Thursday, August 24, 2023

H C Verma solutions, The Nucleus, Chapter-46, OBJECTIVE-I, Concepts of Physics, Part-II

OBJECTIVE-I


       1. The mass of a neutral carbon atom in the ground state is
(a) exact 12 u
(b) Less than 12 u
(c) More than 12 u
(d) depends on the form of carbon such as graphite or charcoal.     

ANSWER: (a).   

EXPLANATION: Unified atomic mass unit is denoted by 'u' which is equal to 1/12th of the mass of a neutral carbon atom in its lowest energy state. So by definition, the mass of a neutral carbon atom is taken exactly as 12 u. Option (a) is correct.   




 

       2. The mass number of a nucleus is equal to  
(a) The number of neutrons in the nucleus
(b) The number of protons in the nucleus
(c) The number of nucleons in the nucleus
(d) None of them.     

ANSWER: (c).   

EXPLANATION: The mass number A is the total number of protons Z and neutrons N (i.e. nucleons) in a nucleus. A =Z +N. Hence option (c) is correct. 




  
       3. As compared to ¹²C atom, ¹⁴C atom has 
(a) two extra protons
(b) two extra protons but no extra electron
(c) two extra neutrons and no extra electron
(d) two extra neutrons and two extra electrons.       

ANSWER: (c).   

EXPLANATION: ¹⁴C is an isotope of carbon. Its nucleus has the same number of protons as ¹²C (otherwise if the number of protons changes, the element will change and it will not remain a C atom). But it has two extra neutrons that's why its mass number is 12+2 =14. In a neutral atom, the number of electrons and the number of protons are the same. Hence ¹⁴C has the same number of electrons as ¹²C. Option (c) is correct.  

 


   
       4. The mass number of a nucleus is 
(a) always less than its atomic number
(b) always more than its atomic number
(c) equal to its atomic number
(d) sometimes more than and sometimes equal to its atomic number. 

ANSWER: (d).   

EXPLANATION: The mass number =Atomic number Z (number of protons) + N (number of neutrons. 
  So the mass number can not be less than its atomic number but may be more than or equal to it. It is because of the fact that in a nucleus Z cannot be zero. But N may be zero as is the case with hydrogen atoms where Z=1, N=0, and A =1. 
   So the mass number is sometimes more than and sometimes equal to its atomic number. Option (d) is correct. 

 


 
       5. The graph of ln (R/Rₒ) versus ln A (R = radius of a nucleus and A = its mass number) is
(a) a straight line 
(b) a parabola
(c) an ellipse
(d) None of them.      

ANSWER: (a).   

EXPLANATION: The average radius R of a nucleus is given as 
R =RₒA1/3
Where Rₒ =1.1 fm.
→R/Rₒ =A1/3
→ln (R/Rₒ) =(1/3) ln A
If y =ln (R/Rₒ) and x =ln A
then, y =(1/3)x
    It is in the form of y =mx, which is an equation of a straight line passing through the origin and having slope =1/3.
    Option (a) is correct.   




  
       6. Let Fₚₚ, Fₚₙ, and Fₙₙ denote the magnitude of the nuclear force by a proton on a proton, by a proton on a neutron, and by a neutron on a neutron respectively. When the separation is 1 fm, 
(a) Fₚₚ > Fₚₙ = Fₙₙ
(b) Fₚₚ = Fₚₙ = Fₙₙ
(c) Fₚₚ > Fₚₙ > Fₙₙ
(d) Fₚₚ < Fₚₙ = Fₙₙ. 

ANSWER: (b).   

EXPLANATION: At the given separation of 1 fm, the close-range attractive nuclear force is predominant which is the same for each nucleon pair and independent of charge. Hence option (b) is correct. 

 


  
       7. Let Fₚₚ, Fₚₙ, and Fₙₙ denote the magnitude of the net force by a proton on a proton, by a proton on a neutron, and by a neutron on a neutron respectively. Neglect gravitational force. When the separation is 1 fm, 
(a) Fₚₚ > Fₚₙ = Fₙₙ
(b) Fₚₚ = Fₚₙ = Fₙₙ
(c) Fₚₚ > Fₚₙ > Fₙₙ
(d) Fₚₚ < Fₚₙ = Fₙₙ.   

