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BCECE Engineering BCECE Engineering Solved Paper-2011

  • question_answer
    If\[\left( \frac{0.51\times {{10}^{-10}}}{4} \right)\]m, is the radius of smallest electron orbit in hydrogen like atom, then this atom is

    A)  H-atom                               

    B)  \[H{{e}^{+}}\]

    C)  \[L{{i}^{2+}}\]                  

    D)         \[B{{e}^{3+}}\]

    Correct Answer: D

    Solution :

    Radius of hydrogen like atom \[\Rightarrow \]               \[r_{n}^{Z}=\frac{{{n}^{2}}}{Z}{{r}_{0}}\] where, \[{{r}_{0}}=0.51\times {{10}^{-10}}\,m\] and        \[r_{n}^{Z}=\frac{0.51\,\times {{10}^{-10}}}{4}m\] At ground state \[n=1,\]therefore \[Z=4\] Hence, it is berillium \[(B{{e}^{3+}}),N=\frac{n(n-1)}{2}\]


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