Thursday, September 21, 2023

H C Verma solutions, The Nucleus, Chapter-46, EXERCISES, Q31 to Q40, Concepts of Physics, Part-II

EXERCISES, Q31 to Q40


       31. A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its 1 cm² window. If the source contains 6.0x10¹⁶ active nuclei and the counter records a rate of 50,000 counts per second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and that the alpha particles fall nearly normally on the window. 

ANSWER: Surface area of a sphere of radius 1 m with the point source at the center, 
=4πr² 
=4π*1² m²
=4π m²     
Number of alpha particles falling on 1 cm², 
=50000 counts/s. 
Total alpha particles emitted by the point source =Total counts on the surface area of the above sphere, 
=4π*10000*50000 counts/s
=2x10⁹π counts per second. 
So the activity of sample A =2x10⁹π per second. 
A =𝜆N
→𝜆 =A/N
    =2x10⁹π/6x10¹⁶ per s
    =1.05x10⁻⁷ per s
Hence the decay constant, 𝜆 =1.05x10⁻⁷ s⁻. 






 
     32. ²³⁸U decays to ²⁰⁶Pb with a half-life of 4.47x10⁹ y. This happens in a number of steps. Can you justify a single half-life for this chain of processes? A sample of rock is found to contain 2.00 mg of ²³⁸U and 0.600 mg of ²⁰⁶Pb. Assuming that all the lead has come from uranium, find the life of the rock.

ANSWER:  The unstable nucleus of ²³⁸U decays to different unstable nuclei in a number of steps and finally decays to the stable nucleus of ²⁰⁶Pb, each step having its own half-life. But this final change of nucleus from U to Pb will have a fixed decay constant and hence a single half-life for this chain of processes can be justified.
 
238 g of ²³⁸U and 206 g of ²⁰⁶Pb  each contain 6.022x10²³ number of atoms. 

    0.600 mg of Pb will contain,
0.6*6.022x10²³/(206*1000) atoms
=1.75x10¹⁸ atoms. 

Similarly, 2.00 mg of the given uranium will contain,
2*6.022x10²³/(238*1000) atoms
=5.06x10¹⁸ atoms.

Hence the original number of uranium atoms,
Nₒ =(1.75+5.06)x10¹⁸ atoms
   =6.81x10¹⁸ atoms.

The present number of uranium atoms,
N =5.06x10¹⁸ atoms. If 𝜆 is the decay constant then,
N =Nₒe-𝜆t
5.06x10¹⁸ =6.81x10¹⁸*e(-0.693/4.47x10⁹)t,
{where t is the life of the rock}
→e(-1.55x10⁻¹⁰)t =5.06/6.81
→t ={ln (5.06/6.81)}/(-1.55x10⁻¹⁰) y.
    =1.92x10⁹ y.
    




 
       33. When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ¹⁴C per gram per minute. A sample from an ancient piece of charcoal shows ¹⁴C activity to be 12.3 disintegrations per gram per minute. How old is this sample? The half-life of ¹⁴C is 5730 y.  

ANSWER: The present activity of an ancient sample of charcoal, 
A =12.3 disintegrations/minute. 

Fresh charcoal's activity, 
Aₒ =15.3 disintegrations/minute
The half-life of charcoal
   t½ =5730 y
       
Decay constant 𝜆 =0.693/5730 per y
              =1.20x10⁻⁴ per y
Suppose the age of the sample =t years
A =Aₒe⁻𝜆t
  e⁻𝜆t =A/Aₒ
→⁻𝜆t =ln (12.3/15.3)
→t =-0.218/(-1.20x10⁻⁴) 
      ≈1800 y.   
 

     




 
       34. Natural water contains a small amount of tritium (³₁H). This isotope beta-decays with a half-life of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things, he finds a sealed bottle of whiskey. On return, he analyses the whiskey and finds that it contains only 1.5 percent of ³₁H radioactivity as compared to a recently purchased bottle marked '8 years old'. Estimate the time of that unsuccessful attempt. 

ANSWER: Given A/Aₒ =1.5% =0.015
But A/Aₒ =e⁻𝜆t
Hence e⁻𝜆t =0.015
→⁻𝜆t =ln (0.015) =-4.2 
→t =4.2/𝜆 
      =4.2*t½/0.693 
      =6.06*12.5 y
      =75.7 y 
But the recently purchased whisky was manufactured 8 years ago. So the above duration is up to that time. Hence from now on, the unsuccessful event was 75.7+8 =83.7 y, hence about 84 y ago.  
     




 
       35. The count rate of nuclear radiation coming from a radioactive sample containing ¹²⁸I varies with time as follows,
Time t (minutes);         0    25  50   75  100
Count rate R(10⁹ s⁻¹): 30  16  8.0  3.8  2.0

(a) Plot ln(Rₒ/R) against t. (b) From the slope of the best straight line through the points, find the decay constant 𝜆. (c) Calculate the half-life t½.  

ANSWER: (a) Given Rₒ =30x10⁹ per s
Hence the variations of ln (Rₒ/R) with time in the given problem will be as follows:

Time  (minutes) :  0     25    50     75   100
  Rₒ/R                :  1   1.88  3.75  7.89  15
ln (Rₒ/R)            :  0   0.63  1.32  2.07   2.71

It can be plotted as follows;
Plot for Q-35(a)

 

(b) The best line is drawn through the points and the slope of this line is, 
m =AB/BO
    =2.70/100 per min
    =0.027 per min.
    
