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BHU PMT BHU PMT Solved Paper-2003

  • question_answer
    The number of\[\alpha \]and \[\beta \]particles emitted in the nuclear reaction\[_{90}T{{h}^{228}}{{\xrightarrow[{}]{{}}}_{83}}B{{i}^{212}}\]are:

    A) \[4\alpha \]and\[1\beta \]                          

    B) \[3\alpha \]and\[7\beta \]

    C) \[8\alpha \]and\[1\beta \]                          

    D) \[4\alpha \]and\[7\beta \]

    Correct Answer: A

    Solution :

                     Number of\[\alpha -\]particles \[=\frac{change\text{ }in\text{ }mass\text{ }number}{4}\]                                 \[=\frac{228-212}{4}\]                                 \[=\frac{16}{4}=4\] Number of\[\beta \]particles \[=2\alpha -\] (change in atomic number) \[=2\times 4-(90-83)\] \[=8-7=1\] \[\therefore \]no. of \[\alpha -\]particles emitted = 4 and\[\beta \]particles = 1


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