chapter 1: Atomic Nucleus

example 1.1;pg no:2

In [1]:
#cal of elelectric field and force
#intiation of all variables
# Chapter 1
print"Example 1.1, Page:2  \n \n"
#Given:
v=1000# potential
d=0.05# distance
q=3.8*10**-9# charge
#solution:
e=v/d;#electric field
f=e;# force
f1=f*q;# force on metal sphere
print"\n The electric field in V/m is =",e
print"\n The force in N/C is =",f
print"\n The force on metal sphere in N is =",f1

Example 1.1, Page:2  
 


 The electric field in V/m is = 20000.0

 The force in N/C is = 20000.0

 The force on metal sphere in N is = 7.6e-05

example 1.2;pg no:2

In [2]:
#cal of potential
#intiation of all variables
# Chapter 1
print"Example 1.2, Page:2  \n \n"
#Given:
energy=2*10**-6
c=2.5*10**-8# velocity of light
#solution:
v=energy/c# potential
print"The potential in V is =",v

Example 1.2, Page:2  
 

The potential in V is = 80.0

example 1.3;pg no:3

In [3]:
#cal of elecrtron and photon wavelength
#intiation of all variables
# Chapter 1
print"Example 1.3, Page:3 \n"
#Given:
energy=10 #in electron volts
m=9.1*10**-31# mass of electron in kg
h=6.626*10**-34# planck's constant J.s
c=3*10**8# speed of light in m/s
#solution (a):
energy1=energy*1.6*10**-19# energy in J
p=(2*m*energy1)**0.5# momentum
wavelength=h/p*(10)**10
print"The wavelength in Angstroms is =",round(wavelength,2)
#solution (b):
wavelength1=h*c/energy1*(10)**10;#photon wavelength
print"The photon wavelength in Angstroms is =",round(wavelength1,2)

Example 1.3, Page:3 

The wavelength in Angstroms is = 3.88
The photon wavelength in Angstroms is = 1242.38

example 1.4;pg no:3

In [4]:
#cal of kinetic energy of an electron
#intiation of all variables
# Chapter 1
print"Example 1.4, Page:3  \n \n"
#Given:
wavelength=10**-10
m=9.1*10**-31
h=6.626*10**-34
#solution:
p=h/wavelength
e=p*p/(2*m) # energy in J
e1=e/(1.6*10**-19)# energy in eV
print"The energy in eV is =",round(e1,2)

Example 1.4, Page:3  
 

The energy in eV is = 150.77

example 1.5;pg no:3

In [5]:
#cal of wavelength of oxygen and nitrogen nucleus
#intiation of all variables
# Chapter 1
print"Example 1.5, Page:3  \n \n"
#Given:
m=1.66*10**-27# 1u=1.66*10^-27 kg
h=6.6262*10**-34#planck's constant in J.s
energy1=120# in Mev for oxygen
energy2=140# in MeV for nitrogen
#solution(a):
p=(2*m*16*energy1*(1.6022*10**-13))**0.5
wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms
print"The wavelength in 10^-5 Angstroms is =",round(wavelength1,2)
#solution (b):
p=(2*m*14*energy2*(1.6022*10**-13))**0.5
wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms
print"The wavelength in 10^-5 Angstroms is =",round(wavelength2,2)
# 1 Angstrom = 10^-10 m

Example 1.5, Page:3  
 

The wavelength in 10^-5 Angstroms is = 0.66
The wavelength in 10^-5 Angstroms is = 0.65

example 1.6;pg no:3

In [6]:
#cal of energy of a gamma photon
#intiation of all variables
# Chapter 1
print"Example 1.6, Page:3  \n \n"
#Given:
wavelength=1.5*10**-10
h=6.62*10**-34
c=3*10**8
#solution:
e=(h*c)/wavelength# energy in J
e1=e/(1.6*10**-19)# energy in eV
print"The energy in eV is =",e1

