Tuesday, September 12, 2023

H C Verma solutions, The Nucleus, Chapter-46, EXERCISES, Q21 to Q30, Concepts of Physics, Part-II

EXERCISES, Q21 to Q30


       21. Radioactive ¹³¹I has a half-life of 8.0 days. A sample containing ¹³¹I has activity 20 µCi at t =0. (a) What is its activity at t =4.0 days? (b) What is its decay constant at t =4.0 days?  

ANSWER: Given t½ =8 days  
                     
The decay constant λ =0.693/t½
                         =0.693/8 per day 
                         =0.087 per day
Aₒ =20 µCi

(a) t =4.0 days
        
Activity at t =4.0 days

A =Aₒe-𝜆t 
  =(20 µCi)*e-(0.087*4.0)
  =20*e-0.348 µCi 
  =14.12 µCi
  ≈14 µCi. 

(b) The decay constant at t =4.0 days will remain the same as it depends only on the half-life of the sample. Hence,
λ =0.087 per day
   =0.087/(24*3600) per second
   =1.01x10⁻⁶ s⁻¹.
  
  




 
       22. The decay constant of ²³⁸U is 4.9x10⁻¹⁸ s⁻¹. (a) What is the average life of ²³⁸U? (b) What is the half-life of ²³⁸U? (c) By what factor does the activity of a ²³⁸U sample decrease in 9x10⁹ years?  

ANSWER: Given the decay constant, 
λ =4.9x10⁻¹⁸ s⁻¹ 
(a) Average life of the given sample of U,
tₐᵥ =1/λ
     =1/4.9x10⁻¹⁸ s
     =2.041x10¹⁷ s
     =6.47x10⁹ years. 
 
(b) Half-life of the given sample, 
t½ =0.693/λ
      =0.693/4.9x10⁻¹⁸ s
      =1.414x10¹⁷ s
      =4.5x10⁹ y. 

(c) Aₒ/A =2t/t½ 
Here t/t½ =9x10⁹/4.5x10⁹ =2 
Hence Aₒ/A =2² =4.      




 
       23. A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes. (a) What is the decay constant of the sample? (b) What is its half-life?  

ANSWER: Given Aₒ =500 per second.
After time t =50 min =3000 s,
Remaining activity, 
A =200 per second.

(a) The decay constant  lambda is given as, 
A =Aₒ*e-λt
→λt =ln (Aₒ/A) =ln (500/200) =0.916 
→λ =0.916/3000 per second 
       =3.05x10⁻⁴ s⁻¹.

(b) Half-life =1/λ
                 =1/3.05x10⁻⁴ s
                 =2272 s ≈ 38 min.   




 
       24. The count rate from a radioactive sample falls from 4.0x10⁶ per second to 1.0x10⁶ per second in 20 hours. What will be the count rate 100 hours after the beginning?   

ANSWER: eλt =Aₒ/A
λt =ln (Aₒ/A)
→λ ={ln (4x10⁶/1x10⁶)}/t
      ={ln 4}/20 per hr
      =0.0693 per hr

If we take t =100 hours from the beginning,
λt =0.0693*100 =6.93
Hence from,
eλt =Aₒ/A
A =Aₒ*e-6.93
   =4x10⁶*e-6.93 per second 
   =3.9x10³ per second.     



 

       25. The half-life of ²²⁶Ra is 1602 y. Calculate the activity of 0.1 g of RaCl₂ in which all the radium is in the form of ²²⁶Ra. Taken atomic weight of Ra is 226 g/mol and that of Cl is 35.5 g/mol.  

ANSWER: Molecular weight of RaCl₂ 
  =226+2*35.5 =297 
Hence 0.1 g of RaCl₂ will contain molecules, 
N =0.1*6.02x10²³/297 molecules
    =2.03x10²⁰ molecules.
Half-life t½ =1602 y

Decay constant 
λ =0.693/1602 per year
   =4.33x10⁻⁴ /y

Activity A =λN
      =4.33x10⁻⁴*2.03x10²⁰/(365*24*3600) s⁻¹
      =2.8x10⁹ disintegrations/s.  




