Manipal Medical Manipal Medical Solved Paper-2008

  • question_answer
    \[^{23}Na\] is 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 :

     \[n/p\]ratio of\[_{13}^{24}Na=13/11\]ie, greater than unity. To achieve stability, it would tend to become unity. This can happen if n decreases or p increases. To do so, a neutron changes into proton and an electron(\[\beta -\]particle) which is emitted. \[_{0}^{1}n\xrightarrow[{}]{{}}\,_{1}^{1}P{{+}_{-1}}{{e}^{0}}({{\beta }^{-}})\]


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