# chapter 3:nuclear models¶

## example 3.1;pg no:38¶

In :
#cal of coulomb barrier for the penetration of Th by alpha particle and proton
#intiation of all variables
# Chapter 3
print"Example 3.1, Page:38  \n \n"
#Given in cgs units
m1=232.;
m2=1.;
m3=4.;
z1=90.;
z2=1.;
z3=2.;
e=4.8*10**-10;# in ergs
c=1.4;# nuclear radius constant

#Formula: E=(z1*z2*e^2)/(r1+r2)
r1=(m1)**(1./3.);
r2=(m2)**(1./3.);
r3=(m3)**(1./3.);
E1=(z1*z2*e*e)/(c*(r1+r2)*10**-13*(1.6*10**-6));
print"The coulomb barrier for the penetration of Th by proton in MeV=",round(E1,2)
E2=(z1*z3*e*e)/(c*(r1+r3)*10**-13*(1.6*10**-6));
print"The coulomb barrier for the penetration of Th by alpha particle in MeV=",round(E2,2)

Example 3.1, Page:38

The coulomb barrier for the penetration of Th by proton in MeV= 12.96
The coulomb barrier for the penetration of Th by alpha particle in MeV= 23.94


## example 3.2;pg no:38¶

In :
#cal of coulomb barrier for the penetration of Th by alpha particle and proton
#intiation of all variables
# Chapter 3
print"Example 3.2, Page38 \n \n"
#Given in cgs units
m1=112;
m2=1;
m3=4;
m4=66;
z1=50;
z2=1;
z3=2;
z4=30;
e=4.8*10**-10;# in ergs
c=1.4;# nuclear radius constant

#Formula: E=(z1*z2*e^2)/(r1+r2)
r1=(m1)**(1./3.);
r2=(m2)**(1./3.);
r3=(m3)**(1./3.);
r4=(m4)**(1./3.);
E1=(z1*z2*e*e)/(c*(r1+r2)*10**-13*(1.6*10**-6));
print"\n The coulomb barrier for the penetration of Th by proton in MeV=",round(E1,2)
E2=(z4*z3*e*e)/(c*(r4+r3)*10**-13*(1.6*10**-6));
print"\n \n The coulomb barrier for the penetration of Th by alpha particle in MeV=",round(E2,2)

Example 3.2, Page38

The coulomb barrier for the penetration of Th by proton in MeV= 8.84

The coulomb barrier for the penetration of Th by alpha particle in MeV= 10.96


## example 3.3;pg no:39¶

In :
#cal of closest distance of approach
#intiation of all variables
# Chapter 3
print"Example 3.3, Page:39  \n \n"
# Given:
E=6;# in MeV
z1=79;
z2=2;
q=4.8*10**-10;
# Solution:

# At the closest distance of approach, the kineic energy of the alpha particle balances the columb barrier energy.

r1=(z1*z2*q*q)/(E*1.6*10**-6);# distance in cm
r=r1*10**13;# distance in fm

print"The closest distance of approach in fm=",round(r,1)

Example 3.3, Page:39

The closest distance of approach in fm= 37.9


## example 3.4;pg no:39¶

In :
#cal of stable nuclied of the isobaric series
#intiation of all variables
# Chapter 3
print"Example 3.4, Page:39  \n \n"
# Given:
A=180.;
# Solution:
z=(40.*A)/(0.6*(A**(2./3.))+80.);
print"The stable nuclied of the isobaric series is Hf atomic no. =",round(z,1)

Example 3.4, Page:39

The stable nuclied of the isobaric series is Hf atomic no. = 72.6


## example 3.5;pg no:39¶

In :
#cal of stable nuclied of the isobaric series is Sr atomic no
#intiation of all variables
# Chapter 3
print"Example 3.5, Page:39  \n \n"
# Given:
A=87.;

# Solution:
z=(40.*A)/(0.6*(A**(2./3.))+80.);
print"The stable nuclied of the isobaric series is Sr atomic no. =",round(z)
# nereast integer is 38
print"\n Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola."

Example 3.5, Page:39

The stable nuclied of the isobaric series is Sr atomic no. = 38.0

Hence the nuclides of z<38 fall on the left of the limb of B vs Z parabola while the nuclides of z>38 fall on the right limb of the parabola.


