Saturday, September 2, 2023

H C Verma solutions, The Nucleus, Chapter-46, EXERCISES, Q11 to Q20, Concepts of Physics, Part-II

EXERCISES, Q11 to Q20


       11. A free neutron beta decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process. 

ANSWER: Given that, t½ =14 minutes 
                                 =14*60 s =840 s.
(a) The decay constant 𝜆 =0.693/t½
            →𝜆 =0.693/840 s⁻¹
                  =8.25x10⁻⁴ s⁻¹.

(b) In a beta decay of a free neutron, a beta particle which is an electron is emitted and the neutron converts to a proton. The energy liberated in this process will be the energy equivalent of the mass that disappeared in the process of decay.
ΔE =Δm*c²
     =[mₙ -(mₑ +mₚ)]c²
   =[1.008665-(0.0005486+1.007276)]*c² 
   =(0.0008404 u)*c² 
   =0.0008404*931 MeV 
   =0.782 MeV 
   =782 keV.  



 
       12. Complete the following decay schemes. 
(a) ²²⁶₈₈Ra → α⁺
(b) ¹⁹₈O → ¹⁹₉F⁺
(c) ²⁵₁₃Al → ²⁵₁₂Mg⁺.  

ANSWER: (a) With an alpha decay, The Ra atom will lose two protons and two neutrons in its nucleus. 
The number of protons now =88-2 =86
Total number of nucleons =226-4 =222
Since the number of protons in a nucleus is the atomic number that is related to a particular element, the new atom with 86 protons in its nucleus is Rn. 
   So the decay scheme is  as follows:
²²⁶₈₈Ra 
      ↓→α⁺   
²²²₈₆Rn 

(b) When an oxygen atom with 8 protons and 11 neutrons decays to an F atom with 9 protons and 10 neutrons, it is clear that in the nucleus a neutron has decayed into a proton and an electron. While this proton remains in the nucleus the electron is emitted as a beta-minus particle with an antineutrino. The decay scheme is as follows:
¹⁹₈O
    ↓→ e⁻ +𝜈̅ 
¹⁹₉F⁺

(c) Here an Al nucleus with 13 protons and 12 neutrons decays to an Mg⁺ nucleus having 12 protons and 13 neutrons. Clearly, the number of nucleons remains the same but a proton in the Al nucleus breaks into a neutron and a positron (e⁺). Also in this process, a neutrino is released that is emitted with the positron. 
   The decay scheme is as follows: 
²⁵₁₂Al
     ↓→e⁺ 𝜈
²⁵₁₃Mg⁺ 
 



 
       13. In the decay ⁶⁴Cu →⁶⁴Ni +e⁺ +𝜈, the maximum kinetic energy carried by the positron is found to be 0.650 MeV.
(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV? 
(b) What is the momentum of this neutrino in kg-m/s?
Use the formula applicable to a photon.   

ANSWER: (a) In beta decay, the released kinetic energy is shared by the beta particle and the neutrino/antineutrino. Here a positron has a maximum kinetic energy of 0.650 MeV, hence in other decays this much kinetic energy will be shared by the positron and neutrino. If the positron has a kinetic energy of 0.150 MeV, the neutrino will have a kinetic energy,
=0.650 -0.150 MeV =0.500 MeV.

(b) Momentum of this neutrino,
  p =E/c
     =0.500*10⁶*1.602x10⁻¹⁹/3x10⁸ kg-m/s
     =2.67x10⁻²² kg-m/s.
 


 

       14. Potassium-40 can decay in three modes. It can decay by ß⁻-emission, ß⁺-emission, or electron capture. (a) Write the equations showing the end products. (b) Find the Q-value in each of the three cases. Atomic masses of ⁴⁰₁₈Ar, ⁴⁰₁₉K, and ⁴⁰₂₀Ca are 39.9624 u, 39.9640 u, and 39.9626 u respectively. 

