# Chapter 1: The Nucleus¶

## Example 1.3.1, Page 31¶

In :
import math

#Variable declaration
h = 6.626e-034;    # Planck's constant, Js
e = 1.602e-019;    # Charge on an electron, C
red_h = h/(2*math.pi*e*1e+06);    # Reduced Planck's constant, MeV
lamda = 5.0e-015;    # de_Broglie wavelength of neutron, m

#Calculations
p = red_h/lamda;    # Momentum of the neutron, MeV-s/m

#Result
print "The momentum of the neutron from de-Broglie relation : %5.3e MeV-s/m"%p

The momentum of the neutron from de-Broglie relation : 1.317e-07 MeV-s/m


## Example 1.3.2, Page 32¶

In :
#Variable declaration
E = [[0,0,0],[0,0,0],[0,0,0]];    # Declare a cell array of empty matrices for nuclides information
E = 'C';    # Assign element 'C' to (1,1) cell
E = 'N';    # Assign element 'N' to (2,1) cell
E = 'O';    # Assign element 'o' to (3,1) cell
E = 6;    # Assign atomic No. 6 to (1,2) cell
E = 7;    # Assign atomic No. 7 to (2,2) cell
E = 8;    # Assign atomic No. 8 to (3,2) cell
E = [12,13,14,16];    # Assign mass numbers for 'C' to (1,3) cell
E = [14,15,16,17];    # Assign mass numbers for 'N' to (2,3) cell
E = [14,15,16,17];    # Assign mass numbers for 'O' to (3,3) cell

#Calculations&Results
# Isotopes
print "\nIsotopes:"
print "\n========="
for i in range(0,3):    # Search for the three elements one-by-one
print "\n(Z = %d)"%(E[i])
for j in range(0,4):
print "\t%s(%d)"%(E[i],E[i][j]),

# Isotones
print "\n\nIsotones:";
print "\n========"
for N in range(6,10):    # Search for the neutron numbers from 6 to 9
print "\n(N = %d)\n"%N;
for i in range(0,3):
for j in range(0,4):
if  E[i][j]- E[i] == N:           # N = A-Z
print "\t%s(%d)"%(E[i],E[i][j]),

# Isobars
print "\n\nIsobars:"
print "\n======="
for A in range(14,18):  # Search for the mass numbers from 14 to 17
print "\n(A = %d)\n"%A
for i in range(1,3):
for j in range(0,3):
if  E[i][j] == A:
print "\t%s(%d)"%(E[i-1],E[i][j]),


Isotopes:

=========

(Z = 6)
C(12) 	C(13) 	C(14) 	C(16)
(Z = 7)
N(14) 	N(15) 	N(16) 	N(17)
(Z = 8)
O(14) 	O(15) 	O(16) 	O(17)

Isotones:

========

(N = 6)

C(12) 	O(14)
(N = 7)

C(13) 	N(14) 	O(15)
(N = 8)

C(14) 	N(15) 	O(16)
(N = 9)

N(16) 	O(17)

Isobars:

=======

(A = 14)

C(14) 	N(14)
(A = 15)

C(15) 	N(15)
(A = 16)

C(16) 	N(16)
(A = 17)



## Example 1.4.1, Page 33¶

In :
#Variable declaration
m = 9.1e-031; # Mass of the electron, Kg
C = 3e+08; # Velocity of the light,m/s

#Calculations
E = m*C**2/1.6e-013; # Energy of the electron at rest, MeV

#Result
print "Energy of the electron at rest : %5.3f MeV"%E

Energy of the electron at rest : 0.512 MeV


## Example 1.4.2, Page 33¶

In :
#Variable declaration
r = 3.46e-015; # Radius of the nucleus, m
r0 = 1.2e-015; # Distance of closest approach of the nucleus, m

#Calculations
A = round((r/r0)**3); # Mass number of the nucleus
if A == 23:
element = "Na";
elif A == 24:
element = "Mg";
elif  A == 27:
element = "Al";
elif  A == 28:
element = "Si";

#Result
print "The mass number of the nucleus is %d and the nucleus is of %s"%(A, element)

The mass number of the nucleus is 24 and the nucleus is of Mg


## Example 1.4.3, Page 34¶

In :
import math

#Variable declaration
m = 40*(1.66e-027); # Mass of the nucleus, kg
r0 = 1.2e-015; # Distance of the closest approach, m
A = 40; # Atomic mass of the nucleus

#Calculations
r = r0*A**(1./3); #Radius of the nucleus, m
V = 4./3*(math.pi*r**3); # Volume of the nucleus, m^3
density = m/V; # Density of the nucleus, kg/m^3

#Result
print "Radius of the nucleus: %3.1e m\nVolume of the nucleus: %5.3e m^3\nDensity of the nucleus: %3.1e kg/m^3"%(r,V,density)

Radius of the nucleus: 4.1e-15 m
Volume of the nucleus: 2.895e-43 m^3
Density of the nucleus: 2.3e+17 kg/m^3


