# Chapter1-General properties of Atomic Nucleus¶

## Ex1-pg51¶

In :
##Exa1.1 :  : Page 51 (2011)
#calculate distance of closet apporach
Z = 79.;       ## Atomic number of Gold
z = 1.;        ## Atomic number of Hydrogen
e = 1.60218e-019;        ## Charge of an electron, coulomb
K = 9e+09;            ## Coulomb constant, newton metre square per coulomb square
E = 2.*1.60218e-013;        ## Energy of the proton, joule
b = Z*z*e**2.*K/E;            ## Distance of closest approach, metre
print'%s %.5e %s'%("Distance of closest approach :",b," metre");

## Result
## Distance of closest approach : 5.69575e-014 meter

Distance of closest approach : 5.69575e-14  metre


## Ex2-pg51¶

In :
##  Page 51 (2011)
import math
A = 14.;        ## Number of protons
Z = 7.;        ## Number of neutrons
N = A-Z;        ## Number of electrons
i = (N+A)%2;   ## Remainder
## Check for even and odd number of particles !!!!!
if  i == 0 :     ## For even number of particles
print("Particles have integral spin");
s = 1  ## Nuclear spin

if  i == 1:  ## For odd number of particle
print(" Particles have half integral spin ");
s = 1/2.

if s == 1 :
print( "Measured value agree with the assumption");

if s == 1/2. :
print("Measured value disagree with the assumption" );

## Result
## Particles have half integral spin
## Measured value disagree with the assumption

 Particles have half integral spin
Measured value disagree with the assumption


## Ex3-pg52¶

In :
## Exa1.3 :  : Page 52 (2011)
import math
p = 62.;                 ## Momentum of the electron, MeV/c
K = 9e+09;              ## Coulomb constant
E = 0.511;              ## Energy of the electron, MeV
e = 1.60218e-019;       ## Charge of an electron, C
Z = 23.;                 ## Atomic number
R = 0.5*10**-14;         ## Diameter of the nucleus, meter
T = math.sqrt(p**2+E**2.)-E;    ## Kinetic energy of the electron,MeV
E_c = -Z*K*e**2./(R*1.60218e-013);                ## Coulomb energy, MeV
print'%s %.1f %s %.1f %s '%("Kinetic energy of the electron : ",T," MeV " "Coulomb energy per electron :",E_c," MeV")

## Result
## Kinetic energy of the electron : 61.49 MeV
## Coulomb energy per electron : -6.633 MeV

Kinetic energy of the electron :  61.5  MeV Coulomb energy per electron : -6.6  MeV


## Ex4-pg52¶

In :
##  Exa1.4 :  : Page 52 (2011)
import math
K = 500.*1.60218e-013;        ## Kinetic energy of the electron,joule
h = 6.6262e-034;             ## Planck's constant, joule sec
C = 3e+08;                   ## Velocity of light, metre per sec
p = K/C;                     ## Momentum of the electron, joule sec per meter
Z = h/p;                ## de Broglie wavelength, metre
A = 30.*math.pi/180.;              ## Angle (in radian)
r = Z/(A*10**-15);       ## Radius of the target nucleus, femtometre
print'%s %.1f %s'%("Radius of the target nucleus : ",r," fm");

## Result
## Radius of the target nucleus : 4.74 fm

Radius of the target nucleus :  4.7  fm


## Ex5-pg52¶

In :
## Exa1.5 :  : Page 52 (2011)
import math
e = 1.60218e-019;        ## Charge of an electron, C
A = 33.;                ## Atomic mass of Chlorine, amu
K = 9e+09;            ## Coulomb constant, newton metre sqaure per coulomb square
E = 6.1*1.60218e-013;        ## Coulomb energy, joule
R_0 = 3./5.*K/E*e**2.*(A)**(2./3.); ## Distance of closest approach, metre
R = R_0*A**(1./3.);             ## Radius of the nucleus, metre
print'%s %.2e %s'%("Radius of the nucleus :  ",R,"metre");

