chapter2 :properties of nucleon and nuclei

example 2.1;pg no:19

In [1]:
#cal of density nucleus
#intiation of all variables
# Chapter 2
print"Example 2.1, Page:19  \n \n"
#Given:
f=19;# atomic mass no. of F
a=197;# atomic mass no.of Au
p=239;# atomic mass no. of Pu
#solution:(a)

m1=f/(6.02*10**23);
Rf=1.4*(f**(1/3))*10**-13;# in cm
V1=1.3333*3.14*(Rf)**3;
df=m1/(V1*10**14);# density in 10^14 g cm^-3
print"\n The density nucleus of F(19) in 10^14 g cm^-3 is =",df

#(b)
m2=a/(6.02*10**23);
Ra=1.4*(a**(1/3))*10**-13;# in cm
V2=1.3333*3.14*(Ra)**3;
da=m2/(V2*10**14);# density in 10^14 g cm^-3
print"\n The density nucleus of Au(197) in 10^14 g cm^-3 is =",da


#(c)
m3=p/(6.02*10**23);
Rp=1.4*(p**(1/3))*10**-13;# in cm
V3=1.3333*3.14*(Rp)**3;
dp=m3/(V3*10**14);# density in 10^14 g cm^-3
print"\n The density nucleus of P(239) in 10^14 g cm^-3 is =",dp
# Note: The density for Au(197) is not calculated correctly in the textbook.

Example 2.1, Page:19  
 


 The density nucleus of F(19) in 10^14 g cm^-3 is = 27.473593965

 The density nucleus of Au(197) in 10^14 g cm^-3 is = 284.857790058

 The density nucleus of P(239) in 10^14 g cm^-3 is = 345.588892507

example 2.2;pg no:20

In [2]:
#cal of Fission
#intiation of all variables
# Chapter 2
print"Example 2.2, Page:20  \n \n"
# Given:
# (a) Be(8)= 2 He(4)
# (b) Kr(80)= 2 Ar(40)
# (c) Cd(108)= 2 Cr(54)


# Solution:
m1=8.0053-2*4.00260;
m2=79.81638-2*39.96238;
m3=107.90418-2*53.93888;


if m1>0:
	print"\n Case (a) Fission is possible since m1=",m1
else:
	print"\n Case (a) Fission is not possiblesince m1=",m1


if m2>0:
    print"\n Case (b) Fission is possible since m2=",m2
else:
    print"\n Case (b) Fission is not possible since m2=",m2

 
if m3>0:
    print"\n Case (c) Fission is possible since m3=",m3
else:
	print"\n Case (c) Fission is not possible since m3=",m3

Example 2.2, Page:20  
 


 Case (a) Fission is possible since m1= 9.99999999998e-05

 Case (b) Fission is not possible since m2= -0.10838

 Case (c) Fission is possible since m3= 0.02642

example 2.3;pg no:21

In [3]:
#cal of  Nuclear spin and Magnetic moment
#intiation of all variables
# Chapter 2
print"Example 2.3, Page:21  \n \n"
# Given:
l=3.;
f=3.;
s=1./2.;

# Solution:
I=l-s;# total angulr momentum
P=(-1)**(l);#nuclear parity
mm=(I-2.293*(I/(I+1)));# in nuclear magneton

print"\n Nuclear spin of the nucleus is ",I
if P>0:
    print" (+) "
else:
    print" (-) "

print"\n Magnetic moment is = in nuclear magneton",round(mm,4)

Example 2.3, Page:21  
 


 Nuclear spin of the nucleus is  2.5
 (-) 

 Magnetic moment is = in nuclear magneton 0.8621

example 2.4;pg no:21

In [4]:
#cal of Nuclear spin and Magnetic moment
#intiation of all variables
# Chapter 2
print"Example 2.4, Page:21  \n \n"
# Given:

l=4.;
g=4.;
s=1./2.;

# Solution:
I=l+s;# total angulr momentum
P=(-1)**(l);#nuclear parity
mm=(I+2.293);# for odd proton and I=l+s in nuclear magneton

print"\n Nuclear spin of the nucleus is",I
if P>0:
    print" (+) "
else:
    print" (-) "
print"\n Magnetic moment is =in nuclear magneton",mm

Example 2.4, Page:21  
 


 Nuclear spin of the nucleus is 4.5
 (+) 

