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

  • question_answer
    Three \[\alpha -\]particles and one \[\beta -\]particle decaying takes place in series from an isotope \[_{88}R{{a}^{238}}\]. Finally the isotope obtained will be                                                          [CPMT 1989; DCE 2000]

    A)            \[_{84}{{X}^{220}}\]          

    B)            \[_{86}{{X}^{222}}\]

    C)            \[_{83}{{X}^{224}}\]          

    D)            \[_{83}{{X}^{215}}\]

    Correct Answer: C

    Solution :

                       By using \[{{n}_{\alpha }}=\frac{A-A'}{4}\] and \[{{n}_{\beta }}=2{{n}_{\alpha }}-Z+Z'\] Þ \[A'=A-4{{n}_{\alpha }}=236-4\times 3=224\] and \[Z'=({{n}_{\beta }}-2{{n}_{\alpha }}+Z)=(1-2\times 3+88)=83\]


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