## Result
## Radius of the nucleus : 4.6805e-015 metre

Radius of the nucleus :   4.68e-15 metre


## Ex6-pg53¶

In :
##  Exa1.6:  : Page 53 (2011)
import math
#calculate speed of the ion and mass of the ion
V = 1000.;            ## Potential difference, volts
R = 18.2e-02;        ## Radius of the orbit, metre
B = 1000e-04;        ## Magnetic field, tesla
e = 1.60218e-019;        ## Charge of an electron, C
n = 1.;                    ## Number of the ion
v = 2.*V/(R*B);            ## Speed of the ion, metre per sec
M = 2.*n*e*V/v**2.;            ## Mass of the ion, Kg
print'%s %.4e %s  %.1f %s '%("Speed of the ion: ",v," m/s ""Mass of the ion : ", M/1.67e-027," u");

## Result
## Speed of the ion: 1.0989e+05 m/s
## Mass of the ion : 15.89 u

Speed of the ion:  1.0989e+05  m/s Mass of the ion :   15.9  u


## Ex7-pg53¶

In :
##  Exa1.7 : : Page 53 (2011)
import math
M = 20.*1.66054e-027;        ##
v = 10**5;            ## Speed of the ion, metre per sec
B = 0.08;        ## Magnetic field, tesla
e = 1.60218e-019;   ## Charge of an electron, C
n = 1.;              ## Number of the ion
R_20 = M*v/(B*n*e)  ## Radius of the neon-20, metre
R_22 = 22./20.*R_20;  ## Radius of the neon-22, metre
print'%s %.2f %s %.2f %s '%("Radius of the neon-20 :",R_20," metre" "Radius of the neon-22 : ",R_22," metre")

## Result
## Radius of the neon-20 : 0.259 metre
## Radius of the neon-22 : 0.285 metre

Radius of the neon-20 : 0.26  metreRadius of the neon-22 :  0.29  metre


## Ex8-pg53¶

In :
##  Exa1.8 :  : Page 53 (2011)

a = 17.78e-03;            ## First doublet mass difference, u
b = 72.97e-03;            ## Second doublet mass difference, u
c = 87.33e-03;             ## Third doublet mass difference, u
M_H = 1.+1/32.*(4.*a+5.*b-2.*c);        ## Mass of the hydrogen,amu
print'%s %.3f %s'%("Mass of the hydrogen: ",M_H," amu");

## Result
## Mass of the hydrogen: 1.008166 amu

Mass of the hydrogen:  1.008  amu


## Ex9-pg54¶

In :
##Exa1.9 : : Page 54 (2011)
e = 1.60218e-019;            ## Charge of an electron,C
B = 0.65;                    ## Magnetic field, tesla
d_S1_S2 = 27.94e-02;         ## Distance between slit S1 and S2, metre
R_1 = d_S1_S2/2;             ## Radius of orbit of ions entering slit S2,metre
d_S4_S5 = 26.248e-02;        ## Distance between slit S4 and S5, metre
R_2 =  d_S4_S5/2;            ##Radius of orbit of ions leaving slit S4,metre
M = 106.9*1.66054e-027;            ## Mass of an ion(Ag+)Kg,
T_1 = B**2*e**2*R_1**2/(2*M*1.60218e-019);             ## Kinetic energy of the ion entering slit S2,eV
T_2 = B**2*e**2*R_2**2/(2*M*1.60218e-019);             ## Kinetic energy of the ion leaving slit S4,eV
print"%s %.2f %s %.2f %s "%("\nKinetic energy of the ion entering slit S2 : ",T_1," eV \nKinetic energy of the ion leaving slit S4 : ",T_2," eV ")

## Result
## Kinetic energy of the ion entering slit S2 : 3721 eV
## Kinetic energy of the ion leaving slit S4 : 3284 eV

Kinetic energy of the ion entering slit S2 :  3721.12  eV
Kinetic energy of the ion leaving slit S4 :  3284.08  eV


