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JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Self Evaluation Test - Nuclei

  • question_answer
    Atomic weight of boron is 10.81 and it has two isotopes \[_{5}{{B}^{10}}\] and\[_{5}{{B}^{11}}\]. Then ratio of \[_{5}{{B}^{10}}{{:}_{5}}{{B}^{11}}\] in nature would be

    A)  19 : 81

    B)  10 : 11

    C)  15 : 16

    D)         81 : 19

    Correct Answer: A

    Solution :

    [a] Let the percentage of \[{{B}^{10}}\] atoms be x, then average atomic weight \[=\frac{10x+11(100-x)}{100}=10.81\Rightarrow x=19\] \[\therefore \,\,\,\,\,\,\,\,\,\frac{{{N}_{{{B}^{10}}}}}{{{N}_{{{B}^{11}}}}}=\frac{19}{81}\]


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