## example 3.7;pg no:40¶

In :
#cal of binding energy for the last proton,neutron
#intiation of all variables
# Chapter 3
print"Example 3.7, Page:40  \n \n"
# Given:
B=11.009305;
C1=12;
C2=11.001433;
p=1.0078;
n=1.0087;
Al=26.981535;
Si1=27.976927;
Si2=26.986705;
# Solution:
m1=(B+p-C1);#(a)
E1=m1*931;# of last proton in C in MeV
print"\n The binding energy for the last proton in 12C in (MeV)=",round(E1,2)

m2=(C2+n-C1);#(b)
E2=m2*931;# of last neutron in C in MeV
print"\n The binding energy for the last neutron in 12C in (MeV)=",round(E2,2)

m3=(Al+p-Si1);#(c)
E3=m3*931;# of last proton in Si in MeV
print"\n The binding energy for the last proton in 28Si in (MeV)=",round(E3,2)

m4=(Si2+n-Si1);#(d)
E4=m4*931;# of last neutron in Si in MeV
print"\n The binding energy for the last neutron in 28Si in (MeV)=",round(E4,2)

# Note: There is a calculation error in the textbook for the (b) part.

Example 3.7, Page:40

The binding energy for the last proton in 12C in (MeV)= 15.92

The binding energy for the last neutron in 12C in (MeV)= 9.43

The binding energy for the last proton in 28Si in (MeV)= 11.55

The binding energy for the last neutron in 28Si in (MeV)= 17.2


## example 3.8;pg no:41¶

In :
#cal of binding energy for N,O
#intiation of all variables
# Chapter 3
print"Example 3.8, Page: 41 \n \n"
# Given:
d=2.014102;
C=12;
a=4.002603;
N=14.003074;
O=15.994915;

# Solution:
m1=(C+d-N);
E1=m1*931;# The binding energy for N(14)
print"\n The binding energy for N(14) in (MeV)=",round(E1,2)

m2=(C+a-O);
E2=m2*931;#The binding energy for O(16)
print"\n The binding energy for O(16) in (MeV)=",round(E2,2)

Example 3.8, Page: 41

The binding energy for N(14) in (MeV)= 10.27

The binding energy for O(16) in (MeV)= 7.16


## example 3.9;pg no:41¶

In :
#cal of binding energy
#intiation of all variables
# Chapter 3
print"Example 3.9, Page:41  \n \n"
# Given:
D=-1.997042;
n=1.0087;
# Solution:
m=(D+2.*n);
E=m*931.;
print"\n The binding energy in (MeV)=",round(E,2)

Example 3.9, Page:41

The binding energy in (MeV)= 18.95


## example 3.10;pg no:41¶

In :
#cal of seperation energy
#intiation of all variables
# Chapter 3
print"Example 3.10, Page:41  \n \n"
# Given:
mH=1.007825;
mn=1.008665;
M1=207.97666;# mass of Pb 208
M2=206.97590;# mass of Pb 207
M3=206.97739;# mass of Tl 207

# Solution:

B1=((82*1.007825+126*1.008665)-207.97666)*931;# binding energy for Pb 208
B2=((82*1.007825+125*1.008665)-206.97590)*931;# binding energy for Pb 207
B3=((81*1.007825+126*1.008665)-206.97739)*931;# binding energy for Tl 207
Sn=B1-B2;# neutron seperation energy
Sp=B1-B3;# proton seperation energy

print"\n The neutron seperation energy in (MeV)=",round(Sn,4)
print"\n The proton seperation energy in (MeV)=",round(Sp,4)

Example 3.10, Page:41

The neutron seperation energy in (MeV)= 7.3596

The proton seperation energy in (MeV)= 7.9647


## example 3.11;pg no:42¶

In :
#cal of seperation energy
#intiation of all variables
# Chapter 3
print"Example 3.11, Page:42  \n \n"
# Given:
mH=1.007825;
mn=1.008665;
M1=22.98977;# mass of Na 23
M2=21.994435;# mass of Na 22
M3=21.991385;# mass of Ne 22
# Solution:

m1=((11.*1.007825+12.*1.008665)-M1);
m2=((11.*1.007825+11.*1.008665)-M2);
m3=((10.*1.007825+12.*1.008665)-M3);
Sn=(m1-m2)*931.;# neutron seperation energy
Sp=(m1-m3)*931.;# proton seperation energy

print" The neutron seperation energy in (MeV)=",round(Sn,2)
print" The proton seperation energy in (MeV)=",round(Sp,2)

# Note: The answers are given in the form of atomic mass units where as in the question its asked for energies.

Example 3.11, Page:42

The neutron seperation energy in (MeV)= 12.41
The proton seperation energy in (MeV)= 8.79