 
       26. The half-life of a radioisotope is 10 h. Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1 Ci.  

ANSWER: Given t½ =10 h. Hence the decay constant, 
𝜆 =0.693/10 =0.0693 per h.
   =1.925x10⁻⁵ s⁻¹
Given Aₒ =1 Ci =3.7x10¹⁰ disintegrations/s.
Number of active nuclei at t =0,
 Nₒ =Aₒ/𝜆
     =3.7x10¹⁰/1.925x10⁻⁵
     =1.922x10¹⁵
Number of active nuclei at t =9 h, 
N =Nₒ e-𝜆t 
  =1.922x10¹⁵*e-1.925x10⁻⁵*9*3600
   =1.030x10¹⁵
Since the half-life of the sample is 10 h, the number of active nuclei at t =10h, 
N½ =Nₒ/2
    =1.922x10¹⁵/2
    =0.961x10¹⁵
Hence the total number of disintegrations in the tenth hour,
=N -N½
=(1.030 -0.961)x10¹⁵ 
=6.9x10¹³.
 




 
       27. The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one-month-old ³²P (t½ =14.3 days) source if it was originally purchased for 800 rupees? 

ANSWER: Given t½ =14.3 days.
Ratio of the original activity remaining,
A/Aₒ =1/2t/t½
        =1/2(30/14.3) =0.234
 Since the rate is proportional to its activity, the second-hand rate after one month will be reduced in the above proportion.
This rate is =0.234x800 rupees 
 ≈187 rupees.    




 

       28. ⁵⁷Co decays to ⁵⁷Fe by ß⁺-emission. The resulting ⁵⁷Fe is in its excited state and comes to the ground state by emitting 𝛾-rays. The half-life of ß⁺-decay is 270 days and that of 𝛾-emission is 10⁻⁸ s. A sample of ⁵⁷Co gives 5.0x10⁹ gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5x10⁹ per second?  

ANSWER: Since the gamma decay from an active nucleus is only after a beta-decay, the number of gamma rays will be equal to the beta-decays. Since the number of beta decay is half after 270 days, the number of gamma rays will also drop to half after 270 days.




 
       29. Carbon (Z =6) with mass number 11 decays to boron (Z =5). (a) Is it a ß⁺-decay or ß⁻-decay? (b) The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon-11 and 10% boron-11 (by the number of atoms) converts itself into a mixture of 10% carbon-11 and 90% boron-11?  

ANSWER: (a) As a result of decay, the atomic number decreases which means the number of protons in a decayed nucleus has reduced by one. Clearly, a proton has changed to a neutron and a positron. The positron is emitted as ß⁺-decay. So it is a ß⁺-decay. 

  (b) t½ =20.3 minutes =1218 s.
Suppose we have a sample of this mixture with a total of 100 atoms. Out of this, only 90 atoms of carbon-11 will decay with time. Let after time t, 80 atoms of carbon-11 decay to boron-11. Now the mixture has 10 carbon-11 atoms and 90 boron-11 atoms. 
Here Nₒ =90, N =10, 
𝜆 =0.693/t½
   =0.693/1218 per second
   =5.69x10⁻⁴ s⁻
Since N =Nₒ*e-𝜆t
→𝜆t = ln (Nₒ/N) =ln (90/10)
→t =2.197/(5.69x10⁻⁴) s
      =3861 s
      =64 min.
 



 

       30. 4x10²³ tritium atoms are contained in a vessel. The half-life of decay of tritium nuclei is 12.3 y. Find (a) the activity of the sample, (b) the number of decays in the next 10 hours, and (c) the number of decays in the next 6.15 y. 

ANSWER: (a) t½ =12.3 y 
                     =12.3*365*24*3600 s  
                     =3.88x10⁸ s
Decay constant 𝜆 =0.693/t½
                             =1.787x10⁻⁹ s⁻
The activity of the sample,
Aₒ =𝜆Nₒ
     =1.787x10⁻⁹*4x10²³ disintergrations/s
     =7.146x10¹⁴ disintegrations/s.

  (b) t =10 hours =36000 s
 𝜆t =1.787x10⁻⁹*36000 =6.433x10⁻⁵
Number of active nuclei after 10 hours,
N =Nₒ*e⁻𝜆t
    =4x10²³*e-6.433x10⁻⁵
    =3.999742x10²³
Hence the number of decays in 10 hours, 
  =Nₒ -N
  =4x10²³ -3.999742x10²³
  =2.57x10¹⁹.

  (c) t =6.15 y =1.94x10⁸ s
𝜆t =1.787x10⁻⁹*1.94x10⁸ =0.347 
N =Nₒ*e-𝜆t
    =4x10²³*e-0.347 
    =2.83x10²³ 
The number of decays in this period,
  =4x10²³ -2.83x10²³
  =1.17x10²³.     
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CHAPTER- 34- Magnetic Field

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