Chapter 12:Nuclear Structure and Reactivity¶

Example 12.1, Page 375¶

In [4]:
#initiation of variable
Z=2;A=4;N=A-Z;         # Given values

#result
print"The following method of representing atoms is followed throughout the chapter\n\t\t x,y,z\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\n\n"
print"The helium can be reperesented as He-- ",Z,A,N;

#part b
Z=50.0;N=66.0;A=Z+N;       # Given values and standard formulae
print"The Tin can be reperesented as Sn-- ",Z,A,N;

#part c
A=235;N=143;Z=A-N;
print"The Uranium can be reperesented as U-- ",Z,A,N;

The following method of representing atoms is followed throughout the chapter
x,y,z
where x=atomic number y=mass number z= Neutron Number S=symbol of the atom

The helium can be reperesented as He--  2 4 2
The Tin can be reperesented as Sn--  50.0 116.0 66.0
The Uranium can be reperesented as U--  92 235 143


Example 12.2, Page 377¶

In [6]:
#initiation of variable
r0=1.2;         #standard value.
A=12.0;
r= r0*A**(1.0/3);

#result
print"The value of mean radius for C in fm is",round(r,3);

#part2
A=70.0;        #given value
r= r0*A**(1.0/3);

#result
print"The value of mean radius for C in fm is",round(r,3);

#part3
A=209;
r= r0*A**(1.0/3);

#result
print"The value of mean radius for C in fm is",round(r,3);

The value of mean radius for C in fm is 2.747
The value of mean radius for C in fm is 4.946
The value of mean radius for C in fm is 7.121


Example 12.3, Page 379¶

In [3]:
#initiation of variable
from math import pi
m=1.67*10**-27; r0=1.2*10**-15; v=4*pi*(r0**3)/3.0     #standard values of mass radius and volume

#calculation
p=m/v;                                               #density

#result
print"Density of typical nucleus in kg/m3 is %.0e" %p;

#part 2
r0=0.01;v=4*pi*(r0**3)/3.0;p=2.0*10**17;              #/hypothetical values
m1=p*v;

#result
print"The mass of the hypothetical nucleus would be in Kg %.0e" %m1;

Density of typical nucleus in kg/m3 is 2e+17
The mass of the hypothetical nucleus would be in Kg 8e+11


Example 12.4, Page 380¶

In [10]:
#initiation of variable
N=30.0;Z=26.0;A=56.0;Mn=1.008665;Mp=1.007825;m=55.934939;c2=931.5; #given values and constants for case-1
B=((N*Mn)+(Z*Mp)-(m))*c2;                                  #binding energy(per nucleon)

#result
print"Binding energy per nucleon for 26,56Fe30 in MeV is",round(B/A,3);

#part 2
N=146.0;Z=92.0;A=238.0;Mn=1.008665;Mp=1.007825;m=238.050785;c2=931.5;      #given values and constants for case-2
B=((N*Mn)+(Z*Mp)-(m))*c2;                             #binding energy(per nucleon)

#result
print"Binding energy per nucleon for 26,56Fe30 in MeV is",round(B/A,3);

Binding nergy per nucleon for 26,56Fe30 in MeV is 8.79
Binding nergy per nucleon for 26,56Fe30 in MeV is 7.57


Example 12.5, Page 382¶

In [4]:
#initiation of variable
from math import exp
t12=2.7*24*3600;              #converting days into seconds
w=0.693/t12;                  #lambda

#result
print"The decay constant in sec is %.2e" %w;

#partb
print"The decay constant is equal to probability of decay in one second hence %.2e" %w;

#partc
m=10**-6;Na=6.023*10**23; M=198.0;     #given values and constants
N=m*Na/M;                        #number of atoms in the sample
Ao=w*N;             #activity

#result
print"The activity was found out to be in Ci is %.2e" %Ao,"=0.244Ci";

#partd
t=7*24*3600.0;         #given time
A=Ao*exp(-w*t);        #activity

#result
print"The activity after one week was found out to be in decays/sec %.1e" %A;

The decay constant in sec is 2.97e-06
The decay constant is equal to probability of decay in one second hence 2.97e-06
The activity was found out to be in Ci is 9.04e+09 =0.244Ci
The activity after one week was found out to be in decays/sec 1.5e+09


Example 12.6, Page 384¶

In [13]:
#initiation  of variable
t1=4.55*10**9;t2=7.04*10**8;       #given values of time at 2 different instants

#calculation
age=t1/t2;
r=2**age;

#result
print "The original rock hence contained",round(r,3),"Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23";

The original rock hence contained 88.222 Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23


Example 12.7, Page 385¶

In [14]:
#initiation of variable
m236Ra=226.025403;
m222Rn=222.017571;
m4He=4.002603;c2=931.5; #mass of various elements and c2=c^2

#calculation
Q=(m236Ra-m222Rn-m4He)*c2;#Q of the reaction
A=226.0
K=((A-4)/A)*Q;                           #kinetic energy

#result
print"The kinetic energy of the alpha particle in Mev is",round(K,3);

The kinetic energy of the alpha particle in Mev is 4.785


Example 12.8, Page 387¶

In [15]:
#initiation of variable
m226Ra=226.025403; #mass of various elements
m212Pb=211.991871;
m14c=14.003242;
c2=931.5;       #value of c^2

#calculation
Q=(m226Ra-m212Pb-m14c)*c2;      #Q of the reaction

#result
print"The value of Q for 14c emission in MeV is",round(Q,3);
print"The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker."

