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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    In a radioactive decay chain, the initial nucleus is \[_{90}^{232}Th.\]At the end there are \[6\alpha -\]particles and \[4\beta -\]particles which are emitted. it the end nucleus is \[_{Z}^{A}X,\] A and Z are given by [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[A=202;\text{ }Z=80\]                         

    B) \[A=200;\text{ }Z=81\]

    C) \[A=208;\text{ }Z=80\]   

    D)               \[A=208;\text{ }Z=82\]

    Correct Answer: D

    Solution :

    The mass number A of the end nucleus is \[232-\left( 6\times 4 \right)=208\]                           The atomic number Z of \[_{Z}^{A}X\]by conservation of charge is \[90-\left( 6\times 2-4\times 1 \right)=82\]


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