Example 1.6, Page:3  
 

The energy in eV is = 8275.0

example 1.7;pg no:4

In [7]:
#cal of threshold frequency,wavelength,energy of photoelectrone
#intiation of all variables
# Chapter 1
print"Example 1.7, Page:4  \n \n"
#Given:
E=5.12*1.6*10**-19# energy in J
h=6.626*10**-34
c=3*10**8
wavelength=200*10**-9
w=2.3# in eV
#solution:
tf=E/h# (part a)
print"\n The threshold frequency in s^-1 is =",round(tf,2)
tl=c/tf*10**10# (part b)
print"\n The threshold wavelength in Angstroms is =",round(tl,2)
e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)
pe=e-w
print"\n The energy of photoelectrone in eV is =",round(pe,2)

Example 1.7, Page:4  
 


 The threshold frequency in s^-1 is = 1.23634168427e+15

 The threshold wavelength in Angstroms is = 2426.51

 The energy of photoelectrone in eV is = 3.91

example 1.8;pg no:4

In [8]:
#cal of velocity of alpha particles,deuteron,proton
#intiation of all variables
# Chapter 1
print"Example 1.8, Page:4  \n \n"
#Given:
e1=1 # in MeV
e2=2 # in MeV
ma=4 # in u(amu)
md=2 # in u(amu)
mp=1 # in u(amu)
# 1u = 1.6*10^-27 Kg
#solution: part a)For alpha particles
v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5
print"\n The velocity of alpha particles for 1 MeV in mega m/s is =",round(v1a/10**6,2)# For 1 MeV
v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5
print"\n The velocity of alpha particles for 2 MeV in mega m/s is =",round(v2a/10**6,2)# For 2 MeV
#solution: part b)For deuteron particles
v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5
print"\n The velocity of deuteron particles for 1 MeV in mega m/s is =",round(v1b/10**6,2) # For 1 MeV
v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5
print"\n The velocity of deuteron particles for 2 MeV in mega m/s is =",round(v2b/10**6,2) # For 2 MeV
#solution: part c)For proton particles
v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5
print"\n The velocity of proton particles for 1 MeV in mega m/s is =",round(v1p/10**6,2) # For 1 MeV
v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5
print"\n The velocity of proton particles for 2 MeV in mega m/s is =",round(v2p/10**6,2) # For 2 MeV

Example 1.8, Page:4  
 


 The velocity of alpha particles for 1 MeV in mega m/s is = 6.94

 The velocity of alpha particles for 2 MeV in mega m/s is = 9.82

 The velocity of deuteron particles for 1 MeV in mega m/s is = 9.82

 The velocity of deuteron particles for 2 MeV in mega m/s is = 13.88

 The velocity of proton particles for 1 MeV in mega m/s is = 13.88

 The velocity of proton particles for 2 MeV in mega m/s is = 19.63

example 1.9;pg no:5

In [9]:
#cal of The energy equivalence
#intiation of all variables
# Chapter 1
print"Example 1.9, Page:5  \n \n"
#Given:
m=1./(6.023*10**23)#mass of 1 atom in g
m1=m*10**-3#mass of 1 atom in Kg
c=3.*10**8# velocity in m/s
#solution:
e=m1*c*c; # energy in J
e1=e/(1.6*10**-13)# energy in MeV
print"The energy in MeV is =",round(e1)

Example 1.9, Page:5  
 

The energy in MeV is = 934.0

example 1.10;pg no:5

In [10]:
#cal of The energy of formation
#intiation of all variables
# Chapter 1
print"Example 1.10, Page:5  \n \n"
#Given:
enthalpy=1278 # enthalpy of combustion in kJ/mol
#solution:
energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)
print"The energy in eV is =",round(energy,2)

Example 1.10, Page:5  
 

The energy in eV is = 13.26

example 1.11;pg no:5

In [11]:
#cal of mean binding energy of helium and oxygen
#intiation of all variables
# Chapter 1
print"Example 1.11, Page:5  \n \n"
#Given:
mh=1.0078
mn=1.0087
ma=4.0026
mo=15.9949
Ah=4.0026 # atomic mass of helium
Ao=15.9949 # atomic mass of oxygen
#solution:
# part (a)
B1=(2*mh+2*mn-ma)*931 # in MeV
Bh=B1/Ah
print"\n The mean binding energy of helium atom in MeV is =",round(Bh,2)
# part (b)
B2=(8*mh+8*mn-mo)*931 # in MeV
Bo=B2/Ao
print"\n The mean binding energy of oxygen atom in MeV is =",round(Bo,2)

