E_ 0 = 0. 15 5;
// End point e ner gy , mega
ele ctr on volt s // En ergy of be ta particle
5 E_be ta = 0. 02 5;
mega electron
// Ene rgy of th e neu tri no ,
6 E_v = E_ 0- E_ bet a;
mega electron
,
volts volts
// Lin ear momentum of ne ut ri no , mega electron volts pe r c
7 p_v = E_ v/c ;
64
8 m = 0. 51 1; 9 M = 14* 1. 66e-2 7; 10 c = 3e +8;
// Mass of an electron , Kg // Mass of car bon 14 ,K g // Vel oci ty of lig ht , m e t r e
pe r sec // Charge of an electron
11 e = 1. 602 18e-1 9;
, co ul om b 12 p_be ta =
beta
sqrt (2*m*E_beta);
// Lin ear m omentum of
pa rt icl e , MeV/c
13 sin_the ta = p_b eta/p_v*si
nd(45);
// Si ne of an gl e
theta 14 p_R = p_b eta*cos d(45)+p_v*
//
sqrt (1-sin_theta^2);
Linea r momemtum of r e c o i l nucleus
, MeV/c
*M*e); // Recoil energy of produc t nucleus , MeV 16 printf ( ” \nThe li n ea r momentum of neu trin o = %4.2 f MeV/ c \nThe li ne ar momentum of bet a pa rt ic le = %6 . 4 f MeV/c \ nThe en er gy of th e rec oi l nuc leu s = %4 .2 f eV” , p_v, p_ be ta, E_ R); 15 E_R = (p_R*1. 6e -13/ 2.99 79e+08)^2/(2
17 18 // Res ult 19 // The li ne ar momentum of
neutri no = 0.13
MeV/c
20 // The / c li ne ar momentum of bet a pa rt ic le = 0.1598 M 21 // T he e ne rg y of th e reco il
eV
nu cl eu s = 1. 20 eV
Scilab code Exa 6.7 Energies during disintergation of Bi210
1 // Scilab co de Exa6 .7 : : Page −242 (2011) 2 clc ; clear ; 3 N = 3. 7e+10 *60 ; // Number of disin tegrat
ion ,
pe r sec 4 H = 0.0 268 *4. 182 ;
// He at pr od uc ed at the
ou tp ut , jou le 5 E = H/(N* 1. 6e-01 3);
partic
// Energy of th e be ta
le , joule 65
// Mas s of Bi sm ut h , MeV // Mas s of polon ium , MeV // End poin t ener gy , MeV // Ra ti o of be ta par ticl e en er gy w it h end poi nt en er gy 10 printf ( ” \nThe en er gy of th e be ta pa rti cl e = %5.3 f 6 7 8 9
M_B i = -1 4. 81 5; M_P o = -1 5. 97 7; E_0 = M_ Bi-M_P o; E_ra tio = E/E_ 0;
MeV \nThe ratio of be t a particle en er gy w i t h end point ene rgy = % 5.3 f ” , E, E_ ra ti o); 11 12 // Res ult 13 // The en er gy of th e be ta part icle = 0.31 6 M eV 14 // T he ratio of be t a particle en er gy w i t h end po in t
energy = 0.272
Scilab code Exa 6.9 The unstable nucleus in the nuclide pair
1 2 3 4 5 6 7 8 9 10
// Scilab
co de Exa6 .9 : : Page
clc ; clear ; M = rand (4,2); M(1, 1) = 7.0 182 *93 1.5 ; M(1, 2) = 7.0 192 *93 1.5 ; M(2, 1) = 13. 007 6*9 31. 5; M(2, 2) = 13. 010 0*9 31. 5; M(3, 1) = 19. 004 5*9 31. 5; M(3, 2) = 19. 008 0*9 31. 5; M(4, 1) = 33. 998 3*9 31. 5;
− 243(2011)
// M ass of lithium , MeV // M ass of beryllium , MeV // M as s of car bon , MeV // M ass of nitrogen , MeV // M ass of fluo rin e , MeV // Ma ss o f ne on , MeV // M ass o f pho sph oro us ,
MeV 11 12 13 14 15 16 17 18
// M as s of sulphur
M(4, 2) = 33. 998 7*9 31. 5; j =1;
// Check th e sta bi li ty
, MeV
!!!!
for i = 1:4 if ro un d (M(i, j+1 )-M(i ,j)) == 1 then printf ( ” \n From pair a : ” ) printf ( ” \n Be ( 4, 7 ) i s u n s t a b l e” els eif rou nd
(M (i, j+ 1)- M( i,j )) = =
66
2
then
);
19 20 21 22 23 24 25 26
printf ( ” \n From pair b : ” ) printf ( ” \n N( 7, 1 3 ) i s u n s t a b l e” ) ; els eif rou nd (M(i, j+1 )-M( i,j)) == 3 then printf ( ” \n From pair c : ” ) printf ( ” \n Ne ( 1 0, 1 9 ) i s u n s t a b l e
”); els eif rou nd (M(i, j+1 )-M( i,j)) == 0 then printf ( ” \n From pair d : ” ) printf ( ” \n P ( 1 5, 3 4 ) i s u n s t a b l e” ); end
27 28 end 29 30 // Res ult 31 // 32 // F rom pair a : 33 // Be ( 4, 7 ) i s u n s t a b l e 34 // F rom pair b : 35 // N( 7, 1 3 ) i s u n s t a b l e 36 // F rom pair c :
37 // Ne (d1 0:, 1 9 ) i s u n s t a b l e 38 // F rom pair 39 // P ( 1 5, 3 4 ) i s u n s t a b l e
Scilab code Exa 6.10 Half life of tritium
1 2 3 4 5 6
// Scilab
co de Exa6 .10 : : Page
−244 (2011)
clc ; clear ; ta u_ 0 = 70 00 ; // Time cons tant , sec M_mo d_s qr = 3; // Nuc lea r mat rix E_ 0 = 0. 01 8; // En er gy of beta sp ec tr um , MeV ft = 0.69 3*tau_0/M_mo d_sq r; // Co mp ar at iv e h alf
life 7 fb = 10 ^( 4. 0* log10 (E_0)+0.78+0.02);
67
//
8 t = 10^( log10 (ft)- log10 (fb));
// Hal f l i f e of H3,
sec 9 printf ( ” \nThe ha lf l i f e of H3 = % 4.2 e sec ” 10 11 // Res ult 12 // The hal f l i f e of H3 = 2.44 e+ 009 sec
, t);
Scilab code Exa 6.11 Degree of forbiddenness of transition
1 // Scilab co de Exa6 .11 : : Page −244 (2011) 2 clc ; clear ; 3 t_p = 33/0.9 2*365* 84800 ; // Par tia l half
l i f e for
be ta emi ssi on , sec 4 E_0 = 0. 51 ; // Kinetic en er gy 5 Z = 55; // At om ic num be r of cesium 6 log_ fb = 4. 0* log10 (E_0)+0.78+0.02*Z-0.005*(Z-1)* log10 (E_0); // Com par iti ve hal f l i f e 7 log_ ft1 = lo g_f b+ log10 (t_p); // For bidd en
tansition 8 // Fo r 8 perc ent be ta minus emissi on 9 t_p = 33/0.0 8*365* 84800 ; // P ar ti al hal
f life ,
sec 10 E_0 = 1. 17 ; // Kinetic en er gy 11 Z = 55; // Ato mi c energy 12 log_ fb = 4. 0* log10 (E_0)+0.78+0.02*Z-0.005*(Z-1)* log10 (E_0); // Com par iti ve hal f l i f e 13 log_ ft2 = lo g_f b+ log10 (t_p); // For bid den
transition 14 // Check th e degr ee of forbiddenness !!!!! 15 if lo g_f t1 <= 10 then 16 printf ( ” \ nFor 92 perc ent be ta emissi on : ” ) 17 printf ( ” \n\tTrans ition is on ce for bid den and parity ch an ge” ) ; 18 end 19 if lo g_f t2 >= 10
then
68
20 21
printf ( ” \ nFor 8 per ce nt be ta emi ssio n : ” ) printf ( ” \n\ t ransition is tw ic e fo rb idd en a nd n o parity ch an ge” ) ;
22 end 23 24 // Res ult 25 26 27 28
/ / For 92 per ce nt be ta emi ssio n : // Tr ans iti on is o n c e fo rb id de n and par ity c h a n g e // Fo r 8 pe rc en t be ta em is si on : // Tr ans iti on is tw ic e fo rb idd en a nd n o par ity change
Scilab code Exa 6.12 Coupling constant and ratio of coupling strengths
for beta transitons 1 // Scilab co de Exa6 .12 : : Page − 244(2011) 2 clc ; clear ; 3 h_ kt = 1.0 54 57e-3 4; // Re du ce d planc k ’ s constant
,
joule sec 4 c = 3e+ 08;
// velocity
of ligh t , m e t r e
pe r sec 5 m_e = 9. 1e-31 ; 6 ft_ O = 316 2.2 8;
// Mass of th e electron , Kg // Co mp ar at iv e ha lf l i f e for
oxygen 7 ft_ n = 117 4.9 0;
// Co mp ar at iv e ha lf
l i f e for
neutron 8 M_f _s qr = 2 // Ma tr ix eleme nt log (2)/(m_e^5*c^4*ft_O* 9 g_ f = sqrt (2*%pi^3*h_kt^7* M_f_sqr)); // Co up li ng co nst ant , joule cubi c
metre 10 C_rati o = (2*ft _O/(ft_ n) -1 )/3;
couplin g strength 11 printf ( ” \nThe value of coupling j o u l e c u b i c m e tr e \ nThe ratio = %5. 3 f ” , g_ f, C_ rat io); 69
// Ra ti o of constant = %6.4 e of cou plin g con st ant
12 13 // Res ult 14 // T he va lue
of cou plin g con sta nt = 1. 396 5e j o u l e c u b i c m e tr e 15 // T he ratio of cou plin g con sta nt = 1.4 61
−062
Scilab code Exa 6.13 Relative capture rate in holmium for 3p to 3s sub-
levels 1 2 3 4
// Scilab
co de Exa6 .13 : : Page
−245 (2011)
clc ; clear ; Q_ EC = 85 0; B_ p = 2. 0;
// Q value for ho lmium 1 61 , keV // B in di ng e ne rg y for p − orbital electron , keV 5 B_ s = 1. 8; // Bi nd in g en er gy for s − orbital electron , keV 6 M_ra tio = 0.05 *(Q_E C -B_p)^2/ (Q_E C -B_s)^2 ; // Ma tr ix rati o 7 Q_ EC = 2. 5; // Q value for ho lmium 1 63 , keV 8 C_ra te = M_ra tio*(Q_ EC-B_s)^2/(Q_
EC-B_p)^2*100
;
// The re la ti ve cap tur e rate in holmium , perc ent 9 printf ( ” \nThe re la ti ve cap tur e rate in holmium 161 = %3.1 f perc ent ” , C_ ra te); 10 11 // Res ult 12 // T he rel ati ve
ca pt ur e rate
in holmium 1 61 = 9.8
percent
Scilab code Exa 6.14 Tritium isotope undergoing beta decay
1 // Scilab co de Exa6 .14 : : Page 2 clc ; clear ;
70
−246 (2011)
// H alf l i f e of
3 t_hal f = 12.5* 365*2 4;
hyd rog en 3 , ho ur 4 lamb da =
// Decay cons tant , per
log (2)/t_half;
hour // Avogadr o ’ s nu mb er ,
5 N_0 = 6. 02 3e+26 ;
per mo le 6 7 8 9
m = 0. 1e-0 3; // Mass of tri tium , Kg dN_by_ dt = lam bda*m*N_ 0/3; // De cay rate , per ho ur H = 21 *4 .1 8; // He at pr od ue d , jou le E = H/d N_ by _d t; // The ave rag e en er gy of
t he be ta partic
le , joule en er gy of be ta par tic les = %4 .2 e jo ul e = %3.1 f keV” , E, E/1 .6e-0 16 );
10 printf ( ” \nThe av er age
11 12 // Res ult 13 // The av er ag e en er gy of be t a partic
les = 6.9 1e
−016
j o u l e = 4 . 3 keV
Scilab code Exa 6.15 Fermi and Gamow Teller selection rule for allowed
beta transitions 1 2 3 4 5 6 7 8 9 10
11
// Scilab
co de Exa6 .15 : : Page
−246 (2011)
clc ; clear ; S = string ( rand (2,1)) S(1, 1) = ’ antip aral lel sp in ’ S(2, 1) = ’ para llel sp in ’ for i = 1:2 if S(i, 1) == ’ antip arall el sp in ’ then printf ( ” \ nF or Fe rm i types : ” ) printf ( ” \n\n T he selection rule s for
al lo we d tran siti ons a r e : \n \td el ta I is z e r o \ n\td el ta p i is p l u s \nThe emit ed neutrin o and electro n hav e %s” ,S(i,1))
elseif
S(i, 1) ==
’ paral lel 71
sp in ’
then
printf ( ” \ nF or Gamow−Teller ty pe s : ” ) printf ( ” \nThe selection rules for al lo we d
12 13
tran plus plus have
siti ons ar e : \n \td el ta I is z e r o , on e and mi nus on e \n\td el ta p i is \ nThe em it ed neutri no and electro n %s” ,S(i,1))
14 end end 15 16 // Ca lc ul at io n of ratio of transit ion prob abil ity 17 M_ F = 1; // M at r ix for F ermi par tic les 18 g_ F = 1; // Co up li ng con st ant of fe rmi
particles // Ma tri x fo r Gamow Tel le r // Coup ling constant of Gamow
19 M_ GT = 5/ 3; 20 g_ GT = 1. 24 ;
Teller = g_ F^2*M_F/(g_G T^2*M_GT); // Ra ti o of transi tion prob abil ity 22 // Calc ulati on of S pa c e ph as e factor 23 e = 1. 6e-1 9; // Charge of an electron , coulomb 24 c = 3e+ 08; // Vel oci ty of lig ht , m e tr e 21 T_prob
sec 9; 25 K = pe 8.r99e+ 26 R_0 = 1. 2e-15 ;
// Coulom b const ant // Di st an c e of closest
approa ch , metre 27 28 29 30 31 32 33
A = 57; // Ma ss nu mb er Z = 28; // Ato mic nu mb er m_n = 1. 67 49e-2 7; // M ass o f neu tro n , K g m_p = 1. 67 26e-2 7; // M ass of pr ot on , K g m_e = 9. 1e-31 ; // Mass of electro n . Kg E_1 = 0. 76 ; // Fi rs t ex ci te d st at e of nic ke l delt a_E = ((3 *e^2*K/(5*R_ 0*A^(1/3 ))*((Z+1)^2-Z^2 )) -( m_ n - m_p ) * c ^2) /1.6 e -13; // Mass differ ence
, mega electr
on volts
34 E_0 = delt a_E -(2*m_e*c^2)/1.6e
en er gy , mega electro
-13;
// End point
n volts
35 P_facto r = (E_ 0 -E_1)^5 /E_0^5;
factor 72
// Sp ac e ph ase
36
printf ( ” \ nThe ratio
of transition probabil ity = %4 . 2 f \nThe spac e ph as e fac tor = % 4.2 f ” , T_ pr ob,
P_factor); 37 38 // Res ult 39 // T he em it ed neutri no and electro
n ha ve
antiparallel sp in // For Gamow −Tel ler t yp e s : / / T he selection rule s for al lo w ed transitions ar e : // del ta I is z e ro , pl us o ne and minus on e // d e l t a p i is p l u s // T he e mi te d neu tri no and electron h a ve par all el spin 45 // T he rati o of transit ion pro babi lity = 0. 39 46 // The sp ac e ph as e factor = 0.62 40 41 42 43 44
73
Chapter 7 Gamma Radiation
Scilab code Exa 7.1 Bragg reflection for first order in a bent crystal spec-
trometer 1 // Scilab co de Exa7 .1 : : Page −292 (2011) 2 clc ; clear ; 3 h = 6. 62 61e-0 34 ; // Pl anc k ’ s constant
,
joule sec 4 C = 2. 99 8e+0 8;
// Ve loc ity
of ligh t ,
me tr e pe r sec // Ra d i u s of focal
5 f =2;
circl e ,
metre // Inter plane r sp ac in g for
6 d = 1. 18e- 01 0;
qua rtz
crystal
, me tr e // Energy of th e
7 E_1 = 1.17* 1.602 2e-013;
gamma rays , jo ul e // Energy of th e
8 E_2 = 1.33* 1.602 2e-013;
gamma rays , jo ul e *d); // Dis tan ce to be moved for obta inin g f i r s t ord er ref lec tio n for two differe nt ene rgi es , m e t r e 10 printf ( ” \nThe distance to be mo ved for obtaining f i r s t order Br ag g re fl ec ti on = % 4.2 e me tr e” , D); 9 D = h*C*f*( 1/E_ 1 -1 /E_2)*1/(2
11
74
12 // Res ult 13 // The distance
to be moved for obtaining first or de r Brag g refl ectio n = 1. 08e −003 metre
Scilab code Exa 7.2 Energy of the gamma rays from magnetic spectro-
graph data 1 // Scilab co de Exa7 .2 : : Page −293 (2011) 2 clc ; clear ; 3 m_ 0 = 9. 10 94e-0 31 ; // Re st m ass of th e elec tron ,
Kg // Ma gn et ic fiel d , tes la me tr e // Cha rge of th e electron ,
4 B_R = 12 50e-0 6; 5 e = 1. 60 22 e-0 19;
coulomb // Vel oci ty of th e ligh t ,
6 C = 3e+ 08;
me tr e pe r sec // Bi nd in g en er gy of th e K
7 E_ k = 0. 08 9;
sh el l electron
8 v = B_ R*e/ (m_ 0* sqrt (1+B_R^2*e^2/(m_0^2*C^2)));
Velo city
−
,MeV //
of th e pho toe lec tro n , m e t r e pe r sec
9 E_pe = m_ 0/(1.602 2e -01 3)*C^2*(1 / sqrt (1-v^2/C^2)-1);
// En er gy of the photoel
ectron
,MeV
10 E_g = E_ pe+E_ k; // En er gy of the g amma rays , M eV 11 printf ( ” \nThe energ y of the gamma ray s = %5. 3 f MeV” E_g); 12 13 // Res ult 14 // The energy of the gamma rays = 0.212 MeV
Scilab code Exa 7.3 Attenuation of beam of X rays in passing through
human tissue 1 // Scilab
co de Exa7 .3 : : Page 75
−292 (2011)
,
2 clc ; clear ; 3 a_ c = 0. 22 1; // At ten uat ion coe ffi cie nt , cmˆ2 /g 4 A = (1- exp (-0.22))*100; // Att enu ati on of b eam of
