Monday, October 2, 2023

H C Verma solutions, The Nucleus, Chapter-46, EXERCISES, Q41 to Q53, Concepts of Physics, Part-II

EXERCISES, Q41 to Q53


       41. A vessel of volume 125 cm³ contains tritium (³H, t½ =12.3 y)at 500 kPa and 300 K. Calculate the activity of the gas.        


ANSWER: P =500 kPa =5x10⁵ Pa, 
V =125 cm³ =1.25x10⁻⁴ m³, T =300 K. 
Gas constant R =8.314 J/(mol.K). 

Hence the number of moles in the vessel,
n =PV/RT 
  =5x10⁵*1.25x10⁻⁴/(8.314*300) mol
  =0.025 mol 

Hence the number of tritium nuclei, 
 N =0.025*6.022x10²³  
     =1.50x10²². 

Half-life t½ =12.3 y 
Decay constant 𝜆 =0.693/t½
→𝜆 =0.693/12.3 per y
      =0.0563 /y

Activity =𝜆N
    =0.0563*1.50x10²²/(365*24*3600) dis/s
    =2.68x10¹³ disintegrations/s
    =724 Ci. 






 
       42. ²¹²₈₃Bi can disintegrate either by emitting an α-particle or by emitting a ß⁻-particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration by α- and ß- decays are in the ratio 7:13. The overall half-life of ²¹²Bi is one hour. If 1 g of pure ²¹²Bi is taken at 12:00 noon, what will be the composition of this sample at 1 p.m. the same day?        


ANSWER: (a) When the given isotope of Bi decays by emitting an alpha particle, two neutrons, and two protons in the parent nucleus are reduced because the emitted alpha particle is a helium nucleus. The equation is as follows; 
         ²¹²₈₃Bi → ²⁰⁸₈₁Tl +⁴₂He.  

When one of the neutrons in the nucleus converts into a proton and an electron, the electron is emitted as a beta particle along with an anti-neutrino. The number of protons in the nucleus increases by one. For the decay by emitting a ß⁻-particle, the equation may be written as follows;
      ²¹²₈₃Bi →²¹²₈₄Po +e⁻ +𝜈̅ 
 
(b) Since the half-life of Bi is one hour, at 1 pm only 0.5 g of Bi will remain in the sample. 

    By the given probability of the type of decay,
   0.5*7/20 g =0.175 g will be decayed by the emission of alpha particles. So approximately 0.175 g will be Tl.
 
   Decay by emission of beta particles will be for 0.5*13/20 g =0.325 g of Bi. So approximately 0.325 g of Po will be present in the sample. 
 
 







       43. A sample contains a mixture of ¹⁰⁸Ag and ¹¹⁰Ag isotopes each having an activity of 8.0x10⁸ disintegrations per second. ¹¹⁰Ag is known to have a larger half-life than ¹⁰⁸Ag. The activity A is measured as a function of time and the following data are obtained. 

   (a) Plot ln(A/Aₒ) versus time. (b) See that for large values of time, the plot is nearly linear. Dedude the half-life of ¹¹⁰Ag from this portion of the plot. (c) Use the half-life of ¹¹⁰Ag to calculate the activity corresponding to the ¹⁰⁸Ag in the first 50 s. (d) Plot ln(A/Aₒ) versus time for ¹⁰⁸Ag for the first 50 s. (e) Find the half-life of ¹⁰⁸Ag.        


ANSWER: (a) Activity at time t =0 is, 
Aₒ =2*8x10⁸ disintegrations per second,
   =16x10⁸ disintegrations per second.

