BVP Medical BVP 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. \[_{0}^{1}n\xrightarrow{{}}_{1}^{1}p{{+}_{-1}}{{e}^{0}}({{\beta }^{-}})\]


You need to login to perform this action.
You will be redirected in 3 sec spinner