Since ln(Rₒ/R) =𝜆t, that is in the form of y =mx, a straight line passing through the origin. The slope of the line will give the value of the decay constant λ. 
Hence 𝜆 =m =0.027 per min.

(c) Hence half-life,
t½ =0.693/𝜆
      =0.693/0.027 min
      ≈ 25 min. 




 
       36. The half-life of ⁴⁰K is 1.3x10⁹ y. A sample of 1 g of pure KCl gives 160 counts/s. Calculate the relative abundance of ⁴⁰K (fraction of ⁴⁰K present) in natural potassium.  

ANSWER: The average atomic mass of K =39. Hence 39+35.5 =74.5 g of KCl will have 6.022x10²³ potassium atoms. 

Hence 1 g of  KCl will have 6.022x10²³/74.5 atoms, 
N'=8.08x10²¹ atoms. 

Given that count rate A =160 counts/s and t½ =1.3x10⁹ y
Decay constant 𝜆 =0.693/1.3x10⁹ per y
                     =5.33x10⁻¹⁰ per y
The number N of active atoms of ⁴⁰K is given as
 A =𝜆N
→N =A/𝜆
     =160*3600*24*365/5.33x10⁻¹⁰
     =9.465x10¹⁸
Hence the relative abundance of ⁴⁰K present in the sample is,
N/N' =9.465x10¹⁸/8.08x10²¹
        =0.0012
        =0.12%.
    



 

       37. ¹⁹⁷₈₀Hg decays to ¹⁹⁷₇₉Au through electron capture with a decay constant of 0.257 per day. (a) What other particle/particles are emitted in the decay? (b) Assume that the electron is captured from the K-shell. Use Moseley's law √v =a(Z-b) with a =4.95x10⁷ s⁻¹⁾² and b =1 to find the wavelength of the Kₐ X-ray emitted following the electron capture.  

ANSWER: (a) In the decay by electron capture, the nucleus captures one of its orbiting electrons and one of the protons converts into a neutron. The atomic number of the new nucleus is reduced by one and a new element is formed. In this process, a neutrino is emitted from the nucleus. 

(b) Assuming that an electron is captured from the K-shell, the frequency of the radiation of Kₐ X-ray is given from Moseley's law, 
   √𝜈 =a(Z -b)
      =4.95x10⁷*(79 -1)
 →𝜈 =1.49x10¹⁹

Hence the wavelength,
  =c/𝜈 
  =3x10⁸/1.49x10¹⁹ m
  =2.0x10⁻¹¹ m
  =20x10⁻¹² m
  =20 pm.    




 
       38. A radioactive isotope is being produced at a constant rate dN/dt =R in an experiment. The isotope has a half-life t½. Show that after a time t >> t½ the number of active nuclei will become constant. 

ANSWER: The rate of production of radioactive isotope =R =constant.

  In a very small time dt, isotopes produced =Rdt. If at t =0, there are N existing isotopes, then after time t remaining active nuclei. 
=Ne-𝜆t +∫(Rdt)e-𝜆t
=Ne-𝜆t +R∫e-𝜆t dt,     
{Between limits t =0 to t =t}
=Ne-𝜆t +(-R/𝜆)[e-𝜆t]₀t
=Ne-𝜆t +(R/𝜆)(1-e-𝜆t)
=R/𝜆 +(N -R/𝜆)e-𝜆t
If t >> t½, then e-𝜆t →0.
In this case, the number of active nuclei will be ≈R/𝜆.
=R*t½/0.693.  
It is a constant.  




 
       39. Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at t =0. Find the number of active nuclei at time t.  

ANSWER: Consider a very small interval of time dt just after t =0. The number of nuclei produced in this time =Rdt.
   We should keep in mind that as soon as radioactive isotopes are produced, they start decaying with their own decay constant. Hence after time t, the remaining active nuclei of Rdt will be 
=(Rdt)e-𝜆t
       Since the nuclei are being produced continuously, The number of active nuclei after time t
=∫(Rdt)e-𝜆t
=R∫e-𝜆t dt,     
{Between limits t =0 to t =t}
=(-R/𝜆)[e-𝜆t]₀t
=(-R/𝜆)(e-𝜆t -1)
=(R/𝜆)(1 -e-𝜆t).






 
       40. In an agricultural experiment, a solution containing 1 mole of radioactive material (t½ =14.3 days) was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruits. If the activity measured was 1 µCi, what percent of activity is transmitted from the root to the fruit in a steady state?   

ANSWER: One mole of active nuclei means the number of nuclei,
=6.022x10²³. 
  After 70 days (=2.9 days), the remaining active nuclei,
 N =Nₒ/(2t/t½)
   =6.022x10²³/(20.2)
   =5.24x10²³
 𝜆 =0.693/t½
    =0.693/14.3
    =0.048 per day
Activity A =𝜆N
→A =0.048*5.24x10²³ per day
       =2.52x10²² per day
       =2.91x10¹⁷ per second
 
  Fruits get 1µCi of nuclei which is equal to,
=3.7x10¹⁰*10⁻⁶
=3.7x10⁴ count per second. 
  Hence the percentage of activity transferred from root to fruits in a steady state,
=3.7x10⁴*100/2.91x10¹⁷
=1.27x10⁻¹¹%.   
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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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