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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    The natural boron of atomic weight 10.81 is found to have two isotopes \[{{B}^{10}}\] and \[{{B}^{11}}\]. The ratio of abundance of isotopes in natural boron should be

    A)  \[11:10\]                            

    B)  \[81:19\]            

    C)         \[10:11\]            

    D)         \[15:16\]

    E)  \[19:81\]

    Correct Answer: E

    Solution :

    Let abundance of\[{{B}^{10}}\]be m% . So, abundance of \[{{B}^{11}}=(100-m)%\] \[\therefore \]  \[10.81=\frac{(10\times m)+11(100-m)}{100}\] Or           \[1081=10m+1100-11m\] Or           \[m=19\] \[\therefore \] Ratio of abundances \[=\frac{19}{100-19}=\frac{19}{81}\]


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