CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • 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)        \[4.29\overset{\text{o}}{\mathop{\text{A}}}\,\]

    E) \[2.38\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: D

    Solution :

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


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