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JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Nucleus (Stability and Reaction)

  • question_answer
    An element \[_{96}{{X}^{227}}\] emits \[4\alpha \] and \[5\beta \] particles to form new element Y. Then atomic number and mass number of Y are                                           [MH CET 2002]

    A)                 93; 211

    B)                 211; 93

    C)                 212; 88 

    D)                 88; 211

    Correct Answer: A

    Solution :

           \[_{96}{{X}^{227}}\to Y+4\alpha +5\beta \]                    On equating mass number                    227 = y + 4 × 4 + 0,  \[y=211\]                    On equating atomic number                                 96 = y + 2 × 4 ? 5,  y = 93.


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