## Example 1.4.4, Page 34¶

In :
import math

#Variable declaration
m = 1.66e-027;    # Mass of a nucleon, kg
A = 235; # Atomic mass of U-235 nucleus
M = A*m;    #Mass of the U-235 nucleus, kg
r0 = 1.2e-015; # Distance of closest approach, m

#Calculations
r = r0*(A)**(1./3); # Radius of the U-235 nucleus
V = 4./3*(math.pi*r**3); # Volume of the U-235 nucleus,m^3
d = M/V; # Density of the U-235 nucleus,kg/m^3

#Result
print "The density of U-235 nucleus : %4.2e kg per metre cube"%d

The density of U-235 nucleus : 2.29e+17 kg per metre cube


## Example 1.4.5, Page 35¶

In :
import math

#Variable declaration
m_O = 2.7e-026; # Mass of O nucleus, kg
r_O = 3e-015; # Radius of O nucleus, m
V_O = 4/3*(math.pi*(r_O)**3); # Volume of O nucleus, metre cube
d_O = m_O/V_O; # Density of O nucleus, kg/metre cube
m_Pb = 3.4e-025; # Mass of Pb nucleus, kg
r_Pb = 7.0e-015; # Radius of Pb nucleus, m

#Calculations
V_Pb = 4./3*(math.pi*(r_Pb)**3); # Volume of Pb nucleus, metre cube
d_Pb = m_Pb/V_Pb; #Density of Pb nucleus,kg/metre cube

#Results
print "The density of oxygen nucleus : %4.2e in kg/metre cube"%d_O
print "The density of Pb nucleus : %4.2e in kg/metre cube"%d_Pb

The density of oxygen nucleus : 3.18e+17 in kg/metre cube
The density of Pb nucleus : 2.37e+17 in kg/metre cube


## Example 1.4.6, Page 35¶

In :
import math

#Variable declaration
E = 5.48*1.6e-013; # Energy of alpha particle, J
e = 1.6e-019; # Charge of an electron, C
Z = 79; # Mas number of Au nucleus,
epsilon_0 = 8.85e-012; # Permittivity of free space,

#Calculations
D = (2*Z*e**2)/(4*math.pi*epsilon_0*E); # Distance of closest approach, m

#Result
print "The distance of closest appproach of alpha particle : %4.2e m"%D

The distance of closest appproach of alpha particle : 4.15e-14 m


## Example 1.4.7, Page 36¶

In :
#Variable declaration
A = 208; # Mass number of Pb-208
r0 = 1.2e-015; # Distance of closest approach, m

#Calculations
r = r0*((A)**(1./3)); # Radius of Pb-208, m

#Result
print "The radius of Pb-208 : %4.2e m"%r

The radius of Pb-208 : 7.11e-15 m


## Example 1.5.1, Page 36¶

In :
#Variable declaration
amu = 931.49; # Atomic mass unit, MeV
M_p = 1.00758; # Mass of proton, amu
M_n = 1.00897;  # Mass of neutron, amu
M_He = 4.0028; # Mass of He nucleus, amu
Z = 2; # Atomic number
N = 2; # Number of neutron

#Calculations
M_defect = Z*M_p+N*M_n-M_He;    # Mass defect, amu
BE_MeV = M_defect*amu; # Binding energy, MeV
BE_J = M_defect*1.49239e-010;    # Binding energy, J

#Results
print "The binding energy (in MeV): %5.2f"%BE_MeV
print "The binding energy (in J): %4.2e"%BE_J

The binding energy (in MeV): 28.22
The binding energy (in J): 4.52e-12


## Example 1.5.2, Page 37¶

In :
#Variable declaration
amu = 1.49239e-010; # Atomic mass unit, J
M_C = 12; # Mass of C-12, amu
M_a = 4.0026;    # Mass of alpha particle, amu

#Calculations
M_3a = 3*M_a; # Mass of 3 alpha particle, amu
D = M_C-M_3a; # Difference in two masses, amu
E = D*amu; # Required energy,J

#Result
print "The energy required to break 3 alpha particles : %4.2e J"%E

The energy required to break 3 alpha particles : -1.16e-12 J


## Example 1.5.3, Page 37¶

In :
#Variable declaration
M_p = 1.007895; # Mass of proton, amu
M_n = 1.008665; # Mass of neutron, amu
M_He = 4.0026; # Mass of He-nucleus, amu
Z = 2; # Number of proton
N = 2; # Number of neutron

#Calculations
D_m = ((Z*M_p)+(N*M_n)-M_He); # Mass defect, amu
amu = 931.49; # Atomic mass unit, MeV
E = D_m*amu; # Required energy, MeV

#Result
print "The energy required to knock out nucleons from the He nucleus = %5.2f MeV"%E