X−ray s in passing
th ro ug h human tis sue ion of beam of X −ra ys in th rou gh human ti ss ue = %d percent ” , ceil (
5 printf ( ” \nThe attenuat
passing A));
6 7 // Res ult 8 // T he attenuat
ion of beam of X −ra ys in pas sin g th rou gh human ti ss ue = 2 0 percent
Scilab code Exa 7.4 Partial half life for gamma emission of Hg195 isomer
1 // Scilab co de Exa7 .4 : : Page −293 (2011) 2 clc ; clear ; 3 al ph a_ k = 45 ; // Rati o be tw ee n dec ay
constants 4 sum _al pha = 0. 08 ;
// Sum of alphas
5 P = 0.3 5*1 /60 ;
// Pro babi lity of t he on , pe r ho ur 6 lambda _g = P*sum_al pha/alpha _k; // D ecay constant of the g amma radiat ions , per ho ur 7 T_g = 1/ (lambda_ g *365 *24); // Partial life time fo r gamma emission , year s 8 printf ( ” \nThe pa rt ia l l i f e ti me for gamma emission = %5.3 f year s ” , T_ g); isomeric
transiti
9 10 // Res ult 11 // The par tia l l i f e ti me for
gamma emiss ion = 11.0 08
years
Scilab code Exa 7.5 Estimating the gamma width from Weisskopf model
76
1 2 3 4
// Scil ab c o d e Exa7 .5:
: Page
−294 (2011)
clc ; clear ; A = 11; E_g = 4. 82 ;
// Mas s num be r of bor on // En er gy of th e gamma radi atio n , mega electron volts 5 W_g = 0.06 75*A^(2/ 3)*E_g^3; // Gamma wid th , mega electron
volts
6 printf ( ” \nThe re qu ir ed gamma width = %5. 2 f MeV” , W_g ); 7 8 // Res ult 9 // The req ui red gamma wid th = 37.3 9 MeV
Scilab code Exa 7.8 K electronic states in indium
1 // Scil ab c o d e Exa7 .8: 2 clc ; clear ; 3 e = 1. 60 22e-1 9;
−295 (2011)
: Page
// Charge of an electron
,
coulomb 4 BR = 23 70e-0 6;
, tesla
// M ag ne ti c f i e l d in an orbit
metre
5 m_0 = 9. 10 94e-3 1; 6 c = 3e+ 08;
// Mass of an electr on , Kg // Ve loc ity of ligh t , m e t r e
pe r sec // velocity
7 v = 1 / sqrt ((m_0/(BR*e))^2+1/c^2);
of th e partic
le , m et r e pe r se c
8 E_ e = m_ 0*c^2*( (1-(v/c)^2)^(-1/2
) -1 )/1.6e-13 ;
En er gy of an electron , MeV 9 E_ b = 0. 02 8; // Binding ener gy , MeV 10 E_ g = E_ e+E _b; // Exci tati on ene rgy , MeV 11 alph a_k = 0. 5; // K co nv er si on c o e f f i c i e n t 12 Z = 49; // Number of protons 13 alph a = 1/1 37 ; // F in e structure con sta nt 14 L = (1/ (1-(Z^3/alpha
^(15/2))))/2;
_k*alpha^4*(2*0
// Angular 77
.511/ 0.392)
momentum
//
15 l = 1 ; // Orb ita l angul ar momentum 16 I = l-1 /2 ; // Pari ty 17 printf ( ” \ nFor K −elect ron state : \ nThe excit atio n
ener gy = % 5. 3 f MeV \nThe an gu la r momentum = % d \ nThe par ity : %3.1 f ” , E_g, ceil (L) , I); 18 // Res ult 19 20 21 22
// // // //
For K −electr on stat e : The excit atio n en er gy = 0.393 M eV The ang ula r momentum = 5 T he par ity : 0.5
Scilab code Exa 7.9
Radioactive lifetime of the lowest energy electric dipole
transition for F17 1 // Scilab co de Exa7 .9 : : Page −295 (2011) 2 clc ; clear ; 3 c = 3e+ 10; // Vel oci ty of lig ht ,
cen tim etr e pe r sec 4 R_0 = 1. 4e-13 ;
ap pr oa ch , centime 5 alph a = 1/1 37 ; 6 A = 17; 7 E_g = 5* 1. 6e-06 ;
// Di st an c e of closest tre // F in e scatterin g co nst ant // Ma ss nu mb er // Ene rgy of gamma
trans ition , er gs 8 h_cut =
// Re du ce d planck
1.054 57162 8e -27 ;
co ns ta nt , ergs
pe r sec
9 lamb da = c/4*R_ 0^2*alph a*(E_g/(h_cu
t*c))^3*A^(2/
3);
// Disintegration co ns ta nt , pe r sec 10 tau = 1/la mb da; // Radioactive lifr \ e ti me , sec 11 printf ( ” \nThe rad ioa cti ve l i f e ti me = % 1.0 e sec ” , tau); 12 13 // Res ult 14 // The radio active
l i f e ti me = 9e 78
−018 sec
Scilab code Exa 7.10 Electric and magnetic multipolarities of gamma rays
from transition between Pb levels 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
// Scilab
co de Exa7 .10 : : Page
−296 (2011)
clc ; clear ; l = 2, 3, 4 printf ( ” \nThe possi ble multipolarities ar e ” ) for l = 2:4 if l == 2 then printf ( ”E%d, ” , l); elseif l == 3 then printf ( ” M%d” , l); elseif l == 4 then printf ( ” and E%d” , l); end end for l = 2:4 if l == 2 then printf ( ” \nThe tr an si ti on E%d domina tes” ,l); end end
// Res ult // The possible mult ipo lari ties // The tra nsi tio n E2 do mi na te s
ar e E2 , M3 a nd E4
Scilab code Exa 7.13 Relative source absorber velocity required to obtain
resonance absorption 1 // Scilab
co de Exa7 .13 : : Page
2 clc ; clear ;
79
−297 (2011)
3 E_0 = 0.014 *1.60 22e-13;
// Ene rg y of the g amma
ra y s , joule // Ma ss nu mb er // Mass of ea ch nu cl eon ,
4 A = 57; 5 m = 1. 67e- 27 ;
Kg // Ve loc ity
6 c = 3e+ 08;
me tr e pe r sec
of ligh t ,
// Number of at om s in the
7 N = 100 0;
lattice // Rala tiv e veloci ty , me tr e pe r sec 9 printf ( ” \nThe rel ati ve sou rce abs orb er velocity .3 f m/s ” , v); 8 v = E_ 0/(A* N*m* c);
10 11 // Res ult 12 // The relati
ve so ur ce ab so rb er velocity = 0.0
= %5
79 m /s
Scilab code Exa 7.14 Estimating the frequency shift of a photon
1 // Scilab co de Exa7 .14 : : Page −297 (2011) 2 clc ; clear ; 3 g = 9.8; // Accel eratio n due to gra vi ty ,
me tr e pe r squ are sec 4 c = 3e+ 08;
// Ve loc ity
of ligh t ,
me tr e pe r sec // Vertical distanc e be t w ee n and abso rbe r , me tr e 6 del ta_ v = g*y/ c^2 ; // Fr eq ue nc y sh if t 7 printf ( ” \nThe requ ired fre que ncy sh ift of th e p h o to n = % 4.2 e ” , del ta _v); 5 y = 20;
source
8 9 // Res ult 10 // T he requi red
fr equ enc y shi ft of th e p h o t o n = 2.1 8
e −015
80
Chapter 8 Beta Decay
Scilab code Exa 8.3 Neutron and proton interacting within the deuteron
1 // Scilab co de Exa8 .3 : : Page −349 (2011) 2 clc ; clear ; 3 b = 1. 9e-1 5; // Width of squ are well
potential
, me tr e // Re duc ed planck ’ s
4 h_ kt = 1.0 545 71e-0 34 ;
co ns ta nt , joule
sec // Ve loc ity
5 c = 3e+ 08;
of ligh t ,
me tr e pe r sec 6 m_n = 1. 67e-2 7; // Mass of a nu cl eo n , Kg 7 V_0 = 40 *1 .6e-1 3; // De pt h , metre 8 E_ B = (V_ 0 -(1 /(m_n*c^2 )*(%pi*h_k t*c/(2*b))^2 ))/1.6e -13; // Bin din g en er gy , mega ele ctro n
volts 9 alp ha =
//
sqrt (m_n*c^2*E_B*1.6e-13)/(h_kt*c);
scatt erin g co efficient
, p e r m et r e // Probabil
10 P = (1 +1 /(alp ha*b) )^-1 ; 11 R_me an = sqrt (b^2/2*(1/3+4/%pi^2+2.5));
ity
// Mean squar e radiu s , me tr e 12 printf ( ” \nThe proba bilit y tha t th e pr ot on moves wit hin th e ra ng e of ne ut ro n = %4.2 f \nTh e mean squ are
radius
of th e deu te ron = %4.2 e me tr e” 81
, P,
R_mean); 13 14 // Res ult 15 // T he probability
th at th e pr ot on moves wi thi n th e ra ng e of ne ut ro n = 0.5 0 16 // T he m ean squ are radius of th e deu te ron = 2.4 2e −015 metre
Scilab code Exa 8.5 Total cross section for np scattering at neutron en-
ergy 1 2 3 4 5 6 7 8
// Scilab
co de Exa8 .5 : : Page
−349 (2011)
clc ; clear ; a_t = 5. 38e-1 5; a_s = -2 3. 7e-15 ; r_ ot = 1. 70e-1 5; r_ os = 2. 40e-1 5; m = 1. 67 48e-2 7; E = 1. 6e-1 3;
9 h_ cu t = 1.0 54 9e-34 ; 10 K_ sq r = m*E/h _c ut^2 ; 11 sig ma = 1/4* (3*4 *%pi*a_t^2/ (a_t^2*K_sq r+(1-1/2*K_sq *a_t*r_ot)^2)+4*%pi*a_s^2/(a_s^2*K_sqr+(1-1/2* K_sqr*a_s*r_os)^2))*1e+028; // To ta l cros s −
se ct ion
fo r n −p scatteri
ng , b a rn
12 printf ( ” \nThe tota l cro ss sec tion for n −p scattering = %5.3 f bar n” , si gm a); 13 14 // Res ult 15 / / T he tota l cro ss sec tio n for n −p scatt ering =
2.911
ba rn
Scilab code Exa 6.8 Beta decayed particle emission of Li8
82
r
1 2 3 4
// Scilab
co de Exa6 .8 : : Page
−243 (2011)
clc ; clear ; l =2; // Or bi ta l ang ula r m omentum q ua nt um nu mb er P = (+ 1)^2 *(-1 )^l; // Pari ty of th e 2.9 MeV leve l
in Be −8 5 M_L i = 7.0 18 2;
// Mass of li thium , MeV
6 M_ Be = 7.9 988 76; // Mass of bery llium , MeV 7 m_ n = 1; // Mas s of neut ron , MeV 8 E_th = (M_Li+m _n-M_Be)*931 .5; // Thre shold ene rgy
, MeV 9 printf ( ” \nThe parity
of th e 2.9 M eV l evel in be −8 = +%d \nThe thre shold en er gy for lith ium 7 ne ut ro n ca pt ur e = %d MeV” ,P, E_ th);
10 11 // Res ult 12 // T he par ity of t he 2.9 M eV level in be −8 = +1 13 // T he thre shold en er gy for lith ium 7 ne ut ro n
capture = 18 MeV
Scilab code Exa 8.8 Possible angular momentum states for the deuterons
in an LS coupling scheme 1 // Scilab co de Exa8 .8 : : Page −351 (2011) 2 clc ; clear ; 3 S =1; // Spin an gu la r momentum( s1+
4 5 6 7 8
−s2 ) , w he r e as s1 is t he sp in of pr ot on a nd s 2 is th e spin of ne ut ro n . m = 2*S+ 1; // S p i n multiplicity j =1; // Total ang ula r momentum printf ( ” \nThe po ss ib le angular momentum st at es wit h th ei r pari tie s a r e a s fo ll ow s : ” ) ; printf ( ” \n %dS%d h a s e v e n p a r i t y ” , m, j); printf ( ” \n
%dP%d h a s odd p a r i t y ”
, j);
83
, m
printf ( ” \n , j);
9
10 S = 0 ; 11 m = 2*S +1 12 printf ( ” \n ;
%dD%d h a s e v e n p a r i t y”
%dP%d h a s odd p a r i t y ”
, m
, m, j)
13 14 // Res ult 15 // The poss ible 16 // 17 // 18 // 19 //
angu lar momentum stat es wi th thei r pari tie s a r e a s fo ll ow s : 3 S1 h a s ev e n p a r i t y 3 P1 h a s odd p a r i t y 3D1 h a s ev e n p a r i t y 1 P1 h a s odd p a r i t y
Scilab code Exa 8.9 States of a two neutron system with given total an-
gular momentum 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
// Scilab
co de Exa8 .9 : : Page
clc ; clear ; printf ( ” \nThe possi ble
−351 (2011)
stat es ar e : ”
);
// For s = 0 s =0; // Spin ang ula r momentum m = 2*s+ 1; // S p i n multiplicity for j = 0:2 // Total ang ula r momentum l = j if l == 0 then printf ( ” \n %dS%d , ” , j,m); elseif l == 2 then printf ( ” %dD%d, ” , j,m); end end
// For s = 1 s =1;
17 m = 2*s+ 1;
84
18 l = 2 19 for j = 0:2 20 if j == 0 then 21 printf ( ” %dP%d, ” , j,m); 22 elseif j ==1 then 23 printf ( ” %dP%d, ” , j,m); 24 25 26 27 28 29 30 31 32 33 34 35
elseif j ==2 then printf ( ”%dP%d and ” , j,m); end end for j = 2 printf ( ” %dF%d” , j,m) end
// Res ult / / Po ss ib le sta te s a r e : / / T he possi ble stat es ar e : // 0 S1 , 2D1, 0P3, 1P3, 2 P3 and
2 F3
Scilab code Exa 8.10 Kinetic energy of the two interacting nucleons in
different frames 1 // Scilab co de Exa8 .10 : : Page −352 (2011) 2 clc ; clear ; 3 r = 2e-0 15 ; // Range of nuclear force ,
metre // Re du ce d value
4 h_ kt = 1. 05 46e-3 4;
co ns ta nt , joule
of Pl an ck ’ s
sec
5 m = 1. 67 4e-2 7; // Mass of ea ch nu cl eon , Kg 6 K = round (2*h_kt^2/(2*m*r^2*1.6023e-13));
//
Kine tic en er gy of e ac h nu cl eo n in cent re of mass frame , mega electro n volts 7 K_ t = 2*K; // To ta l kin eti c en er gy , mega ele ctr on volt s 8 K_ in c = 2*K_ t;
// Ki ne tic e ne rg y of th e inciden 85
t
nu cl eo n , mega electro
n volts en er gy of ea ch nuc leo n = %d MeV\nThe tot al ki ne ti c ene rgy = %d MeV \nThe kin eti c en er gy of th e incident nuc le on = %d MeV”
9 printf ( ” \nThe kin etic
K, K_t, K_ in c); 10 11 // Res ult 12 //
86
,
Chapter 9 Nuclear Models
Scilab code Exa 9.1 Estimating the Fermi energies for neutrons and pro-
tons 1 // Scilab co de Exa9 .1 2 clc ; clear ; 3 h_ cu t = 1. 05 4e-03 4;
: : Page
−389 (2011)
// Re du ce d Pla nc k ’ s constant
,
joule sec 4 rh o = 2e+ 04 4; 5 6 7 8 9
// Dens it y of th e nuclear ma tt er , kg per me tr e cu be V = 23 8/r ho ; // V olume of the nuclear ma tt er , met re cub e // For neu tro n N = 238- 92; // Number of neutrons M = 1.6 74 82e-02 7; // M ass of a neu tr on , kg e = 1. 60 2e-01 9; // E nerg y equivalen t of 1 eV, J/ eV
10 E_ f = (3 *%pi^2)^(2/
3)*h_cu t ^2/ (2*M)*(N/V)^(
// Ferm i ene rgy of ne ut ro n , eV 11 printf ( ” \nThe Fe rm i energy of neutron
= %5.2 f MeV”
E_f/1e+006); 12 // For pro ton 13 N = 92; // Number of protons 14 M = 1.6 74 82e-02 7;
// M ass of a pr ot on , kg 87
2/3 )/e; ,
// E nerg y equivalen
15 e = 1. 60 2e-01 9;
t of 1 eV, J/
eV 16 E_ f = (3 *%pi^2)^(2/
3)*h_cu t ^2/ (2*M)*(N/V)^(
// Ferm i ene rgy of ne ut ro n , eV 17 printf ( ” \nThe Fe rm i energy of proton
2/3 )/e;
= %5.2 f MeV”
,
E_f/1e+006); 18 19 // Res ult 20 // Th e Fe rmi ene rgy 21 // Th e Fe rmi ene rgy
of neu tro n = 48.92 MeV of pro ton = 35.96 MeV
Scilab code Exa 9.3 General propeties of a neutron star
1 // Scilab co de Exa9 .3 : : Page −390 (2011) 2 clc ; clear ; 3 h_ cu t = 1.0 54 5e-34 ; // Re du ce d Pl anc k ’ s constant
,
joule sec // Gravitational co nst ant , newton squar e me tr e per squar e Kg
4 G = 6. 6e-1 1;
5 m = 10 ^3 0; 6 m_n = 1. 67e-2 7; 7 R = (9 *%pi/4)^(2/ ^(1/3);
// Mass of th e sta r , Kg // Mass of t he ne ut ro n , Kg 3)*h_cut^2/
(G*(m_n)^3
)*(m_n/m)
// Ra di us of th e ne ut ro n sta r ,
metre 8 printf ( ” \nThe radius of th e ne ut ro n star = %3.1 e metre” , R); 9 10 // Res ult 11 // The radius of th e ne ut ro n star = 1.6 e+ 004 me tr e
Scilab code Exa 9.4 Stability of the isobar using the liquid drop model
1 // Scilab
co de Exa9 .4 : : Page 88
−391 (2011)
2 clc ; clear ; 3 A = 77; // Mass number of th e isoto pes 4 Z = round (A/((0.015*A^(2/3))+2)); // At omi c
number of stable
isotope !!!!!
