# Chapter2-Atomic and Nuclear Physics¶

## Ex1-pg8¶

In [1]:
## 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,'');

 Number of deuterium atoms =  9.90e+20


## Ex2-pg9¶

In [2]:
## 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,'');

 Atomic Weight of Oxygen =  16.00


## Ex3-pg12¶

In [3]:
## 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');

 Rest mass energy of electron =  8.19e-07  ergs

Expressing the result in joules

Rest mass energy of electron =  8.19e-14  joules

Expressing the result in MeV

Rest mass energy of electron =  0.51  MeV



## Ex4-pg12¶

In [4]:
## 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');

 Energy equivalent of one amu =  931.52  MeV



## Ex5-pg16¶

In [5]:
## 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');

The ionization energy of K shell electron in Lead atom is 88keV
From Planck''s law of photoelectric effect
Energy = (h*c)/lambda

Wavelength of radiation =  1.410e-11  m



## Ex6-pg25¶

In [6]:
## 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');

 Theoretical maximum activity =  7.47  curie (Ci)

Time to reach 80 percent of maximum activity =  150.49  hours



## Ex7-pg29¶

In [7]:
## 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');

The reactants are Nitrogen and neutron
One of the known product is Hydrogen

For unknown element the atomic number is  6.00  and atomic mass number is  14.00

The element corresponds to Carbon-14


## Ex8-pg29¶

In [8]:
## 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");

The reaction is Tritium(d,n)Helium-4

Q value for the reaction =  17.59  MeV

The reaction is exothermic.



## Ex9-pg33¶

In [9]:
## 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");


The binding Energy of the last neutron in Carbon-13 atom =  4.95  MeV


## Ex10-pg36¶

In [10]:
## 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.

Using mass equation
Mass =  99582.07  =  106.91 f u
Actual mass = 106.905092 u

Binding Energy =  915.49  MeV 8.56 MeV/nucleon



## Ex11-pg39¶

In [11]:
## 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");

 Most probable energy of air molecules =  0.01  eV

Average energy of air molecules =  0.04  eV



## Ex12-pg40¶

In [12]:
## 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");

Atom density of sodium =  2.54e+22  atoms/cm^3



## Ex13-pg41¶

In [13]:
## 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");

 Atom density of Sodium and Chlorine =  2.2360e+22  molecules/cm^3



## Ex14-pg42¶

In [14]:
## 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");

Atom density of water =  3.3427e+22  molecules/cm^3

Atom density of Hydrogen(H) =  6.6854e+22  atoms/cm^3

Atom density of Oxygen(O)=  3.3427E+22  atoms/cm^3

Atom density of Deuterium(H-2)=  1.0028E+19  atoms/cm^3



## Ex15-pg43¶

In [15]:
## 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");

Amount of Uranium-235 in the reactor =  300.00  kg

Atom density of Uranium-235 =  9.7871e+21  atoms/cm^3

Atom density of Uranium-238 =  3.8654e+22  atoms/cm^3



## Ex16-pg43¶

In [16]:
## 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");

Atom density of Uranium-235 =  7.1110e+21  atoms/cm^3