VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    A radioactive element X emits \[3\alpha \],\[1\beta \] and \[1\gamma \]-particles and forms \[_{76}{{Y}^{235}}\]. Element X is

    A)  \[_{81}{{X}^{247}}\]         

    B)  \[_{80}{{X}^{247}}\]

    C)  \[_{81}{{X}^{246}}\]       

    D)  \[_{80}{{X}^{246}}\]

    Correct Answer: A

    Solution :

     The complete nuclear reaction of the given transformation is \[_{Z}{{X}^{A}}{{\xrightarrow{{}}}_{76}}{{Y}^{235}}+\underset{\alpha -particle}{\mathop{{{3}_{Z}}H{{e}^{4}}}}\,+\underset{\beta -particle}{\mathop{_{-1}{{e}^{0}}}}\,\]         \[+\underset{\gamma -particle}{\mathop{\gamma }}\,\] On comparing mass number, we get \[A=235+12+0\] \[=247\] On comparing atomic number, we get \[Z=76+6-1\] \[=81\] Thus, element X is \[_{81}{{X}^{247}}\].


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