5 // Check th e sta bi li ty 6 if Z == 34 then 7
printf ( ” \ nSe( %d, %d) is
8 9
); elseif Z == 33 then printf ( ” \ nAs( % d,% d) is
Br(% d,% d) are
Br(% d,% d) are
sta ble \ nSe (%d,% d) an d uns tab le ” , Z, A, Z+1, A, Z+2, A
); elseif Z == 35 then printf ( ” \ nBr( % d,% d) is
10 11
As(% d,% d) are 12 13 14 15 16
stable \nAs (%d,% d) and uns tab le ” , Z, A, Z-1, A, Z+1, A
sta ble \ nSe (%d,% d) an d uns tab le ” ,Z,A,Z-2,A,Z-1,A);
end
// Res ult // Se( 34 ,7 7) is sta ble // As (33 ,77) and Br(35 ,77)
are unstabl
e
Scilab code Exa 9.5 Energy difference between neutron shells
1 // Scilab co de Exa9 .5 2 clc ; clear ; 3 m_ 40 = 39. 962 589 ;
: : Page
−391 (2011)
// Mass of cal ciu m 40 ,
at omi c mass unit 4 m_ 41 = 40. 962 275 ;
// Mass of cal ciu m 41 ,
at omi c mass unit 5 m_ 39 = 38. 970 691 ;
// Mass of calc ium 39 ,
at omi c mass unit 6 m_n = 1.0 086 65;
// Mass of the ne ut ro n ,
at omi c mass unit 7 BE_1d = (m_39+m _n-m_40)*931.
5;
89
// Bind ing
en er gy of 1d 3/ 2 ne ut ro n , mega electro n volts 5; // Bind ing en er gy of 1 f 7/ 2 ne ut ro n , mega electro n volts 9 de lt a = BE _1 d -BE_ 1f; // E nergy diff eren ce be tw een ne ut ro n shel ls , mega electro n volts 10 printf ( ” \nThe en er gy dif fer enc e be tw een ne ut ro n 8 BE_1f = (m_40+m _n-m_41)*931.
s h e l l s = % 4.2 f MeV” , de lt a);
11 12 // Res ult 13 // T he en er gy diff erenc
e b et w e en ne ut ro n she ll s =
7. 25 MeV
Scilab code Exa 9.7 Angular frequency of the nuclei
1 // Scilab co de Exa9 .7 2 clc ; clear ; 3 h_ cu t = 1.0 54 5e-34 ;
co ns ta nt , joule
: : Page
−392 (2011)
// Re du ce d Plan ck ’ s
sec // Di st an ce of closest
4 R = 1. 2e-1 5;
approa ch , metre 5 6 7 8 9
// Mass of t he nu cl eo n , Kg
m = 1.6 74 82e-2 7;
// For O −17 for A = 17:60 // Mas s nu mbe rs if A == 17 then omeg a_O = 5*3^ (1/3 )*h_cut*17^ (-1/3)/(2^(7
// Ang ul ar frequency 10 // For Ni −60
/3)*m*R^2 );
of ox yg en
11 elseif A == 60 then 12 omeg a_Ni = 5*3^ (1/3 )* h_c ut*60^(-1/3 ) /(2^ (7/3 )*m*R ^2); // An gu la r fr equ enc y of nickel 13 end 14 end 15 printf ( ” \nThe angular frequency for ox yg en 17 = %4.2
e \ nThe angu lar
fre quen cy for
omeg a_O , omega_Ni);
90
nickel
60 = %4.2 e”
,
16 17 // Res ult 18 // Th e angular 19 // The ang ular
frequency for freq uenc y for
ox yg en 17 = 2.43 e+ 022 nickel 60 = 1.60 e+ 022
Scilab code Exa 9.9 Angular momenta and parities
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
// Scilab
co de Exa9 .9 : : Page
// Ele me nt s allo cated = ’ Carbo n ’ = ’ Bor on ’ = ’ Oxygen ’ = ’ Zi nc ’ = ’ Nitr ogen ’ = 6; // Number of pro ton in ca rb on nuclei Z( 2, 1) = 5; // Number of pro ton in bo ro n nuclei Z( 3, 1) = 8; // Number of proto n in ox yg en nuclei Z(4, 1) = 30 ; // Number of pr ot on in zinc nuclei Z( 5, 1) = 7; // Number of pro ton in nitrogen nuclei N( 1, 1) = 6; // Mas s num be r of carbon N( 2, 1) = 6; // Mass nu mb er of boro n N( 3, 1) = 9; // Mass nu mb er of oxy gen N(4, 1) = 37 ; // Mass number of zinc N( 5, 1) = 9; // Mass number of nitro gem E(1, 1) E(2, 1) E(3, 1) E(4, 1) E(5, 1) Z( 1, 1)
17 18 19 20 21 22 for i = 1:5 23
−393 (2011)
clc ; clear ; Z = rand (5,1); N = rand (5,1); E = string ( rand (5,1));
if Z( i, 1) == 8
then
91
24
elseif
25 26
27 28 29 30 31 32 33 34 35 36 37 end
printf ( ” \nThe ang ula r momentum i s 5/2 and t he par ity is +1 for %s ” , E(i, 1) ); Z( i, 1) == 5 then printf ( ” \nThe ang ula r momentum i s 3/2 and t he par ity is −1 for %s” , E(i, 1))
; end if Z( i,1) == N(i, 1) then printf ( ” \nThe angul ar mometum i s 0 an d the par ity is +1 for %s” , E(i, 1)); end if N( i,1)- Z(i, 1) == 2 then printf ( ” \nThe angul ar momentum i s 2 an d the pa ri ty is −1 for %s” , E(i, 1)); end if N( i,1)- Z(i, 1) == 7 then printf ( ” \nThe ang ula r momentum i s 5/2 an d t h e pa ri ty is −1 for %s” , E(i, 1)); end
38 39 // Res ult 40 // The ang ular m ometum is 0 and th e parity 41 // 42 // 43 // 44 //
Carbon T he angular m for Bor on The angular m fo r Oxyge n T he angular m for Z in c T he angular m for Nit rog en
is +1 for −1
omentum is 3/ 2 and the parity
is
omentum is 5/ 2 and the parity
is +1
omentum is 5/ 2 and the parity
is
omentum is 2 and the parity
92
is
−1 −1
Scilab code Exa 9.11 Quadrupole and magnetic moment of ground state
of nuclides 1 // Scilab co de Exa9 .11 2 clc ; clear ; 3 R_0 = 1. 2e-01 5; 4 5 6 7 8 9 10 11 12 13 14 15 16
: : Page
// Di st an ce of closest
approa ch , metre // M ass n umber of th e nuclei N = rand (4,1) N(1, 1) = 17 ; N(2, 1) = 33 ; N(3, 1) = 63 ; N(4,1 ) = 20 9; for i = 1:4
−394 (2011)
ar e allocated
b e lo w :
// for ox yge n // for su lp hu r // for co pp er // for bi sm ut h
if N( i,1 ) == 17 then printf ( ” \n For Ox ygen : ” ) I = 5/2; // Total ang ula r momentum l = 2; // Orb ita l angul ar momentum mu = -1 .9 1; // for odd ne ut ro n and I
= l +1/2 17
Q = -3/ 5*( 2*I-1)/(2// *I+2)*(R_0*N( i,1)^(1/ ^2*10^28; Q uadrupol e moment of 3))
oxy gen , barn 18
19 20 21 22
printf ( ” \n
The v a l u e o f m a g n e t i c moment i s : %4.2 f \n The v a l u e o f quadrup ole moment is : %6.4 f ba rn” , mu,
Q); elseif N( i,1 ) == 33 then printf ( ” \n\n Fo r S ul p hu r : ” ) I = 3/2; // Total ang ula r momentum l = 2; // Orb ital ang ul ar
momentum 23
// for
mu = 1. 91 *I/(I+ 1);
ne ut ro n and I = l 24
odd
−1/2
Q = -3/ 5*( 2*I-1)/(2 *I+2)*(R_0*N( i,1)^(1/3 )) ^2*10^28; // Quadrup ole moment of
sulph ur , ba rn 93
25
printf ( ” \n
The v a l u e o f m a g n e t i c moment i s : %5.3 f \n The v a l u e o f quadrup ole moment is : %6.4 f ba rn” , mu,
26 27
Q); elseif N( i,1 ) == 63 then printf ( ” \n\n For Copper : ” )
28 29
I = 3/2; l = 1;
30
mu = I+2 .2 9;
// Total ang ula r momentum // Orb ital ang ul ar
momentum // for
odd pr ot ons
an d I = l +1/2 31
Q = -3/ 5*( 2*I-1)/(2 *I+2)*(R_0*N( i,1)^(1/3 )) ^2*10^28; // Quadrupole momentum of
copp er , bar n 32
printf ( ” \n
The v a l u e o f m a g n e t i c moment i s : %4.2 f \n The v a l u e o f quadrup ole moment is : %6.4 f ba rn” , mu,
33 34 35
Q); elseif N( i,1 ) == 20 9 then printf ( ” \n\n For Bismuth : ” ) I = 9/2; // Total an gul ar m omentum
36 37
l // Orb lar odd momentum mu= =5;I-2.2 9*I/ (I+1 ); ita l angu // for pr ot on s
38
Q = -3/ 5*( 2*I-1)/(2 *I+2)*(R_0*N( i,1)^(1/3 )) ^2*10^28; // Quadrupole momentum of
and I = l
−1/2
bism uth , barn 39
printf ( ” \n
The v a l u e o f m a g n e t i c moment i s : %4.2 f \n The v a l u e o f quadrup ole moment is : %5.3 f ba rn” , mu,
Q); 40 end 41 end 42 43 // Res ult 44 // Fo r Oxygen : 45 // The v a l u e o f m a g n e t i c moment is : 46 // The v a l u e o f q u a d r u p o l e moment is
94
−1.91 :
−0.0 326 barn 47 48 // For Su lp hu r : 49 // The v a l u e o f m a g n e t i c moment is : 1 . 1 4 6 50 // The v a l u e o f q u a d r u p o l e moment is :
−0.0 356 barn 51 52 // For Cop pe r : 53 // The v a l u e o f m a g n e t i c moment is : 3 . 7 9 54 // The v a l u e o f q u a d r u p o l e moment is :
−0.0 547 barn 55 56 // Fo r Bi sm ut h : 57 // The v a l u e o f m a g n e t i c moment is : 2 . 6 3 58 // The v a l u e o f q u a d r u p o l e moment is :
−0. 22 1 bar n
Scilab code Exa 9.12 Kinetic energy of iron nucleus
1 // Scilab co de Exa9 .12 : : Page −395 (2011) 2 clc ; clear ; 3 h_cut = 1.054 57162 8e -34 ; // Red ue d planck ’ s
co ns ta nt , joule
sec // Di st an ce of closest
4 a = 1e-0 14 ;
approa ch , metre 5 m = 1. 67e- 27 ; // Mass of e ac h nu cl eon , Kg 6 KE = 14*%p i^2*h_cu t ^2/ (2*m*a^2*1 .6e-13);
Kinetic en er gy of iron nuc le us , MeV 7 printf ( ” \nThe kinetic en er gy of iron nuclei MeV”, KE) ; 8 9 // Res ult 10 // T he kineti
c en er gy of iron
95
nuclei = 28.
//
= %5.2 f
76 M eV
Scilab code Exa 9.14 Electric quadrupole moment of scandium
1 // Scilab co de Exa9 .14 : : Page 2 clc ; clear ; 3 R_0 = 1. 2e-15 ; // Dis ta nc e of
−396 (2011) clos est
a p p r o a ch ,
metre 4 5 6 7
j = 7/2; // Total ang ula r momentum A = 41; // Mas s nu mb er of Sc and ium Z = 20; // At om ic nu mb er of Ca lci um Q_S c = -( 2*j-1 )/(2*j+2)*(R_ 0*A^(1/3 ))^2;
//
El ect ri c quad rupo le of Scand ium nuc leu s , Sq . m 8 Q_ Ca = Z/( A-1)^ 2* abs (Q_Sc); // Elect ric qua drup ole of calc ium nuc leu s , Sq . m 9 printf ( ” \nThe e l e c t r i c quadr upole of scand ium nucleus = %4.2 e square me tr e \ nThe e l e c t r i c qua drup ole of calc ium nucleus = %4.2 e squar e metre” , Q_ Sc, Q_ Ca) ; 10 11 // Res ult 12 // The e l e c t r i c quadrup ole
of sc an di um nucleus = −1.14e −029 square me tr e 13 // The e l e c t r i c quadrup ole of calcium nucleus = 1.43 e −031 square me tr e
Scilab code Exa 9.16 Energy of lowest lying tungsten states
1 // Scilab co de Exa9 .16 : : Page 2 clc ; clear ; 3 h_cut_sqr _upon_ 2f = 0.01 667;
va lu e , joule 4 for I = 4:6 5 if I == 4
−398 (2011)
sq ua re pe r sec cu b e then
96
// A cons tant
E = I*(I+1 )*h_cut_sqr_ upon_2 f; printf ( ” \nThe en er gy for 4+ tun gst en state = %5 . 3 f MeV” , E); elseif I == 6 then E = I*(I+1 )*h_cut_sqr_ upon_2 f; printf ( ” \ nThe en er gy for 6+ tun gst en state =
6 7 8 9 10 11 12 13 14 15 16
end
%5 . 3 f MeV” , E);
end
// Res ult // The en er gy for 4+ tun gst en state // The en er gy for 6+ tun gst en state
97
= 0.333 M eV = 0.700 M eV
Chapter 10 Nuclear Reactions
Scilab code Exa 10.1 Q value for the formation of P30 in the ground state
1 // Scilab co de Exa10 .1 2 clc ; clear ; 3 M = 47 .6 68 ;
: : Page
−455 (2011)
// To ta l mass of react ion ,
MeV 4 E = 44 .3 59 ; // Total ener gy , MeV 5 Q = M-E ; // Q−val ue , MeV 6 printf ( ” \nThe Q −va lu e for th e fo rm at ion of P30 = %5 . 3 f MeV” , Q); 7 8 // Res ult 9 // The Q −val ue for th e for mat ion of P30 = 3.309 M eV
Scilab code Exa 10.2 Q value of the reaction and atomic mass of the resid-
ual nucleus 1 // Scilab co de Exa10 .2 2 clc ; clear ; 3 E_x = 7. 70 ;
: : Page
−455 (2011)
/ / Energy of th e al ph a partic 98
le , MeV
4 5 6 7 8 9
E_y = m_ x = m_ y = M_ X = M_ Y = th et a
4. 44 ; // En er gy of the pr oto n , MeV 4. 0; // Ma ss n umber of alp ha pa rti cle 1. 0; // M ass n umber of pro tiu m ion 14; // M ass n umber of nitrogen nucleus 17; // Ma ss nu mber of ox yg en nucleus = 90* 3.1 4/1 80; // Ang le be tw ee n incident
beam
dire ctio n and em itt ed pr ot on , degr ee
10 A_x = 4.0 026 033 ; / / Atomic mass of al ph a particle , u 11 A_X = 14. 003 074 2; // Atomic mass of nitrogen nuc le us
, u 12 A_y = 1.0 078 252 ; // Atomic ma ss of pr ot on , u 13 Q = ((E_ y*(1+m_y/M_ Y))-(E_x*( 1- m_ x/M_Y))-2/M_ Y* sqrt ((m_x*m_y*E_x*E_y))* cos (theta))/931.5; // Q−
va lue , u 14 A_ Y = A_ x+A_ X-A_ y-Q; // Atomic mass of O −17, u 15 printf ( ” \nThe Q −va lu e of th e reaction = %9.7 f u \ nThe at omi c mass of the O −17 = %1 0. 7 f u” , Q, A_Y) ; 16 17 // Res ult 18 // The Q −va lu e of t h e rea ct ion = − 0.0 012 755 u 19 // at omma ic ss mass = 16.9991278 20 // The Atomic of of th e th O e O−17 −17 : 16. 99 91 27 8 u
u
Scilab code Exa 10.3 Kinetic energy of the neutrons emitted at given an-
gle to the incident beam 1 // Scilab co de Exa10 .3 : : Page −455 (2011) 2 clc ; clear ; 3 m_p = 1.0 072 76; // Atomic mass of th e p ro to n ,
u 4 m_H = 3.0 160 49;
// Atom ic mass of the
tri tium , u 5 m_ He = 3.0 160 29;
// Atom ic mass of the He
io n , u 99
// Atomic mass of the
6 m_n = 1.0 086 65;
emit ted ne ut ro n , u // Q−va lu e in
7 Q = (m_p+m _H-m_H e -m_n)*931 .5;
MeV // Kine tic
8 E_ p = 3;
en er gy of
the proton , MeV 9 th et a = 30* 3.1 4/1 80; // an gl e , radi an cos (theta); 10 u = sqrt (m_p*m_n*E_p)/(m_He+m_n)*
// 11 v = ((m_H e*Q)+E_p*(m
_H e -m_p))/(m_
He+m_n);
// // Kine tic energy of the emitted neu tro n ,MeV 13 printf ( ” \nThe kin etic en er gy of th e em itt ed ne ut ro n = %5 . 3 f MeV” , E_ n); 12 E_n = (u + sqrt (u^2+v))^2;
14 15 // Res ult 16 // T he kineti
c en er gy of th e em it te d ne ut ro n = 1. 445
MeV
Scilab code Exa 10.4 Estimating the temperature of nuclear fusion reac-
tion 1 // Scilab co de Exa10 .4 2 clc ; clear ; 3 r_mi n = 4e-0 15 ;
: : Page
−456 (2011)
// Distanc e be tw ee n two
deut rons , me tr e 4 k = 1.3 806 504e-0 23 ;
// B olt zm ann ’ s co nsta nt ,
Jo ul e pe r kelvin 5 alph a = 1/1 37 ; // Fi ne structure con sta nt 6 h_red = 1.05 45716 8e -03 4; // Re duc ed planck ’ s
co nst ant , Jou le sec // Vel oci ty of lig ht ,
7 C = 3e+ 08;
me te r pe r sec ond 8 T = alp ha*h_red*C/
(r_min*k);
100
9 printf ( ” \nThe te mp er at ur e in th e fusion = %3. 1 e K” , T); 10 11 // Res ult 12 // T he te mp er at ur e in th e fusion reaction
reaction
is
is = 4.2 e
+009 K
Scilab code Exa 10.5 Excitation energy of the compound nucleus
1 2 3 4 5 6
// Scilab
co de Exa11 .5 : : Page
−456 (2011)
clc ; clear ; E_0 = 4. 99 ; // En er gy of the pr oto n , MeV m_ p = 1; // Mass number of the proton m_ F = 19; // Mass number of th e flo uri ne E = E_ 0/(1 +m_p/m _F); // En ergy of th e
re la ti ve mo ti on , MeV 7 A_F = 18. 998 405 ;
// Atom ic mass of the
flu orin e , amu 8 A_H = 1.0 072 76;
//
Atomic mass of th e
proton , amu 9 A_ Ne = 19. 992 440 ;
// Atom ic mass of the neon
, amu .5; // Bind ing ene rgy of the abs orb ed pr ot on , MeV 11 E_ex c = E+de l_ E; // Exci tati on en er gy of the co mp ou nd nuc leu s , MeV 12 printf ( ” \nThe exc ita tio n ene rgy of the co mpound nu cl eu s = %6. 3 f MeV” , E_ ex c); 10 del_E = (A_F+A _H-A_Ne)*931
13 14 // Res ult 15 // T he excit atio n en er gy of th e compound nucl eus =
17 .0 74 MeV
101
Scilab code Exa 10.6 Excitation energy and parity for compound nucleus
1 2 3 4
// Scilab
co de Exa10 .6 : : Page
clc ; clear ; E_ d = 0. 6; m_ d = 2;
−457 (2011)
// En er gy of the deu tro n , MeV // Mass number of the deutr on
5 m_L i = 19 ; // Mass number of the Lit hiu m 6 E = E_ d/(1 +m_d/m _L i); // Energy of th e
re la ti ve mo ti on , MeV // At om ic ma ss of th e Li th iu m , amu A_ d = 2. 01 5; // Atomic mass of th e de ut ro n , amu A_B e = 8. 00 8; // Atomic mass of the Beryllium , amu del_E = (A_Li+A _d-A_Be)*931. 5; // Bind ing ene rgy of the abs orb ed pr ot on , MeV E_ex c = E+de l_ E; // Exci tati on en er gy of the co mp ou nd nuc leu s , MeV l_ f = 2; // or bi ta l angular momentum of two al ph a par ticl e
