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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    When Uranium is bombarded with neurons, undergoes fission. The fission reaction can be written as:   \[_{92}{{\bigcup }^{235}}{{+}_{0}}{{n}^{1}}\to 56\operatorname{B}{{a}^{141}}{{+}_{36}}{{\operatorname{Kr}}^{92}}+3x\,+Q(\text{energy)}\]                 where there particles named \[x\] are produced and energy Q is released. What is the name of the particle \[x\]?           JEE Main Online Paper (Held On 09 April 2013)        

    A)                 electron                                              

    B)                 \[\alpha \]-particle                                         

    C)                 neutron                              

    D)                        neutrino                

    Correct Answer: C

    Solution :

                    The fission of \[_{92}{{U}^{235}}\] is represented by                 \[_{92}{{U}^{235}}{{+}_{0}}{{n}^{1}}{{\to }_{56}}B{{a}^{141}}{{+}_{36}}K{{r}^{92}}{{+}_{3}}{{n}^{1}}+Q\]                 The name of the particle x is neutron \[{{(}_{0}}{{n}^{1}})\].                


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