Q 1 :    

Which of the following nuclear fragments corresponding to nuclear fission between neutron (n01) and uranium isotope (U92235) is correct?        [2024]

  • B56144a+K3689r+3n01

     

  • B56144a+K3689r+4n01

     

  • S51153b+N4199b+3n01

     

  • X56140e+S3894r+3n01

     

(A)        After balancing mass number and atomic number,the original fission reaction is given by,

              U92235+n01B56144a+K3689r+3n01

 



Q 2 :    

In a hypothetical fission reaction

 

X23692Y14156+Z9236+3R

 

The identity of emitted particles (R) is                   [2024]

  • γ-radiations

     

  • Electron

     

  • Neutron

     

  • Proton

     

(C)          Z in LHS = 92

               Z in RHS = 56 + 36 = 92

               A in LHS = 236

               A in RHS = 141 + 92 = 233

 



Q 3 :    

The energy released in the fusion of 2 kg of hydrogen deep in the sun is EH and the energy released in the fission of 2 kg of U235 is EU. The ratio EHEU is approximately       

            

(Consider the fusion reaction as 4H11+2e-H24e+2v+6γ+26.7 MeV, energy released in the fission reaction of U235 is 200 MeV per fission nucleus  and NA=6.023×1023)                  [2024]

  • 9.13

     

  • 15.04

     

  • 7.62

     

  • 25.6

     

(C)            In each fusion reaction, 4H11 nucleus is used.

                  Energy released per Nuclei of H11=26.74 MeV

                           Energy released by 2 kg hydrogen,

                  EH=20001×NA×26.74MeV

                           Energy released by 2 Kg Uranium,

                   EU=2000235×NA×200MeV

                   So,EHEU=235×26.74×200=7.84

                       Approximately close to 7.62

 



Q 4 :    

The explosive in a Hydrogen bomb is a mixture of H2,H311 and Li63 in some condensed form. The chain reaction is given by 

 

Li6+n103He4+H312

 

H2+H31He4+n1021

 

During the explosion the energy released is approximately [Given : M(Li) = 6.01690 amu, M (H21) = 2.01471 amu, M(He42) = 4.00388 amu, and 1 amu = 931.5 MeV]                       [2024]

  • 28.12 MeV

     

  • 12.64 MeV

     

  • 16.48 MeV

     

  • 22.22 MeV

     

(D)            Li63+n10He42+H31

                  H21+H31He42+n10

                  Li63+H212(He4)2

                  Q=Δmc2

                  Q=[M(Li)+M(H12)-2×M(He4)]×931.5 MeV

                  Q=[6.01690+2.01471-2×4.00388]×931.5 MeV

                   Q22.22 MeV



Q 5 :    

The disintegration energy Q for the nuclear fission of U235C140e+Z94r+n is _________ MeV.

Given atomic masses of U235:235.0439 u; Ce:139.905 u Z94r:93.9063 u; n:1.0086 u, Value of c2=931 MeV.                [2024]



(208)            U235C140e+Z94r+n

                     Disintegration energy Q=Δmc2

                      Q=(mR-mP)·c2

                      mR=235.0439u

                     mP=139.9054u+93.9063u+1.0086u=234.8203u

                     Q=(235.0439u-234.8203u)c2=0.2236×931MeV

                        208.17MeV208MeV