In [1]:

```
## Exa8.3 : : Page-349 (2011)
#find The probability that the proton moves within the range of neutron
import math
b = 1.9e-15; ## Width of square well potential, metre
h_kt = 1.054571e-034; ## Reduced planck's constant, joule sec
c = 3e+08; ## Velocity of light, metre per sec
m_n = 1.67e-27; ## Mass of a nucleon , Kg
V_0 = 40*1.6e-13; ## Depth, metre
E_B = (V_0-(1/(m_n*c**2)*(math.pi*h_kt*c/(2*b))**2))/1.6e-13; ## Binding energy, mega electron volts
alpha = math.sqrt(m_n*c**2*E_B*1.6e-13)/(h_kt*c); ## scattering co efficient, per metre
P = (1+1/(alpha*b))**-1.; ## Probability
R_mean = math.sqrt (b**2./2.*(1./3.+4./math.pi**2.+2.5)); ## Mean square radius, metre
print'%s %.2f %s %.2e %s'%("\nThe probability that the proton moves within the range of neutron = ",P," \n" "The mean square radius of the deuteron = ",R_mean," metre")
## Result
## The probability that the proton moves within the range of neutron = 0.50
## The mean square radius of the deuteron = 2.42e-015 metre
```

In [2]:

```
## Exa8.5 : : Page-349 (2011)
#find The total cross section for n-p scattering
import math
a_t = 5.38e-15;
a_s = -23.7e-15;
r_ot = 1.70e-15;
r_os = 2.40e-15;
m = 1.6748e-27;
E = 1.6e-13;
h_cut = 1.0549e-34;
K_sqr = m*E/h_cut**2;
sigma = 1/4.*(3.*4*math.pi*a_t**2./(a_t**2.*K_sqr+(1.-1/2.*K_sqr*a_t*r_ot)**2)+4*math.pi*a_s**2/(a_s**2*K_sqr+(1-1./2.*K_sqr*a_s*r_os)**2))*1e+028; ## Total cross-section for n-p scattering, barn
print'%s %.2f %s'%("\nThe total cross section for n-p scattering = ",sigma," barn");
## Result
## The total cross section for n-p scattering = 2.911 barn
```

In [22]:

```
## Exa8.8 : : Page-351 (2011)
#find The possible angular momentum states with their parities are as follows
import math
S = 1.; ## Spin angular momentum(s1+-s2), whereas s1 is the spin of proton and s2 is the spin of neutron.
m = 2.*S+1.; ## Spin multiplicity
j = 1.; ## Total angular momentum
print("\nThe possible angular momentum states with their parities are as follows : ");
print'%s %.2f %s %.2f %s '%("\n ",m, " " and "S has even parity ",j,"");
print'%s %.2f %s %.2f %s '%("\n ",m," " and "P has odd parity ", j,"");
print'%s %.2f %s %.2f %s'%("\n ",m, " " and "S has odd parity ",j,"");
S = 0.;
m = 2.*S+1.
print(m)
print'%s %.2f %s %.2f %s '%("\n ",m," " and "P has odd parity ", j,"");
## Result
## The possible angular momentum states with their parities are as follows :
## 3S1 has even parity
## 3P1 has odd parity
## 3D1 has even parity
## 1P1 has odd parity
```

In [10]:

```
# Exa8.9 : : Page-351 (2011)
#find The possible states are
print("\nThe possible states are : ");
#For s = 0
s = 0; # Spin angular momentum
m = 2*s+1; # Spin multiplicity
for j in range(0,3): # Total angular momentum
l = j
if l == 0:
print"%s %.1f %s %.d %s "%("",j,""and "S",m,"")
elif l == 2:
print"%s %.1f %s %.d %s "%("",j,""and "D",m,"")
#For s = 1
s = 1;
m = 2*s+1;
l = 2
for j in range(0,3):
if j == 0:
print"%s %.1f %s %.d %s "%("",j,""and "P",m,"")
elif j ==1:
print"%s %.d %s %.d %s "%("",j,""and "P",m,"")
elif j ==2:
print"%s %.d %s %.d %s "%("",j,""and "P",m,"")
for j in range(2,3):
print"%s %.d %s %.d %s "%("",j,"" and "F",m,"")
#Result
#Possible states are :
# The possible states are :
# 0S1, 2D1, 0P3, 1P3, 2P3 and 2F3
```

In [4]:

```
## Exa8.10 : : Page-352 (2011)
#find The kinetic energy of each nucleon and The total kinetic energy
import math
r = 2e-015; ## Range of nuclear force, metre
h_kt = 1.0546e-34; ## Reduced value of Planck's constant, joule sec
m = 1.674e-27; ## Mass of each nucleon, Kg
K = round (2*h_kt**2./(2*m*r**2*1.6023e-13)); ## Kinetic energy of each nucleon in centre of mass frame, mega electron volts
K_t = 2.*K; ## Total kinetic energy, mega electron volts
K_inc = 2.*K_t; ## Kinetic energy of the incident nucleon, mega electron volts
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe kinetic energy of each nucleon = ",K," MeV" and "The total kinetic energy =",K_t," MeV"and "The kinetic energy of the incident nucleon =",K_inc," MeV")
## Result
##
```