## example 3.12;pg no:43¶

In :
#cal of excited level density,level spacing and nuclear temperature
#intiation of all variables
# Chapter 3
print"Example 3.12, Page:43  \n \n"
# Given:
C=0.3;# in MeV^-1
a=2.0;# in MeV
E=8; # in MeV
import math
# Solution:
d=C*(math.exp(2*((2*8)**(0.5))));# excited level density
s=(1/d)*1000;# level spacing
nT=(E/a)**(0.5);# nuclear temperature
print"\n The excited level density in (MeV)=",round(d,2)
print"\n The level spacing in (keV)=",round(s,2)
print"\n The nuclear temperature in (MeV)=",round(nT,2)

Example 3.12, Page:43

The excited level density in (MeV)= 894.29

The level spacing in (keV)= 1.12

The nuclear temperature in (MeV)= 2.0


## example 3.13;pg no:43¶

In :
#cal of expression for 1st, 2nd, and 3rd excited states
#intiation of all variables
# Chapter 3
print"Example 3.13, Page:43  \n \n"
# Given:
I0=3./2.;# ground state of spin

# Solution:
I1=I0+1.;
I2=I0+2.;
I3=I0+3.;
K=1.;# Assumed as some constant
# Formula: E=(h^2/(2*I))*((I*(I+1))-I0*(I0+1))
# Consider   K=(h^2/(2*I))=1

E1=K*((I1*(I1+1.))-(I0*(I0+1.)));# For 1 excited state

E2=K*((I2*(I2+1.))-(I0*(I0+1.)));# For 2 excited state

E3=K*((I3*(I3+1.))-(I0*(I0+1.)));# For 3 excited state

print"The expression for 1st, 2nd, and 3rd excited states are K times respectively.",E1,E2,E3

Example 3.13, Page:43

The expression for 1st, 2nd, and 3rd excited states are K times respectively. 5.0 12.0 21.0


## example 3.14;pg no:44¶

In :
#cal of energy of state 4,6,8,10
#intiation of all variables
# Chapter 3
print"Example 3.14, Page:44  \n \n"
# Given:
E2=44;# in keV

# Solution:
E4=E2*((4.*5.)/(2.*3.));# for part (a)
E6=E2*((6.*7.)/(2.*3.));# for part (b)
E8=E2*((8.*9.)/(2.*3.));# for part (c)
E10=E2*((10.*11.)/(2.*3.));# for part (d)

print"\n The energy of state 4 (+) in (keV)=",round(E4,2)
print"\n The energy of state 6 (+) in (keV)=",round(E6,2)
print"\n The energy of state 8 (+) in (keV)=",round(E8,2)
print"\n The energy of state 10 (+) in (keV)=",round(E10,2)

Example 3.14, Page:44

The energy of state 4 (+) in (keV)= 146.67

The energy of state 6 (+) in (keV)= 308.0

The energy of state 8 (+) in (keV)= 528.0

The energy of state 10 (+) in (keV)= 806.67


## example 3.15;pg no:44¶

In :
#cal of energy
#intiation of all variables
# Chapter 3
print"Example 3.15, Page:44  \n \n"
# Given:
E2=44;# in keV
En=525;# in keV

# Solution:
n=(En)/E2;
#
print"n=",n
print"\n For the required level of energy 525 keV nearest even integer is = & spin is (+)",round(n+1,2)

Example 3.15, Page:44

n= 11

For the required level of energy 525 keV nearest even integer is = & spin is (+) 12.0


## example 3.16;pg no:45¶

In :
#cal of  energy
#intiation of all variables
# Chapter 3
print"Example 3.16, Page:45  \n \n"
# Given:
E2=44.;# in keV
En1=146.;# in keV
En2=304.;# in keV
En3=514.;# in keV
# Solution:
n1=(En1)/E2;
n2=(En2)/E2;
n3=(En3)/E2;
print"n1=",n1
print"\n For the required level of energy 146 keV nearest even integer is =& spin is (+)",round(n1+1.,2)
print"\n \n n2=",n2
print"\n For the required level of energy 304 keV nearest even integer is =& spin is (+)",round(n2,2)
print"\n \n n3=",n3
print"\n For the required level of energy 514 keV nearest even integer is =& spin is (+)",round(n3+1.,2)

#Note: In the last part (c) the answer given in the textbook is 8(+). But the correct answer is 12(+)

Example 3.16, Page:45

n1= 3.31818181818

For the required level of energy 146 keV nearest even integer is =& spin is (+) 4.32

n2= 6.90909090909

For the required level of energy 304 keV nearest even integer is =& spin is (+) 6.91

n3= 11.6818181818

For the required level of energy 514 keV nearest even integer is =& spin is (+) 12.68

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