Example 1.11, Page:5  
 


 The mean binding energy of helium atom in MeV is = 7.07

 The mean binding energy of oxygen atom in MeV is = 7.98

example 1.12;pg no:6

In [12]:
#cal of mean binding energy of Be atom
#intiation of all variables
# Chapter 1
print"Example 1.12, Page:6 \n"
#Given:
mh=1.0078;
mn=1.0087;
ABe=8.0053; # atomic mass of beryllium
#solution:
B1=(4*mh+4*mn-ABe)*931; # in MeV
Bh=B1/ABe;
print"\n The mean binding energy of Be atom in MeV is =",round(Bh,3)
print"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas"

Example 1.12, Page:6 


 The mean binding energy of Be atom in MeV is = 7.059
From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas

example 1.13;pg no:6

In [13]:
#cal of amount of coal
#intiation of all variables
# Chapter 1
print"Example 1.13, Page:6 \n"
#Given:
e=200; # in Mev
m=0.235; # weight of uranium atom in Kg
enthalpy=393.5; # in KJ/mol
Na=6.02*10**23;
#solution:
e1=e*1.6*10**-19*10**6;
atoms=Na/m;
e2=atoms*e1;#energy released in J
m1=(e2*12)/(393.5*1000*1000);# in Kg
m2=m1/1000;# in tons
print"The amount of coal required in Kg is =", round(m2/1000,2)

Example 1.13, Page:6 

The amount of coal required in Kg is = 2.5

example 1.14;pg no:7

In [14]:
#cal of The energy releases
#intiation of all variables
# Chapter 1
print"Example 1.14, Page:7  \n \n"
#Given:
H1=241.8; # in KJ/mol
H2=887.2; # in KJ/mol
# 1 KJ/mol = 0.0104 eV/atom
#solution: part (a)
e1=H1*0.0104;
print"The energy release in part (a) in eV/molecule is =",round(e1,2)
#solution: part (b)
e2=H2*0.0104;
print"The energy release in part (b) in eV/molecule is =",round(e2,2)

Example 1.14, Page:7  
 

The energy release in part (a) in eV/molecule is = 2.51
The energy release in part (b) in eV/molecule is = 9.23

example 1.15;pg no:7

In [15]:
#cal of The energy releases
#intiation of all variables
# Chapter 1
print"Example 1.15, Page:7 \n"
#Given:
H1=4.1; # in eV/molecule
H2=17.4; # in eV/molecule
H3=200;# in MeV/atom of U
# 1 eV/atom  = 96.32 KJ/mol
#solution: part (a)
e1=H1*96.32;
print"The energy release in part (a) in KJ/mol of carbondioxide is =",round(e1,1)
#solution: part (b)
e2=H2*96.32;
print"The energy release in part (b) in KJ/mol of alumina is =",round(e2,1)
#solution: part (c)
e3=H3*1000*96.32;# in MJ/atom of U(235)
print"The energy release in part (c) in MJ/atom of U(235) is =",round(e3/10**6,3)

Example 1.15, Page:7 

The energy release in part (a) in KJ/mol of carbondioxide is = 394.9
The energy release in part (b) in KJ/mol of alumina is = 1676.0
The energy release in part (c) in MJ/atom of U(235) is = 19.264

example 1.16;pg no:7

In [16]:
#cal of The rate of energy release
#intiation of all variables
# Chapter 1
print"Example 1.16, Page:7  \n \n"
#Given:
e=200.; #MeV/ atom of U
# 1 eV = 1.6*10^-19 J
Na=6.023*10**23;
M=0.235; # mass in Kg
#solution:
e1=e*1.6*10**-19*10**6;
A=Na/M;
e2=A*e1; # energy released in MJ/day
e3=e2/(24.*3600.);
print"\n The rate of energy release in MW is=",round(e3/10**6,2)