ANSWER: (d).   

EXPLANATION: The net force between a nucleon pair at the given distance will be the sum of nuclear force and coulomb force. The coulomb force here will be only between a proton-proton pair because only a proton has a positive charge while the neutron is neutral. So the net force between the proton-neutron pair and the neutron-neutron pair will be equal. The coulomb force between a proton-proton pair will be repulsive while the nuclear force between them is attractive. So the net force between a proton-proton pair will be less than the proton-neutron pair or neutron-neutron pair.
    Option (d) is correct.  




 
       8. Two protons are kept at a separation of 10 nm. Let Fₙ and Fₑ be the nuclear force and the electromagnetic force between them. 
(a) Fₑ = Fₙ,
(b) Fₑ >> Fₙ,
(c) Fₑ << Fₙ,
(d) Fₑ and Fₙ differ only slightly.    

ANSWER: (b).   

EXPLANATION: Nuclear force is only appreciable when separation is of the order of fm. The nm is a very large unit compared to the fm. The nuclear force is a close-range force that reduces drastically with an increase in separation and becomes negligible when the distance is increased up to 10 fm. So at a separation of 10 nm, Fₙ will be nearly zero while Fₑ will have appreciable value. Thus Fₑ >> Fₙ. 
    Option (b) is correct. 




  
       9. As the mass number A increases, the binding energy per nucleon in the nucleus 
(a) increases 
(b) decreases
(c) remains the same  
(d) varies in a way that depends on the actual value of A.        

ANSWER: (d).   

EXPLANATION: The binding energy per nucleon in the nucleus depends upon the number of surrounding nucleons around an inside nucleon and the surface area of a nucleus. In lighter nuclei surrounding nucleons inside a nucleus increase with A, so initially the B.E. per nucleon increases with A up to the point when the surrounding nucleons around a nucleon inside are full. Thereafter with a continuous increase in surface area, the B.E. per nucleon starts to decrease slowly. This behavior cannot be explained with options (a), (b), or (c). So option (d) is correct.




  
       10. Which of the following is a wrong description of the binding energy of a nucleus? 
(a) It is the energy required to break a nucleus into its constituent nucleons.  
(b) It is the energy made available when free nucleons combine to form a nucleus. 
(c) It is the sum of the rest mass energies of its nucleons minus the rest mass energy of the nucleus. 
(d) It is the sum of the kinetic energy of all the nucleons in the nucleus.     

ANSWER: (d).   

EXPLANATION: Options (a), (b), and (c) give the correct description of binding energy. The energies described in these options are exactly the same. The kinetic energies of nucleons have nothing to do with the binding energy of a nucleus. So option (d) is the answer.  

 


  
       11. In one average life,  
(a) Half the active nuclei decay 
(b) Less than half the active nuclei decay
(c) More than half the active nuclei decay
(d) All the nuclei decay.    

ANSWER: (c).   

EXPLANATION: The average life of a nuclei is =Half-life/0.693 
=1.44*Half-life. 

   So the average life of a nuclei is more than its half-life. Option (c) is correct.    





   
       12. In radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?  
(a) proton   
(b) neutron
(c) electron
(d) photon.    

ANSWER: (d).   

EXPLANATION: If a proton is emitted in a radioactive decay, atomic number as well as mass number changes. If a neutron is emitted, the atomic number remains unchanged but the mass number changes. If an electron is emitted, the mass number remains unchanged but the atomic number changes.  
    With the given condition in the problem, options (a), (b), and (c) are not correct. If a photon is emitted neither atomic number nor mass number changes as is the case with gamma radiation. Hence option (d) is correct. 



  

       13. During negative beta decay,  
(a) An atomic electron is ejected 
(b) An electron that is already present within the nucleus is ejected 
(c) A neutron in the nucleus decays emitting an electron 
(d) a proton in the nucleus decays emitting an electron.         

ANSWER: (c).   