7 A_L i = 6. 01 7; 8 9 10 11 12
13 P = co (-1 )^2; mp)^l_ ou ndf*(+1 nuc leu s
// Pa ri ty of t he of the co mpound nuc leu s = % 6. 3 f MeV \ nThe par ity of the compound nucleus = %d” , E_ ex c , P) ;
14 printf ( ” \nThe exc ita tio n ene rgy
15 16 // Res ult 17 // T he excit atio n en er gy of th e compound nucl eus =
22 .8 99 MeV 18 // The parity
of th e compound nucle us = 1
Scilab code Exa 10.7 Cross section for neutron induced fission
1 // Scilab
co de Exa10 .7 : : Page 102
−457 (2011)
2 clc ; clear ; 3 la mbd a = 1e-0 16 ;
// Disintegration
co ns ta nt ,
pe r sec // Ne ut ro n flux , neutr ons
4 ph i = 10 ^1 1;
pe r squ are c m per 5 sigma =
sec
5*lamb da/(phi*10
se ct ion , mill i ba rn s section f i s s i o n = %d m i l l i barns”
6 printf ( ” \nThe cross 7 8 // Res ult 9 // The cross
milli
section ba r ns
for
// Cr oss
^-27);
for
ne ut ro n in du ce d
, si gm a);
ne ut ro n in du ce d f i s s i o n = 5
Scilab code Exa 10.8 Irradiance of neutron beam with the thin sheet of
Co59 1 2 3 4
// Scilab
co de Exa10 .8 : : Page
clc ; clear ; N_ 0 = 6. 022 52e+0 26 ; rho = 8.9 *10 ^3;
−59, K g per cubic 5 M = 59; 6 si gm a = 30 e-02 8;
square
// Av og ad ro ’ s const ant // Nu cl ea r dens ity of Co me tr e // Ma ss nu mb er // Cr os s sec tion , pe r
met re
// Ne ut ro n flux , neutr ons pe r sq ua re m e t r e pe r sec d = 0. 04e- 02 ; // Thi ckn ess of Co −59 sheet , me tr e t = 3* 60 *6 0; // To ta l reaction time , sec t_hal f = 5.2*3 65*864 00; // Hal f l i f e of Co −60, sec la mbd a = 0. 69 3/t_h al f; // Disintegration
7 ph i = 10 ^1 6; 8 9 10 11
−457 (2011)
co nst ant , pe r sec 103
12 N_nuc lei =
//
round (N_0*rho/M*sigma*phi*d*t);
Number of nucl ei of Co −60 produc ed = lam bda*N_nuc lei; // I n i t i a l activity , de ca ys pe r sec 14 printf ( ” \nThe number of nu cl ei of Co60 pro duc ed = %5 . 2 e \nThe i n i t i a l ac ti vi ty per Sq . me tr e = % 1.0 g 13 Init_act ivity
dec ays per sec ” , N_n ucl ei , Init_ activ ity); 15 16 // Res ult 17 // The number of nu cle i of Co60 pro duc ed = 1.18 e+ 019 18 // The i n i t i a l ac ti vi ty per Sq . me tr e = 5e+ 010 de ca ys pe r sec
Scilab code Exa 10.9 Bombardment of protons on Fe54 target
1 // Scilab co de Exa10 .9 2 clc ; clear ; 3 d = 0.1;
Kg per squre
: : Page
// Thi ckn ess of Fe
9 10
11 12 13
−54 sheet ,
me tr e
4 M = 54; 5 m = 1. 66e- 02 7; 6 n = d/ (M *m ); 7 8
−458 (2011)
// Mas s num be r of Fe // Mass of the pr ot on , Kg // Number of nuclei in uni t ar ea of th e ta rg et , nuclei pe r sq ua re m e t r e ds = 10 ^-5 ; // Area , me tr e square r = 0.1; // Dis tan ce be tw ee n detector and target foil , m e t r e d_omeg a =ds/r^2 ; // Solid an gl e , steradi an d_sigm a = 1.3e-03 *10 ^ -3* 10^-28; // Differ entia l cross sec ti on , sq ua re m e t r e p er nuclei P = d_ si gm a*n; // Prob ablit y , eve nt per pro ton I = 10^-7 ; // Cur ren t , am pe re e = 1. 6e-1 9; // Ch ar ge of the pr ot on , C 104
// Number of proton s per sec ond beam , pr ot on pe r sec 15 dN = P* N; // Number of events detected pe r se co nd , eve nts pe r sec 16 printf ( ” \nThe number of events detected = %d events per sec ” , dN) ; 14 N = I/e;
in th e incident
17 18 // Res ult 19 // T he nu mber of eve nts
detec ted = 9 0 even ts pe r sec
Scilab code Exa 10.10 Fractional attenuation of neutron beam on passing
through nickel sheet 1 2 3 4
// Scilab
co de Exa10 .10
clc ; clear ; N_ 0 = 6. 022 52e+2 6; si gm a = 3. 5e-28 ;
: : Page
−458 (2011)
// Av og ad ro ’ s cons tant // Cr os s sec tion , sq uar e
metre // Nuc lear
5 rho = 8. 9e+03 ;
densit y , Kg
pe r cubi c me tr e 6 M = 58; 7 summat ion = rh o/M*N_0*sig
section
ma;
, pe r me tr e // Thi ck nes s of
8 x = 0. 01e- 02 ;
nickel
// Ma ss nu mb er // Ma cr osc opi c cross
she et , me tr e
exp (summation*x/2.3026); // Fractional attenuat ion of ne ut ro n beam o n passing th ro ug h nickel she et 10 printf ( ” \nThe fra cti on al attenuat ion of ne ut ro n b eam on passing th ro ug h nickel sheet = % 6.4 f ” , 9 I0_rat io_I =
I0_ratio_I); 11 12 // Res ult 13 // T he frac tion al
pass ing
atte nuat ion
th ro ug h nickel
of ne ut ro n beam on
she et = 1. 001 4 105
14 // W rong an sw er given
Scilab code Exa 10.11
in the tex tboo k
Scattering contribution to the resonance
1 // Scilab co de Exa10 .11 : : Page 2 clc ; clear ; 3 lamb da = sqrt (1.45e-021/(4*%pi));
−458 (2011) //
Wave leng th , metre 4 W_r ati o = 2. 3e-07 ; 5 sig ma = W_r atio*(4 *%pi)*lambd
// Width rat io a ^2*1 0^28 ;
// Scatteri
ng cont ribu tion , bar n contribu tion e = % 4.2 f bar ns” , si gm a);
6 printf ( ” \nThe scattering
resonanc
7 8 // Res ult 9 // The scattering
3.33
cont ribu tion
to th e
t o t he re so na nc e =
ba rn s
Scilab code Exa 10.12 Estimating the relative probabilities interactions
in the indium 1 // Scilab co de Exa10 .12 2 clc ; clear ; 3 si gm a = 2. 8e-02 4;
: : Page
−458 (2011)
// Cr os s section
, me tr e
square 4 la mbd a = 2. 4e-11 ;
// de Brogl ie wa ve le ng th ,
metre da^2; // Rela tive pro bab ili ties of (n , n) and (n , y) in i n di u m 6 printf ( ” \nThe relati ve proba bilit ies of (n ,n) and (n , y) in ind ium = % 5.3 f ” , R_ pr ob); 5 R_pro b = %p i*sigma/lamb
7 8 // Res ult
106
9 // The relati
ve proba bilit ies indium = 0.015
of (n ,n) and (n , y) in
Scilab code Exa 10.13 Peak cross section during neutron capture
1 // Scilab co de Exa10 .13 2 clc ; clear ; 3 h = 6. 62 5e-3 4;
: : Page
−459 (2011)
// Pla nck ’ s constant
,
joule sec 4 5 6 7
m_n E = w_ y w_ n
= 1. 67e-2 7; // Mass of neu tro n , Kg 4. 90 6; // En ergy , jou le = 0. 12 4; // radiati on width , eV = 0. 00 7*E^(1 /2 ); // Probabil ity
of e la st ic emi ssio n of ne ut ro n , eV // Total ang ula r momentum // To ta l angu lar momentum in the com po un d st at e
8 I =3; 9 I_ c = 2;
10 sig ma = ((h^2 )*(2*I_c+1 )*w_y*w_n)*10^ *1.602e-019*(2*I+1)*(w_y+w_n)^2);
section
28/(2 * %p i*m_n*E
// Cro ss
, ba rn s
11 printf ( ” \nThe cross section of ne ut ro n ca pt ur e = %5 .3 e bar ns” , si gm a); 12 13 // Res ult 14 // The cross section of ne ut ro n cap tur e = 3.755 e+ 004
barns
Scilab code Exa 10.14 Angle at which differential cross section is maxi-
mumat a givem l value 1 // Scilab co de Exa10 .14 2 clc ; clear ; 3 R =5;
: : Page
−459 (2011)
// Ra di us , fe mto me tr e 107
// The va lu e of k for
4 k_d = 0. 98 ;
deutron // The v al ue of k for trit on // Ang les at wh ic h d i f f e r e t i a l cr os s se ct io n i s maximum, degree 7 // Use of for lo op for ang le s calculation ( in de gr ee ) 5 k_p = 0. 82 ; 6 the ta = rand (1,5);
8 for l = 0:4 th et a = round (( acos ((k_d^2+k_p^2)/(2*k_d*k_p)-l 9 ^2/(2*k_d*k_p*R^2)))*180/3.14); 10 printf ( ” \ nF or l = %d” , l); 11 printf ( ” , the value of th et a max = % d degree” , ceil (theta)); 12 end 13 14 // Res ult 15 // For l = 0 , th e va lue of th e ta ma x = 0 de gre e 16 // For l = 1 , th e va lue of th e ta ma x = 8 de gre e 17 // For l = 2 , th e valu e of th et a m ax = 2 4 degr ee 18 // For l = 3 , th e valu e of th et a m ax = 3 8 degr ee 19 // For l = 4 , th e valu e of th et a m ax = 5 2 degr ee
Scilab code Exa 10.15
1 2 3 4 5 6
// Scilab
Estimating the angular momentum transfer
co de Exa10 .15
: : Page
−459 (2011)
clc ; clear ; k_d = 2. 02e+3 0; // The va lu e of k for de ut ro n k_t = 2. 02e+3 0; // The v al ue of k for trit on th et a = 23* 3.1 4/1 80; // An gl e , radi ams q = sqrt (k_d+k_t-2*k_t* cos (theta))*10^-15;
// th e va lu e of q in f e m t o m e tr e // Di st anc e of clos est a p p r o a ch , femt o met re 8 A = 90; // Mas s num be r of Zr −90 9 z = 4.3 0; // De ut ro n size , fe mt o me tr e 7 R_ 0 = 1. 2;
10 R = R_ 0*A^(1 /3 )+1/2 *z;
// R ad i us of th e 108
nucleu s , fe mto me tr e 11 l = round (q*R);
// Orb ital
ang ul ar
momentum // To ta l angu lar
12 I = l+ 1/2
momentum 13 printf ( ” \nThe to ta l angular
momentum tr an sf er = %3.1
f ” , I);
14 15 // Res ult 16 // The tot al
angular
momentum tra ns fer = 4.5
109
Chapter 11 Particle Accelerators
Scilab code Exa 11.1 Optimum number of stages and ripple voltage in
Cockcroft Walton accelerator 1 2 3 4 5 6 7 8 9
// Scilab
co de Exa11 .1 : : Page
− 535(2011)
clc ; clear ; V_0 = 10 ^5 ; // Acce lerati ng vo lt age , volts C = 0. 02e- 00 6; // Cap aci tan ce , farad I = 4* 1e-0 03 ; // Cur ren t , am pe re f = 200; // Frequ enc y , cycles pe r sec n = sqrt (V_0*f*C/I); // Number of p ar ti cle s delta_ V = I*n*(n+1 )/(4*f*C); printf ( ” \nThe op timum number of sta ges in the acce lera tor = %d” , n); 10 printf ( ” \nThe rip ple voltage = %4.1 f kV” , del ta_ V/1e +003); 11 12 // Res ult 13 // T he op timum number of stages in th e acce lera tor =
10 14 // T he ripple
volt age = 27.5 k
110
V
Scilab code Exa 11.2 Charging current and potential of an electrostatic
generator 1 // Scilab co de Exa11 .2 : : Page −536 (2011) 2 clc ; clear ; 3 s = 15; // Spe ed , me tr e per sec 4 w = 0.3; 5 E = 3e+ 06;
// Width of th e electrode , m et re // Breakdown strengt h , volt s per
metre // A bs ol ut e perm itiv ity of free pe r me tr e C = 11 1e-1 2; // Cap ac ita nce , farad i = round (2*eps*E*s*w*10^6); // C ur re nt , mic ro ampere V = i/ C*1 0^- 12 ; // Rate of rise of elect rode pot ent ial , mega volts pe r se c printf ( ” \nThe charging curren t = %d mi cr o −ampere \ nThe rat e of rise of electr ode potenti al = %4. 2 f MV/ s e c ” , i, V);
6 eps = 8. 85e-1 2;
sp ac e , farad
7 8 9 10
11 12 // Res ult 13 // current mi cro 14 // TThe he char ra te ging of rise of =ele23ctr 9ode pote −ampere nti al = 2. 15 MV
/ se c
Scilab code Exa 11.3 Linear proton accelerator
1 // Scilab co de Exa11 .3 : : Page −536 (2011) 2 clc ; clear ; 3 f = 20 0* 10 ^6 ; // F r e qu en c y of th e acceler
ator
, cy cl e p e r se c 4 M = 1. 67 24e-2 7; 5 E = 45. 3*1 .6e-1 3;
// Mass of t he pr ot on , Kg // Accelerating en er gy ,
joule 6 L_ f =
round
(1/f* sqrt (2*E/M)*100);
111
// Le ng th of
t h e final
drift
tube , cen ti m e t r e // Len gt h of th e f i r s t dr if t tu be , me tr e 8 K_E = (1/ 2*M*L_1^2 *f^2)/1.6e-1 3; // Kine tic en er gy of th e injec ted pr ot on , MeV 9 E_ in c = E/1. 6e-1 3-K_E; // Increase in e ne r gy , 7 L_1 = 5. 35 *1 0^-2;
MeV
// Ch ar ge of the pr ot on ,
10 q = 1. 6e-1 9;
C 11 V = 1. 49e+ 06 ;
// Accel eratin g vo lt age ,
volts 12 N = E_ in c*1. 6e-13 /(q*V );
// Number of dr if t
protons 13 L = 1/f* sqrt (2*q*V/M)* integrate ( ’nˆ(1/2) ’ , ’ n ’ , 0, N ); // To ta l len gth of th e accelera tor , m e tr e 14 printf ( ” \nThe len gth of th e fin al dri ft tu be = %d cm
\ nThe kinetic en er gy of th e injected pr ot ons = %4 . 2 f MeV \ nThe total le ngt h of t he accelerator = %3 .1 f me tr e” , L_f, K_E, L) ; 15 16 // Res ult 17 // T he ng th ofenter h egyfinal t u b ed= pr47otcon m s = 0. 60 18 // T he lekinetic of thdrift e injecte
MeV 19 // T he total
le ng th of t he accelerator = 9
.2 m e t r e
Scilab code Exa 11.5 Energy and the frequency of deuterons accelerated
in cyclotron 1 2 3 4 5
// Scilab
co de Exa11 .5 : : Page
−536 (2011)
clc ; clear ; B = 1.4; // M a gn e t ic field , tesla R = 88e-0 02 ; // Ra di us of th e orbit , me tr e q = 1. 60 23e-0 19 ; // Cha rge of th e
deut ron , C 112
// Mass of th e
6 M_d = 2.014 102*1 .66e-27;
deutron , Kg // Mass of th e He
7 M_He = 4.002 603*1 .66e-27;
ion , Kg 8 E = B^2*R^2 *q^2/(2*M_
d*1.6e-13 );
th e em er gi ng de ut ro n , mega electro
// En er gy og n volts
9 f = B*q/ (2 *%pi*M _d)*1 0^-6 ;
// Fr eq ue nc y of pe r sec 10 B_He = 2*%p i*M_He*f*10^6 /(2*q); // Magnetic field required for He(++) ions , we be r per squar e me tr e 11 B_c ha nge = B-B_ He; // Chang e in magn eti c fie ld , tes la 12 printf ( ” \nThe ene rgy of the eme rgi ng deu tro n = %4.1 f MeV\nThe freq uenc y of th e de e voltage = %5.2 f MHz\nThe ch ang e in magne tic f i e l d = % 4.2 f te sl a ” , th e de ut ro n vo lt age , mega cycles
E, f, B_ ch an ge) ; 13 14 15 16 17
// // // //
Res ult The ene rgy of the eme rg ing deu tro n = 36.4 The freq uenc y of th e de e voltage = 10.68 M The ch an ge in mag ne tic f i e l d = 0.01 te sl a
MeV Hz
Scilab code Exa 11.6 Protons extracted from a cyclotron
1 // Scilab co de Exa11 .6 : : Page −537 (2011) 2 clc ; clear ; 3 K_ E = 7.5 *1. 602 3e-13 ; // Kinetic en er gy ,
joule 4 r = 0.5 1;
// Ra di us of th e pr ot on
’ s orbit , me tr e 5 E = 5* 10 ^6 ;
// Ele ctr ic fie ld , vo lts
per me tr e 6 m = 1. 67e- 27 ; 7 q = 1. 60 23e-1 9;
// Mass of the pr ot on , Kg // Char ge of th e
prot on , C 113
// Velo city
8 v = sqrt (2*K_E/m);
of th e pr ot on ,
me tr e pe r sec // The effec tive in m ag n e t ic field , tesla 10 B = m* v/ (q* r); // Total magnetic f pr od uc ed , te sl a 9 B_ re d = E/v ;
re du ct ion
ield
11 r_c ha nge = r*B_ re d/B; 12
// The ch an ge in orbit radius , me tr e printf ( ” \ nThe ef fe ct iv e reducti on in ma gn et ic f i e l d = %5.3 f tes la \ nThe c ha n ge in orbit radi us = %5.3 f me tr e ” , B_ re d , r_ cha ng e);
13 14 // Res ult 15 // The ef fe ct iv e reduct ion
tesla 16 // T he c ha ng e in orbit
in ma gn et ic f i e l d = 0.132
radiu s = 0.087
metre
Scilab code Exa 11.7 Energy of the electrons in a betatron
1 2 3 4 5 6 7
// Scilab
co de Exa11 .7 : : Page
−537 (2011)
clc ; clear ; B = 0.4; // M a gn e t ic field , tesla e = 1. 62 03e-1 9; // Charg e of an elec tron , C R = 30* 2. 54e-0 2; // Ra di us , met re c = 3e+ 08; // Cap aci tan ce , farad E = B*e* R*c/1 .6e-1 3; // The en er gy of th e
elec tron , mega electron volts // Freq ue nc y , cycles pe r sec // Total number of revolutions 10 Avg_E_p er_re v = E*1e+006 /N; // Av er age energy ga in ed pe r re vol uti on , elect ron vol t 11 printf ( ” \nThe ene rgy of the ele ctro n = %4.1 f MeV \ nThe ave rag e en er gy gai ned pe r revolution = %6.2 f 8 f = 50; 9 N = c/(4 *2 *%pi*f *R);
eV” , E, Avg _E_ per _re v); 114
12 13 // Res ult 14 // T he en er gy of th e electro n = 92.6 MeV 15 // The av er ag e en er gy ga in ed pe r revolution = 295.
