##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
## 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
## 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
## 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
## Exa1.5 : : Page 52 (2011)
import math
#calculate radius of the nucleus
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
## 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
## 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
## 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
##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
## 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
##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
## 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
## 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)
## 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
## 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
## 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.