EXPLANATION:  During a negative beta decay, a neutron converts into a proton and an electron. This electron is emitted while the proton remains in the nucleus. It increases the atomic number of the nucleus but the mass number remains the same. Option (c) is correct.

 



   
       14. A freshly prepared radioactive source of half-life 2 h emits radiation of intensity that is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is 
(a) 6 h
(b) 12 h
(c) 24 h
() 128 h.     

ANSWER: (b).   

EXPLANATION: For safe working with the source, the radiation activity should be 1/64 of the initial activity. So, 
A/Aₒ = 1/2t/t½
→1/2t/t½ =1/64 =1/2⁶
→t/t½ =6
→t =6*t½ =6*2 h =12 h. 
    Option (b) is correct. 





   
       15. The decay constant of a radioactive sample is λ. The half-life and the average life of the sample are respectively 
(a) 1/λ and (ln 2/λ)
(b) (ln 2/λ) and 1/λ
(c) λ(ln 2) and 1/λ
(d) λ/(ln 2) and 1/λ.   

ANSWER: (b).   

EXPLANATION: The half-life is given as, 
t½ =(ln 2)/λ
While the average life is given as
tₐᵥ =1/λ

   Option (b) is correct.
 

 




  
       16. An α-particle is bombarded on ¹⁴N. As a result, a ¹⁷O nucleus is formed and a particle is emitted. This particle is a 
(a) neutron
(b) proton
(c) electron
(d) positron.    

ANSWER: (b).   

EXPLANATION: An α-particle has two protons and two neutrons. When it is absorbed by a ¹⁴N nucleus which has seven protons and seven neutrons, the new temporary nucleus has nine protons and nine neutrons and the mass number becomes 18. But the new nucleus formed is given that of oxygen which has 8 protons only. So clearly a proton is emitted and the mass number also becomes 17.  
    Option (b) is correct. 

 



  
       17. Ten grams of ⁵⁷Co kept in an open container beta-decays with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly 
(a) 10 g 
(b) 5 g
(c) 2.5 g
(d)1.25 g.     

ANSWER: (a).   

EXPLANATION: In beta decay, an electron from the nucleus is emitted. The nucleus is composed of protons and neutrons, each of them is about 1840 times more in weight than an electron. A ⁵⁷Co nucleus has a total of 57 protons and neutrons. So in comparison to a Co nucleus, the weight of an electron is negligible. So after 540 days of beta decay, the residual weight of the Co in the container will be almost the same i.e. 10 g. 
    Option (a) is correct.   

 




  
       18. Free ²³⁸U nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x in time t after the decay. If a decay takes place when the train is moving at a uniform speed v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger will be 
(a) x +vt
(b) x -vt
(c) x
(d) depends on the direction of the train.     

ANSWER: (c).   

EXPLANATION: Since the frame of reference is moving with uniform velocity, objects will appear to an observer moving with a velocity that is equal to the object's velocity minus the frame's velocity. 
   Suppose the velocity of the alpha particle =u and that of the recoiling nucleus = -u' when the train is at rest. So the relative velocity of the alpha particle relative to the recoiling nucleus is u -(-u') =u +u'. It is the velocity by which both particles will appear to separate from each other to the observer.
     Now consider the train moving with a uniform velocity v. The observer will see the velocity of the alpha particle =u -v and the velocity of the recoiling nucleus =-u' -v. So both particles will appear to the observer to separate from each other with a velocity equal to 
=(u -v) -(-u' -v) 
=u +u' 
   It is the same when the train is at rest. Hence in time t after the decay, the passenger will measure the same distance x when the train is moving with a uniform velocity. 

 



  



       19. During a nuclear fission reaction,
(a) A heavy nucleus breaks into two fragments by itself 
(b) A light nucleus bombarded by thermal neutrons breaks up 
(c) A heavy nucleus bombarded by thermal neutrons breaks up
(d) Two light nuclei combine to give a heavier nucleus and possibly other products.       

ANSWER: (c)   

EXPLANATION:  When a heavy nucleus absorbs a bombarded thermal neutron and becomes unstable, it breaks up. This nuclear reaction is called a fission reaction. 
      Option (c) is correct.    
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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

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Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion


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Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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