2. Questions Bank
  3. 12th Class
  4. Chemistry
  5. Nuclear Chemistry / नाभिकीय रसायन

12th Class Chemistry Nuclear Chemistry / नाभिकीय रसायन Question Bank Critical Thinking Nuclear chemistry

  • question_answer
    \[_{11}^{23}Na\] is the more stable isotope of Na. Find out the process by which \[_{11}^{24}Na\] can undergo radioactive decay  [IIT Screening 2003]

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

    B)            \[\alpha \]emission

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

    D)            K electron capture

    Correct Answer: A

    Solution :

           \[_{11}^{23}Na\to \frac{n}{p}\text{ratio}=12/11\]                    \[_{11}^{24}Na\to \frac{n}{p}\text{ratio}=23/11\]                    so decrease in \[\frac{n}{p}\]ratio gives out \[\beta \]-particle                    \[n\to p+e\,({{\beta }^{-}})\].


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