2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A nucleus with \[Z=92\] emits the following in a sequence: \[\alpha ,\alpha ,{{\beta }^{-}},{{\beta }^{-}},\alpha ,\alpha ,\alpha ,\alpha ;{{\beta }^{-}},{{\beta }^{-}},\alpha ,{{\beta }^{+}},{{\beta }^{+}},\alpha \]. The Z of the resulting nucleus is     AIEEE  Solved  Paper-2003

    A) 76          

    B)       78       

    C)       82          

    D)       74

    Correct Answer: B

    Solution :

                    Since, 8 \[\alpha \]-particles, \[4{{\beta }^{-}}\] -particles and \[2{{\beta }^{+}}\] -particles are emitted, so new atomic number                     \[Z'=Z-8\times 2+4\times 1-2\times 1\]                                 \[=92-16+4-2\]                                 \[=92-14=78\]   


You need to login to perform this action.
You will be redirected in 3 sec spinner