# Chapter8-Nuclear Forces¶

## Ex3-pg349¶

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

The probability that the proton moves within the range of neutron =  0.50
The mean square radius of the deuteron =   2.42e-15  metre


## Ex5-pg349¶

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

The total cross section for n-p scattering =  2.91  barn


## Ex8-pg351¶

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

The possible angular momentum states with their parities are as follows :

3.00 S has even parity  1.00

3.00 P has odd parity   1.00

3.00 S has odd parity  1.00
1.0

1.00 P has odd parity    1.00


## Ex9-pg351¶

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

The possible states are :
0.0  1
2.0  1
0.0  3
1  3
2  3
2  3


## Ex10-pg352¶

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
##

The kinetic energy of each nucleon =  10.00 The total kinetic energy = 20.00 The kinetic energy of the incident nucleon = 40.00  MeV