## Example 2.1
import math
#determine
## Given data
atom_h = 6.6*10**24; ## Number of atoms in Hydrogen
## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium
isoab_H2 = 0.015; ## Isotropic abundance of deuterium
## Calculation
totatom_d=(isoab_H2*atom_h)/100.;
## Result
print"%s %.2e %s "%('\n Number of deuterium atoms = ',totatom_d,'');
## Example 2.2
import math
#determine
## Given data
## Using the data given in the example 2.2
atwt_O16 = 15.99492; ## Atomic weight of O-16 isotope
isoab_O16 = 99.759; ## Abundance of O-16 isotope
atwt_O17 = 16.99913; ## Atomic weight of O-17 isotope
isoab_O17 = 0.037; ## Abundance of O-17 isotope
atwt_O18 = 17.99916; ## Atomic weight of O-18 isotope
isoab_O18 = 0.204; ## Abundance of O-18 isotope
## Calculation
atwt_O=(isoab_O16*atwt_O16 + isoab_O17*atwt_O17 + isoab_O18*atwt_O18)/100.;
## Result
print"%s %.2f %s "%('\n Atomic Weight of Oxygen = ',atwt_O,'');
## Example 2.3
import math
#determine
## Given data
me = 9.1095*10**(-28); ## Mass of electron in grams
c = 2.9979*10**10; ## Speed of light in vacuum in cm/sec
## Calculation
rest_mass = me*c**2;
## Result
print"%s %.2e %s "%('\n Rest mass energy of electron = ',rest_mass,' ergs\n');
print('Expressing the result in joules')
## 1 Joule = 10^(-7)ergs
rest_mass_j = rest_mass*10**(-7);
print"%s %.2e %s "%('\n Rest mass energy of electron = ',rest_mass_j,' joules\n');
print('Expressing the result in MeV')
## 1 MeV = 1.6022*10^(-13)joules
rest_mass_mev = rest_mass_j/(1.6022*10**(-13));
print"%s %.2f %s "%('\n Rest mass energy of electron = ',rest_mass_mev,' MeV\n');
## Example 2.4
import math
#Calculate the
## From the result of Example 2.3
## Rest mass energy of electron = 0.5110 MeV
rest_mass_mev = 0.5110;
me = 9.1095*10**(-28); ## Mass of electron in grams
## From standard data table
## 1 amu = 1.6606*10^(-24)g
amu = 1.6606*10**(-24);
## Calculation
en_eq = (amu/me)*rest_mass_mev;
## Result
print'%s %.2f %s'%('\n Energy equivalent of one amu = ',en_eq,' MeV\n');
## Example 2.5
import math
#Calculate the
## From the standard data table
h = 6.626*10**(-34); ## Planck's constant in J-s
c = 3*10**8; ## Speed of light in vacuum in m/sec
## Given data
print('The ionization energy of K shell electron in Lead atom is 88keV');
E = 88*10**3; ## Ionization energy in keV
## Expressing the result in joules by using 1 eV = 1.6022*10^(-19) J
E = E*1.6022*10**(-19);
print("From Planck\''s law of photoelectric effect \n Energy = (h*c)/lambda\n");
## Calculation
lambd = (h*c)/E;
## Result
print'%s %.3e %s'%('\n Wavelength of radiation = ',lambd,' m\n');
## Example 2.6
import math
#Calculate the
## Given data
T12 = 64.8; ## Half life = 64.8 hour
lambd = 0.693/T12; ## Decay constant in hour^(-1)
t = 12.; ## Analysis time of gold sample in hours
alpha = 0.9; ## Activity of gold sample after analysis time
## 1.
## Calculation
R = alpha/(1-math.exp(-lambd*t));
## Result
print'%s %.2f %s'%('\n Theoretical maximum activity = ',R,' curie (Ci) \n');
