Tuesday, August 29, 2023

H C Verma solutions, The Nucleus, Chapter-46, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

EXERCISES, Q1 to Q10


       1. Assume that the mass of a nucleus is approximately given by M =Amₚ where A is the mass number. Estimate the density of the matter in kg/m³ inside a nucleus. What is the specific gravity of nuclear matter?     

ANSWER: mₚ =1.007276 u
Given the mass of the nucleus,
M =Amₚ

The radius of a nucleus is
R = Rₒ∛A, where Rₒ =1.1 fm.

The volume of the nucleus,
V =(4/3)πR³
   =(4/3)πRₒ³A

Hence the density of the nucleus
D =M/V
   =3Amₚ/4πRₒ³A
   =3mₚ/4πRₒ³
=3*1.007276*1.66x10⁻²⁷/4π*(1.1x10⁻¹⁵)³ kg/m³
  =3.0x10¹⁷ kg/m³.
(Because of 1 u ≈1.66x10⁻²⁷ kg. You can calculate it from the given data where 1 u =931 MeV/c²)
1 u =931*10⁶*1.602x10⁻¹⁹/(3x10⁸)² kg 
      =1.66x10⁻²⁷ kg.
 
The specific gravity of the nuclear matter,
 =Density of the matter/Density of water
 =3.0x10¹⁷ kg/m³/1x10³ kg/m³
 =3.0x10¹⁴.     






 
       2. A neutron star has a density equal to that of nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is 4.0x10³⁰ kg (twice the mass of the sun).   

ANSWER: The volume of the neutron star, 
V = Mass/Density 

Given, M =4.0x10³⁰ kg,
Density =density of nuclear matter
      =3.0x10¹⁷ kg/m³
(Calculated in the previous problem). 

Hence the volume,
V =4.0x10³⁰/3.0x10¹⁷ m³
   =(4/3)x10¹³ m³.
But volume of a sphere =(4/3)πR³  
If we take R = radius of the neutron star then,
 (4/3)πR³ =(4/3)x10¹³ 
→πR³ =1x10¹³ 
→R =∛(1x10¹³/π) 
       =14710 m 
       ≈15 km.          




 

       3. Calculate the mass of an α-particle. Its binding energy is 28.2 MeV.    

ANSWER: An α-particle is made of two protons and two neutrons. Hence, 
2mₚ +2mₙ =mₐ +binding energy
where mₐ =mass of the α-particle. 

Given binding energy =28.2 MeV
In terms of mass B.E. =28.2 Mev/c² 
Hence, 
mₐ =2*1.007276 u+2*1.008665 u -28.2 Mev/c²
   =4.031882 u -28.2 MeV/c² 
  =4.031882 u -(28.2/931) u  
  =4.0016 u.
 




 
       4. How much energy is released in the following reaction?
           ⁷Li + p = α + α
The atomic mass of ⁷Li =7.0160 u and that of ⁴He =4.0026 u.    

ANSWER: The mass of a proton =1.007276 u.
The sum of the rest mass of Li and p on the left side,
LHS =(7.0160 +1.007276) u
      =8.023276 u. 
 
Given that the mass of an α-particle   
  =4.0026 
   
So the sum of the rest mass of two α-particles on the RHS is,  
  =2*4.0026 u
  =8.0052 u.
Since the final product on the RHS has a lesser mass than the LHS, this difference in mass is released as energy. 
So the mass equivalent of energy released is 
=8.023276 -8.0052 u 
=0.018076 u
=0.018076*931 MeV/c² 
=16.83 MeV/c² 
The energy equivalent of this released energy,
=16.83 MeV.               





 
       5. Find the binding energy per nucleon of ¹⁹⁷₇₉Au if its atomic mass is 196.96 u.   

ANSWER: The number of protons =79, 
The number of neutrons =197 -79 =118. 
The sum of the rest mass of the individual protons and neutrons, 
M =79x1.007276+118x1.008665 u 
    =198.597274 u 
But the rest mass of the Au atom is =196.96 u.
Binding energy =difference in mass in energy units
            =198.597274 -196.96 u 
            =1.637274 u
            =1.637274*931 MeV
            =1524.3 MeV. 
Total number of nucleons =197, 
Hence the binding energy per nucleon, 
     =1524.3/197 = 7.74 MeV.       