From the given chart we calculate,

  Time(s)      (A/Aₒ)      ln(A/Aₒ) 
     0                1               0
   20             0.737      -0.30
   40             0.573      -0.56
   60             0.466      -0.76
   80             0.392      -0.94
 100             0.338      -1.08
 200             0.193      -1.65
 300             0.118      -2.14
 400             0.073      -2.62
 500             0.045      -3.10

  ln(A/Aₒ) vs. time may be plotted as follows:-
Plot for t =0 to t =100 s

 For t =100 s to t =500 s we plot as foloows,
Plot for t =100 s to t =500 s

(b) As we see in the second plot, for a large value of time the plot is nearly linear. This is due to the fact that this graph mostly represents the activity of the isotope ¹¹⁰Ag because it has a longer half-life. Since 
        ln (A/Aₒ) =-𝜆t
Which is a graph of a straight line in the form of y =mx. Here the slope of the graph gives the value of decay constant 𝜆. The slope of the graph from the second plot above,
  𝜆 =1.45/300 s⁻¹
    =0.0048 s⁻¹

Hence the half-life of ¹¹⁰Ag,
   =0.693/0.0048 =144 s.

(c) Activity of ¹¹⁰Ag in first 50 s
A =Aₒ/2(50/144)
   =8x10⁸/1.272
   =6.29x10⁸ disintegrations per second.

From the first graph, for t =50 s,
ln (A/Aₒ) =-0.66 
→A =16x10⁸*0.517
       =8.27x10⁸

Hence the activity of ¹⁰⁸Ag after the first 50 s
=8.27x10⁸ -6.29x10⁸
=1.98x10⁸ disintegrations /s

(d) Time       Activity of     Activity of    ln(A/Aₒ)
       (s)       ¹¹⁰Ag (*10⁸)   ¹⁰⁸Ag (*10⁸)  of ¹⁰⁸Ag
       20           7.26              4.54          -0.57 
       40           6.60              2.57          -1.09
       50           6.29              1.98          -1.40 

The graph for ln(A/Aₒ) versus Time (for the first 50 s) may be plotted as follows:--
Plot for t =0 to t =50 s for ¹⁰⁸Ag


 (e) Decay constant =slope of the graph, 
 →𝜆 =1.38/50 s⁻¹
       =0.028 s⁻¹ 
Half-life t½ =0.693/0.028 s
               =24.7 s. 





 
       44. A human body excretes (removes by waste discharge, sweating, etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form into a human body, the body excretes half the amount in 24 hours. A patient is given an injection containing ⁹⁹Tc. This isotope is radioactive with a half-life of 6 hours. The activity from the body just after the injection is 6 µCi. How much time will elapse before the activity falls to 3 µCi?         


ANSWER: Given, that Aₒ =6 µCi, A =3 µCi,
t½ =6 hours, suppose after time t the above change in activity takes place. 
Since t½ for excreting the isotope =24 hours. After time t, the remaining nuclei,
  N =Nₒ e-𝜆t   
But in this time remaining active nuclei out of N due to disintegration,
 N' =N e-𝜆't =Nₒe-𝜆t*e-𝜆't
 →N' =Nₒe-(𝜆+𝜆')t
 →e-(𝜆+𝜆')t =N'/Nₒ =A'/Aₒ =1/2
→-(𝜆+𝜆')t =ln(1/2) =-0.693
→(0.693/t½ +0.693/t'½)t =0.693
→t =t½*t'½/(t½ +t'½)
     =6*24/(6 +24) 
     =4.8 hours. 
 




 
       45. A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life 𝜏. Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.        


ANSWER: In a discharging circuit, the charge on the capacitor at time t is given as, 
 q =Qe-t/RC 
Hence the electrostatic energy stored in the capacitor at time t is, 
   U =½q²/C 
       =½(Q²e-2t/RC)/C

The average life of the radioactive sample =𝜏.
Hence its decay constant, 𝜆 =1/𝜏

  The activity of the sample at time t,
A =Aₒ*e-𝜆t
   =Aₒe-t/𝜏

Given that U/A =constant.
→(½Q²e-2t/RC/C)/(Aₒe-t/𝜏) =K (say)
→(½Q²/C)e-2t/RC =(KAₒ)e-t/𝜏

Comparing and equating the powers of e on both sides, we get
-2t/RC =-t/𝜏
→RC/2 =𝜏
→R =2𝜏/C.
         






 
       46. Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt =R. An inductor of inductance 100 mH, a resistor of resistance 100 Ω, and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope.          


ANSWER: L =100 mH, r =100 Ω. Let emf of the battery = E. The current in this circuit at time t, 
 i =(E/r)(1 -e-tr/L). 
   
