PROBLEM ON NUCLEAR PHYSICS

                  

Q/1.  The measured mass of deuteron atom(² ₁H) , Hydrogen atom(¹ ₁H)  , proton and neutron is 2.01649 u ,1.00782 u , 1.00727 u and 1.00866 u . Find the binding energy of the deuteron nucleus (unit MeV /nucleon).

 Ans:  Here A=2, Z=1, N=1  

B E = [ZM_H + NM_N + M(¹₁H)]✖ 931.5 MeV

         = [1✖ 1.00782 +1✖ 1.00866 -2.01649] ✖931.5 MeV

          = [0.00238]✖ 931.5 MeV

          =2.224 MeV

 

Q/2. The binding energy of the neon isotope (²⁰₁₀Ne) is 160.647 MeV. Find its atomic mass?

 Ans: Here A=10, Z=10, N=10  

             M (^A _ZX) = [ZM_H + NM_N] – B/(931.5 MeV/u)

             M (^A _ZNe) =[10 ✖1.00782 + 10 ✖ 1.00866] -160.647 /931.5 MeV/u

                               = 19.992u

 

Q/3.  (a) Find the energy needed to remove a neutron from the nucleus of the calcium isotope         (⁴²₂₀Ca) .

(b) Find the energy needed to remove a proton from this nucleus.

 (c) Why are these energies different?

     Given: Atomic masses of ⁴²₂₀Ca = 41.958622u, ⁴¹₂₀Ca= 40.962278u, ⁴¹ ₁₉K = 40.961825u, and mass of ¹₀n =1.008665u, ¹₁P =1.007276u .

Ans: (a) ⁴²₂₀Ca ➡ ⁴¹₂₀Ca +¹₀n ;

 Total mass of the  ⁴¹₂₀Ca & ¹₀n = 41.970943u

 Mass defect Dm=41.970943 -41.958622= 0.012321u

     So, B.E. of missing neutron = Dm ✖931.5MeV = 11.48MeV

   (b) ⁴²₂₀Ca ➡ ⁴¹ ₁₉K +¹₁P;

Total mass of the ⁴¹ ₁₉K & ¹₁P41.919101u

    Mass defect Dm=  41.919101 - 41.958622= 0.010479u

      So, B.E. of missing protron =  D m✖ 931.5MeV = 10.27MeV 

 (c) The neutron was acted upon only by attractive nuclear forces whereas the proton was also acted upon by repulsive electric forces that decrease its binding energy.

 

 Q/4.  Half life of P is 14.3 days. If you have 1.00 g of P today, then what would be the amount remaining in 10 days?

Ans:   t_1/2 =0.693/λ

         λ= 0.693/14.3= 0.04847 per day

    N= N₀e^-λt   ⇒N = (1)e^-0.04847*10

         ⇒N = 0.616g

                                        

Q/5.  A radioactive nucleus has a half life of 100 years. If the number of nuclei t = 0 is N₀

 , then find the number of nuclei that have decayed in 300 years?

 Ans :   Number of nuclei present after 300 year  N=N₀(1/2)^(T/T1/2)   

             N' = N₀ - N₀ (1/2)³ = (7/8) N₀  

 

   Q/6. The atomic ratio between the uranium isotopes  ²³⁸U and ²³⁴U in a mineral sample is found to be 1.8 ´ 10⁴ . The half  life of  ²³⁸U is 4.5 ´10⁹ years, then find the half life of ²³⁴U .

    Ans: N_Aλ_A=N_B λ_B

        Þ N_A/N_B =λ_A/λ_B =T_1/2A/T_1/2B                     

       Þ T_1/2B  = (N_A/N_B ) T_1/2A  =4.5´10⁹ /1.8 ´ 10⁴ = 2.5 ´ 10⁵

 

Q/7: A radioactive sample contains 1.00 mg of radon ²²²Rn, whose atomic mass is 222 u . The half life of the radon is3.8 day . Then find the activity of the radon?

 

 Ans: Decay constant    λ = 0.693/ T_1/2  = 0.693/(3.8✖24✖60✖60)=2.1✖10⁵years

          Number of atoms in 1.00 mg is  N=  1✖10^-6kg/{(222u)✖1.66✖10^-27kg/u}

          N= [ 1✖10^-6kg/{(222u)✖1.66✖10^-27kg/u}]✖6.023✖10²³ = 2.7✖10¹⁸atoms

         Hence, activity R= Nλ =2.1✖10^-6 ✖2.7✖10¹⁸ =5.7✖10¹² decay/sec