In [2]:

```
#pg 239
#calculate the nuclear density
#Given:
import math
mp = 1.67*10**-27 ; # proton mass in kg
r0 = 1.2*10**-15; # constant in m
a0 = 0.5*10**-10; # atomic dimensions in m
#rho_nucleus = nuclear mass/ nuclear volume
#calculations
rho_nucleus = (3*mp)/(4*math.pi*r0**3); # nuclear density in kg/m**3
#ratio = rho_nucleus/rho_atom = (a0/r0)**3
ratio = a0**3/r0**3;
#results
print"Nuclear density is",round(rho_nucleus*10**-17,1),"x 10^17 kg/m**3"
print"Nuclear density is",round(ratio*10**-13,0),"x 10^13 times Atomic density."
```

In [3]:

```
#pg 241
#calculate the rest mass of pion
#Given :
h = 1.05*10**-34; #planck's constant in Js
m = 9.1*10**-31; #electron rest mass in kg
c = 3*10**8; #Speed of light in m/s
b = 1.7*10**-15; # range of nuclear force in m
#calculations
m_pi = h/(b*c); # rest mass of a pion in kg
t = m_pi/m; # times the rest mass of an electron
#results
print "Rest mass of a pion is",round(t,0), "times the rest mass of an electron"
print 'textbook answer is 220 , because approximate value for m_pi was considered.'
```

In [4]:

```
#pg 242
#calculate the Binding energy
#Given :
mp = 1.007276470 ; # proton mass in u
mn = 1.008665012; # neutron mass in u
md = 2.013553215; # deuteron mass in u
#E = ( mp + mn - md)*c**2
# 1 u * c**2 = 931.5 MeV , where 1 u = 1.66*10**-27 kg and c = 3*10**8 m/s
#calculations
E = (mp + mn - md)*931.5; # Binding energy in MeV
#results
print "Binding energy :",round(E,3),"MeV"
```

In [5]:

```
#pg 243
#calculate the Average binding energy per nucleon
#Given :
m_alpha = 4.001506106; # mass of an alpha particle in u
mp = 1.007276470 ; # proton mass in u
mn = 1.008665012; # neutron mass in u
#E = ( 2*mp + 2*mn - m_alpha)*c**2
# 1 u * c**2 = 931.5 MeV , where 1 u = 1.66*10**-27 kg and c = 3*10**8 m/s
#calculations
E = (2*mp + 2*mn - m_alpha)*931.5; # Binding energy in MeV
#resuts
print "Average binding energy per nucleon :",round(E/4.,3)," MeV"
```

In [7]:

```
#pg 249
#calculate the Q value
#Given :
Mn = 14.00753; #mass of Nitrogen 14 in u
Mo = 17.0045; # mass of Oxygen 17 in u
m_alpha = 4.00387; # mass of alpha particle in u
mp = 1.00184; # mass of proton in u
#Q = (m_alpha + Mn - Mo - mp)*c^2
## 1 u * c^2 = 931.5 MeV , where 1 u = 1.66*10^-27 kg and c = 3*10^8 m/s
#calculations
Q = (m_alpha + Mn - Mo - mp)*931.5 ;# Q value in MeV
#results
print "Q value is (MeV) = ",round(Q,1)
```

In [8]:

```
#pg 250
#calculate the angle of ejection
#Given :
import math
Q = 4.;# in MeV
Ex = 2.; # in MeV
Ey = 5.; # in MeV
mx = 4.; # in u
my = 1.; # in u
My =13.; # in u
#calculations
theta = math.acos(( (Ey*(1 + (my/My))) - (Ex*(1 - (mx/My))) - Q )/((2/My)*math.sqrt(mx*Ex*my*Ey)))*57.3; # angle of ejection in degrees
#resutls
print"Angle of ejection is (degrees) = ",round(theta,0)
```

In [9]:

```
#pg 251
#calculate the energy for electron and nucleon
#Given :
h = 6.625*10**-34 ; #planck's constant in Js
me = 9.1*10**-31 ; #electron mass in kg
mn = 1.67*10**-27;# a nucleon mass in kg
#calculations
#(a)For electron
L1 = 1; # in A
#E = (n**2*h**2)/(8*m*L**2) , here n value is not given , so let us calculate the remaining part (neglecting n**2 in the formula)
#L1 = 1*10**-10 m , 1A = 1.0*10**-10 m
E1 = h**2/(8*me*(L1*10**-10)**2); # energy in J
#(b)For nucleon
L2 = 1; # in fm
#E = (n**2*h**2)/(8*m*L**2) , here n value is not given , so let us calculate the remaining part (neglecting n**2 in the formula)
#L2 = 1*10**-15 m , 1 fm = 1.0*10**-15 m
E2 = h**2/(8*mn*(L2*10**-15)**2);#energy in J
#results
print"Energy for an electron :",round(E1*10**17,1)," x 10**-17 x n^2 J"
print"Energy for a nucleon :",round(E2*10**11,2)," x 10**-11 x n^2 J"
```

