Chapter15-Atomic Nucleus And Nuclear Energy

Ex1-pg-pg427

In [15]:
import math
##Example 15.1
##calculation of binding energy per nucleon

##given values
Mp=1.00814;##mass of proton in amu
Mn=1.008665;##mass of nucleon in amu
M=7.01822;##mass of Lithium nucleus in amu
amu=931.;##amu in MeV
n=7-3;##no of neutrons in lithium nucleus

##calculation
ET=(3*Mp+4*Mn-M)*amu;##total binding energy in MeV
E=ET/7.;##7 1s the mass number
print'%s %.2f %s'%('Binding energy per nucleon in MeV is',E,'');

Binding energy per nucleon in MeV is 5.43

Ex2-pg427

In [14]:
import math
##Example 15.2
##calculation of energy 

##given values
M1=15.00001;##atomic mass of N15 in amu
M2=15.0030;##atomic mass of O15 in amu
M3=15.9949;##atomic mass of O16 in amu
amu=931.4;##amu in MeV
mp=1.0072766;##restmass of proton
mn=1.0086654;##restmass of neutron

##calculation
Q1=(M3-mp-M1)*amu;
print'%s %.2f %s'%('energy required to remove one proton from O16 is',Q1,'');
Q2=(M3-mn-M2)*amu;
print'%s %.2f %s'%('energy required to remove one neutron from O16 is',Q2,'');

energy required to remove one proton from O16 is -11.54 
energy required to remove one neutron from O16 is -15.62

Ex3-pg428

In [13]:
import math
##Example 15.3
##calculation of binding energy 

##given values
Mp=1.00758;##mass of proton in amu
Mn=1.00897;##mass of nucleon in amu
M=4.0028;##mass of Helium nucleus in amu
amu=931.4;##amu in MeV

##calculation
E1=(2*Mp+2*Mn-M)*amu;##total binding energy
print'%s %.2f %s'%('Binding energy in MeV is',E1,'');
E2=E1*10**6*1.6*10**-19;
print'%s %.3e %s'%('binding energy in Joule is',E2,'');

Binding energy in MeV is 28.22 
binding energy in Joule is 4.515e-12

Ex4-pg435

In [12]:
import math
##Example 15.4
##calculation of amount of unchanged material

##given values
T=2;##half life in years
k=.6931/T;##decay constant
M=4.0028;##mass of Helium nucleus in amu
amu=931.4;##amu in MeV
No=1.;##initial amount in g

##calculation
N=No*(math.e**(-k*2*T));
print'%s %.2f %s'%('amount of material remaining unchanged after four years(in gram) is',N,'');

amount of material remaining unchanged after four years(in gram) is 0.25

Ex5-pg435

In [11]:
import math
##Example 15.5
##calculation of amount of halflife

##given values
t=5.;##time period in years
amu=931.4;##amu in MeV
No=5.;##initial amount in g
N=5.-(10.5*10**-3);##amount present after 5 years


##calculation
k=math.log(N/No)/t;##decay constant
T=-.693/k;
print'%s %.2f %s'%('halflife in years is',T,'');

halflife in years is 1648.27

Ex6-pg435

In [10]:
import math
##Example 15.6
##calculation of activity

##given values
t=28.;##half life in years
m=10**-3;##mass of sample
M=90.;##atomic mass of strontium
NA=6.02*10**26;##avogadro's number


##calculation
n=m*NA/M;##no of nuclei in 1 mg sample
k=.693/(t*365*24.*60.*60.);##decay constant
A=k*n;
print'%s %.3e %s'%('activity of sample(in disintegrations per second) is',A,'');

activity of sample(in disintegrations per second) is 5.250e+12

Ex7-pg439

In [9]:
import math
##Example 15.7
##calculation of age of mineral

##given values
t=4.5*10**9.;##half life in years
M1=238.;##atomic mass of Uranium in g
m=.093;##mass of lead in 1 g of uranium in g
NA=6.02*10**26;##avogadro's number
M2=206.;##atomic mass of lead in g

##calculation
n=NA/M1;##no of nuclei in 1 g of uranium sample
n1=m*NA/M2;##no of nuclei in m mass of lead
c=n1/n;
k=.693/t;##decay constant
T=(1/k)*math.log(1+c);
print'%s %.3e %s'%('age of mineral in years is',T,'');

age of mineral in years is 6.627e+08

Ex8-pg440

In [8]:
import math
##Example 15.8
##calculation of age of wooden piece

##given values
t=5730.;##half life of C14 in years
M1=50.;##mass of wooden piece in g
A1=320.;##activity of wooden piece (disintegration per minute per g)
A2=12.;##activity of living tree

