#Rate of consumption of U-235 per year
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
amu = 1.6e-027 # Mass of a nucleon, kg
P_out = 250e+06 # Output power of nuclear reactor, J/s
E = 200e+06*e # Energy released per fission of U-235, J
n = P_out/E # Number of fissions per second
m = 235*amu # Mass of a nucleon, kg
m_sec = m*n # Consumption per second of U-235, kg
m_year = m_sec*365*24*60*60 # Consumption per year of U-235, kg
print "The rate of consumption of U-235 per year = %5.2f kg"% m_year
#Rate of fission of U-235
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
E1 = 32e+06 # Energy released per second, J
E2 = 200e+06 # Energy released per fission, J
N = E1/E2 # Number of atoms undergoing fission per second
print "The number of atoms undergoing fission per second = %1.0e"%(N/e)
#Binding energy of helium nucleus
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
amu = 931 # Energy equivalent of 1 amu, MeV
m = 2*1.007825+2*1.008665-4.002603 # Mass difference in formation of He, amu
E = m*amu # Energy equivalent of mass difference for He nucleus, MeV
print "The minimum energy required to break He nucleus = %5.2f MeV"% E
#Energy released during fusion of deuterium nuclei
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
amu = 931.5 # Energy equivalent of 1 amu, MeV
M_H = 2.014102 # Mass of hydrogen nucleus, amu
M_He = 4.002603 # Mass of helium nucleus, amu
m = 2*M_H-M_He # Mass difference, amu
E = m*amu # Energy released during fusion of deuterium nuclei, MeV
print "The energy released during fusion of deuterium nuclei = %6.3f MeV"%E
#Energy required to break one gram mole of helium
amu = 931.5 # Energy equivalent of 1 amu, MeV
mp = 1.007825 # Mass of proton, amu
mn = 1.008665 # Mass of neutron, amu
M_He = 4.002603 # Mass of helium nucleus, amu
N = 6.023e+023 # Avogadro's number, g/mol
m = 2*mp+2*mn-M_He # Mass difference, amu
E1 = m*amu # Energy required to break one atom of He, MeV
E = N*E1 # Energy required to break one gram mole of He, MeV
print "The energy required to break one gram mole of He = %5.3e MeV"% E
#Energy liberated during production of alpha particles
amu = 931 # Energy equivalent of 1 amu, MeV
mp = 1.007825 # Mass of proton, amu
M_Li = 7.016005 # Mass of lithium nucleus, amu
M_He = 4.002604 # Mass of helium nucleus, amu
dm = M_Li+mp-2*M_He # Mass difference, amu
U = dm*amu # Energy liberated during production of two alpha particles, MeV
print "The energy liberated during production of two alpha particles = %5.2f MeV"%U
# Result
# The energy liberated during production of two alpha particles = 17.34 MeV
#Kinetic energy of neutrons
d = 2.2 # Binding energy of deuterium, MeV
H3 = 8.5 # Binding energy of tritium, MeV
He4 = 28.3 # Binding energy of helium, MeV
KE = He4-d-H3 # Kinetic energy of the neutron, MeV
print "The kinetic energy of the neutron = %4.1f MeV"% KE
#Consumption rate of U-235
N = 6.023e+026 # Avogadro's number, No. of atoms per kg
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
P = 100e+06 # Average power generation, J/s
U = P*365*24*60*60 # Energy required in one year, J
U1 = 180e+06*e # Energy produced by one atom fission of U-235
n = U/U1 # Number of atoms required to produce energy in one year
M = n*235/N # Mass of U-235 required per year, kg
print "The rate of consumption of U-235 per year = %7.4f kg"% M
#Minimum disintegraton energy of nucleus
mn = 1.008665 # Mass of neutron, amu
mp = 1.007276 # Mass of proton, amu
amu = 931 # Energy equivalent of 1 amu, MeV
BE = 2.21 # Binding energy of deutron nucleus, MeV
E = BE/amu # Binding energy of deutron nucleus, amu
M_D = mp+mn-E # Mass of deuterium nucleus, amu