## 2.
## Calculation
## The expression to calculate 80 percent of maximum activity is \n 0.8R = R*(1-exp(-lambda*t))
t = -math.log(0.2)/lambd;
## Result
print'%s %.2f %s'%('\n Time to reach 80 percent of maximum activity = ',t,' hours \n');
## Example 2.7
import math
#Calculate the
print('The reactants are Nitrogen and neutron')
## The total atomic number of reactants
Z_reactant = 7.+0.;
## The total atomic mass number of reactants
A_reactant = 14.+1.;
print('One of the known product is Hydrogen')
Z_H = 1.; ## The atomic number of Hydrogen
A_H = 1.; ## The atomic mass number of Hydrogen
## The atomic number of unknown element
Z_unknown = Z_reactant-Z_H;
## The atomic mass number of unknown element
A_unknown = A_reactant-A_H;
## Result
print'%s %.2f %s %.2f %s '%(" \n For unknown element the atomic number is ",Z_unknown," and atomic mass number is ",A_unknown," \n");
## From periodic table
print('The element corresponds to Carbon-14');
## Example 2.8
import math
#Calculate the
print('The reaction is Tritium(d,n)Helium-4');
## Using standard data table of mass in amu
M_H3 = 3.016049; ## Atomic mass of Tritium
M_He4 = 4.002604; ## Atomic mass of Helium
M_d = 2.014102; ## Atomic mass of Deuterium
M_n = 1.008665; ## Atomic mass of neutron
## Calculation of total mass of reactants
tot_reac = M_H3+M_d;
## Calculation of total mass of products
tot_prod = M_He4+M_n;
## Calculation
Q = tot_reac-tot_prod;
## Expressing in MeV by using 1 amu = 931.5 MeV
Q_mev = Q*931.5;
## Result
print'%s %.2f %s'%(" \n Q value for the reaction = ",Q_mev," MeV");
if Q_mev > 0:
print("\n The reaction is exothermic. \n");
else:
print("\n The reaction is endothermic. \n");
## Example 2.9
import math
#Calculate the
## Using standard data table of mass in amu
M_C12 = 12.; ## Atomic mass of Carbon-12
M_n = 1.00866; ## Atomic mass of neutron
M_C13 = 13.00335; ## Atomic mass of Carbon-13
##If one neutron is removed from carbon-13, carbon-12 is obtained
tot = M_C12+M_n;
dm = tot-M_C13; ## Mass defect
## Converting to energy equivalent from mass by using 1 amu = 931.5 MeV
Es = dm*931.5;
## Result
print'%s %.2f %s'%(" \n The binding Energy of the last neutron in Carbon-13 atom = ",Es," MeV");
## Example 2.10
import math
#calculate the
## Using standard data table of mass and the coefficients of mass equation for Silver-107
N = 60.; ## Number of neutrons
Z = 47.; ## Atomic number
A = 107.; ## Atomic mass number
## The coefficients used in mass equation are
alpha = 15.56;
bet = 17.23;
gam = 0.697;
zeta = 23.285;
mn = 939.573; ## Mass of neutron in terms of energy
mH = 938.791; ## Mass of proton in terms of energy
## Calculation
print('Using mass equation');
M = (N*mn)+(Z*mH)-(alpha*A)+(bet*(A**(2/3.)))+(gam*Z**2/A**(1/3.))+(zeta*(A-2*Z)**2/A);
## Expressing in amu by using 1 amu = 931.5 MeV
M_amu = M/931.5;
print'%s %.2f %s %.2f %s '%(" Mass = ",M," MeV" and " = ",M_amu,"f u" );
print('Actual mass = 106.905092 u');
## Calculation
BE = (alpha*A)-(bet*(A**(2/3.)))-(gam*Z**2/A**(1/3.))-((zeta*(A-(2.*Z))**2)/A);
## Result
print'%s %.2f %s %.2f %s '%("\n Binding Energy = ",BE," MeV" or"" ,BE/107,"MeV/nucleon \n");
## The value is different from the answer given in the textbook. The textbook answer is wrong.
## Example 2.11
import math
#calculate the
## Given data
T_C = 38.; ## Given temeperature in celsius
##The temperature in Kelvin
T_K = T_C+273.15;
T_0 = 293.61; ## The temperature in kelvin equivalent to 0 deg celsius
kT = 0.0253; ## The term 'kT' in eV at temperature T0
## Calculation
Ep = 0.5*kT*(T_K/T_0);
Ebar = 3*Ep;
## Result
print'%s %.2f %s'%(" Most probable energy of air molecules = ",Ep," eV \n");
print'%s %.2f %s'%(" Average energy of air molecules = ",Ebar," eV \n");
## Example 2.12
import math
#calculate the
## Given data
rho = 0.97; ## Density of Sodium in gram/cm^3
## From standard data table
NA = 0.6022*10**24; ## Avagodro number
M = 22.99; ## Atomic weight of Sodium
## Calculation
N = rho*NA/M;
## Result
print'%s %.2e %s'%("Atom density of sodium = ",N," atoms/cm^3 \n");
## Example 2.13
import math
#calculate the
## Given data
rho_NaCl = 2.17; ## Density of Sodium Chloride(NaCl) in gram/cm^3
## From standard data table
NA = 0.6022*10**24; ## Avogodro number
M_Na = 22.99; ## Atomic weight of Sodium(Na)
M_Cl = 35.453; ## Atomic weight of Chlorine(Cl)
M_NaCl = M_Na+M_Cl; ## Molecular weight of Sodium Chloride(NaCl)
## Calculation
N = rho_NaCl*NA/M_NaCl;
## As in NaCl, there is one atom of Na and Cl
N_Na = N;
N_Cl = N;
## Result
print'%s %.4e %s'%(" Atom density of Sodium and Chlorine = ",N," molecules/cm^3 \n");
