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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Radioactivity

  • question_answer
    What is the respective number of \[\alpha \]and \[\beta \] particles emitted in the following radioactive decay            \[_{90}{{X}^{200}}{{\to }_{80}}{{Y}^{168}}\] [CBSE PMT 1995; Pb. PMT 2004]

    A)            6 and 8                                    

    B)            8 and 8

    C)            6 and 6                                    

    D)            8 and 6

    Correct Answer: D

    Solution :

                       \[{{n}_{\alpha }}=\frac{A-A'}{4}=\frac{200-168}{4}=8\] \[{{n}_{\beta }}=2{{n}_{a}}-Z+Z'=2\times 8-90+80=6\]


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