 Magnetic moment is =in nuclear magneton 6.793

example 2.5;pg no:22

In [5]:
#cal of  Magnetic moment
#intiation of all variables
# Chapter 2
print"Example 2.5, Page:22  \n \n"
# Given:(a) Cs(55) (b) Kr(36)

# Solution: Part(a)
print" The odd nucleon is 55 proton in level 2 d 5/2 (+). Hence spin is 5/2 and parity (+)"
mm=((5./2.)+2.293);
print"\n Magnetic moment is =in nuclear magneton",mm
# Solution: Part(b)
print"\n \n The odd nucleon is 47 neutron in level 1 g 9/2 (+). Hence spin is 9/2 and parity (+)"
print"Magnetic moment for neutron is m=-1.913 in nuclear magneton"

Example 2.5, Page:22  
 

 The odd nucleon is 55 proton in level 2 d 5/2 (+). Hence spin is 5/2 and parity (+)

 Magnetic moment is =in nuclear magneton 4.793

 
 The odd nucleon is 47 neutron in level 1 g 9/2 (+). Hence spin is 9/2 and parity (+)
Magnetic moment for neutron is m=-1.913 in nuclear magneton

example 2.7;pg no:22

In [6]:
#cal of The nuclear g factor for P
#intiation of all variables
# Chapter 2
print"Example 2.7, Page:22  \n \n"
# Given:
h=6.6262*10**-34;# in J.s
f=17.24*10**6;# in Hz/T
m=5.05*10**-27;# in J/T
# Solution:

E=h*f;
g=E/(m)
print"The nuclear g factor for P is =",round(g,3)

Example 2.7, Page:22  
 

The nuclear g factor for P is = 2.262

example 2.8;pg no:23

In [7]:
#cal of The NMR frequency
#intiation of all variables
# Chapter 2
print"Example 2.8, Page:23  \n \n"
# Given:

h=6.6262*10**-34;# in J.s
f=17.24*10**6;# in Hz/T
m=5.05*10**-27;# in J/T
g=1.405;

# Solution:

E=g*m;
f=E/(h*10**6);# NMR frequency

print"The NMR frequency in MHz",round(f,2)

Example 2.8, Page:23  
 

The NMR frequency in MHz 10.71

example 2.9;pg no:23

In [8]:
#cal of Magnetic field required
#intiation of all variables
# Chapter 2
print"Example 2.9, Page:23 "
# Given:

h=6.6262*10**-34;# in J.s
f=30.256*10**6;# in Hz/T
m=5.05*10**-27;# in J/T
g1=5.585;
g2=1.405;

# Solution:
H1=(h*f)/(g1*m);
H2=(h*f)/(g2*m);
print"Magnetic field required for a proton in T=",round(H1,3)
print"Magnetic field required for C 13 in T=",round(H2,3)

Example 2.9, Page:23 
Magnetic field required for a proton in T= 0.711
Magnetic field required for C 13 in T= 2.826

example 2.10;pg no:23

In [9]:
#cal of Magnetic field required
#intiation of all variables
# Chapter 2
print"Example 2.10, Page:23"
# Given:

h=6.62*10**-34;# in J.s
f=9.302*10**9;# in Hz/T
m=9.2741*10**-24;# in J/T
g1=2.0025;


# Solution:

H1=(h*f)/(g1*m);
print"Magnetic field required for a proton in T=",round(H1,4)

Example 2.10, Page:23
Magnetic field required for a proton in T= 0.3316

example 2.11;pg no:24

In [10]:
#cal of frequency needed
#intiation of all variables
# Chapter 2
print"Example 2.11, Page:24  \n \n"
# Given:
mf=1.201;# In T
h=6.6262*10**-34;# in J.s
m=9.2741*10**-24;# in J/T
g=2.0025;
# Solution:
v=(g*m*mf)/(h*10**9);
print"The frequency needed to bring in resonance in GHz=",round(v,2)

Example 2.11, Page:24  
 

The frequency needed to bring in resonance in GHz= 33.66

example 2.12;pg no:24

In [11]:
#cal of The frequency needed to bring proton and electron
#intiation of all variables
# Chapter 2
print"Example 2.12, Page:24  \n \n"
# Given:
mf=1.5;# In T
h=6.6262*10**-34;# in J.s
mb=9.2741*10**-24;# in J/T
mn=5.0504*10**-27;#in J/T
ge=2.002;
gp=5.5854;
# Solution: Part(a)
v1=(gp*mn*mf)/(h*10**6);
print"\n The frequency needed to bring proton spin resonance is=MHz",round(v1,2)
# Solution: Part(b)
v2=(ge*mb*mf)/(h*10**10);
print"\n The frequency needed to bring electron spin resonance in GHz",round(v2,2)