eV 16 // Not e : Wrong an sw er is given 17 //
57
in the text book
A ve r a g e en er gy ga in ed pe r revo luti on : 29 5. 57 ele ctr on volt s
Scilab code Exa 11.8 Electrons accelerated into betatron
1 2 3 4
// Scilab
co de Exa11 .8 : : Page
clc ; clear ; R = 0.3 5; N = 10 0e+06/ 480 ;
−537 (2011)
// Or bi t rad ius , me tr e // Total number of
revolutions // Dis ta nc e trave rsed
5 L = 2*%p i*R *N;
th e electron 6 t = 2e-0 6; 7 e = 1. 62 03e-1 9;
by
, me tr e // Pu lse dur at ion , sec // Cha rge of an electron
,
C 8 9 10 11 12 13
n = 3e+ 09; // Number of ele ctr on s f = 180; // fr eq uen cy , hertz I_p = n*e /t ; // Peak current , am pe re I_ av g = n*e *f; // Av er age current , am pe re ta u = t* f; // Duty cycle printf ( ” \ nThe pe ak cur ren t = %3.1 e am pe re \ nThe
average curren t = %4.2 e am pe re = %3. 1 e” , I_p, I_ av g , ta u); 14 15 16 17 18
// // // //
\ nThe du ty cyc le
Res ult The pe ak current = 2.4 e −004 ampe re The ave rag e curren t = 8.75 e −008 ampe re The dut y cycle = 3.6 e −00 4
115
Scilab code Exa 11.9 Deuterons accelerated in synchrocyclotron
1 2 3 4
// Scilab
co de Exa11 .9 : : Page
clc ; clear ; q = 1. 60 23e-1 9; B_ 0 = 1. 5;
−538 (2011)
// Charg e of an elec tron , C // Mag net ic f i e l d at the
ce nt re , tesla // Mass of th e
5 m_d = 2.014 102*1 .66e-27;
deutron , Kg B_0*q/( 2*%pi*m_ d*10^6); // Maximum fr equ en cy of th e d e e vo lt age , mega cycles pe r sec B_p rim e = 1.4 310 ; // Mag net ic f i e l d at the pe ri phe ry of th e dee , tesla f_pr ime = 10 ^7 ; // Fr equ enc y , cycles pe r sec c = 3e+ 08; // Vel oci ty of th e ligh t , me tr e pe r sec M = B_pr ime*q/(2 *%pi*f_pri me*1.66e-27 ); //
6 f_max = 7 8 9 10
Relati
visti c mass , u
.5; // Kine tic e ne rg y of th e partic le , mega electr on volts printf ( ” \ nThe m aximum frequency of the dee volt age = %5. 2 f MHz \ nThe kinetic en er gy of th e de ut er on = %5 . 1 f MeV” , f_ ma x , K_ E);
11 K_E = (M-m_d/1. 66e-27)*931 12
13 14 // Res ult 15 // T he m aximum frequency
of the de e voltage = 11.44 MHz 16 // T he kin eti c en er gy of th e de ute ron = 17 1.6 M eV
Scilab code Exa 11.10 Electrons accelerated in electron synchrotron
116
1 2 3 4 5 6
// Scilab
co de Exa11 .10
clc ; clear ; e = 1. 60 23e-1 9; E = 70 *1 .6e-1 3; R = 0.2 8; c = 3e+ 08;
: : Page
// Charg e // Energy , // Rad iu s of // Vel oci ty
pe r sec
−538 (2011) of an elec electron th e orb it of lig ht ,
tron , C volts , metre m e tr e
// Mag net ic f i e l d int ens ity ,
7 B = E/ (e* R* c);
tesla // Fr eq ue nc y , cyc le
8 f = e*B* c^2/ (2 *%pi*E );
pe r sec 88.5* (0.07 ) ^4*1 0^3/ (R); // En er gy rad iat ed by an ele ctr on , elect ron volts printf ( ” \ nThe frequency of the appl ied e l e c t r i c f i e l d = % 5.3 e cycl es per sec \nThe magnetic f i e l d int ensi ty = % 4.3 f tes la \ nThe energy radiated by th e electr on = %3.1 f eV” , f, B,
9 del_E = 10
del_E); 11 12 // Res ult 13 // The frequency
of the appli ed e l e c t r i c f i e l d =
1.705ma e+gn008 pe rens secity = 0.83 2 tes la 14 // The et iccycles f i e l d int 15 // T he en er gy radi ated by th e electron = 7.6 eV
Scilab code Exa 11.11 Kinetic energy of the accelerated nitrogen ion
1 // Scilab co de Exa11 .11 : : Page −538 (2011) 2 clc ; clear ; 3 E =3; // Energy of pr ot on sy nch rot ron , giga
ele ctr on volt s // Relat ivist ic e n e r g y ,
4 m_0 _c _sq = 0. 93 8;
mega electron 5 P_ p =
volts // Momentum o f
sqrt (E^2-m_0_c_sq^2);
t he p r o t o n , gi ga electr
on volts 117
pe r c
// Momentum of the gi ga ele ctr on volt s
N(14) ions
6 P_n = 6*P_ p; 7 T_ n = 8
sqrt (P_n^2+(0.938*14)
,
^2)-0.938 *14;
//
Kin et ic en er gy of th e accel erate d nit rog en io n printf ( ” \ nThe kinetic en er gy of th e accelerated ni tro gen ion = %4.2 f MeV” , T_ n);
9 10 // Res ult 11 // T he kinetic
en er gy of th e accel erate d nit rog en
ion = 8.43 MeV
Scilab code Exa 11.12 Maximum magnetic flux density and frequency of
proton in cosmotron proton synchrotron 1 2 3 4 5 6 7
// Scilab
co de Exa11 .12
clc ; clear ; e = 1. 6e-1 9; R = 9. 14 4; m_p = 1. 67e-0 27 ; E = 3.6 *1 .6e-1 3; L = 3. 04 8;
section
−539 (2011)
: : Page
// Charg e of an elec tron , C // Ra di us , met re // Mass of t he pr ot on , Kg // Energy , jou le // Le ng th of th e on e sync hrot ron
, me tr e // Kine tic
8 T =3;
e ne rg y , gig a electron
volts // Ve loc ity
9 c = 3e+ 08;
of t he lig ht , m e tr e
pe r sec 10 m_0 _c _sq = 0. 93 8;
// Rel ati vis tic
en er gy , m ega
ele ctr on volt s 11 B = round
// web per squ are
( sqrt (2*m_p*E)/(R*e)*10^4);
Maximum magnetic metre
f i e l d dens ity ,
12 v = B*1 0^-4 *e*R/m _p;
// Ve loc ity
of t he
pr ot on , me tr e per sec // Fr eq ue nc y of th e
13 f_ c = v/(2 *%pi*R* 10 ^6 );
circular
or bi t , mega cycles 118
pe r se c
14 f_0 = 2*%p i*R*f_c*10^3
fr eq ue nc y , kilo
/(2*%pi*R+4
cycles
*L);
// Re du ce d
pe r se c
// , web pe r sq ua re m e t r e
15 B_m = 3. 33 * sqrt (T*(T+2*m_0_c_sq))/R;
Relati
visti c field
16 f_ 0 = c^2*e*R*B* 1e-004/(( 2*%pi*R+4*L)* (T+m_0_c_ sq)*e *1e+015); // Maximum frequency of the 17
accelerati
ng vo lt age , mega cycles pe r sec fl ux den si ty = %5.3 f weber /Sq . m\ nThe maximum frequ ency of the acc ele rat ing voltage = %4.2 f MHz” , B_m, f_ 0);
printf ( ” \ nThe maximum magnetic
18 19 // Res ult 20 // The m aximum magn etic
flu x dens ity = 1.3 93 we be r/ Sq .m 21 // The m aximum freq uenc y of th e acce lera ting voltage = 0. 09 MHz 22 // A nswer is giv en wr on gl y in th e te xt boo k
Scilab code Exa 11.13 Energy of the single proton in the colliding beam
1 // Scilab co de Exa11 .13 : : Page −539 (2011) 2 clc ; clear ; 3 E_c = 30e+ 00 9; // Energy of th e pr ot on
accel erato r , GeV // Relat ivist ic e n e r g y
4 m_0_c_ sq = 0.938 *10^6 ;
, GeV 5 E_p = (4 *E_c^2-2*m_0_
c_sq^2)/( 2*m_0_c_ sq) ;
En er gy of the pr oto n , GeV 6 printf ( ” \nThe energy of the proto n = %5.2 e GeV” /1e+009); 7 8 // Res ult 9 // The energy of the proto n = 1.92 e+ 006 G eV 10 // W rong an sw er given in the tex tboo k
119
// , E_p
Scilab code Exa 11.14 Energy of the electron during boson production
1 // Scilab co de Exa11 .14 : : Page −539 (2011) 2 clc ; clear ; 3 M_ z = 92; // Mass of t he boson , giga elect
ron
volts // Ene rgy of t he electron
4 E_e = M_ z/2 ;
, giga
ele ctr on volt s 5 c = 3e+ 08;
// Ve loc ity
of t he lig ht , m e tr e
per sec ond 1.6e-019* 1e+009); // Ma ss of el ec tr on , gi ga ele ctr on volt s 7 E_e_p lus = M_ z^2/(2*m_e); // Thr esh old ene rgy for t h e po si tr on , gi ga ele ctr on vol ts 8 printf ( ” \ nThe ene rgy of the ele ctr on = %d GeV \ nThe threshold ene rgy of the positr on = %4.2 e G eV” , 6 m_e = 9. 1e-31*c^2/(
E_ e, E_e_ plus); 9 10 // Res ult 11 // The en er gy of th e electro n = 46 G eV 12 // The threshold en er gy of th e positron
GeV
120
= 8.27 e+ 006
Chapter 12 Neutrons
Scilab code Exa 12.1 Maximum activity induced in 100 mg of Cu foil
// Scilab
co de Exa12 .1 : : Page
−573 (2011)
1 2 3 4 5
clc ; clear ; N_0 = 6. 23e+2 3; // Av og ad ro ’ s num be r , per mol e m = 0.1; // Mass of co pp er foil , Kg ph i = 10 ^1 2; // Neu tr on flux den sit y , pe r
6 7 8 9
a_6 3 = 0. 69 1; // Ab un da nc e of Cu −63 a_6 5 = 0. 30 9; // Ab un da nc e of Cu −65 W_ m = 63 .5 7; // Molecular we igh t , gram si gm a_ 63 = 4. 5e-2 4; // Ac ti vat ion cros s secti
sq ua re cen tim etr e sec
on
for Cu −63, sq ua re centi
me t r e 10 sig ma _65 = 2. 3e-24 ; // Act ivat ion cross sectio n for C u −65, sq ua re centi m e t re // p e r se c 12 A_65 = ph i*sigma_6 5*m*a_65/W_ m*N_0; // Activ ity for C u −65, disintegr ations p e r se c 13 printf ( ” \nThe activ ity for Cu −63 is = %4.3 e disintegr ations p e r se c \ nThe activi ty for Cu −65 is = %4.2 e disi ntegr atio ns pe r sec ” , A_ 63, A_ 65) ; 11 A_63 = ph i*sigma_6 3*m*a_63/W_
Activ ity
m*N_0;
for C u −63, disintegr
14
121
ations
15 // Res ult 16 // T he activity
for disintegr ations 17 // T he activity for disintegr ations
C u per C u per
−63 is = 3.047 e+ 009 se c −65 is = 6.97 e+ 008 se c
Scilab code Exa 12.2 Energy loss during neutron scattering
1 2 3 4 5
// Scilab
co de Exa12 .2 : : Page
−573 (2011)
clc ; clear ; A_ Be = 9; // Mass number of beryll ium A_ U = 23 8; // Mass nu mb er of ur an iu m E_los_Be = (1-((A_B e -1)^2/(A_Be+1) ^2))*100;
Energ y loss 6 E_lo s_U =
for
//
ber yll ium
round ((1-((A_U-1)^2/(A_U+1)^2))*100);
// Energy los s for uranium 7 printf ( ” \nThe en er gy loss for ber ylli um is = %d percent \nThe en er gy los s for uranium is = %d perc ent ” , E_l os_ Be , E_lo s_U); 8 9 // Check for greater en er gy loss 10 if E_los_ Be >= E_lo s_U then 11 printf ( ” \ nThe en er gy loss is beryll ium ” ) ; 12 else 13 printf ( ” \ nThe ene rg y loss is ”); 14 end 15 16 // Res ult 17 // T he e ne rg y loss for be ry lli um 18 // T he en er gy loss for ura nium is 19 // T he en e rg y loss is grea ter for
122
!!!! grea ter
for
greater
for
uranium
is = 3 6 pe rc en t = 2 per ce nt be ry ll ium
Scilab code Exa 12.3 Energy loss of neutron during collision with carbon
1 2 3 4
// Scilab
co de Exa12 .3 : : Page
−574 (2011)
clc ; clear ; A = 12; // Mass nu mb er of Ca rb on alpha = (A-1)^2/ (A+1)^2; // Scatteri
ng
coefficient = 1/2 *(1-alpha)*100 ; // En ergy loss of neutron 6 printf ( ” \nThe en er gy los s of ne ut ro n = %5.3 f perc ent ” ,E_loss) 5 E_loss
7 8 // Res ult 9 // T he en er gy loss
of ne ut ro n = 14 .20 1 per cen t
Scilab code Exa 12.4 Number of collisions for neutron loss
1 2 3 4
// Scilab
co de Exa12 .4 : : Page
clc ; clear ; zet a = 0. 20 9; E_c ha nge = 10 0/ 1;
−574 (2011)
// Mo de rat ed assembly // Change in en er gy of th e
neutron // The rmal ene rgy of the
5 E_t her mal = 0. 02 5;
ne ut ro n , electron
volts
6 E_n = 2* 10^ 6;
// En erg y of the ne ut ro n ,
ele ctr on volt s // Number of c o l l i s i o n s of neutr ons to lo ss 99 percent of thei r ene rgie s
7 n = 1/ze ta * log (E_change);
8 n_t her mal = 1/ze ta* log (E_n/E_thermal);
Number of c o l l i s i o n s of neutron s to reach energies 123
// thermal
9 printf ( ” \nThe num ber of
c o l l i s i o n s of neutron s to loss 99 pe rc en t of thei r ene rgie s = %d \ nThe number of c o l l i s i o n s of neutrons to reach therm al ener gie s = %d” ,n,n_thermal)
10 11 // Res ult 12 // The number of
c o l l i s i o n s of neutr ons to lo ss 99 pe rc en t of thei r ene rgi es = 22 13 // The num be r of c o l l i s i o n s of neutron s to reach th er ma l energies =87
Scilab code Exa 12.5 Average distance travelled by a neutron
1 // Scilab co de Exa12 .5 : : Page −574 (2011) 2 clc ; clear ; 3 L =1; // For simpl icity as su me t he rm al di ffus ion
len gth
to be un it y , uni t ( ’ x ∗ exp( −x/L) ’ , ’ x ’ , 0, 10 0) ; // A v er a ge distanc e travell ed by th e ne ut ro n , uni t 5 x_r ms = sqrt ( integrate ( ’xˆ2 ∗ exp( −x/L) ’ , ’ x ’ , 0, 100) ); // R oot mean squ are of th e distance trvel led by th e n e ut r on , uni t 6 printf ( ” \nThe av er age distan ce travell ed by th e neutron = %d ∗L” , x_ ba r); 7 printf ( ” \nThe root m ean squ are distance trav ell ed by the neutron = % 5.3 fL = % 5.3 fx ba r ” , x_ rm s , x_ rm s 4 x_b ar =
integrate
); 8 9 // Res ult 10 // T he av er ag e distanc
e travell ed by th e ne ut ro n = 1∗L 11 // T he roo t mean sq ua re distanc e travell ed by th e neu tro n = 1.414L = 1.414 x bar
124
Scilab code Exa 12.6 Neutron flux through water tank
1 // Scilab co de Exa12 .6 : : Page −574 (2011) 2 clc ; clear ; 3 Q = 5e+ 08; // Rat e at wh ic h neutr ons
pr od uc e , ne ut ro ns pe r sec 4 r = 20; // Dis tan ce from th e sou rc e , centi me tr e 5 // For wa te r 6 lam bda _wt r = 0. 45 ; // Tr an sp or t m ean fre e pa th , centi me tr e 7 L_ wt r = 2. 73 ; // Thermal dif fus io n le ngt h , centi me tr e 8 phi_wt r = 3*Q/( 4*%pi*lambd
// Neu tr on flux for cen time tre pe r sec 9 // Fo r he av y wat er 10 lam bda _h_ wtr = 2. 40 ;
pat h , cent i me tr e
11 L_h_ wtr = 17 1;
le ngt h , centi
a_wt r*r)*
exp (-r/L_wtr);
wat er , neu tro ns pe r squ are
//
Tr an spo rt mean free
// Thermal di ffu sio n
me tr e
12 phi_h_w tr = 3*Q/(4 *%pi*lambda_h L_h_wtr); // Ne utr on flux
_wtr*r)*
exp (-r/
for he avy wa ter , ne ut ro ns pe r sq ua re cen tim etr e pe r sec 13 printf ( ” \nThe neutr on flu x thr ough wat er = % 5.3 e neu tro ns pe r squ are c m per sec \nThe neu tro n flux th rou gh he av y wa te r = % 5.3 e neutr ons per squar e cm per sec ” , ph i_w tr , phi_ h_wt r); 14 15 // Res ult 16 // The neu tro n flux
th rou gh wa te r = 8.730 e+ 003 neu tro ns pe r squ are c m per sec 17 // The neu tro n flux th rou gh he av y wa te r = 2.212 e+ 006 neu tro ns pe r squ are c m per 125
