Chapter 12: Applied Nuclear Physics

Example 12.1, Page 12.31

In [1]:
from math import log

# Given 
E = 1.14 # energy of gamma radiation in Mev
l = 0.2 # length of aluminium in meter
p = 0.03 # reduce in intensity in beam
d = 2700 # density of aluminium in kg/m^3 

#Calculations
mu = (1 / l) * log(1 / p)
k = mu / d
x = 0.693 / mu

#Result
print "Mass absorption coeffiecient of Al for this radiation is %f m^2/kg\nHalf value thickness is %.4f meter"%(k,x)

Mass absorption coeffiecient of Al for this radiation is 0.006494 m^2/kg
Half value thickness is 0.0395 meter

Example 12.2, Page 12.32

In [2]:
from math import *

# Given that
E = 1.1 # energy of gamma radiation in Mev
l = 0.25 # length of aluminium in meter
p = 0.02 # reduce in intensity in beam
d = 2700 # density of aluminium in kg/m^3 

#Calculations
mu = (1 / l) * log(1 / p)
k = mu / d
x = 0.693 / mu

#Result
print "Mass attenuation coefficient of Al for this radiation is %.1e m^2/kg\nHalf value thickness is %.4f meter"%(k,x)

Mass attenuation coefficient of Al for this radiation is 5.8e-03 m^2/kg
Half value thickness is 0.0443 meter

Example 12.3, Page 12.32

In [3]:
from math import *

# Given 
t = 15 # half-life for Na(23) in hours
r = 93.75 # percentage fraction of sample which decayed 

#Calculations
lamda = 0.693 / t
T = (1 / lamda) * (log(100 / (100 - r)))

#Result
print "Time taken for 93.75 per decay of sample is %d hours"%T

Time taken for 93.75 per decay of sample is 60 hours

Example 12.4, Page 12.33

In [4]:
from math import *

# Given 
t = 4 # half-life of radioactive element in years
r = 1. / 64 # ratio of mass of element present in specimen to the initial mass of element 

#Calculations
lamda = 0.693 / t
T = (1 / lamda) * log(1 / r)

#Result
print "Time after which element present in specimen reduce to 1/64 of its original value is %d years"%T

Time after which element present in specimen reduce to 1/64 of its original value is 24 years

Example 12.5, Page 12.33

In [5]:
from math import *

# Given 
t = 15 # half-life of radioactive element in years
r = 0.025 # ratio of mass of element present in specimen to the intial mass of element 

#Calculations
lamda = 0.693 / t
T = (1 / lamda) * log(1 / r)

#result
print "Period in which 2.5 percent of the initial quantity left over is %.2f years"%T

Period in which 2.5 percent of the initial quantity left over is 79.85 years

Example 12.6, Page 12.33

In [7]:
from math import *

# Given 
t = 3.8 # half-life for radon in days
r = 60. # percentage fraction of sample which decayed 

#Calculations
lamda = 0.693 / t
T = (1 / lamda) * (log(100 / (100 - r)))

#Result
print "Time taken for 60 percent decay of sample is %.3f days"%T

Time taken for 60 percent decay of sample is 5.024 days

Example 12.7, Page 12.34

In [8]:
# Given 
lamda = 4.28e-4 # decay constant in per year

#Calculations
T = 0.693 / lamda
t = 1 / lamda

#Result
print "Half life time is %.2f years\nMean life time is %.2f years"%(T,t)

Half life time is 1619.16 years
Mean life time is 2336.45 years

Example 12.8, Page 12.34

In [9]:
from math import *

# Given 
t = 30 # time in years
r = 1. / 64 # ratio of final mass of element to the intial mass of element 

#Calculations
lamda = log(1 / r) / t
T = 0.693 / lamda

#Result
print "Half life of radioactive material is %d years"%(ceil(T))

Half life of radioactive material is 5 years

Example 12.9, Page 12.34

In [10]:
# Given 
t = 2.1 # half life in minute
r = 60 # percentage fraction of sample which decayed 