Example 2.12, Page:24  
 


 The frequency needed to bring proton spin resonance is=MHz 63.86

 The frequency needed to bring electron spin resonance in GHz 4.2

example 2.13;pg no:24

In [12]:
#cal of Magnetic field required
#intiation of all variables
# Chapter 2
print"Example 2.13, Page:24  \n \n"
# Given:
h=6.6262*10**-34;# in J.s
f=40.2*10**6;# in Hz/T
m=5.05*10**-27;# in J/T
g1=5.256;


# Solution:
H1=(h*f)/(g1*m);
print"\n Magnetic field required for causing resonance in T=",round(H1,3)

Example 2.13, Page:24  
 


 Magnetic field required for causing resonance in T= 1.004

example 2.14;pg no:25

In [13]:
#cal of ratio of NMR frequencies
#intiation of all variables
# Chapter 2
print"Example 2.14, Page: 25 \n \n"
#Given:
mf=1.0;# In T
h=6.6262*10**-34;# in J.s
mn=5.0504*10**-27;#in J/T
gB=5.4;
gN=4.01;
# Solution:
v1=(gB*mn*mf)/(h*10**6);

v2=(gN*mn*mf)/(h*10**6);

print"\n The ratio of NMR frequencies of B/N is =",round(v1/v2,2)

Example 2.14, Page: 25 
 


 The ratio of NMR frequencies of B/N is = 1.35

example 2.15;pg no:25

In [14]:
#cal of recoil energy
#intiation of all variables
# Chapter 2
print"Example 2.15, Page:25  \n \n"
# Given:
E=14.4*10**-3;# in MeV
m=57;
# Solution:
Er=(536*(E)**2)/(m*10**-3);
print"The recoil energy in (meV)=",round(Er,2)

Example 2.15, Page:25  
 

The recoil energy in (meV)= 1.95

example 2.16;pg no:25

In [15]:
#cal of energy emtted by the nucleus
#intiation of all variables
# Chapter 2
print"Example 2.16, Page:25  \n \n"
import math
# Given:
Er=2.551;# in meV
m=119;# atomic wt of Sn

# Solution:
E=math.sqrt(2.551*10**-3*119/536);# energy emitted by nucleus
print"The energy emtted by the nucleus in (KeV)=",round(E*10**3,2)

Example 2.16, Page:25  
 

The energy emtted by the nucleus in (KeV)= 23.8

example 2.17;pg no:26

In [16]:
#cal of doppler shift frequency
#intiation of all variables
# Chapter 2
print"Example 2.17, Page:26  \n \n"
# Given:
l=10**-10;# in m
m=100;# in u
h=6.6262*10**-34;# in J.s


# Solution:
v=h/(m*l*1.67*10**-27);# velocity
f=v/l;# frequency

print"The doppler shift frequency in 10^11 Hz=",round(f/10**11,2)

Example 2.17, Page:26  
 

The doppler shift frequency in 10^11 Hz= 3.97

example 2.18;pg no:26

In [17]:
#cal of frequency
#intiation of all variables
# Chapter 2
print"Example 2.18, Page:26  \n \n"
#Given:
# 1 ev=8065 cm^-1
E=14.4*10**3;# in eV
# Solution:
f1=E*8065;# frequency in cm^-1
print"The frequency in 10**8 cm^-1 is =",round(f1/10**8,3)
fr=f1*3*10**8*100;
print"The frequency in 10^12 MHz is =",round(fr/10**18,2)

Example 2.18, Page:26  
 

The frequency in 10**8 cm^-1 is = 1.161
The frequency in 10^12 MHz is = 3.48

example 2.19;pg no:26

In [18]:
#cal of shift in frequency between the source and the sample
#intiation of all variables
# Chapter 2
print"Example 2.19, Page:26  \n \n"
# Given:
#Given:
# 1 ev=8065 cm^-1
E=14.4*10**3;# in eV
v1=2.2*10**-3;# in m/s
# Solution:
f1=E*8065;# frequency in cm^-1
fr=f1*3*10**8*100;
fr1=(fr*v1)/(3*10**8);
print"The shift in frequency between the source and the sample in MHz=",round(fr1/10**6,2)

