Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    The radius of the first Bohr orbit of hydrogen atom is\[0.529\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]. The radius of the third orbit of\[{{H}^{+}}\]will be

    A) \[8.46\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]                       

    B) \[0.705\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[1.59\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]       

    D)        \[4.79\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

    According to Bohr model, radius of hydrogen atom                 \[({{r}_{n}})=\frac{0.529\times {{n}^{2}}}{Z}\overset{\text{o}}{\mathop{\text{A}}}\,\] (where,\[n=\]number of orbit, \[Z=\]atomic number)                 \[{{r}_{3}}=\frac{0.529\times {{(3)}^{2}}}{1}=4.761\,\,\overset{\text{o}}{\mathop{\text{A}}}\,\]


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