#Calculation
lamda = 0.693 / t

#Result
print "Decay constant is %.2f per minute"%lamda

Decay constant is 0.33 per minute

Example 12.10, Page 12.35

In [11]:
from math import *

# Given 
t = 2.7 # half-life of Au(198) in days
m = 1e-6 # mass of sample in gm
T = 8 * 86400 # time in seconds

#Calculations
lamda = 0.693 / (t * 86400)
N = (m * 6.023e23) / 198 # by the formula (N = mass*Avogadro number/molar mass)
A_ = lamda * N
A = A_ * (1 / exp(lamda * T))

#Result
print "Activity of sample is %.4e decays/sec"%A

Activity of sample is 1.1595e+09 decays/sec

Example 12.11, Page 12.35

In [12]:
# Given 
n = 3 # no. of half lives

#Calculation
f = (1. / 2)**n

#Result
print "Fraction of sample left after %d half lives is %.3f  "%(n,f)

Fraction of sample left after 3 half lives is 0.125

Example 12.12, Page 12.35

In [13]:
# Given 
t = 2 # life period of radioactive substance in years
T = 4 # time in years
m = 10. # mass of substance in mg

#Calculation
N = m / T # in mg

#Result
print "Substance remained unchanged after 4 years is %.1f mg"%N

Substance remained unchanged after 4 years is 2.5 mg

Example 12.13, Page 12.36

In [15]:
from math import *

# Given that
m = 1. # initial mass of radium in gm
m_ = 0.0021 # final mass of radium in gm
t = 5. # time for decay from m to m_ in years

#Calculations
lamda = log(m / (1 - m_)) / t
T = 0.693 / lamda
T_ = 1. / lamda

#Result
print "Decay constant is %.4f per year\nHalf life of sample is %.1f years\nAverage life of sample is %.3f years"%(lamda,T,T_)
#Incorrect answer in the textbook

Decay constant is 0.0004 per year
Half life of sample is 1648.3 years
Average life of sample is 2378.452 years

Example 12.14, Page 12.36

In [16]:
from math import *

# Given 
t = 10. # time in days
r = 15. # percentage fraction of sample which remain 

#Calculations
lamda = log(100. / 15) / t
T = 0.693 / lamda

#Result
print "Half life of sample is %.3f days"%T

Half life of sample is 3.653 days

Example 12.15, Page 12.36

In [17]:
from math import *

# Given 
t = 12.3 # half life in year
T = 50 # time in year 

#Calculations
lamda = 0.693 / t
f = 1 / exp(lamda * T)

#Result
print "Fraction of radioactive isotope remained is %.2f "%f

Fraction of radioactive isotope remained is 0.06

Example 12.16, Page 12.37

In [18]:
# Given 
R = 1 # radioactivity of Pb(214) in curie
t = 26.8 # half life in minute 

#calculations
lamda = 0.693 / (t * 60)
R = 1 * 3.7e10 # in disintegration per sec
m = (R * 214) / (6.023e23 * lamda)

#Result
print "Mass of Pb(214)is %.2e gm"%m

Mass of Pb(214)is 3.05e-08 gm

Example 12.17, Page 12.37

In [19]:
# Given 
R = 1e6 # radioactivity of Pb(214) in disintegrations per sec
t = 26.8 # half life in minute 

#Calculations
lamda = 0.693 / (t * 60)
m = (R * 214) / (6.023e23 * lamda)

#Result
print "Mass of Pb(214) is %.2e gm"%m

Mass of Pb(214) is 8.24e-13 gm

Example 12.18, Page 12.37

In [21]:
# Given 
m = 1. # mass of Ra(226) in gm 
R = 1. # radioactivity of Ra(226) in curie

#Calculations
r = R * 3.7e10 # in disintegrations per sec
N = 6.023e23 * m / 226
lamda = r * 226 / 6.023e23
T = 1. / lamda
T_ = 0.693 / lamda

#Result
print "Mean life of radium is %.2e year\nHalf life of radium is %.2e year"%(T,T_)

Mean life of radium is 7.20e+10 year
Half life of radium is 4.99e+10 year

Example 12.19, Page 12.38

In [23]:
# Given 
m = 0.0001 # mass of Sr(90) in gm
t = 28. # half life of Sr(90) in year
t_ = 9. # time in sec

#Calculations
lamda = 0.693 / (t * 86400 * 365)
N_ = 6.023e23 * m / 90
n = N_ * lamda * t_

#Result
print "Activity of Sr is %.2e disintegration/sec."%n
#Incorrect answer in the textbook

Activity of Sr is 4.73e+09 disintegration/sec.