sec
Scilab code Exa 12.7 Diffusion length and neutron flux for thermal neu-
trons 1 // Scilab co de Exa12 .7 : : Page −575 (2011) 2 clc ; clear ; 3 k = 1. 38e- 23 ; // Bo lt zma nn cons tant , jo ul es
pe r kelvin 4 5 6 7
T = 323; // Te mp er at ur e , kel vin E = (k*T )/1 .6e-1 9; // T hermal e ne rg y , jo ul es sig ma_ 0 = 13 .2e-2 8; // Cr os s section , squ are me tr e E_ 0 = 0. 02 5; // En er gy of the ne ut ro n ,
ele ctr on volt s // Absorpt ion sec tion , sq ua re m e tr e 9 t_ha lf = 2. 25 ; // Ha lf life , ho ur s 10 la mbd a = 0. 69 /t_ha lf; // Decay cons tant , per hour 11 N_0 = 6. 02 3e+02 6; // Avogadr o ’ s nu mb er , 8 sig ma_ a = sig ma_ 0* sqrt (E_0/E);
cross
per
12 m_M n = 55 ; 13 w = 0. 1e-0 3;
// Mas s num be r of ma ng ne se // Weigh t of ma ng nes e fo il ,
Kg // Ac ti vi ty , disintegrations
14 A = 200;
pe r sec 15 N = N_ 0*w/m _M n;
in the
// Number of ma ng ne se nu cle i
foil
16 x1 = 1.5 ; // Ba se , met re 17 x2 = 2.0 ; // He igh t , me tr e 18 phi = A/(N*sigm a_a*0.41 6); // Ne ut ro n flux ,
neu tro ns pe r squ are me tr e pe r sec // Fo r si mp li ci ty as su me i n i t i a l ne ut ro n flux to be un it y , neu tro ns/S q .m −s e c 20 phi 2 = 1/ 2*ph i1; // Gi ven neut ro n flu x , neu tro ns/ 19 ph i1 = 1;
Sq .m−s e c 126
21 L1 = 1/
// Th er mal gi ve n ne ut ro n fl ux , m
log (phi1/phi2)/(x2-x1);
diffusion
le ng th for
22 L = sqrt (1/((1/L1)^2+(%pi/x1)^2+(%pi/x2)^2));
// Diffusion
le ng th , m e t r e = %3.2 e neutr ons per squ are me tre pe r sec \nThe dif fus io n length =
23 printf ( ” \nThe neu tro n flux
%4
.2 f me tr e” , phi, L) ;
24 25 // Res ult 26 // T he neu tro n flux
= 3.51 e+ 008 neutr ons per squar e me tr e pe r sec 27 // T he diff usio n len gth = 0 .38 me t r e 28 // No te : t he difussion le ng th is sol ve d w r o n gl y in th e test book
Scilab code Exa 12.8 Diffusion length for thermal neutrons in graphite
1 // Scilab co de Exa12 .8 2 clc ; clear ; 3 N_0 = 6. 02 3e+02 6;
: : Page
− 575(2011)
// A vog adr o ’ s n um be r , per
mole 4 rho = 1. 62e+0 3;
// De ns it y , kg pe r cubi c
metre 5 sig ma_ a = 3. 2e-31 ;
section
// Abs or pt ion
cross
, squ are me tr e
6 sig ma_ s = 4. 8e-28 ;
// Sc att er ed cros s se ct ion
, squ are me tr e 7 A = 12; // Ma ss nu mb er 8 lambda _a = A/(N_0*rh o*sigma_ a);
// Absorpt ion
mean fr ee pa th , me tr e 9 lamb da_t r = A/(N_0*rh o*sigma_ s *(1-2/ (3*A)));
// Tr ansp ort mean fre e pat h , me tr e // Diffu sion
10 L = sqrt (lambda_a*lambda_tr/3);
length
for
th er ma l ne ut ro n
11 printf ( ” \nThe diff usio n len gth
127
for
th er ma l ne ut ro n =
%5.3 f me tr e ” ,L ) 12 13 // Res ult 14 // T he diff usio n len gth
for
th er ma l ne ut ro n = 0 .59 0
metre
Scilab code Exa 12.9 Neutron age and slowing down length of neutrons
in graphite and beryllium 1 // Scilab co de Exa12 .9 : : Page −575 (2011) 2 clc ; clear ; 3 E_0 = 2e+ 06 ; // Av era ge en er gy of th e ne ut ro n
, el ec tr on vo lt s // Thermal ene rgy of the ne ut ro n , electron volts 5 // Fo r graphite 6 A =12 // Ma ss nu mb er 7 sigm a_g = 33 .5 ; // The v al ue of si gm a for graphite 4 E = 0. 02 5;
8 tau _0 = 1/( 6*sigma_ g^2)*(A+2/3 )/(1 -2/ (3*A))* log (E_0/ E); // A ge of ne ut ro n for grap hit e , Sq .m 9 L_ f = sqrt (tau_0); // Slo wi ng down len gth of
neu tro n th rou gh graphite 10 11 12 13 14 15 16
, m
printf ( ” \ nF or Graphite , A = % d” , A); printf ( ” \ nNeutron age = %d Sq . cm” , tau _0*1e+004 ); printf ( ” \ nSlowing down leng th = %5.3 f m” , L_ f);
// For beryll ium
A = 9 // Ma ss nu mb er si gm a_ b = 57 ; // The v al ue of s ig m a for ber yll ium tau _0 = 1/( 6*sigma_ b^2)*(A+2/3 )/(1 -2/ (3*A))* log (E_0/ E); // A ge of ne ut ro n for ber yll ium , Sq .m 17 L_ f = sqrt (tau_0); // Slo wi ng down len gth of
neu tro n th rou gh graphite 18 printf ( ” \n \ nF or Beryllium , 19 printf ( ” \ nNeutron
, m A = %d” , A);
age = %d Sq . cm” , tau _0*1e+004 ); 128
20 21 22 23 24 25
printf ( ” \ nSlowing
// // // //
down leng th = %3.1 e m” , L_ f);
Res ult Fo r Gra phi te , A = 12 Ne ut ro n age = 36 2 Sq .c m Sl ow in g down lengt h = 0.190 m
26 27 // For Ber yll ium , A = 9 28 // Ne ut ro n age = 97 Sq .c m 29 // Slo win g down length = 9.9 e
−002 m
Scilab code Exa 12.10 Energy of the neutrons reflected from the crystal
1 2 3 4 5 6
// Scilab
co de Exa12 .10
clc ; clear ; th et a = 3. 5*%p i/180 ; d = 2. 3e-1 0; n =1; h = 6. 62 56e-3 4;
: : Page
−576 (2011)
// Reflection an gl e , rad ian // Lattice spa ci ng , me tr e // Fo r f i r s t order // P la nc k ’ s const ant , jou le
sec 7 m = 1. 67 48e-2 7; // Mass of t he ne ut ro n , Kg 8 E = n^2 *h^2/ (8 *m*d^2 * sin (theta)^2*1.6023e-19);
// Energ y of th e ne ut ro ns , electron
volts
9 printf ( ” \nThe ene rgy of the neutr ons = %4.2 f eV” ; 10 11 // Res ult 12 // T he en er gy of th e neu tro ns = 1.04 eV
129
, E)
Chapter 13 Nuclear Fission and Fusion
Scilab code Exa 13.1 Fission rate and energy released during fission of
U235 1 // Scilab co de Exa13 .1 2 clc ; clear ; 3 E = 200 *1. 602 3e-13 ;
fission
: : Page
−600 (2011)
// Energ y released p
er
, joul e
4 E_ t = 2;
// Total
po we r pr od uc ed ,
watt // Fis sio n rate , f i s s i o n s per
5 R_fi ss = E_ t/E;
sec 6 m = 0.5; 7 M = 235; 8 N_0 = 6. 02 3e+26 ;
// Mas s of ur an iu m , Kg // Mas s num be r of ur an iu m // A vog adr o ’ s n um be r , per
mole 9 N = m/ M*N _0 // Number of ur an iu m nu cl ei 10 E_ re l = N*E/4 .0 8* 10 ^-3; // En er gy released ,
kilo calorie s 11 printf ( ” \nThe rate
f i s s i o n s per
se c
of f i s s i o n of U −235 = %4. 2 e \ nE ne rg y rel eas ed = % e kcal ” ,
R_ fis s , E_r el); 12 13 // Res ult
130
14 // The rate
of f i s s i o n of U −235 = 6 . 2 4 e+010 pe r sec 15 // En erg y rele ase d = 1.006535 e+ 010 kcal
fissions
Scilab code Exa 13.2 Number of free neutrons in the reactor
1 // Scilab co de Exa13 .2 2 clc ; clear ; 3 E = 200 *1 .6e-1 3;
6 7 8
−600 (2011)
// Energ y released pe r p e r ne ut ro n t = 10^-3 ; // Time , sec P = E/t; // Power pr odu ce d by on e fre e ne ut ro n , wa tt per neu tro n P_l = 10 ^9 ; // Power level , wa tt N = P_l/P ; // Number of free neu tro ns in th e reac tor , neu tro ns printf ( ” \nThe n umber of free neu tro ns in th e reactor = % 5.3 e neutrons” , N); fission
4 5
: : Page
, joules
9 10 // Res ult 11 // T he n umber of
free ne ut ro ns in th e reactor = 3.12 5 e+ 016 neutron s
Scilab code Exa 13.3 Number of neutrons released per absorption
1 2 3 4
// Scilab
co de Exa13 .3 : : Page
clc ; clear ; N_ 0_ 23 5 = 1; N_ 0_ 23 8 = 20 ;
−600 (2011)
// Number of ur an iu m 23 5 per 23 8 // Number of ur an iu m 23 8 fo r
one ur an iu m 235 5 sig ma_ a_2 35 = 68 3;
// A bs or pt io n cross section
ur an ium 235 , bar n 131
for
6 sig ma_ a_2 38 = 2. 73 ;
// Abs or pt io n cross
secti on for
ur an ium 238 , bar n 7 sig ma_ f_2 35 = 58 3; // Fiss ion cros s sec tio n , b a r n 8 sigm a_a = (N_0_2 35*sigma_a_ 235+N_0_2 38*sigma_a_ 238) /(N_0_235+N_0_238); // Asor ption cros s se c , ba rn 9 sigm a_f = N_0_ 235*sigma_f _235/(N_0_ 235+N_0_2 38);
// Fis ssio n cr oss
10 v = 2.4 3; 11 et a = v*sig ma_ f/sigm a_ a;
sec tio n
// A v e ra g e number of ne ut ro n released pe r abso rpt ion 12 printf ( ” \nThe aver age number of neutr ons rel eas ed per absor ption = % 5.3 f ” , et a); 13 14 // Res ult 15 // T he ave rag e number of neu tro ns releas
ed pe r
absorption = 1.921
Scilab code Exa 13.4 Excitation energy for uranium isotopes
1 // Scilab co de Exa13 .4 : : Page − 600(2011) 2 clc ; clear ; 3 a_v = 14 .0 ; // Volume binding ene rgy
cons tant on volts a_s = 13 .0 ; // Surfac e bin ding en er gy co nst ant , mega electro n volts a_ c = 0. 58 3; // Cou lo mb const ant , mega ele ctr on volt s a_a = 19 .3 ; // As ym me tr ic constant , mega ele ctr on volt s a_p = 33 .5 ; // Pairing ene rgy con stan t , mega ele ctr on volt s Z = 92; // Ato mic nu mb er // For U −236 A = 235; // Ma ss nu mb er , mega electr
4 5 6 7 8 9 10
11 E_ex c_23 6 = a_ v*(A+1-A)-a_
s*((A+1)^(2/
132
3)-A^(2/3 ))-
a_c*(Z^2/(A+1)^(1/3)-Z^2/A^(1/3))-a_a*((A+1-2*Z) ^2/(A+1)-(A-2*Z)^2/A)+a_p*(A+1)^(-3/4);
Excitation volts 12 // For U −239
en er gy for
uranium 23 6 , mega electro
// n
// Ma ss nu mb er
13 A = 238;
14 E_ex c_23 9 = a_ v*(A+1-A)-a_ s*((A+1)^(2/ 3)-A^(2/3 ))a_c*(Z^2/(A+1)^(1/3)-Z^2/A^(1/3))-a_a*((A+1-2*Z) ^2/(A+1)-(A-2*Z)^2/A)+a_p*((A+1)^(-3/4)-A^(-3/4)) ; // Excitation ene rg y for ur anium 2 39 15 // Now calculate th e rat e of sp on ta ne ou s fiss ion ing
for U −235 16 N_ 0 = 6. 022 14e+2 3;
// Av og ad ro ’ s consta nt ,
per mo le 17 M = 235; // Ma ss nu mb er 18 t_ hal f = 3e+17 *3. 15e+7 ; // Ha l f life , ye ar s 19 la mbd a = 0. 69 3/t_h al f; // Decay cons tant , per
year // Mass of ur an iu m
20 N = N_0/M ;
235 , Kg // Rate of
21 dN _d t = N*la mbd a*360 0;
sp on(ta ne ou s excit fis sioatio ni ng oferura 23 5 , pe r236 hou=r %3 22 printf ” \nThe n en gy nium for uranium . 1 f MeV \ nThe excit atio n en er gy for uranium 23 9 = %3 . 1 f MeV \ nThe rat e of sp on ta ne ou s fiss ioni ng of ur an iu m 23 5 = %4.2 f per ho ur” , E_exc _236 , E_exc _239 , dN_dt); 23 24 25 26 27
// // // //
Res ult The excit The excit T he rate = 0.68
atio n en er gy for uranium 236 = 6.8 M eV atio n en er gy for uranium 239 = 5.9 M eV of sp on ta ne ou s fis sio ni ng of uranium 23 5 pe r h o u r
Scilab code Exa 13.5 Total energy released in fusion reaction
133
1 2 3 4 5
// Scilab
co de Exa13 .5 : : Page
−601 (2011)
clc ; clear ; a = 10^ 5; // Area of th e la ke , sq ua re mi le d = 1/2 0; // Depth of th e la ke , m ile V = a*d*( 1. 6e+03 )^3; // V olume of th e la ke , cub ic
metre 6 rho = 10 ^3 ;
cubic
// De nsi ty of wat er , kg pe r
me tr e
// To ta l mass of wa te r in
7 M_wa ter = V*rh o;
the lak e , Kg // A vo ga dr o ’ s constant
8 N_ 0 = 6. 022 14e+2 6;
, per
mole // Milecular mass of wa te r // Number of mole cules of
9 A = 18; 10 N = M_w at er*N_ 0/A;
wa te r , molecule
s
11 abund_ det = 0.0 156e-02 ; // Abu nd an ce of dete rium 12 N_ d = N*2 *abund _de t; // Number of deterium at om s 13 E_p er_ det = 43 /6 ; // Energ y released pe r
det eri um atom , mega elec tro n vol ts // To ta l e ne rg y released du ri ng fus ion , mega electron volt
14 E_ t = N_ d*E_pe r_d et;
15 printf E_ t); en er gy released %4 . 2(e” \nThe MeV” , total 16 17 // Res ult 18 // To ta l en er gy releas
ed dur ing
fusion
du ri ng fusion =
= 1.53 e+ 039
MeV
Scilab code Exa 13.6
Maximum temperature attained by thermonuclear
device 1 // Scilab co de Exa13 .6 : : Page −601 (2011) 2 clc ; clear ; 3 r = 1/2; // Ra di us of th e tube , me tr e 4 a = %p i*r^ 2;
// Area of th e to ru s , s qu ar e 134
metre 5 V = 3*%p i*a ;
// Volume of th e to rus , cub ic
metre // Pre ssu re of th e g a s , me tr e // Cap ac ita nce , farad
6 P = 10 ^-5*1 3. 6e+3*9 .81 ;
ne wt on per square 7 C = 12 00e- 6; 8 9 10 11 12
v = 4e +4; // T_ro om = 29 3; // N_k = P*V/T _ro om; // E = 1/ 2*C *v^ 2; // T_ k = 1/ 6*E/ (N_k*1 0);
po te nt ia l , vo lts Room tem pera ture , kelvi n From gas equati on Energy sto re d , joules // Te mp er at ur e attained b y the rmo nuc lea r de vic e , kelvin 13 printf ( ” \nThe tem per atur e attained by ther monuc lear dev ice = % 4.2 e K” , T_ k); 14 15 // Res ult 16 // T he tem pera ture
attained
by ther monuc lear
device
= 4. 75 e+0 05 K
Scilab code Exa 13.7 Energy radiated and the temperature of the sun
1 // Scilab co de Exa13 .7 : : Page −601 (2011) 2 clc ; clear ; 3 G = 6. 67e- 11 ; // Gravitational co nst ant ,
newton squar e m per
squar e kg
4 r = 7e+ 08; // Ra di us of th e sun , m et re 5 M_0 = 2e+ 30 ; // Mass of t he sun , kg 6 E_ re l = 3/ 5*G*M _0^2 /r; // En ergy releas ed by
th e sun , joule // En er gy rel eas ed when sun di am et er shr ink by 10 pe rc en t , joule 8 R = 8. 31 4; // Univ ers al ga s co ns ta nt , joule pe r kelv in p e r kel vin pe r mole 9 T = E_ re l/(M_ 0*R); // Te mp er at ur e of the su n , 7 E_dia_sh rink_ 10 = E_r el/9;
kelvin 135
10 printf ( ” \nThe ene rgy
rel eas ed by the su n = % 4.2 e j o u l e \nThe en er gy releas ed when sun dia me te r is shrin ked by 10 perc ent = % 4.2 e joule \ nThe te mpe rat ure of th e su n = %4.2 e kelvin ” ,E_rel,
E_dia_shr ink_10 , T); 11 12 // Res ult 13 // The en er gy releas 14 // T he en er gy releas
by 10 percent = 15 // Th e tem pera ture
ed by th e sun = 2.29 e+ 041 joule ed when sun dia me te r is shrin ked 2.54 e+ 040 jou le of the su n = 1.38 e+ 010 kelvin
Scilab code Exa 13.8 Estimating the Q value for symmetric fission of a
nucleus 1 2 3 4
// Scilab