print "The mass of deuterium nucleus = %8.6f amu"% M_D
#Rate of fission of U-235
N = 6.023e+026 # Avogadro's number, No. of atoms per kg
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
P = 1 # Average power generation, J/s
U = P*365*24*60*60 # Energy required in one year, J
U1 = 200e+06*e # Energy produced by one atom fission of U-235
n = U/U1 # Number of atoms undergoing fission per year
M = n/N # Mass of U-235 required per year, kg
print "The rate of fission of U-235 per year = %5.3e kg"% M
#Energy released during fission of U-235
N = 6.023e+023 # Avogadro's number
A = 235 # Mass number of U-235
n = N/235 # Number of atoms in 1g of U-235
E = 200 # Energy produced by fission of 1 U-235 atom, MeV
U = n*E # Energy produced by fission of 1g of U-235 atoms, MeV
print "The energy produced by fission of 1g of U-235 atoms = %5.3e MeV"% U
#Minimum energy of gamma photon for pair production
c = 3.0e+08 # Speed of light, m/s
me = 9.1e-031 # Mass of electron, kg
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
mp = me # Mass of positron, kg
U = (me+mp)*c**2/(e*1e+06) # Energy of gamma-ray photon, MeV
print "The energy of gamma-ray photon = %5.3f MeV"%U
from sympy import symbols, solve
#Uranium atom undergoing fission in a reactor
P_out = 800e+06 # Output power of the reactor, J/s
E1 = P_out*24*60*60 # Energy required one day, J
eta = 0.25 # Efiiciency of reactor
N = symbols('N') # Declare N as the variable
E2 = N*200e+06*1.6e-019*eta # Useful energy produced by N atoms in a day, J
N=solve(E2-E1, N)[0] # Number of U-235 atoms consumed in one day
m = N*235/6.023e+026 # Mass of uranium consumption in one day, kg
print "The number of U-235 atoms consumed in one day = %4.2e atoms"% N
print "The mass of uranium consumption in one day = %4.2f kg"% m
from sympy import symbols, solve
#Amount of uranium fuel required for one day operation
e = 1.6e-019 # Energy equivalent of 1 eV, J/eV
eta = 0.20 # Efficiency of the nuclear reactor
E1 = 100e+06*24*60*60 # Average energy required per day, J
m = symbols('m') # Suppose amount of fuel required be m kg
n = m*6.023e+026/235 # Number of uranium atoms
E = 200e+06*e # Energy released per fission of U-235, J
U = E*n # Total energy released by fission of U-235, J
E2 = U*eta # Useful energy produced by n atoms in a day, J
m = solve(E2-E1)[0]
print "The mass of uranium fuel required for one day operation = %6.4f kg/day"% m
from __future__ import division
#Binding energy of Fe using Weizsaecker formula
amu = 931.5 # Energy equivalent of 1 amu, MeV
A = 56 # Mass number of Fe
Z = 26 # Atomic number of Fe
av = 15.7 # Binding energy per nucleon due to volume effect, MeV
As = 17.8 # Surface energy constant, MeV
ac = 0.711 # Coulomb energy constant, MeV
aa = 23.7 # asymmetric energy constant, MeV
ap = 11.18 # Pairing energy constant, MeV
BE = av*A - As*A**(2/3) - ac*Z**2*A**(-1/3)-aa*(A-2*Z)**2*A**(-1)+ap*A**(-1/2) # Weizsaecker Semiempirical mass formula
M_Fe = 55.939395 # Atomic mass of Fe-56
mp = 1.007825 # Mass of proton, amu
mn = 1.008665 # Mass of neutron, amu
E_B = (Z*mp+(A-Z)*mn-M_Fe)*amu # Binding energy of Fe-56, MeV
print "The binding energy of Fe-56 using Weizsaecker formula = %6.2f MeV"% BE
print "The binding energy of Fe-56 using mass defect = %6.2f MeV"%E_B
print "The result of the semi empirical formula agrees with the experimental value within %3.1f percent"% abs((BE-E_B)/BE*100)
# Result
# The binding energy of Fe-56 using Weizsaecker formula = 487.75 MeV
# The binding energy of Fe-56 using mass defect = 488.11 MeV
# The result of the semi empirical formula agrees with the experimental value within 0.1 percent