Example 12.20, Page 12.38

In [24]:
# Given 
t = 1600 # the half life of radium(226) in year
t1 = 3.8 # the half life of radon(222) in days
m = 1 # mass of Ra(226) in gm

#Calculation
m_ = (222 * t1 * m) / (226 * 365 * t) # by the formula N1*t = N2*t1

#Result
print "Mass of radon is %.2e gm"%m_

Mass of radon is 6.39e-06 gm

Example 12.21, Page 12.39

In [25]:
# Given 
m1 = 4.002603 # mass of He(4) in a.m.u.
m2 = 3.016056 # mass of H(3) in a.m.u.
m3 = 1.007276 # mass of H(1) in a.m.u.

#Calculations
k = m2 + m3 - m1
E = k * 931

#Result
print "Energy of gamma ray photon is %.3f MeV"%E

Energy of gamma ray photon is 19.299 MeV

Example 12.22, Page 12.39

In [1]:
# Given 
E = 3 # kinetic energy of proton in Mev
m1 = 1.007276 # mass of H(1) in a.m.u.
m2 = 3.016056 # mass of H(3) in a.m.u.
m3 = 1.008665 # mass of neutron in a.m.u.
m4 = 3.016036 # mass of He(3) in a.m.u.

#calculations
k = m1 + m2 - m3 - m4
E = k * 931.5

#Result
print "Q value of reaction is %.4f MeV"%E

Q value of reaction is -1.2752 MeV

Example 12.23, Page 12.40

In [2]:
# Given 
E = 200 # energy released per fission in Mev
m = 0.01 # mass of U(235) in gm
n = 235 # atomic no of sample
N_0=6.023e23 # Avogadro constant

#calculations
E_ = E * 1.6e-13
k = E_ * N_0 * m / n
H = k / 4.168

#Result
print "Heat produce by complete disintegration is %.3e cal\nEnergy released is %.1e J"%(H,E_)

Heat produce by complete disintegration is 1.968e+08 cal
Energy released is 3.2e-11 J

Example 12.24, Page 12.40

In [3]:
# Given 
E = 200 # energy released per fission in Mev
m = 1 # mass of U(235) in kg

#Calculations
E_ = E * 1.6e-13
k = E_ * 6.023e26 * m / 235

#Result
print "Energy released  by fission of 1 kg of U(235)is %.2e J"%k

Energy released  by fission of 1 kg of U(235)is 8.20e+13 J

Example 12.25, Page 12.40

In [4]:
# Given 
P = 1e9 # power required for enlighten the city in watt
e = 30 # percentage efficiency of nuclear reactor
E = 3.2e-11 # energy released per fission in J

#Calculations
E_ = E * 30 / 100
N = P / E_
N_ = N * 24 * 3600
m = N_ * 235 / 6.023e26 

#Result
print "Amount of fuel required per day is %.2f kg"%m

Amount of fuel required per day is 3.51 kg

Example 12.26, Page 12.41

In [5]:
# Given 
E = 200 # energy released per fission of U(235)in Mev
m = 3.7 # mass of U(235) consumed in a day in kg
e = 20 # percentage efficiency of reactor

#Calculations
E_ = E * 1.6e-13
N = m * 6.023e26 / 235
H = E_ * e / 100
k = H * N / (24 * 3600)

#Result
print "Power output of reactor is %.3f GW"%(k * 1e-9)

Power output of reactor is 0.702 GW

Example 12.27, Page 12.42

In [6]:
# Given 
m1 = 4.00260 # mass of He(4) in a.m.u.
m2 = 0.00055 # mass of electron in a.m.u.
m3 = 12 # mass of C(12) in a.m.u.