ANSWER: (a) Potassium has atomic number =19, hence there will be 19 protons and 21 electrons in the nucleus of Potassium-40. 
    For the decay by ß⁻-emission
In this decay, a neutron in the nucleus of the potassium converts into a proton and an electron. The proton remains in the nucleus while the electron emits as ß⁻-radiation with an antineutrino. The number of nucleons in the nucleus remains the same but now there are 20 protons and 20 neutrons. Its new atomic number is 20, so it is now a calcium atom. The equation for this beta decay may be written as follows:
⁴⁰₁₉K →⁴⁰₂₀Ca +e⁻ +𝜈̅

     For the decay by ß⁺-emission
 In this decay, a proton converts into a neutron and a positron. The positron (e⁺) gets emitted as ß⁺-radiation with a neutrino. The nucleus has now 18 protons and 22 neutrons, so its atomic number is 18, and its mass number remains 40. It is now an Ar atom. The equation for this ß⁺-decay may be written as follows:
⁴⁰₁₉K →⁴⁰₁₈Ar +e⁺ +𝜈

    For the decay by electron capture:
In this mode of decay, an orbiting electron of the atom is captured by a proton of the nucleus to convert into a neutron. A neutrino is produced in this process and emitted. Again the daughter nucleus has a total of 18 protons and 22 neutrons. Its atomic number is 18 and its mass number is 40. So it is now an Ar atom. The equation may be written as follows:
⁴⁰₁₉K +e⁻ →⁴⁰₁₈Ar +𝜈

(b) Q-values

Q-value for the beta-minus decay:
Q =Uᵢ -Uⱼ
   =[m(⁴⁰₁₉K) -m(⁴⁰₂₀Ca)]c² 
   =[(39.964 -39.9626) u]c²*931MeV/c²
   =1.3034 MeV.

Q-value for the ₑ⁺ (beta plus) decay:
Q =Uᵢ -Uⱼ
    =[m(⁴⁰₁₉K) -m(⁴⁰₁₈Ar) -2mₑ]c²
 =[(39.964-39.9624)c²*931 MeV/c²-2*511 keV] 
    =[1.4896 MeV -1022 keV
    =1489.6 -1022 keV 
    =467.6 keV
    =0.4676 MeV.

Q-value for the electron capture process:
Q =Uᵢ -Uⱼ
    =[m(⁴⁰₁₉K) -m(⁴⁰₁₈Ar) ]c²*931 MeV/c²
    =(39.964 -39.9624)*931 MeV
    =1.4896 MeV
    ≈1.490 MeV.
 




       15. Lithium (Z =3) has two stable isotopes ⁶Li and ⁷Li. When neutrons are bombarded on the Li sample, electrons and α-particles are ejected. Write down the nuclear processes taking place. 

ANSWER: When a neutron is absorbed by the nucleus of a ⁶Li atom,  it changes to another stable isotope ⁷Li. So the nuclear process is:
    ⁶₃Li +n → ⁷₃Li

Further absorption of neutron by the ⁷₃Li converts it into ⁸₃Li. So,
    ⁷₃Li +n → ⁸₃Li

Since the neutron in the new nucleus has a greater proportion, this atom becomes unstable. To get stability, one of the neutrons in the nucleus converts to a proton and an electron. The converted proton remains in the nucleus while the electron is emitted as ß-radiation along with an antineutrino. The atomic number Z of the daughter nucleus is 4, hence it is a Be nucleus now. The process can be written as follows:
  ⁸₃Li →⁸₄Be +e⁻ +𝜈̅

To achieve further stability, the Be nucleus decays into two alpha particles as follows:
   ⁸₄Be →⁴₂He +⁴₂He.
 



  
       16. The masses of ¹¹C and ¹¹B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the ß⁺-decay of ¹¹C to ¹¹B. 

ANSWER: The energy released in this process is the Q-value of the process.
Q-value =[m(¹¹C) -m(¹¹B) -2mₑ]c²
      =[(11.0114 -11.0093 -2*0.0005486)u]c²*931 Mev/c² 
  =(0.0021 -0.0010972)*931 MeV 
  =0.0010028*931 MeV 
  =0.9336 MeV 
  =933.6 keV. 
This released energy is shared by the emitted positron and the neutrino. The maximum energy a positron will have when the neutrino shares zero energy. Then whole of the energy will be taken by the positron. Hence the maximum energy a positron can have =933.6 keV.  



