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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    In a certain hypothetical radioactive decay process, species A decays into species B and species B decays into species C according to the reactions: \[A\xrightarrow{{}}2B+particles+energy\] \[B\xrightarrow{{}}3C+particles+energy\] The decay constant for species is \[{{\lambda }_{1}}=1{{\sec }^{-1}}\] and that for the species B is \[{{\lambda }_{2}}=100{{\sec }^{-1}}.\] Initially \[{{10}^{4}}\]moles of species of A were present while there was none of B and C. It was found that species B reaches its maximum number of moles of B.

    A)  1

    B)  2

    C)  5

    D)  4

    Correct Answer: B

    Solution :

    [b] \[\frac{d{{N}_{A}}}{dt}=-{{\lambda }_{1}}{{N}_{A}},\,\,\,\frac{d{{N}_{B}}}{dt}=2{{\lambda }_{1}}{{N}_{A}}-{{\lambda }_{2}}{{N}_{B}},\] \[{{N}_{B}}=\] maximum \[\Rightarrow \,\,\frac{d{{N}_{B}}}{dt}=0\] \[\Rightarrow \,\,\,\,\,\,\,\,2{{\lambda }_{1}}{{N}_{A}}={{\lambda }_{2}}{{N}_{{{B}_{\max }}}}\Rightarrow {{N}_{{{B}_{\max }}}}=\frac{2{{\lambda }_{1}}}{{{\lambda }_{2}}}{{N}_{A}}\] \[\Rightarrow \,\,\,\,\,\,\,\,{{N}_{{{B}_{\max }}}}=\frac{2{{\lambda }_{1}}}{{{\lambda }_{2}}}{{N}_{0}}{{e}^{-{{\lambda }_{{{1}^{t}}=2}}}}\]


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