#Calculations
delta_m = 3 * m1 - m3
E = delta_m * 931

#Result
print "Energy produce by each reaction is %.4f MeV"%E

Energy produce by each reaction is 7.2618 MeV

Example 12.28, Page 12.42

In [9]:
# Given 
P = 5e7 # power in watt
e = 33. # percentage efficiency of nuclear reactor
m1 = 2.01478 # mass of H(2) in a.m.u.
m2 = 4.00388 # mass of He(4) in a.m.u.

#calculations
m = (2 * m1) - m2
E = m * 931 * 1.6e-13
E_O = (E * e) /(2 * 100)
N = P / E_O
k = N * m1 / 6.023e26
M = k * 24 * 3600 

#Result
print "Mass of deuterium consumed per day is %.4f kg"%M

Mass of deuterium consumed per day is 0.0229 kg

Example 12.29, Page 12.43

In [10]:
from math import *

# Given
d = 1.8 # diameter in meter
B = 0.8 # magnetic field in tesla
m = 6.68e-27 # mass of He(4) in kg
e = 1.6e-19 # charge on an electron in Coulomb

#Calculations
r = d / 2
E = (B**2 * (2 * e)**2 * r**2) / (2 * m * 1.6e-19 * 10**6) 
f = B * 2 * e / (2 * pi * m)
N = f / 2

#Result
print "Energy  is %.2f MeV\nNumber of revolution made by particle to obtain above energy is %.2e per sec"%(E,N)

Energy  is 24.83 MeV
Number of revolution made by particle to obtain above energy is 3.05e+06 per sec

Example 12.30, Page 12.43

In [11]:
from math import *

# Given
f = 12e6 # oscillator frequency of cyclotron in Hz
r = 0.53 # radius of dee in meter
e = 1.6e-19 # charge on an electron in Coulomb

#Calculations
B = (2 * pi * f * 2 * 1.67e-27) / e

#Result
print "Value of magnetic induction needed to accelerate deuteron in it is %.3f T"%B

Value of magnetic induction needed to accelerate deuteron in it is 1.574 T

Example 12.31, Page 12.44

In [12]:
from math import *

# Given 
f = 10e6 # frequency of applied Emf in Hz
r = 0.32 # radius in meter
m = 3.32e-27 # mass of deuteron in kg
e = 1.6e-19 # charge on an electron in Coulomb

#Calculations
B = (2 * pi * f * m) / e
v = (e * B * r) / m

#Result
print "Flux density of the magnetic field is %.3f T\nVelocity of the deuterons emerging out of the cyclotron is %.2e m/sec "%(B, v)

Flux density of the magnetic field is 1.304 T
Velocity of the deuterons emerging out of the cyclotron is 2.01e+07 m/sec

Example 12.32, Page 12.44

In [13]:
from math import *

# Given 
f = 60 # operating frequency in Hz
d = 1.6 # diameter in meter
B = 0.5 # magnetic field at the orbit in tesla
e = 1.6e-19 # charge on an electron in Coulomb

#Calculations
r = d / 2
w = 2 * pi * f
E = 4 * e * w * r**2 * B
E_ = 3e8 * r * B / 1e6

#Result
print "Energy gained per turn is %.1f eV\nFinal energy is %d MeV"%(E / e,E_)

Energy gained per turn is 482.5 eV
Final energy is 120 MeV

Example 12.33, Page 12.45

In [14]:
# Given 
E = 70 # energy of betatron synchrotron in Mev
r = 0.28 # radius in meter
e = 1.6e-19 # charge on an electron in C

#Calculations
E_ = E * 1.6e-13
B = E_ / (3e8 * e * r)