Example 2.19, Page:26  
 

The shift in frequency between the source and the sample in MHz= 25.55

example 2.20;pg no:27

In [19]:
#cal of recoil velocity and doppler shift for an atom
#intiation of all variables
# Chapter 2
print"Example 2.20, Page:27  \n \n"
# Given:
E=1.6*14.4*10**3*10**-19;# energy in J
c=3*10**8;# in m/s
m=57*1.6*10**-27;
M=10**-4;
h=6.6262*10**-34;# in J.s
# Solution:
p=E/c;
v=p/m;
v1=(v*m)/(M);
v2=(v*m)/(M*10**-20);
f1=(E*v)/(h*c);
f2=(E*v1)/(h*c*10**-10);
print"The recoil velocity of free atom is =(m/s)",round(v,2)
print"The recoil velocity of atom that is part of crystal in 10^-20 (m/s)",round(v2,2)
print"The doppler shift for free atom in 10^10 Hz=",round(f1/10**10,2)
print"The doppler shift of atom that is part of crystal in 10^-10 Hz is =",round(f2,2)

Example 2.20, Page:27  
 

The recoil velocity of free atom is =(m/s) 84.21
The recoil velocity of atom that is part of crystal in 10^-20 (m/s) 7.68
The doppler shift for free atom in 10^10 Hz= 97.6
The doppler shift of atom that is part of crystal in 10^-10 Hz is = 8.9

example 2.21;pg no:28

In [20]:
#cal of Ellipticity
#intiation of all variables
# Chapter 2
print"Example 2.21, Page:28  \n \n"
# Given:
A=175.;
R=1.4*10**-15*((A)**(1./3.));
# Soluiton:
#Part a
sqrBMinusSqrA = (5.9 * (10** (-28))) * 5. /(2. * 71.);
BMinusA = sqrBMinusSqrA / (2. * R);
ellipticity = 2 * (BMinusA) / (2. * R);
print"Ellipticity is =",round(ellipticity,2)
#Part B
b = (BMinusA + (2 * R)) /2;
a = (-BMinusA + (2 * R)) /2;
print"b/a is =",round(b/a,3)

Example 2.21, Page:28  
 

Ellipticity is = 0.17
b/a is = 1.185

example 2.22;pg no:28

In [21]:
#cal of Ellipticity
#intiation of all variables
# Chapter 2
print"Example 2.22, Page:  \n \n"
# Given:
A1=176.;
A2=233.;
R1=1.4*10**-15*((A1)**(1./3.));
R2=1.4*10**-15*((A2)**(1./3.));
# Soluiton:
#Part a
sqrBMinusSqrA = (5. * 7. * (10** (-28))) /(2. * 71.);
BMinusA = sqrBMinusSqrA / (2. * R1);
ellipticity = 2. * (BMinusA) / (2. * R1);
print"Ellipticity is =",round(ellipticity,2)
b = (BMinusA + (2. * R1)) /2.;
a = (-BMinusA + (2. * R1)) /2.;
print"b/a is =",round(b/a,3)
#Part B
sqrBMinusSqrA = -(5. * 3. * (10** (-28))) /(2. * 91.);
BMinusA = sqrBMinusSqrA / (2. * R2);
ellipticity = 2. * (BMinusA) / (2. * R2);
print"Ellipticity is =",round(ellipticity,2)
b = (BMinusA + (2. * R2)) /2.;
a = (-BMinusA + (2. * R2)) /2.;
print"b/a is =",round(b/a,3)

Example 2.22, Page:  
 

Ellipticity is = 0.2
b/a is = 1.222
Ellipticity is = -0.06
b/a is = 0.946

example 2.23;pg no:29

In [22]:
#cal of Quadrapole moment
#intiation of all variables
# Chapter 2
print"Example 2.23, Page:29  \n \n"
# Given:
e = 0.03;
A=75.;
R=1.4*10**-15*((A)**(1./3.));
# Soluiton:
BPlusA = 2. * R;
BMinusA = e * R;
sqrBMinusSqrA = BPlusA * BMinusA;
BMinusA = sqrBMinusSqrA / (2. * R);
Q = (2.*33.*sqrBMinusSqrA)/5.;
q1=Q/10**-28;
print"Quadrapole moment in barns=",round(q1,4)

Example 2.23, Page:29  
 

Quadrapole moment in barns= 0.2761