##calculation
k=.693/t;##decay constant
A=A1/M1;##activity after death

T=(1/k)*math.log(A2/A);
print'%s %.2f %s'%('age of mineral in years is',T,'');

age of mineral in years is 5197.59

Ex9-pg443

In [7]:
import math
##Example 15.9
##calculation of energy released

##given values
M1=10.016125;##atomic mass of Boron in amu
M2=13.007440;##atomic mass of C13 in amu
M3=4.003874;##atomic mass of Helium in amu
mp=1.008146;##mass of proton in amu
amu=931.;##amu in MeV

##calculation
Q=(M1+M3-(M2+mp))*amu;##total binding energy in M
print'%s %.2f %s'%('Binding energy per nucleon in MeV is',Q,'');

Binding energy per nucleon in MeV is 4.11

Ex10-pg444

In [4]:
import math
##Example 15.10
##calculation of crosssection

##given values
t=.01*10**-3;##thickness in m
n=10**13.;##no of protons bombarding target per s
NA=6.02*10**26.;##avogadro's number
M=7.;##atomic mass of lithium in kg
d=500.;##density of lithium in kg/m**3
n0=10**8.;##no of neutrons produced per s
##calculation
n1=d*NA/M;##no of target nuclei per unit volume
n2=n1*t;##no of target nuclei per area
A=n0/(n*n2);
print'%s %.3e %s'%('crosssection(in m^2) for this reaction is',A,'');

crosssection(in m^2) for this reaction is 2.326e-29

Ex11-pg450

In [5]:
import math
##Example 15.11
##calculation of final energy 

##given values
B=.4;##max magnetic field in Wb/m**2
c=3*10**8.;
e=1.6*10**-19.;
d=1.52;##diametre in m
r=d/2.;

##calculation
E=B*e*r*c;##E=pc,p=mv=Ber
print'%s %.3e %s'%('final energy of e(in J) is',E,'');
E1=(E/e)/10**6;
print'%s %.2f %s'%('final energy of e (in MeV) is',E1,'');

final energy of e(in J) is 1.459e-11 
final energy of e (in MeV) is 91.20

Ex12-pg459

In [4]:
import math
##Example 15.12
##calculation of amount of fuel

##given values
P=100*10**6.;##power required by city
M=235.;##atomic mass of Uranium in g
e=20/100.;##conversion efficiency
NA=6.02*10**26.;##avogadros number
E=200*10**6*1.6*10**-19;##energy released per fission
t=8.64*10**4.;##day in seconds


##calculation
E1=P*t;##energy requirement
m=E1*M/(NA*e*E);##no of nuclei N=NA*m/M,energy released by m kg is N*E,energy requirement=e*N*E
print'%s %.2f %s'%('amount of fuel(in kg) required is',m,'');

amount of fuel(in kg) required is 0.53

Ex13-pg459

In [3]:
import math
##Example 15.13
##calculation of power output

##given values
M=235.;##atomic mass of Uranium in kg
e=5/100.;##reactor efficiency
m=25/1000.;##amount of uranium consumed per day in kg
E=200*10**6*1.6*10**-19;##energy released per fission
t=8.64*10**4.;##day in seconds
NA=6.02*10**26.;##avogadros number

##calculation
n=NA*m/M;##no of nuclei in 25g
E1=n*E;##energy produced by n nuclei
E2=E1*e;##energy converted to power
P=E2/t;##power output in Watt
print'%s %.2f %s'%('power output in MW is',P/10**6,'');

power output in MW is 1.19

Ex14-pg460

In [2]:
import math
##Example 15.14
##calculation of power developed

##given values
M=235.;##atomic mass of Uranium in kg
m=20.4;##amount of uranium consumed per day in kg
E=200*10**6*1.6*10**-19;##energy released per fission
t=3600*1000.;##time of operation
NA=6.02*10**26;##avogadros number

##calculation
n=NA*m/M;##no of nuclei in 20.4kg
E1=n*E;##energy produced by n nuclei
P=E1/t;##in Watt
print'%s %.2f %s'%('power developed in MW is',P/10**6,'');

power developed in MW is 464.52

Ex15-pg464

In [1]:
import math
##Example 15.15
##calculation of amount of dueterium consumed

##given values
M1=2.01478;##atomic mass of Hydrogen in amu
M2=4.00388;##atomic mass of Helium in amu
amu=931.;##amu in MeV
e=30/100.;##efficiency
P=50*10**6.;##output power
NA=6.026*10**26.;##avogadro number
t=8.64*10**4.;##seconds in a day

##calculation
Q=(2*M1-M2)*amu;##energy released in a D-D reaction in MeV
O=Q*e*10**6/2.;##actual output per dueterium atom in eV
n=P/(O*1.6*10**-19);##no of D atoms required
m=n*M1/NA;##equivalent mass of D required per s
X=m*t;

print'%s %.2f %s'%('Deuterium requirement per day in kg is',X,'');

Deuterium requirement per day in kg is 0.03