Example 1.16, Page:7  
 


 The rate of energy release in MW is= 949.25

example 1.17;pg no:8

In [17]:
#cal of The mass loss
#intiation of all variables
# Chapter 1
print"Example 1.17, Page:8  \n \n"
#Given:
e=26.03; # in MeV
#solution:
loss=e/931; #in atomic mass units (u)
# 1 u = 1.66*10^-27 Kg
m=(loss*1.66*10**-27)/(1*10**-27);
print"\n The mass loss in 10^-27 Kg/He formed is =",round(m,4)

Example 1.17, Page:8  
 


 The mass loss in 10^-27 Kg/He formed is = 0.0464

example 1.18;pg no:8

In [18]:
#cal of The energy loss
#intiation of all variables
# Chapter 1
print"Example 1.18, Page:8  \n"
#Given:
mh=1.007825;
mt=3.016049;
md=2.014102;
#solution:
m1=(mh+mt-2*md);
e=(-m1)*931; # in MeV
print"\n The energy loss in MeV is =",round(-e,4)

Example 1.18, Page:8  


 The energy loss in MeV is = -4.0312

example 1.19;pg no:8

In [19]:
#cal of  mean binding energy of tritium and nickel atom
#intiation of all variables
# Chapter 1
print"Example 1.19, Page:8 \n"
#Given:
mh=1.007825;
mn=1.008665;
mt=3.016049; # atomic mass of Tritium
mNi=59.93528; # atomic mass of Nickel
#solution:
# part (a)
B1=(1*mh+2*mn-mt)*931; # in MeV
Bh=B1/mt;
print"The mean binding energy of tritium atom in MeV is =",round(Bh,3)
# part (b)
B2=(28*mh+32*mn-mNi)*931; # in MeV
Bo=B2/mNi;
print"The mean binding energy of nickel atom in MeV is =",round(Bo,3)

Example 1.19, Page:8 

The mean binding energy of tritium atom in MeV is = 2.811
The mean binding energy of nickel atom in MeV is = 8.716

example 1.20;pg no:9

In [20]:
#cal of mean binding energy of Cl
#intiation of all variables
# Chapter 1
print"Example 1.20, Page:9  \n"
#Given:
mh=1.00783;
mn=1.00867;
m35=34.96885; # atomic mass of Cl (35)
m37=36.96590; # atomic mass of Cl (37)
#solution:
B1=(17*mh+18*mn-m35)*931; # in MeV
Bh=B1/m35;
print"The mean binding energy of Cl (35) atom in MeV is =",round(Bh,4)
B2=(17*mh+20*mn-m37)*931; # in MeV
Bo=B2/m37;
print"The mean binding energy of Cl (37) atom in MeV is =",round(Bo,4)
Bi=Bo-Bh;
print"The increase in mean binding energy of Cl atom in MeV is =",round(Bi,2)
# NOTE: The answer depends upon how much precise value you take for atomic masses.

Example 1.20, Page:9  

The mean binding energy of Cl (35) atom in MeV is = 8.5281
The mean binding energy of Cl (37) atom in MeV is = 8.5784
The increase in mean binding energy of Cl atom in MeV is = 0.05

example 1.21;pg no:9

In [21]:
#cal of mean binding energy of Na
#intiation of all variables
# Chapter 1
print"Example 1.21, Page:9  \n \n"
#Given:
mh=1.0078;
mn=1.0087;
m22=21.99431;# atomic mass of Na 22
m23=22.9898;# atomic mass of Na 23
m24=23.9909;# atomic mass of Na 24
#solution:
# part (a)
B1=((11*mh+11*mn)-m22)*931; # in MeV
Bh=B1/m22;
print"\n The mean binding energy of Na(22) in MeV is =",round(Bh,4)
# part (b)
B2=((11*mh+12*mn)-m23)*931; # in MeV
Bo=B2/m23;
print"\n The mean binding energy of Na(23)in MeV is =",round(Bo,4)
# part (c)
B3=((11*mh+13*mn)-m24)*931; # in MeV
Bs=B3/m24;
print"\n The mean binding energy of Na(24) in MeV is =",round(Bs,4)

Example 1.21, Page:9  
 


 The mean binding energy of Na(22) in MeV is = 7.9236

 The mean binding energy of Na(23)in MeV is = 8.1154

 The mean binding energy of Na(24) in MeV is = 8.0717
In [ ]: