CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    Radius of the first orbit of the electron in a hydrogen atom is \[0.53\overset{\text{o}}{\mathop{\text{A}}}\,\]. So, the radius of the third orbit will be:

    A) \[2.12\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[4.77\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C)        \[1.06\overset{\text{o}}{\mathop{\text{A}}}\,\]

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

    E)  \[0.18\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: B

    Solution :

    \[{{r}_{n}}\propto {{n}^{2}}\Rightarrow \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{{{1}^{2}}}{{{3}^{2}}}\] \[\Rightarrow \,{{r}_{2}}=9{{r}_{1}}=9\times 0.53=4.77\overset{\text{o}}{\mathop{\text{A}}}\,\]


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