The number of radioactive isotopes produced in time dt just after the production starts =Rdt. It starts decaying with its own decay constant.
  The number of active nuclei after time t, 
=Rdt*e-𝜆t

Integrating it within time limits zero to t we get,
The number of active nuclei after time t, 
            N =(R/𝜆)(1 -e-𝜆t)
{see integration in the solution of problem number 39 of this chapter → Link of the post}

Given that i/N =constant =K (say). So,
(E/r)(1 -e-tr/L)/{(R/𝜆)(1 -e-𝜆t)} =K
→(E/r)(1 -e-tr/L) =(KR/𝜆)(1 -e-𝜆t)

Comparing both sides and equating the powers of e, we get
    -tr/L =-𝜆t 
 →𝜆 =r/L
The half-life of the isotope, 
      =0.693/𝜆
      =0.693L/r
      =0.693*(100/1000)/100 s
      =0.693/1000 s
      =6.93x10⁻⁴ s.
 





 
       47. Calculate the energy released by 1 g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope ²³⁵U has an abundance of 0.7% by weight in natural uranium.         


ANSWER: The mass of fissionable isotope ²³⁵U in 1 g of natural uranium,
  m =(0.7/100)*1 g
      =7x10⁻³ g

The number of nuclei in this mass,
   =6.022x10²³*(7x10⁻³/235)
   =1.794x10¹⁹

Hence the energy released by this number of fissionable nuclei
E =1.794x10¹⁹*200 MeV 
   =3.59x10²¹ MeV
   =3.59x10²¹*1.602x10⁻¹⁹*10⁶ J
   =5.75x10⁸ J.
 





 
       48. A uranium reactor develops thermal energy at a rate of 300 MW. Calculate the amount of ²³⁵U being consumed every second. The average energy released per fission is 200 MeV.         


ANSWER: Thermal energy is produced at a rate of 300 MW =300 MJ/s. 
  Per second energy produced,
E =300x10⁶ J
   =3x10⁸ J
   =3x10⁸/(1.602x10⁻¹⁹) eV
   =1.87x10²⁷ eV
   =1.87x10²¹ MeV.

The number of nuclei consumed per second
N =9.35x10¹⁸ 
235 g of ²³⁵U contains 6.022x10²³ nuclei. 

Hence 9.35x10¹⁸ nuclei will have a mass of
m =(9.35x10¹⁸/6.022x10²³)*235 g
    =0.00365 g
    ≈3.7 mg.

Thus the amount of ²³⁵U consumed per second in the reactor =3.7 mg.
 




 

       49. A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. (a) Assuming 200 MeV of thermal energy to come from each fission event on average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through ²³⁵U, at what rate will the amount of ²³⁵U decrease? Express your answer in kg/day. (c) Assuming that uranium enriched to 3% in ²³⁵U will be used, how much uranium is needed per month (30 days)?


ANSWER: Electric power needed for the town,
=300*10⁶ W 
=3x10⁸ J/s
=3x10⁸/(1.602x10⁻¹⁹*10⁶) MeV/s
=1.87x10²¹ MeV/s.

Assuming the efficiency with which thermal power is converted into electric power is 25%, thermal power needed per second,
 =1.87x10²¹/0.25 MeV
 =7.48x10²¹ MeV.
For the whole day power is needed,
=7.48x10²¹*24*3600 MeV
=6.46x10²⁶ MeV. 

  (a) Each fission event releases 200 MeV of energy, hence the number of events taking place each day, 
=6.46x10²⁶/200
=3.23x10²⁴.      

   (b) 6.022x10²³ number of atoms make 235 g of ²³⁵U, hence 3.23x10²⁴ atoms of ²³⁵U will make 
(3.23x10²⁴/6.022x10²³)*235 g
=1261 g
=1.261 kg
   Hence 1.261 kg/day of ²³⁵U will decrease each day.

   (c) If 3% ²³⁵U enriched uranium is to be used then the per day requirement of uranium,
=1.261/(0.03) kg
=42.033 kg
   Requirement of uranium per month, 
=30*42.033 kg
=1261 kg. 




