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JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

  • question_answer
    The radius of hydrogen atom in its ground state is \[5.3\times {{10}^{-11}}m\]. After collision with an electron it is found to have a radius of \[21.2\times {{10}^{-11}}m\]. What is the principal quantum number n of the final state of the atom [CBSE PMT 1994; CPMT 2001; MH CET 2000]

    A)            n = 4                                        

    B)            n = 2

    C)            n = 16                                      

    D)            n = 3

    Correct Answer: B

    Solution :

                       \[r\propto {{n}^{2}}\ i.e.\frac{{{r}_{f}}}{{{r}_{i}}}={{\left( \frac{{{n}_{f}}}{{{n}_{i}}} \right)}^{2}}\] Þ \[\frac{21.2\times {{10}^{-11}}}{5.3\times {{10}^{-11}}}={{\left( \frac{n}{1} \right)}^{2}}\] Þ \[{{n}^{2}}=4\] Þ n = 2


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