co de Exa13 .8 : : Page
clc ; clear ; A_ 0 = 24 0; A_ 1 = 12 0;
−602 (2011)
// Mass number of pare nt nucleus // Mass number of dau ght er nucleus
5 B_ 12 0 = 8. 5;
// Bi nd in g en er gy of da ug ht er
6 B_ 24 0 = 7. 6;
//
nucleus
B in di n g e ne rg y of pa re nt
nucleus 7 Q = 2*A _1*B _1 20-A_0*B_
9 10 // Res ult 11 // The estimate
// Est ima te d Q −value ,
24 0;
mega electron volts 8 printf ( ” \nThe estimated
d Q −value
Q−val ue i s = %d MeV” , Q);
is = 21 6 M eV
Scilab code Exa 13.9 Estimating the asymmetric binding energy term
1 // Scilab
co de Exa13 .9 : : Page 136
−602 (2011)
2 clc ; clear ; 3 E = 31. 7; // Energy , MeV 4 a_ a = 5/ 9* 2^ (-2/3 )*E; // As ym me tr ic bindi ng
en er gy term , mega electro
n volts
5 printf ( ” \nThe asy mme tri c binding MeV”, a_ a); 6 7 // Res ult 8 // The asy mme tri c binding
energy
137
energy
te rm = % 4.1 f
te rm = 11.1
MeV
Chapter 15 Nuclear Fission Reactors
Scilab code Exa 15.1 Estimation of the leakage factor for thermal reactor
1 2 3 4
// Scilab
co de Exa15 .1 : : Page
clc ; clear ; N_ 0_ 23 5 = 1; N_ 0_ c = 10 ^5 ;
−652 (2011)
// Number of ura niu m at om // Number of graphite at oms per
uraniu m at om 5 sig ma_ a_2 35 = 69 8;
// Ab so rp ti on cross section
fo r ur an iu m , barns // Abs or pt ion gra phi te , ba rn s
6 sig ma_ a_c = 0. 00 3;
section
for
7 f = N_0_ 235*sigma_ sigm a_a_ c ); 8 eta = 2. 08 ;
a_235/(N_0_
235*sigma_
cross a_23 5+ N_0 _c*
// Thermal utili zati on factor // N umber of fa st f i s s i o n neutron
produced 9 k_ in f = et a*f ; // Multi plicat ion factor 10 L_m = 0. 54 ; // Mate rial le ngt h , me tr e 11 L_ sq r = ((L _m)^ 2* (1-f)); // diffusion le ng th ,
metre 12 tau = 0. 036 4; 13 B_ sq r = 3. 27 ; 14 k_e ff = round
);
// Age of th e ne ut ro n // Ge ome tri cal buckl ing (k_inf* exp (-tau*B_sqr)/(1+L_sqr*B_sqr)
// Effe ctive multip
lica tion facto 138
r
15 16 17 18 19 20
N_ lf = k_ ef f/k_i nf; // Non leakag e fact or lf = (1-N _l f)*10 0; // Le ak ag e factor , perc ent printf ( ” \n To ta l leakage facto r = %4.1 f perc ent ” ,lf)
// Res ult / / T o t a l le ak ag e fac tor =
31 .3 pe rc en t
Scilab code Exa 15.2 Neutron multiplication factor of uranium reactor
1 // Scilab co de Exa15 .2 : : Page −652 (2011) 2 clc ; clear ; 3 N_ m = 50; // Number of molecules of he av y
wa te r pe r ur an iu m molec ule 4 N_ u = 1; // Number of ur an iu m molecul 5 sig ma_ a_u = 7. 68 ; // Ab so rp ti on cross
for
es section
ur an iu m , barns
6 sig ma_ s_u = 8. 3;
// Sc att er ed cross
secti on
fo r ur an iu m , barns // Ab so rp ti on cross section he av y wa te r , bar ns 8 sig ma_ s_D = 10 .6 ; // Sc att er ed cros s sectio n fo r ur an iu m , barns 7 sigma_ a_D = 0.00 092;
for
9 f = N_ u*sigma_a _u/(N_u*sigma 10 11 12 13 14 15
_a_u+N_m*sigm
a_a_ D );
// Thermal uti liza tion factor zet a = 0. 57 0; // Averag e num ber of c o l l i s i o n s N_ 0 = N_ u*139 /14 0; // Number of U −238 atoms per unit vol um e sigma_ s = N_m/N_ 0*sigma_s _D; // Scat ter ed cross section , ba rn s sigm a_a_e ff = 3.8 5*(sigma _s/N_0)^0.4 15; // Effective ab sor pt io n cros s se ct ion , ba r ns p = exp (-sigma_a_eff/sigma_s); // Reso nanc e esc ape probablity ep s = 1; // Fas t f i s s i o n fa ct or
16 eta = 1. 34 ;
// N umber of 139
f a s t fission
neutr on pro duc ed // Effective multiplicat ion factor 18 printf ( ” \ nN eut ro n multiplicati on factor 17 k_ in f = ep s*et a*p*f;
= %4.1 f ”
,
k_inf); 19 20 // Res ult 21 // Ne ut ro n multiplicati
on factor
=
1.2
Scilab code Exa 15.3 Multiplication factor for uranium graphite moder-
ated assembly 1 2 3 4
// Scilab
6 7 8 9 10 11 12
−652 (2011)
// Fo r graphite // Ab sor pt io n cross gra phi te , ba rn s sig ma_ s_g = 4. 8; // Sc att er ed cross secti on for gra phi te , ba rn s zet a = 0. 15 8; // Averag e num ber of c o l l i s i o n s N_ m = 50; // Number of molec ules of graphite per ur an iu m molecule // Fo r ur an iu m sigm a_f = 59 0; // Fissio ning cro ss se ct ion , barns sig ma_ a_u = 69 8; // Ab so rp ti on cross section for U −235, bar ns sig ma_ a_2 38 = 2. 75 ; // Ab sor pt io n cross se ct ion fo r U −23 8, bar ns v = 2.4 6; // Number of fa st neutr ons emitted N_u = 1 // Number of ura niu m ato ms sig ma_ a_g = 0.0 032 ;
section
5
co de Exa15 .3 : : Page
clc ; clear ;
for
13 14 f = N_ u*sigma_a _u/(N_u*sigma
_a_u+N_m*sigm
// Thermal uti liza tion 15 N_0 = N_ u*(75 /7 6);
// Number of U 140
a_a_ g );
factor −238 atoms
per unit
vol um e //
16 sigma_ s = N_m*76 /75*sigma _s_g/N_u;
Scat ter ed cross
sec tion , ba rn s
17 sigm a_ef f = 3.8 5*(sigma _s/N_0)^0.41
Effective
//
5;
cros s sec tio n , ba rn s // Res onan ce
18 p = exp (-sigma_eff/sigma_s);
esc ape
probability , ba rn s // Fas t f i s s i o n fa ct or // N umber of fa st f i s s i o n neutron produced 21 k_ in f = ep s*et a*p*f; // Multi plicat ion factor 22 printf ( ” \nThe requ ired multiplicati on factor = %3.1 f ” , k_ in f); 19 ep s = 1; 20 eta = 1. 34 ;
23 24 // Res ult 25 // T he req uir ed
multiplica
tion
factor = 1.
1
Scilab code Exa 15.4 Ratio of number of uranium atoms to graphite atoms
1 // Scilab co de Exa15 .4 : : Page 2 clc ; clear ; 3 eta = 2. 07 ; // N umber of
−653 (2011) fa st
f i s s i o n neutron
produced 4 x = 1/ (e ta -1 ); 5 sig ma_ a_u = 68 7;
// A bs or pt io n cros s secti on for
ur ani um , barns 6 sig ma_ a_g = 0.0 045 ;
// Ab so rp ti on cros s secti on for
graphit e , bar ns a_u; // Ra ti o of number of ur an iu m at om s to grap hite at om s 8 printf ( ” \nThe rat io of number of ur an iu m ato ms to grap hite at om s = %4.2 e ” , N_ rat io); 7 N_rati o = x*sigma_ a_g/sigma_
9 10 // Res ult 11 // T he rati o of number of uranium atoms to graphite
141
atoms = 6.12e
−006
Scilab code Exa 15.5 Multiplication factor for LOPO nuclear reactor
1 2 3 4 5
// Scilab
co de Exa15 .5 : : Page
−653 (2011)
clc ; clear ; f = 0. 75 4; sig ma_ s_o = 4. 2;
// Thermal util izat ion factor // Sc att er ed cross secti on for ox yge n , bar ns sigma _s_ H = 20 ; // Sca tt er ed cros s secti on for hy dr og en , bar ns N_O = 87 9.2 5; // Number of oxy gen at om s N_23 8 = 14. 19 ; // Number of ura niu m ato ms N_H = 15 73 ; // Number of hydr ogen at om s
6 7 8 9 sigm a_s = N_O/N_ 238*sigma _s_o+N_H/N_
// Scat ter ed cross 10 N_ 0 = 14 .1 9; // Number volume 11 zeta _o = 0. 12 0; // Number 12 zet a_ H = 1; // Number hydrogen
238*sigma _s_H;
sec tion , ba rn s of U −238 pe r unit of c o l l i s i o n fo r oxyge n of c o l l i s i o n f o r
13 sigm a_ef f = (N_0/(zet a_o*sigma _s_o*N_O+zet a_H* sigm a_s_ H*N_H )); // Effe ctive cro ss
section
, ba rn s // Res onan ce
14 p = exp (-sigma_eff/sigma_s);
esc ape
probablity
15 eta = 2. 08 ;
// Nu mber of
f i s s i o n neut ron
produce d . 16 ep s = 1; // Fi ssi on fac tor 17 K_ in f = ep s*et a*p*f; // Multip licat ion factor 18 printf ( ” \nThe mult ipli catio n facto r for L OPO reactor = % 3.1 f ” , K_ in f); 19 20 // Res ult 21 // The multiplicati
on factor
for L OPO reactor = 142
1.6
Scilab code Exa 15.6 Control poison required to maintain the criticality
of U235 1 2 3 4
// Scilab
−654 (2011)
co de Exa15 .6 : : Page
clc ; clear ; r = 35; // R ad i us of th e rea cto r , centi m e t r e B_ sq r = (%p i/r)^ 2; // Geome tri cal buck lin g , pe r
squ are 5 D = 0. 22 0;
centi
me tr e // Dif fus ion
coefficient
, cen ti
metre // Rate o f abso rpt ion of the rma l neutr ons 7 v = 2.5; // Number of fas t neu tro ns emi tt ed 8 ta u = 50; // Age of th e ne ut ro n 9 sig ma_ f = 0. 048 ; // Rate of f i s s i o n 6 sig ma_ a_f = 0. 05 7;
10 sigm a_a_ c = -1 /( 1+tau*B_s qr)*(-v*sigm B_sqr*D+tau*B_sqr*sigma_a_f);
Co nt ro ll ed cros s secti on controlled
11 printf ( ” \nThe requ ired .4 f ” , sig ma_ a_c); 12 13 // Res ult 14 // T he req uir ed controlle
a_f+sigma_
cross
section =
d cros s secti on = 0. 02 73
Scilab code Exa 15.7 Dimensions of a reactor
1 2 3 4
// Scilab
co de Exa15 .7 : : Page
−655 (2011)
clc ; clear ; B_s qr = 65 ; // Geo met ric al buckl ing a = sqrt (3*%pi^2/B_sqr)*100; // Si de of t h e
cubical
re act or , centi
a_f+
//
metre 143
%6
/ / R a di us of th e cubical reac tor , centi me tr e 6 printf ( ” \nThe side of th e cubical reactor = %4.1 f cm \ nThe c r i t i c a l radi us of the rea cto r = % d cm” , a, 5 R = round (%pi/ sqrt (B_sqr)*100);
R); 7 8 // Res ult 9 // T he side of th e cubi cal reac tor = 6 7. 5 cm 10 // The c r i t i c a l radius of the react or = 39 cm
Scilab code Exa 15.8 Critical volume of the spherical reactor
1 // Scilab co de Exa15 .8 2 clc ; clear ; 3 sig ma_ a_u = 69 8;
: : Page
−655 (2011)
// Ab so rp ti on cross
section
fo r ur an iu m , barns // Ab sor pt ion cross heav y wat er , ba rn s N_m = 10 ^5 ; // Number of at om s of he av y wat er N_ u = 1; // Number of at om s of ur ani um f = sigm a_a_ u/( sigm a_a_ u+ sigm a_a_ M*N_m/N_u); // Thermal uti liza tion factor eta = 2. 08 ; // N umber of fa st f i s s i o n neutron produced k_ in f = et a*f ; // Multi plicat ion factor L_m_ sqr = 1. 70 ; // Mate rial le ngt h , me tr e L_ sq r = L_ m_s qr*( 1-f); // Diffusion le ng th , m e t r e B_sqr = 1.819 /0.30 381* exp ( -1/12) -1/0.3038 ; // Geome tric al buck ling , per squar e me tr e V_c = 12 0/ (B_sq r* sqrt (B_sqr)); // Volu me of th e reac tor , cubi c me tr e printf ( ” \nThe c r i t i c a l vo lu me of the re ac to r = % 4.1 f cubic me tr e” , V_ c);
4 sigma_ a_M = 0.00 092;
sectio
5 6 7 8 9 10 11 12 13 14
n for
15 16 // Res ult
144
17 // The c r i t i c a l vol um e of the
react or = 36.4
metre
145
cubic
Chapter 16 Chemical and Biological Effects of Radiation
Scilab code Exa 16.1 Radiation dosimetry
1 2 3 4 5 6
// Scilab
co de Exa16 .1 : : Page
−672 (2011)
clc ; clear ; R_ d = 25; // Ra di at io n d os e , mill i r ad R_ c_g y = 25 e-03 ; / / Dose in centi gray R_S v = 25 *1 0^-2 // D ose in milli sieverts printf ( ” \ n25 mrad = %2. 0 e cGy = % 4. 2 f mSv” , R_ c_ gy, R_Sv);
7 8 // Resul ts 9 // 25 mrad = 3e
−002 cGy = 0. 25 mSv
Scilab code Exa 16.2 Conversion of becquerel into curie
1 // Scilab co de Exa16 .2 : : Page 2 clc ; clear ; 3 BC_con v = 100 *1e+009/3 .7e+10;
curie
co nv er si on , milli
curie 146
−673 (2011) // Becque rel
4 printf ( ” \ n100 mega becq uere l = %3.1 f mil li BC_conv) 5 6 // Resul ts 7 // 1 00 mega bec que re l = 2.7 milli curie
curie
”
Scilab code Exa 16.4 Amount of liver dose for a liver scan
1 2 3 4 5
// Scilab
co de Exa16 .4 : : Page
−673 (2011)
clc ; clear ; A = 80 *1 0^ 6; // Acti vity , becquerel t_ hal f = 6* 360 0; // H a l f life , s N = A*t_ ha lf/0. 693 ; // Number of s urvivi
ng
radionuclei 6 E_re leas ed = 0.9 *N*(140e+03
)*1.6e-19;
// En er gy
rele ased , joule // Mass of liver
7 m_ l = 1. 8;
average
of
man, Kg
8 liv_do se = E_re lease d *10 ^2/m_l;
centigray 9 printf ( ” \nThe requiresd
// Liver d
li ve r do se = %3.1 f cGy”
liv_dose); 10 11 // Res ult 12 // T he requires
d li ver
os e ,
do se = 2.8
147
cGy
,
,
Chapter 18 Elementary Particles
Scilab code Exa 18.1 Root mean square radius of charge distribution
1 2 3 4
// Scilab
co de Exa18 .1 : : Page
−770 (2011)
clc ; clear ; m_ sq r = 0. 71 ; // Fo r prot on , (GeV/c −square )ˆ2 R_r ms = sqrt (12)/( sqrt (m_sqr)*5.1); // Ro ot mean
squar e radiu s , fe mt o me tr e mean squ are radius of ch arg e dis tri but ion : %4.2 f ferm i ” , R_ rm s);
5 printf ( ” \nThe root 6 7 // Res ult 8 // T he root
mean squ are radius distribution : 0. 81 fe rm i
of ch arg e
Scilab code Exa 18.3 Isospin of the strange particles
1 // Scilab co de Ex18 .3 2 clc ; clear ; 3 p = rand (1,2); 4 pi_mi nus =
rand (1,2);
: : Page
−763 (2011)
// pro ton // pi m inu s me so n 148
5 6 7 8 9 10
pi_p lus = rand (1,2); // pi plus meson n = rand (1,2); // neutr on lamd a_0 = rand (1,2); // la md a hype ron K_ 0 = rand (1,2); // K zero (Ka ons) K_pl us = rand (1,2); // K plus (Ka ons) sigma_ plus = rand (1,2); // hype ron
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
sigma_m inus = rand (1,2) ksi_m inus = rand (1,2);
26 27 28 29 30 31 32 33 34 35 36
// hype ron // hype ron // All oca te t he va lu e of Isospins (T and T 3) p(1,1 ) = 1/ 2; p(1,2 ) = 1/ 2; pi_ mi nus(1, 1) = 1; pi_ mi nus(1, 2) = -1; pi_ plu s(1,1 ) = 1; pi_ plu s(1,2 ) = +1 ; n(1,1 ) = 1/ 2; n(1,2 ) = -1 /2 ; lam bd a_0(1, 1) = 0; lam bd a_0(1, 2) = 0; K_0(1,1) = pi_m inus(1,1 )+p(1,1); K_ 0(1,2 ) = pi_ min us(1,2 )+p(1, 2) ; K_pl us(1,1) ) == p(1,1)+p( K_plus(1,2 p(1,2)+p( 1,1)-lamb 1,2)-lamb da_0(1,1 da_0(1,2 )-p(1,1); )-p(1,2) ; sigm a_pl us(1,1) = pi_ plus(1,1 )+p(1,1)-K_p lus(1,1 ); sigm a_pl us(1,2) = pi_p lus(1,2)+p( 1,2)-K_pl us(1,2); sigm a_min us(1,1) = pi_m inus(1,1 )+p(1,1)-K_p lus(1,1) ; sigm a_min us(1,2) = pi_m inus(1,2 )+p(1,2)-K_p lus(1,2) ; ksi_ minu s(1,1) = pi_p lus(1,1)+n( 1,1)-K_pl us(1,1)K_plus(1,1); ksi_ minu s(1,2) = pi_p lus(1,2)+n( 1,2)-K_pl us(1,2)K_plus(1,2); printf ( ” \n R e a c t i o n I \n piminus +p . . . . . . > l a m bd a 0 + K 0” ) ; printf ( ” \n T he v al ue of T for K 0 is : %3. 1 f ” ,K_0 (1,1)); printf ( ” \n T he v al ue of T3 for K 0 is : %3. 1 f ” ,K_0
