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NEET Sample Paper NEET Sample Test Paper-68

  • 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

    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 \[227=y+4\times 4+0,\,\,y=\,\,211\] On equating atomic number \[96=y+2\times 4-5,\,\,y\,\,=\,\,93\]


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