The energy required to knock out nucleons from the He nucleus = 28.43 MeV


## Example 1.5.4, Page 38¶

In :
#Variable declaration
M_Fe = 55.934939; # Mass of Fe-56, amu
M_p = 1.007825; # Mass of proton, amu
M_n = 1.008665; # Mass of neutron, amu
Z = 26; # Atomic number of Fe-56
N = 30; # Number of neutron in Fe-56
amu = 931.49; # Atomic mass unit, MeV

#Calculations
BE = ((Z*M_p)+(N*M_n)-M_Fe)*amu; # Binding energy of Fe-56, MeV

#Result
print "The binding energy of Fe-56 : %6.4f MeV"%BE

The binding energy of Fe-56 : 492.2561 MeV


## Example 1.5.5, Page 38¶

In :
#Variable declaration
amu = 931.49; # Atomic mass unit, MeV
M_p = 1.007825; # Mass of proton, amu
M_n = 1.008663;  # Mass of neutron, amu
A = 2;       # Mass number of deutron, amu
M_D = 2.014103;    # Mass of deuteron nucleus, amu

#Calculations
M_Defect = (M_p+M_n-M_D)*amu;     # Mass defect of the nucleus, MeV
P_fraction = (M_D - A)/A;     # Packing fraction of nucleus

#Results
print "Mass defect      %4.2f MeV\n Packing fraction    %7.5f"%(M_Defect,P_fraction);

Mass defect      2.22 MeV
Packing fraction    0.00705


## Example 1.5.6, Page 38¶

In :
#Variable declaration
m_p = 1.007825; # Mass of proton, amu
m_n = 1.008665; # Mass of neutron, amu
m_He = 4.002634; # Mass of He-4 nucleus, amu
amu = 931.47; # Atomic mass unit, MeV
A = 4 # Mass number of He-4 nucleus

#Calculations
BE = (2*m_p+2*m_n-m_He)*amu; # Binding energy of He-4 nucleus, MeV
Av_BE = BE/A; # Average binding energy or binding energy per nucleon, MeV

#Result
print "The binding energy per nucleon : %4.2f MeV"%Av_BE

The binding energy per nucleon : 7.07 MeV


## Example 1.6.1, Page 39¶

In :
#Variable declaration
l1 = 1;    # Orbital qunatum number for p-state nucleon
l2 = 2;    # Orbital qunatum number for d-state nucleon

#Calculations&Result
# Display the value of L within the for loop
print "The possible L values will be"
for i in range(abs(l1-l2),abs(l1+l2+1)):        # Coupling of l-orbitals
print "\t %1d"%i,

The possible L values will be
1 	 2 	 3


## Example 1.6.2, Page 40¶

In :
import numpy
#Variable declaration
# Get the l value from the user
l = 3;    # Orbital qunatum number for f-state proton
s = 0.5;    # Magnitude of spin quantum number

#Calculations&Result
# Display the value of j within the for loop

print "The j values will be between"
for i in numpy.arange((l-s),(l+s+1)):        # l-s Coupling
print "\t %3.1f"%i,

The j values will be between
2.5 	 3.5


## Example 1.11.1, Page 40¶

In :
#Variable declaration
V = 1000; # Potential difference, volts
R = 0.122; # Radius of the circular path, m
B = 1500e-04; # Magnetic field, tesla
e = 1.602e-019; # Charge of the electron, C
amu = 1.673e-027; # Atomic mass unit, kg

#Calculations
v = (2*V)/(R*B); # Speed of the ion, m/s
M = 2*e*V/v**2; # Mass of the ion, kg
A = M/amu; # Mass number

#Result
print "    Speed   >  %5.3e m/s  \n    Mass    >  %5.3e kg  \n    Mass number   >  %5.2f "%(v, M, A)

    Speed   >  1.093e+05 m/s
Mass    >  2.682e-26 kg
Mass number   >  16.03


## Example 1.11.2, Page 41¶

In :
import numpy

#Variable declaration
amu = 1.673e-027; # Atomic mass unit, kg
E = 5e+04; # Electric field, V/m
B1 = 0.4; # Magnetic field, tesla
v = E/B1; # Velocity of ions, m/s
B = 0.8; # Magnetic field, tesla
e = 1.602e-019; #charge of electron,C

#Calculations
m_Ar = [36,38,40]    # Masses of three isoptopes of Ar, amu

r_Ar = [0,0,0];    # Array of radii of three Ar ions, mm
for i in range(len(r_Ar)):
r_Ar[i] = (m_Ar[i]*amu*v)/(B*e)*1e+03; # Radius of Ar ion orbit, mm

d1 = 2*(r_Ar-r_Ar);    # Distance b/w first and second line, mm
d2 = 2*(r_Ar-r_Ar);    # Distance b/w second and third line, mm

#Results
print "The distance between successive lines due to three different isotopes : %3.1f mm and %3.1f mm"%(d1,d2)

The distance between successive lines due to three different isotopes : 6.5 mm and 6.5 mm