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MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    When a radioactive isotope\[_{88}R{{a}^{228}}\]decay in series by the emission of 3a -particles and a \[\beta -\]particle, the isotope finally formed is

    A)  \[_{82}{{X}^{216}}\]                      

    B)  \[_{83}{{X}^{215}}\] 

    C)  \[_{88}{{X}^{216}}\]                      

    D)  \[_{83}{{X}^{216}}\]

    Correct Answer: D

    Solution :

                    \[_{88}R{{a}^{228}}{{\xrightarrow[{}]{3\alpha }}_{82}}{{Y}^{216}}{{\xrightarrow[{}]{1\beta }}_{83}}{{X}^{216}}\]


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