 
       17. ²²⁸Th emits an alpha particle to reduce ²²⁴Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay: 
         ²²⁸Th → ²²⁴Ra* + α
       ²²⁴Ra* → ²²⁴Ra + 𝜸 (217 keV).
The atomic mass of ²²⁸Th is 228.028726 u, that of ²²⁴Ra is 224.020196 and that of ⁴He is 4.00260 u. 

ANSWER: The reduction in the rest mass of the products appears as the kinetic energy of the alpha particle. 
ΔM =m(²²⁸Th) -{m(²²⁴Ra*) +m(⁴He)}
      =228.028726 u-(224.020196 u+217 keV+4.00260) u
      =0.00593 u -217 keV
The energy equivalent of 0.00593 u is,
  =0.00593c²*931 MeV/c²
  =5.52083 Mev        
Hence the kinetic energy of the alpha particle,
ΔE =ΔM (interms of energy)
      =5.52083 MeV -217 keV
      =5.52083 -0.217 MeV
      =5.304 MeV.
 



 
       18. Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme: 
     ¹²N → ¹²C* +e⁺ +𝜈 
    ¹²C* → ¹²C +𝛾 (4.43 MeV) 
The atomic mass of ¹²N is 12.018613 u. 

ANSWER: m(¹²C*) =m(¹²C) +4.43 MeV
                          =12*931 MeV +4.43 MeV 
                          =11176.43 MeV 
m(¹²N) =12.018613*931 MeV 
           =11189.329 MeV.
Q-value of the decay scheme,
=m(¹²N)-m(¹²C*) -2mₑ 
=(11189.329 -11176.43) MeV -2*511 keV
=12.899 -1.022 MeV 
=11.877 MeV
≈11.88 MeV. 
This much energy will be shared by the ß-particle and the neutrino. For the maximum kinetic energy of the ß-particle, we take the energy of the neutrino =zero. The total of the above energy released will be taken by the ß-particle.  




 
       19. The decay constant of ¹⁹⁷₈₀Hg (electron capture to ¹⁹⁷₇₉Au) is 1.8x10⁻⁴ s⁻¹. 
(a) What is the half-life?
(b) What is the average life?
(c) How much time will it take to convert 25% of this isotope of mercury into gold?    

ANSWER: (a) Decay constant given,
 λ =1.8x10⁻⁴ s⁻¹
Hence the half-life,
t½ =0.693/λ
     =0.693/(1.8x10⁻⁴ s⁻¹)
     =3850 s
     ≈64 min.

(b) Average life = 1/λ
             =1/(1.8x10⁻⁴ s⁻¹)
             =5555 s
             ≈92 min.

(c) 25% of the original isotope decays means only 75% of the original nuclei remain active after time t in the above decay.
N/Nₒ =3/4.
Let the time taken in this decay =t, then
N/Nₒ =e-λt =3/4
→-λt =ln (3/4)
→t = -0.2877/(-1.8x10⁻⁴ s⁻¹)
     ≈1600 s.
 



 
       20. The half-life of ¹⁹⁸Au is 2.7 days. (a) Find the activity of a sample containing 1.0 µg of ¹⁹⁸Au. (b) What will be the activity after 7 days? Take the atomic weight of ¹⁹⁸Au to be 198 g/mol. 

ANSWER: 198 g of this Au will contain a 6.02x10²³ number of atoms. 
Number of atoms in 1.0µg,
N =1.0x10⁻⁶*6.02x10²³/198
    =3.04x10¹⁵
Half-life (given) =2.7 days =2.354x10⁵ s
Hence decay constant,
λ =0.693/2.354x10⁵ s⁻¹
   =2.944x10⁻⁶ s⁻¹
(a) Hence the activity of the sample,
Aₒ =λN
   =2.944x10⁻⁶*3.04x10¹⁵ disintegration/s
   =8.95x10⁹ Bq
   =8.95x10⁹/3.7x10¹⁰ Ci
   =0.242 Ci

(b) Activity at time t is,
 A =Aₒ/2t/t½
    =0.242/(27/2.7)   
    =0.040 Ci. 
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