#Result
print "Magnitude of magnetic field is %.2f T"%B

Magnitude of magnetic field is 0.83 T

Example 12.34, Page 12.45

In [15]:
# Given 
E = 4.18 # energy of alpha particle in Mev
n = 12 # no. of particle enter the chamber per sec
E_ = 40 # required energy of an ion pair in ev
e = 1.6e-19 # charge on an electron in C

#Calculations
R = n * E * 10**6 # in eV
N = R / E_
i = N * e

#Result
print "The current produced = %.e Amp"%i

The current produced = 2e-13 Amp

Example 12.35, Page 12.46

In [17]:
# Given
n = 10**8 # no. of electron per discharge counted by GM counter
r = 500. # counting rate in counts per minutes
e = 1.6e-19 # charge on an electron in C

#Calculations
N = r / 60
i = N * n * e 

#Result
print "Average current in the circuit = %.2e Amp"%i

Average current in the circuit = 1.33e-10 Amp

Example 12.36, Page 12.46

In [18]:
from math import *

# Given 
E = 10 # energy of electron in kev
B = 5e-5 # magnetic field of earth in tesla
e = 1.6e-19 # charge on an electron in C

#Calculations
f = e * B / 9.1e-31
E_ = E * 1.6e-16
v = sqrt((2 * E_) / 9.1e-31)
r = v / f

#Result
print "Frequency of cyclotron = %.3e per sec\nLarmour radius = %.3f meter"%(f,r)

Frequency of cyclotron = 8.791e+06 per sec
Larmour radius = 6.745 meter

Example 12.37, Page 12.46

In [19]:
# Given 
B = 5e-9 # magnetic field in tesla
v = 3e5 # velocity of proton stream in m/sec
e = 1.6e-19 # charge on an electron in C

#calculations
r = (1.67e-27 * v) / (e * B)

#Result
print "Larmour radius is %.2e meter"%r

Larmour radius is 6.26e+05 meter

Example 12.38, Page 12.46

In [21]:
from math import *

# Given 
E = 1. # energy of He+ in kev
r = 0.188 # Larmour radius in meter
e = 1.6e-19 # charge on an electron in C

#Calculations
E_ = E * 1.6e-16
v = sqrt((2 * E_) / (4 * 1.67e-27))
B = (4 * 1.67e-27 * v) / (e * r)

#Result
print "Magnetic field is %.3e tesla"%B

Magnetic field is 4.861e-02 tesla

Example 12.39, Page 12.47

In [22]:
from math import *

# Given 
E = 3.5 # energy of He++ ash particle in Mev
B = 8 # magnetic field in tesla
e = 1.6e-19 # charge on an electron in C

#Calculations
E_ = E * 1.6e-13
v = sqrt(2 * E_ / (4 * 1.67e-27))
r = (4 * 1.67e-27 * v) / (2 * e * B)

#Result
print "Larmour radius is %.2e meter"%r

Larmour radius is 3.38e-02 meter

Example 12.40, Page 12.47

In [23]:
from math import *

# Given 
d = 1e12 # electron density in number per m^3
E = 0.1 # thermal energy in eV
e = 1.6e-19 # charge on an electron in C

#Calculations
lamda = sqrt((8.85e-12 * E * e) / (d * e * e))
omega = sqrt(d * e**2 / (9.1e-31 * 8.85e-12))
f = omega / (2 * pi)

#Results
print "Debye length is %.2e meter\nPlasma frequency is %.2f MHz"%(lamda,f / 1e6)

Debye length is 2.35e-03 meter
Plasma frequency is 8.97 MHz

Example 12.41, Page 12.48

In [25]:
from math import *

# Given 
d = 1e16 # density in per m^3
E = 2 # thermal energy in eV
e = 1.6e-19 # charge on an electron in C

#Calculations
lamda = sqrt((8.85e-12 * E * e) / (d * e * e))
omega = sqrt(d * e**2 / (9.1e-31 * 8.85e-12))
f = omega / (2 * pi)

#Result
print "Debye length is %.4e meter\nPlasma frequency is %.3e Hz"%(lamda,f)

Debye length is 1.0518e-04 meter
Plasma frequency is 8.973e+08 Hz