## Example 2.14
import math
#calculate the
## Given data
rho = 1.; ## Density of water in gram/cm^3
## 1.
M_H = 1.00797; ## Atomic weight of Hydrogen(H)
M_O = 15.9994; ## Atomic weight of Oxygen(O)
## As in water, there is two atoms of Hydrogen(H) and one atom of Oxygen(O)
M = (2*M_H)+M_O; ## Molecular weight of water
## From standard data table
NA = 0.6022*10**24; ## Avagodro number
## Calculation
N = rho*NA/M;
## Result
print'%s %.4e %s'%("Atom density of water = ",N," molecules/cm^3 \n");
## 2.
## As in water, there is two atoms of Hydrogen(H) and one atom of Oxygen(O)
N_H = 2*N; ## Atom density of Hydrogen
N_O = N; ## Atom density of Oxygen
## Result
print'%s %.4e %s'%("Atom density of Hydrogen(H) = ",N_H," atoms/cm^3 \n");
print'%s %.4E %s'%("Atom density of Oxygen(O)= ",N_O," atoms/cm^3 \n");
## 3.
## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium
isoab_H2 = 0.015;
## Calculation
N_H2 = isoab_H2*N_H/100.;
## Result
print'%s %.4E %s'%("Atom density of Deuterium(H-2)= ",N_H2," atoms/cm^3 \n");
## Example 2.15
import math
#calculate the
## Given data
rho = 19.1; ## Density of Uranium-235 in gram/cm^3
wt = 1500.; ## Weight of uranium rods in a reactor in kg
nr = 0.2; ## Enrichment(w/o) of Uranium-235
## 1.
## As Enrichment is 20(w/o)
wt_U235 = nr*wt; ## Amount of Uranium-235
## Result
print'%s %.2f %s'%("Amount of Uranium-235 in the reactor = ",wt_U235," kg \n");
## 2.
## From standard data table
NA = 0.6022*10**24; ## Avagodro number
M_U235 = 235.0439; ## Atomic weight of Uranium-235
M_U238 = 238.0508; ## Atomic weight of Uranium-238
## Calculation
N_U235 = nr*rho*NA/M_U235; ## Atom density of Uranium-235
N_U238 = (1.-nr)*rho*NA/M_U238; ## Atom density of Uranium-238
## Result
print'%s %.4e %s'%("Atom density of Uranium-235 = ",N_U235," atoms/cm^3 \n");
print'%s %.4e %s'%("Atom density of Uranium-238 = ",N_U238," atoms/cm^3 \n");
## Example 2.16
import math
#calculate the
## Given data
rho_UO2 = 10.5; ## Density of UO2 pellets in gram/cm^3
nr = 0.3; ## Enrichment(w/o) of Uranium-235
## From standard data table
M_U235 = 235.0439; ## Atomic weight of Uranium-235
M_U238 = 238.0508; ## Atomic weight of Uranium-238
M_O = 15.999; ## Atomic weight of Oxygen
NA = 0.6022*10**24; ## Avogodro number
M = 1./((nr/M_U235)+((1.-nr)/M_U238));
M_UO2 = M+(2.*M_O); ## Molecular weight of UO2
nr_U = M/M_UO2*100.; ## The percent(w/o) of Uranium in UO2 pellet
rho_U = nr_U*rho_UO2/100. ## Density of Uranium in g/cm^3
rho_U235 = nr*rho_U ## Density of Uranium-235 in g/cm^3
## Calculation
N_U235=rho_U235*NA/M_U235;
## Result
print'%s %.4e %s'%("Atom density of Uranium-235 = ",N_U235," atoms/cm^3 \n");