149
(1,2)); 37 printf ( ” \n
R e a c t i o n II \n la m b d a 0 + K p lu s” ) ; 38 printf ( ” \n T he v al ue of T for K_plus(1,1)); 39 printf ( ” \n T he va lu e of T3 for
p i p l u s +p
−>
K p l u s is : %3. 1 f ” K pl us is : %3. 1 f ”
, ,
K_plus(1,2)); 40 printf ( ” \n R e a c t i o n III \n p i p l u s +n −> lambda 0 + K plus” ); 41 printf ( ” \n T he v al ue of T for K p l u s is : %3. 1 f ” , K_plus(1,1)); 42 printf ( ” \n T he va lu e of T3 for K pl us is : %3. 1 f ” , K_plus(1,2)); 43 printf ( ” \n Re ac ti on VI \n pi minus +p −> si gm a m in us + K p lu s” ) ; 44 printf ( ” \n T he va lu e of T for si gm a mi nu s is : %3.1 f ” ,sigma_minus(1,1)); 45 printf ( ” \n T he va lu e of T3 for si gm a mi nu s is : %3.1 f ” ,sigma_minus(1,2)); 46 printf ( ” \n R e a c t i on V \n p i p l u s +p −> si gm a p lu s + K pl us” ) ; 47 printf ( ” \n T he va lu e of T for si gma pl us is : %3. 1 f ” ,sigma_plus(1,1)); 48 printf ( ” \n T he va lu e of T3 for si gm a pl us is : %3. 1 f ” ,sigma_plus(1,2)); 49 printf ( ” \n Re ac ti on VI \n p i p l u s +n −> ks i mi nu s + K p lu s + K p l us” ); 50 printf ( ” \n T he va lu e of T for Ks i m in u s is : %3. 1 f ” ,ksi_minus(1,1)); 51 printf ( ” \n T he va lu e of T3 for Ks i mi nu s is : %3.1 f ” ,ksi_minus(1,2)); 52 53 // Res ult 54 // 55 // R e ac t i on I 56 // pi minus + p −> l a m b da 0 + K 0 57 // T he v a lu e of T fo r K 0 is : 1. 5 58 // T he va lu e of T 3 for K 0 is : −0.5
150
59 // R e ac t i on II 60 // p i p l u s +p −> l am b d a 0 + K p lu s 61 / / T he va lu e of T fo r K p l u s is : 0. 5 62 / / T he va lu e of T 3 for K p l u s is : 0. 5 63 // R e ac t i on III 64 // p i p l u s +n −> l am b d a 0 + K p lu s 65 / / T he va lu e of T fo r K p l u s is : 0. 5 66 / / T he va lu e of T 3 for K p l u s is : 0. 5 67 // R ea ct io n V I 68 // pi minus + p −> s ig ma mi nu s + K p lu s 69 // T he va lu e of T for s ig ma m i nu s is : 1.0 70 // T he va lu e of T 3 for si gm a m in us is : −1.0 71 // R ea ct io n V 72 // p i p l u s +p −> s ig ma pl us + K p lu s 73 / / T he va lu e of T fo r si gm a pl us is : 1. 0 74 / / T he va lu e of T 3 for si gm a pl us is : 1. 0 75 // R ea ct io n V I 76 // p i pl us +n −> k si mi nu s + K p l u s +
K plus 77 // T he va lu e of T for K s i m i n u s is : 0. 5 78 // T he va lu e of T 3 for Ks i m in u s is :
−0.5
Scilab code Exa 18.4 Allowed and forbidden reactions under conservation
laws 1 2 3 4 5 6 7 8 9
// Scilab
co de Exa18 .4 : : Page
clc ; clear ; p = rand (1,3); pi_mi nus = rand (1,3); pi_p lus = rand (1,3); pi_ 0 = rand (1,3); n = rand (1,3); lambd a_0 = rand (1,3); K_ 0 = rand (1,3);
10 K_pl us =
rand (1,3);
−764 (2011)
// pro ton // pi mi nu s meson // pi plus meson // pi zero meson // neu tro n // la mb da zero hy pe ron // k zero meson // k plus 151
meson
11 K_0_ bar = rand (1,3); // an ti parti cle of k z e r o 12 sigma_ plus = rand (1,3); // sig ma hy pe ron 13 // N ow in th e follow ing ste ps we allocate d t he va lu e
of charge (Q) , bary on nu mb er(B) nu mb er (S )
an d st ra ng en es s
14 p( 1, 1) = 1; 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
p( 1, 2) = 1; p( 1, 3) = 0; pi_ mi nus(1, 1) = -1; pi_ mi nus(1, 2) = 0; pi_ mi nus(1, 3) = 0; pi_ plu s(1,1 ) = 1; pi_ plu s(1,2 ) = 0; pi_ plu s(1,3 ) = 0; n( 1, 1) = 0; n( 1, 2) = 1; n( 1, 3) = 0; lam bd a_0(1, 1) = 0; lam bd a_0(1, 2) = 1; lam bd a_0(1, 3) = -1; K_0( 1,1 ) =0 ;
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
K_ 0(1, 0(1, 3) 2) == 1; 0 ; K_ K_pl us(1, 1) = 1; K_ pl us( 1,2 ) = 0 ; K_pl us(1, 3) = 1; sig ma_ plu s(1,1 ) = 1; sig ma _p lu s(1, 2) = 1; sigma_ plus(1,3) = -1 ; K_0 _ba r(1,1 ) = 0; K_0 _ba r(1,2 ) = 0; K_0 _ba r(1,3 ) = -1 ; pi_0( 1, 1) = 0 ; pi_0( 1, 2) = 0 ; pi_0( 1, 3) = 0 ; j =0; k =0; printf ( ” \n R e a c t i o n I
\n
p i p l u s +n 152
......
>
la mb da 0 + K plu s”
47 for i = 1:3 48 if pi_pl us(1,i)+n( (1,i) then 49 j = j+ 1; 50 else
1,i)
)
== lamb da_0(1,i)+K_
51 printf ( ” \n Re ac ti on I is for bid den ” ) if i == 1 then 52 53 printf ( ” \n De lt a Q is n ot zer o ” ) 54 elseif i == 2 then 55 printf ( ” \n De lt a B is n o t ze ro ” ) 56 elseif i == 3 then 57 printf ( ” \n De lt a S is n o t ze ro ” ) 58 end 59 end 60 end 61 62 if j==3 then printf ( ” \n R e a c t io n I is al lo w ed ” ); 63 64 printf ( ” \n D e l t a Q is ze ro \n D e lt a B is De lt a S is ze ro ” ) 65 printf end 66 ( ” \n
......
>
R e a c t i o n II K 0 + K pl us”
\n
plu s
ze ro
p i p l u s +n
)
67 j = 0 ; 68 for i = 1:3 69 if pi_p lus(1,i)+n( 1,i) == K_0( 1,i)+K_plu then 70 j = j+ 1; else 71 72 printf ( ” \n Re ac ti on II is fo rb idd en ” ) 73 if i == 1 then 74 printf ( ” \n De lt a Q is n ot zer o ” ) 75 elseif i == 2 then 76 printf ( ” \n De lt a B is n o t ze ro ” ) 77 elseif i == 3 then 78 printf ( ” \n De lt a S is n o t ze ro ” ) 79 end
153
s(1,i)
\n
80 end 81 end 82 83 if j==3 then 84 printf ( ” \n Re a c t io n II is 85 printf ( ” \n D e l t a Q is ze ro
De lt a S is
86 end 87 j = 0 ; 88 printf ( ” \n
......
>
al lo we d ” ) ; \n D e lt a B is ze ro
\n
ze ro ” )
R e a c t i o n III \n K 0 ba r + sumis on plus ”
p i p l u s +n )
89 for i = 1:3 90 if pi_p lus(1,i)+n( 1,i) == K_0_ bar(1,i)+ sigma_plus(1,i) then 91 j = j+ 1; 92 else 93 printf ( ” \n Re ac ti on III is fo rb id de n ” ) 94 if i == 1 then printf ( ” \n De lt a Q is n ot zer o ” ) 95 96 elseif i == 2 then 97 printf ( ” \n De lt a B is n o t ze ro ” ) 98 elseif 99 printf i( ”== \n 3De then lt a S is n o t ze ro ” ) 100 end 101 end 102 end 103 104 if j==3 then 105 printf ( ” \n Re a c t io n III is al lo we d ” ) ; printf ( ” \n D e l t a Q is ze ro 106 \n D e lt a B is ze ro \n De lt a S is ze ro ” ) 107 end 108 j = 0 ; 109 printf ( ” \n Re ac ti on IV \n p i p l u s +n . . . . . . > pi m in us + p” ) 110 for i = 1:3 111 if pi_p lus(1,i)+n( 1,i) == pi_m inus(1,i)+p( 1,i) then
154
112 113 114 115 116 117
j = j+ 1; else printf ( ” \n Re ac ti on IV is if i == 1 then printf ( ” \n De lt a Q is elseif i == 2 then
for bidd en ” ) n ot zer o ”
)
118 printf ( ” \n De lt a B is n o t ze ro ” ) elseif i == 3 then 119 120 printf ( ” \n De lt a S is n o t ze ro ” ) 121 end 122 end 123 end 124 125 if j==3 then 126 printf ( ” \n Re ac ti on I V is al lo we d ” ) ; 127 printf ( ” \n D e l t a Q is ze ro \n D e lt a B is ze ro De lt a S is ze ro ” ) 128 end 129 j = 0 ; 130 printf ( ” \n R e a c t i on V \n piminus +p . . . . . . > l a m bd a 0 + K 0” ) 131 for iif= 1:3 132 pi_mi nus(1,i)+p( 1,i) == lamb da_0( ) then 133 j = j+ 1; 134 else 135 printf ( ” \n Re ac ti on V is for bid den ” 136 if i == 1 then 137 printf ( ” \n De lt a Q is n ot zer o ” elseif i == 2 then 138 139 printf ( ” \n De lt a B is n o t ze ro ” 140 elseif i == 3 then 141 printf ( ” \n De lt a S is n o t ze ro ” 142 end 143 end 144 end 145 146 if j==3 then
155
\n
1,i)+K_0(1,i
) ) ) )
147 148
printf ( ” \n Re ac ti on V is al lo we d ” ) ; printf ( ” \n D e l t a Q is ze ro \n D e lt a B is De lt a S is ze ro ” )
149 end 150 j = 0 ; 151 printf ( ” \n
......
>
Re ac ti on VI l a m bd a 0
\n
\n
p i p l u s +n
+ K pl us”
152 for i = 1:3 153 if pi_mi nus(1,i)+p( 1,i) i ) then 154 j = j+ 1; 155 else 156 printf ( ” \n Re ac ti on VI 157 if i == 1 then 158 printf ( ” \n De lt a Q 159 elseif i == 2 then 160 printf ( ” \n De lt a B 161 elseif i == 3 then printf ( ” \n De lt a S 162 163 end 164 end
ze ro
) == pi_ 0(1,i)+lambd
a_0(1,
is for bidd en ” ) is n ot zer o ”
);
is n o t ze ro ”
)
is n o t ze ro ”
)
165 end 166 167 if j==3 then 168 printf ( ” \n Re ac ti on V I is 169 printf ( ” \n D e l t a Q is ze ro De lt a S is ze ro ” ) ; 170 end 171 172 // Res ult 173 // Re ac tio n I 174 // p i p l u s + n ...... 175 / / Re ac ti on I is al lo we d 176 // De lt a Q is ze ro 177 // De lt a B is ze ro 178 / / D e lt a S is ze ro 179 // R e ac t i on II 180 // p i p l u s + n ......
156
al lo we d ” ) ; \n D e lt a B is ze ro
>
l am b d a 0
>
K0
\n
+ K p lu s
+ K plus
181 182 183 184 185 186
// // // // // //
R e ac t i on II is De lt a B is n o t Re ac ti on II is D e lt a S is n o t R e ac t i on III pi plus
fo rb id de n ze ro fo rb id de n ze ro
187 188 189 190 191 192 193 194 195 196 197 198 199 200
// // // // // // // // // // // // // //
sumison plus Re ac ti on III is fo rb idd en D e lt a S is n o t ze ro R ea ct io n I V p i p l u s + n ...... Re ac ti on IV is for bid den De lt a Q is n o t ze ro R ea ct io n V p i m i n u s + p ...... Re ac ti on V i s al low ed De lt a Q is ze ro De lt a B is ze ro D e lt a S is ze ro R ea ct io n V I p i p l u s + n ......
+
n ......
>
K 0 bar
>
pi minus
la m b d a 0
>
>
+
l am b d a 0
+p
+ K0
+ K p lu s
201 // // Re forrobid den 202 D e ac lt ati Son isVIn oist ze
Scilab code Exa 18.9 Decay of sigma particle
1 // Scilab co de Ex18 .9 : : Page −766 (2011) 2 clc ; clear ; 3 h_c ros s = 6. 62e-0 22 ; // Red uec ed planck
constant
4 p_w idt h = 0. 88* 35;
decay ,
’s
, MeV sec // Partia
l wi d t h of t he
MeV
5 ta u = h_ cro ss/p_w id th; 6 T_ pi = 1;
// Life t i m e of si gma , se c // Is os pi n of p i p lu s
particle 157
7 T_ la mb da = 0;
// Isospin
of lambda zer o
particle = T_ pi+T _l am bd a; // Isosp in of s i g m a particle 9 printf ( ” \nThe lif eti me of si g m a par ticl e = %4.2 e s nThe reaction is str ong \nThe isos pin of si gm a 8 T_si gm a
10 11 12 13 14
particle
\
is : % d” ,ta u, T_si gma);
// Res ult // T he lifetime of s i g m a particle = 2.1 5e // T he reac tion is st ro ng / / T he iso spi n of s i g m a parti cle is : 1
−023 s
Scilab code Exa 18.10 Estimation of the mean life of tau plus
1 // Scilab co de Exa18 .10 : : Page −767 (2011) 2 clc ; clear ; 3 m_ me w = 10 6; // Mass of mew lepton , m ega
ele ctr on volt s p e r sq ua re c 4 m_ ta u = 17 84 ;
// Mass of t au lep ton , m ega ele ctr on volt s p e r sq ua re c 5 tau _me w = 2. 2e-06 ; // Mean l i f e of mew lepton , sec 6 R = 16 /1 00 ; // Br an ch in g facto r 7 tau_plu s = R*(m_me w/m_tau)^5*tau_ mew; // Mean l i f e for ta u pl us , sec 8 printf ( ” \nThe mean l i f e for ta u plus : %3.1 e sec ” , tau_plus); 9 10 // Res ult 11 // The mean l i f e for
ta u plus
: 2.6 e
−013 sec
Scilab code Exa 18.13 Possible electric charge for a baryon and a meson
158
1 // Scilab co de Exa18 .13 : : Page 2 clc ; clear ; 3 function s = sy mb ol(v al) 4 if val == 2 then 5 s = ’++’ ; 6 elseif v al == 1 then 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
− 768(2011)
s = ’+ ’ ; elseif v al == 0 then s = ’0 ’; elseif va l == -1 then s = ’− ’ ; end endfunction B =1; // Bary on nu mb er S =0; // Stra ngen ess qua ntum num be r Q = rand (1,4) // Ch ar ge I3 = 3/2 ; printf ( ” \nThe possible ch ar ge states ar e” ) ; for i = 0:1:3 Q = I3+( B+S )/2 ; syprintf m = sy(mbo l(Q); , sy m); ” %s” I3 = I3 - 1; end printf ( ” respectively
// Res ult // The possible respectively
” );
ch ar ge states
ar e ++ + 0
−
Scilab code Exa 18.15 Branching ratio for resonant decay
1 // Scilab
co de Exa18 .15
: : Page
2 clc ; clear ;
159
−768 (2011)
3 I_ 1 = 3/ 2; 4 I_ 2 = 1/ 2; 5 delta_r atio
// Isospi n for del ta (1 23 2) // Is os pi n fo r de lt a 0 =
sqrt ((2/3)^2)/
7 8 // Res ult 9 // T he br an ch in g ratio
for
//
sqrt ((1/3)^2);
Br anc hin g ra tio 6 printf ( ” \nThe br an ch in g ratio for = 1 /2 is %d” , delt a_rat io);
a re son anc e w it h I
a re son anc e w it h I = 1 /2
is 2
Scilab code Exa 18.16 Ratio of cross section for reactions
1 2 3 4
// Scilab
co de Exa18 .16
: : Page
−768 (2011)
clc ; clear ; phi = 45 *%pi/1 80 ; // Pha se diff erenc e Cross_sec _ratio = 1/4* (5+4 * cos (phi))/(1- cos (phi));
// C ro s s secti on ratio section ratio
5 printf ( ” \nThe cross
: %4.2 f ”
,
Cross_sec_ratio); 6 7 // Res ult
Scilab code Exa 18.18 Root mean square radius of charge distribution
1 2 3 4
// Scilab
co de Exa18 .18
: : Page
−770 (2011)
clc ; clear ; m_ sq r = 0. 71 ; // Fo r prot on , (GeV/c −square )ˆ2 R_r ms = sqrt (12)/( sqrt (m_sqr)*5.1); // Ro ot mean
squar e radiu s , fe mt o me tr e mean squ are radius of ch arg e dis tri but ion : %4.2 f ferm i ” , R_ rm s);
5 printf ( ” \nThe root 6
160
7 // Res ult 8 // T he root
mean squ are radius distribution : 0. 81 fe rm i
161
of ch arg e