The value of Q for 14c emission in MeV is 28.215
The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.


Example 12.9, Page 389¶

In [16]:
#initiation of variable
m23Ne=22.994465; #mass of various elements
m23Na=22.989768;
c2=931.5;          #value of c^2

#calculation
Q=(m23Ne-m23Na)*c2;     #Q of the reaction

#result
print "Hence the maximum kinetic energy of the emitted electrons in MeV is",round(Q,3);

Hence the maximum kinetic energy of the emitted electrons in MeV is 4.375


Example 12.10, Page 390¶

In [17]:
#initiation of variable
m40K=39.963999;         #mass of various particles
m40Ca=39.962591;
c2=931.5;                    #value of c^2 in MeV

#calculation
Qb1=(m40K-m40Ca)*c2;       #Q value of the reaction

#result
print"The Q value for -VE beta emission in Mev in",round(Qb1,3);

#partb
m40K=39.963999;          #mass of various particles
m40Ar=39.962384;
me=0.000549;
Qb2=(m40K-m40Ar-2*me)*c2;         #Q value of the reaction

#result
print"The Q value for +VE beta emission in Mev in",round(Qb2,3);

#partc
m40K=39.963999;
m40Ar=39.962384;

#calculation
Qec=(m40K-m40Ar)*c2;

#result
print"The Q value for +VE beta emission in Mev in",round(Qec,3);

The Q value for -VE beta emission in Mev in 1.312
The Q value for +VE beta emission in Mev in 0.482
The Q value for +VE beta emission in Mev in 1.504


Example 12.11, Page 392¶

In [18]:
#initiation of variable
Mg=12.000000; #mass of the carbon atom in amu
c2=931.5;
Eg=4.43;     #given energy of gamma ray
Mex=Mg+(Eg/c2);     #mass in excited state
Me=0.000549;        #mass of an electron

#calculation
Q=(12.018613-Mex-2*Me)*c2;     #Q of the particle

#result
print"The maximum value of kinetic energy is in MeV",round(Q,3);

The maximum value of kinetic energy is in MeV 11.885


Example 12.12, Page 393¶

In [5]:
#initiation of variable
m238U=238.050786; #mass of various quantities
m206Pb=205.974455;
m4He=4.002603;
c2=931.5;     #constants

#calculation
Q=(m238U-m206Pb-8*m4He)*c2;
t12=(4.5)*10**9*(3.16*10**7);    #half life years to seconds conversion
w=0.693/t12;                 # lambeda
NoD=(Na/238)*w;      #number of decays
E=NoD*Q*(1.6*10**-19)*10**6;        #rate of liberation of energy,converting MeV to eV

#result
print"Rate of energy liberation in W is %.0e" %E;

Rate of energy liberation in W is 1e-07


Example 12.13, Page 395¶

In [11]:
#initiation of variable
from math import log
R=0.5;t12=4.5*10**9;                 #value of radius and half-life
t1=(t12/0.693)*log(1+(1/R));        #age of rock 1
R=1.0;
t2=(t12/0.693)*log(1+(1/R));       #age of rock-2
R=2.0
t3=(t12/0.693)*log(1+(1/R));       #age of rock 3

#result
print"The ages of rock samples in years respectively are %.1e"%t1," %.1e" %t2,"  %.1e" %t3;

The ages of rock samples in years respectively are 7.1e+09  4.5e+09   2.6e+09


Example 12.14, Page 397¶

In [15]:
#initiation of variable
from math import log
P=2.0*10**14; V=2.0*10**-14; R=8.314; T=295.0;Na=6.023*10**23;   #various constants and given values

#calculation
n=P*V/(R*T);       #ideal gas law
N=Na*n;f=10**-12         #avagadaro's number and fraction of carbon molecules
t12=5730*3.16*(10**7);   #half life
A=(0.693/t12)*N*f;    #activity
D1w=A*7*24*60*60;       #decays per second

#result
print"the no. of decays per second is %.2e" %A
print"The no of decays per week is ",round(D1w);

#partb
c1=1420.0;             #concentration at instant 1
c2=D1w;              #concentration at instant 2
t12y=5730;          #half life
t=t12y*log(c2/c1)/0.693;            #age of the sample

#result
print"Age of the sample in years is",round(t,3);
print"the answer in the book is wrong"

the no. of decays per second is 3.76e-03
The no of decays pers week is  2274.0
Age of the sample in years is 3892.57