VMMC VMMC Medical Solved Paper-2008

  • question_answer
    \[{{\,}^{\text{23}}}\text{Na}\] is more stable isotope of Na. Find out  the process by which \[_{\text{11}}^{\text{24}}\text{Na}\] 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 \[_{\text{13}}^{\text{24}}\text{Na =13/11}\] i.e., 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 (\[-\]particle) which is emitted. \[_{0}^{1}n\xrightarrow{{}}_{1}^{1}p{{+}_{-1}}{{e}^{0}}({{\beta }^{-}})\]


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