In [10]:

```
#pg 260
#calculate the energy released and tonnes of coal
#Given :
Na = 6.023*10**23 ; # Avogadro constant in atoms/mole
LE = 200.; # liberated energy in MeV
mm = 235.; # molar mass of U 235 in gm/mole
#calculations
# 1 eV = 1.6*10**-19 J , 1 MeV = 1.0*10**6 eV
RE = (Na*LE*1.6*10**-19*10**6)/mm ; #released energy in J
# 1 cal = 4.187 J
EC = RE/4.187 ; # energy in cal
#Burning 1 kg of coal releases 7000 K cal of energy
Q1 = EC/(7000*10**3); # Quantity of Coal in Kg
#Exploding 1 kg of TNT releases 1000 cal of energy
Q2 = EC/1000; # Quantity of TNT in kg
#results
print"Energy released :",round(EC*10**-10,0),"x 10^10 cal"
print round(Q1*10**-3,0)," tonnes of Coal"
print round(Q2*10**-3,0)," tonnes of TNT"
print 'Results obtained differ from those in textbook , because approximate values were considered in textbook.'
```

In [12]:

```
#pg 260
#calculate the tonnes of coal and electric power required
#Given :
Na = 6.023*10**23 ; # Avogadro constant atoms/mole
LE = 200.; # liberated energy in MeV
mm = 235.*10**-3; # molar mass of U 235 in gm/mole
p = 30./100. ; # conversion efficiency
#calculations
# 1 eV = 1.6*10**-19 J , 1 MeV = 1.0*10**6 eV
RE = (Na*LE*1.6*10**-19*10**6)/mm ; #released energy in J per day
# 1 day = 24 hrs * 60 mins * 60 sec
P = RE/(24.*60*60); # Power output in W per day
# 1 cal = 4.187 J
EC = RE/4.187 ; # energy in cal
#Burning 1 kg of coal releases 7000 K cal of energy
Q1 = EC/(7000.*10**3); # Quantity of Coal in Kg per day
EP = p*P ; # electric power in W
#results
print" tonnes of Coal = ",round(Q1*10**-3,0)
print" Electric power for 30 percent conversion efficiency (kW) = ",round(EP*10**-3,0)
print' Results obtained differ from those in textbook , because approximate values were considered in textbook.'
```

In [13]:

```
#pg 264
#calculate how long ago tree died
#Given :
import math
T_half = 5730.; # carbon 14 half life in years
Na = 6.023*10**23; # Avogadro constant in nuclei/mole
M = 25.;# charcoal mass in gm
mm = 12.;# molar mass of carbon 12 in gm/mole
a = 250. ; # disinitegrations per minute (Carbon 14 activity)
# 1 year = 525949 minutes
#calculations
lambd = 0.693/(T_half*525949);# disinitegrations per minute per nucleus
N0_1 = (Na/mm)*M ; # Number of nuclei (Carbon 12)
# Carbon 14 to Carbon 12 ratio = 1.3*10**-12
N0_2 = 1.3*10**-12*N0_1 ; # Number of nuclei (Carbon 14)
R0 = N0_2*lambd ; # disinitegrations per minute per nucleus
a0 = R0 ; # initial activity
t = math.log(a0/a)/lambd ;
# 1 year = 525949 minutes
#results
print "The tree died (years ago) = ",t/525949.
print 'Result obtained differs from the textbook, because R0 value obtained here is 375.1025, where as in textbook it is 374.'
```

In [15]:

```
#pg 265
#calculate the Original amount and Remaining amount
#Given :
import math
T_half = 8. ; # iodine 131 haf life in days
lambd = 0.693/T_half ; # decay constant in decays/day
N0 = 20. ; # mass in mg
t = 48.; # time in days
#calculations
N = N0*math.exp(-lambd*t); # in mg
#results
print"Original amount (mg) = ",N0
print"Remaining amount after 48 days (mg) = ",round(N,3)
```

In [16]:

```
#pg 268
#calculate the equivalent dose
#Given :
RBE = 0.7 ; #RBE factor for cobalt 60 gamma rays
dose = 1000 ; # dose in rad
#calculations
e = RBE*dose; # equivalent dose in rem
#results
print"Equivalent dose is (rem) = ",e
```