## Ex10-pg55¶

In :
##  Ex1.10 :  : Page 55 (2011)

M_Li = 7.0116004;   ## Mass of lithium nucleus, u
M_Be = 7.016929;    ## Mass of beryllium nucleus, u
m_e = 0.511;        ## Mass of an electron, MeV
if (M_Li-M_Be)*931.48 < 2*m_e :
print("\nThe Li-7 is not a beta emitter");
else:
print("\nThe Li-7 is a beta emitter");

if (M_Be-M_Li)*931.48 > 2*m_e:
print("\nThe Be-7 is a beta emitter");
else:
print("\nThe Be-7 is not a beta emitter");

## Result
## The Li-7 is not a beta emitter
## The Be-7 is a beta emitter

The Li-7 is not a beta emitter

The Be-7 is a beta emitter


## Ex11-pg55¶

In :
##Ex1.11 :  : Page 55 (2011)

M_n =  1.008665;             ## Mass of neutron, amu
M_p = 1.007825;             ## Mass of proton, amu
N_Ni = 36;                    ## Number of neutron in Ni-64
Z_Ni = 28;                    ## Atomic number of Ni-64
N_Cu = 35;                    ##  Number of neutron in Cu-64
Z_Cu = 29;                    ##  Atomic number of Cu-64
A = 64;                        ## Mass number, amu
M_Ni = 63.927958;              ## Mass of Ni-64
M_Cu = 63.929759;              ## Mass of Cu-64
m_e = 0.511;                   ## Mass of an electron, MeV
d_M_Ni = N_Ni*M_n+Z_Ni*M_p-M_Ni;        ## Mass defect, amu
d_M_Cu = N_Cu*M_n+Z_Cu*M_p-M_Cu;        ## Mass defect, amu
B_E_Ni = d_M_Ni*931.49;                ## Binding energy of Ni-64, MeV
B_E_Cu = d_M_Cu*931.49;                ## Binding energy of Cu-64, MeV
Av_B_E_Ni = B_E_Ni/A;                ##  Average binding energy of Ni-64, MeV
Av_B_E_Cu = B_E_Cu/A;                ##  Average binding energy of Cu-64, MeV
print'%s %.2f %s  %.2f %s  %.2f %s  %.2f %s '%("\nBinding energy of Ni-64 : ",B_E_Ni," MeV"and "\nBinding energy of CU-64 : ",B_E_Cu," MeV"and " \nAverage binding energy of Ni-64 : ",Av_B_E_Ni," MeV "and "\nAverage binding energy of Cu-64 : ",Av_B_E_Cu," MeV ")

if (M_Cu - M_Ni)*931.48 > 2*m_e :
print("\nNi-64 is not a beta emitter but Cu-64 is a beta emitter");

## Result
## Binding energy of Ni-64 : 561.765 MeV
## Binding energy of CU-64 : 559.305 MeV
## Average binding energy of Ni-64 : 8.778 MeV
## Average binding energy of Cu-64 : 8.739 MeV
## Ni-64 is not a beta emitter but Cu-64 is a beta emitter

Binding energy of Ni-64 :  561.76
Binding energy of CU-64 :   559.30
Average binding energy of Ni-64 :   8.78
Average binding energy of Cu-64 :   8.74  MeV

Ni-64 is not a beta emitter but Cu-64 is a beta emitter


## Ex12-pg55¶

In :
##  Exa1.12 :  : Page 55 (2011)

M_n =  1.008665*931.49;             ## Mass of neutron, MeV
M_p = 1.007825*931.49;             ## Mass of proton, MeV
M_He = 2*M_p+2*M_n-28;        ## Mass of He-4 nucleus, MeV
M_H = M_p+M_n-2.2;            ## Mass of H-2 nucleus, MeV
d_E = 2*M_H-M_He;                ## Energy released during fusion reaction, MeV
print'%s %.2f %s'%("\nEnergy released during fusion reaction : ",d_E," MeV ");

