KVPY Sample Paper KVPY Stream-SX Model Paper-4

  • question_answer
    \[^{23}Na\]is the more stable isotope of \[Na\]. Find out the process by which \[_{^{11}}^{24}Na\] can undergo radioactive decay

    A) \[{{\beta }^{-}}\]-emission    

    B) \[\alpha \]-emission

    C) \[{{\beta }^{+}}\]-emission                  

    D) K-electron capture

    Correct Answer: A

    Solution :

    [a] In stable isotope of \[Na,\] there are 11 protons and 12 neutrons. In the given radioactive isotope of sodium \[(N{{a}^{24}}),\] there are 13 neutrons, one neutron is more than that required for stability. A neutron rich isotope always decay by \[{{\beta }^{-}}\text{-}\]emission as \[{{{}_{0}}^{{{n}^{1}}}}\xrightarrow{{}}{}_{-1}{{\beta }^{0}}+{}_{1}{{H}^{1}}\]        


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