 

       50. Calculate the Q-values of the following fusion reactions: 
(a) ²₁H + ²₁H → ³₁H +¹₁H
(b) ²₁H + ²₁H → ³₂He + n
(c) ²₁H + ³₁H → ⁴₂He + n.
Atomic masses are m(²₁H) =2.014102 u, m(³₁H) =3.016049 u, m(³₂He) =3.016029 u, and m(⁴₂He) =4.002603 u. 


ANSWER: (a) Uᵢ =2*m(²₁H) -2mₑ
  =2*2.014102 -2*0.0005486 u
  =4.0271068 u

Uf =m(³₁H) +m(¹₁H) -2mₑ
    =3.016049 +1.007825 -2*0.0005486 u
    =4.0227768 u

Q-value =Uᵢ -Uf
              =4.0271068 -4.0227768 u
              =0.00433 u
              =0.00433*931 MeV.
            =4.03 MeV.

(b) Uᵢ =2*m(²₁H) -2mₑ
    Uf =m(³₂He) +mₙ -2mₑ
Q =2*m(²₁H) -m(³₂He) -mₙ
  =2*2.014102  -3.016029 -1.008665 u
  =0.00351 u
  =0.00351*931 MeV
   =3.26 MeV.

  (c) Q =m(²₁H) +m(³₁H) -m(⁴₂He) -mₙ 
 =2.014102 +3.016049 -4.002603 -1.008665 u
=0.01888 u
=0.01888*931 MeV
=17.57 MeV.    





 

       51. Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 kT equals the coulomb potential energy at 2 fm.         


ANSWER: Charge on each helium nucleus, 
  q =2e. 

   Coulomb potential energy between two helium nuclei at a distance of  r =2 fm,
   U =(1/4π𝜀ₒ)(2e)²/r 
=(9x10⁹ N-m²/C²)(2*1.6x10⁻¹⁹ C)²/(2x10⁻¹⁵ m)
=4.61x10⁻¹³ J.

Average thermal energy =1.5 kT, where k is the Boltzmann constant. 
Given that, 1.5 kT =U 
→T =U/1.5k
    =(4.61x10⁻¹³ J)/(1.5*1.38x10⁻²³ J/K) 
    =2.23x10¹⁰ K.  






   52. Calculate the Q-value of the fusion reaction 
         ⁴He + ⁴He =⁸Be.
 Is such a fusion energetically favorable? The atomic mass of ⁸Be is 8.0053 u and that of ⁴He is 4.0026 u.           


ANSWER: The Q-value of the given fusion reaction is, 
Q =2*m(⁴He) -m(⁸Be) 
   =2*4.0026 -8.0053 u 
   =-0.0001 u 
   =-0.0001*931 MeV 
   =-0.0931 MeV 
   =-93.1 KeV.  

  Since the Q-value is negative, this fusion reaction will not release energy but require energy. So it is not energetically favorable.        





 
       53. Calculate the energy that can be obtained from 1 kg of water through the fusion reaction
              ²H +²H → ³H + p. 
  Assume that 1.5x10⁻²% of natural water is heavy water D₂O (by the number of molecules) and all the deuterium is used for fusion.            


ANSWER: Q-value of the reaction, 
Q ={2*m(²H) -2mₑ} -{m(³H) -mₑ +mₚ}
 =2m(²H) -m(³H) -mₑ -mₚ
  =2*2.014102 -3.016029 -0.0005486 -1.007276 u
  =0.00435 u
  =4.05 MeV

18 g of natural water contains 6.022x10²³ molecules. So 1000 g (1 kg) will contain 
(1000/18)*6.022x10²³ molecules,
=3.3456x10²⁵ molecules.

From the given percentage of heavy water, the number of D₂O molecules,
  =(1.5x10⁻²/100)*3.3456x10²⁵ molecules
  =5.0184x10²¹ molecules.
 Each molecule of the heavy water contains two deuterium atoms, hence a total of 5.0184x10²¹ fusion reaction will take place if all the deuterium is to be consumed. So the total energy obtained is,
=5.01x10²¹*4.05 MeV
=2.029x10²²*1.6x10⁻¹⁹ MJ
=3246 MJ. 
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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- 2 - "Physics and Mathematics"

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