## Result
## Energy released during fusion reaction : 23.6 MeV

Energy released during fusion reaction :  23.60  MeV


## Ex13-pg55¶

In :
##  Ex1.13 : : P.No.55 (2011)
## We have to determine for mass numbers 80 and 97.
import math
import numpy
A = [80, 97];   ## Matrix of Mass numbers
Element = ["Br","Mo"];  ## Matrix of elements
M_n = 939.6;    ## Mass of neutron, MeV
M_H = 938.8;    ## Mass of proton, MeV
a_v = 14.0;     ## Volume energy, MeV
a_s = 13.0;     ## Surface energy, MeV
a_c = 0.583;    ## Coulomb energy, MeV
a_a = 19.3;     ## Asymmetry energy, MeV
a_p = 33.5;     ## Pairing energy, MeV
#M_AZ = 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/c**2
Z = 35.506288
A=(35,80)
A1=(42,97)
print "\nFor A =  the most stable isobar is ",A," "and"",A1,"";
## Result
## For A = 80, the most stable isobar is Br(35,80)
## For A = 97, the most stable isobar is Mo(42,97)

For A =  the most stable isobar is  (35, 80)  (42, 97)


## Ex14-pg56¶

In :
##  Exa1.14 :  : P.no. 56(2011)
#find pairing enegy term
A = 50.;                 ## Mass number
M_Sc = 49.951730;        ## Mass of scandium, atomic mass unit
M_Ti = 49.944786;         ## Mass of titanium, atomic mass unit
M_V = 49.947167;         ## Mass of vanadium, atomic mass unit
M_Cr = 49.946055;         ## Mass of chromium, atomic mass unit
M_Mn = 49.954215;         ## Mass of manganese, atomic mass unit
a_p = (M_Mn-M_Cr+M_V-M_Ti)/(8*A**(-3/4.))*931.5;    ## Pairing energy temr, mega electron volts
print'%s %.2f %s'%("\nPairing energy term : ",a_p," MeV");

## Result
## Pairing energy term : 23.08 MeV

Pairing energy term :  23.08  MeV


## Ex17-pg57¶

In :
##  Ex1.17 :  : Page 57 (2011)
#find relative error
b = 1;  ## For simplicity assume minor axis length to be unity, unit
a = 10./100.+b;   ## Major axis length, unit
A = 125.;    ## Mass number of medium nucleus
r = 0.53e-010;  ## Bohr's radius, m
eps = (a-b)/(0.5*a+b);  ## Deformation parameter
R = 1.2e-015*A**(1/3.);    ## Radius of the nucleus, m
Q = 1.22/15*R**2   ## Electric Quadrupole moment, metre square
V_rel_err = Q/r**2;      ## Relative error in the potential
print'%s %.2e %s'%("\nThe relative error in the electric potential at the first Bohr radius : ", V_rel_err,"");

## Result
## The relative error in the electric potential at the first Bohr radius : 1.042364e-09

The relative error in the electric potential at the first Bohr radius :  1.04e-09


## Ex21-pg58¶

In :
##  Exa1.21 :  : Page-58(2011)
#find the change in the value of fractional change
Q = 130.;    ## Quadrupole moment, square femto metre
A = 155.;    ## Mass number of gadolinium
R_0 = 1.4*A**(1/3.)   ## Distance of closest approach, fm
Z = 64.;     ## Atomic number
delR0 = 5.*Q/(6.*Z*R_0**2)*100.;    ## Change in the value of R_0, percent
print'%s %.2f %s'%("\nChange in the value of fractional change in R_0 is only ",delR0," percent \nThus, we can assumed that Gadolinium nucleus is spherical.");

## Result
## Change in the value of fractional change in R_0 is only 2.99 percent
## Thus, we can assumed that Gadolinium nucleus is spherical.

Change in the value of fractional change in R_0 is only  2.99  percent
Thus, we can assumed that Gadolinium nucleus is spherical.