 

       6. (a) Calculate the energy released if ²³⁸U emits an α-particle.
(b) Calculate the energy to be supplied to ²³⁸U if two protons and two neutrons are to be emitted one by one.
The atomic mass of ²³⁸U, ²³⁴Th, and ⁴He are 238.0508 u, 234.04363 u, and 4.00260 u respectively.   

ANSWER: (a) Energy released in mass equivalent,
 =mass of ²³⁸U -(mass of ²³⁴Th +mass of ⁴He),
 =238.0508 -(234.04363 +4.0026) u.     
 =238.0508 -238.04623 u 
 =0.00457 u
In terms of energy, the released energy, 
 =0.00457*931 MeV 
 =4.255 MeV.  

(b) Rest mass of two protons and two neutrons separately,
 =2*1.007276 +2*1.008665 u 
 =4.031882 u 
So the sum of the masses of the products, 
=234.04363 +4.031882 u
=238.075512 u
      Since the mass of the products is more than the mass of the parent nucleus 238.0508 u, the difference in mass has to be supplied in the form of energy for this process to happen.    Required energy E =238.075512 -238.0508 u
→E =0.024712 u
       =0.024712*931 MeV (in energy units)
       =23.00 MeV.    



 

       7. Find the energy liberated in the reaction 
         ²²³Ra → ²⁰⁹Pb + ¹⁴C
The atomic masses needed are as follows:
²²³Ra →223.018 u
²⁰⁹Pb →208.981 u
¹⁴C → 14.003 u.    

ANSWER: Atomic mass of Ra =223.018 u,
The sum of the atomic masses of products Pb and C, 
=208.981 +14.003 u 
=222.984 u.
 Reduction in mass =223.018 -222.984 u
                                =0.034 u. 
This much mass is converted into energy that is liberated. The energy equivalent of this mass, 
   =0.034*931 MeV 
   =31.65 MeV.          




 

       8. Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is        
            ΔE =(MZ-1,N +MH -MZ,N)c²  
where MZ,N =mass of an atom with Z protons and N neutrons in the nucleus and MH =mass of a hydrogen atom. This energy is known as proton-separation energy.

ANSWER: The energy supplied to separate a proton from a nucleus appears as an increased mass of the products. With the separation of a proton from a nucleus with Z protons and N neutrons, the daughter nucleus has now Z-1 protons and N neutrons. The proton's mass is approximately equal to the mass of a hydrogen atom. Hence the energy supplied,
=Increase in mass of the products 
=(Massof daughter nucleus+mass of the proton) -mass of the parent nucleus
=(MZ-1,N +MH) -MZ,N
=MZ-1,N +MH -MZ,N
The energy equivalent of this mass is given from E=mc², hence the energy needed to separate a proton from a nucleus,
ΔE =(MZ-1,N +MH -MZ,N)c².    






       9. Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons in terms of masses MZ,N, MZ,N-1, and the mass of the neutron.   

ANSWER: Increase in the sum of the mass of the products, 
ΔM=(MZ,N-1 +mN)- MZ,N 
This increase in mass is due to the conversion of the supplied energy to separate a neutron.
      The increase in mass ΔM is written as energy according to the equation ΔE =ΔM*c²
    Hence the required energy,
ΔE =ΔMc²
    =(MZ,N-1 +mN -MZ,N)c².
 
       




 

       10. ³²P beta decays to ³²S. Find the sum of the energy of the antineutrino and the kinetic energy of the ß-particle. Neglect the recoil of the daughter nucleus. The atomic mass of ³²P =31.974 u and that of ³²S =31.972 u.   

ANSWER: As we see, the mass of the atom decreases with beta decay. We take the mass of a beta particle as negligible in comparison to a proton or neutron. So this decrease in mass will appear as kinetic energy of the beta particle and antineutrino.
ΔE =ΔM*c²
   =[m(³²P) -m(³²S)]*c²
   =[31.974 u -31.972 u]*c²
   =[0.002 u]*c²
   =[0.002 u]*c²*931 MeV/c²
   =1.86 MeV.  
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Links to the Chapters






CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion









CHAPTER- 11 - Gravitation





CHAPTER- 10 - Rotational Mechanics






CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 7 - Circular Motion

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion


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CHAPTER- 2 - "Physics and Mathematics"

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