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VMMC VMMC Medical Solved Paper-2011

  • question_answer
    An electron in a hydrogen atom makes a transition          \[{{n}_{1}}\to {{n}_{2}}\]where \[{{n}_{1}}\] and \[{{n}_{2}}\] are principal quantum numbers to be valid. The time period of the electron in initial state is eight times that in the final state. Then the ratio of \[{{n}_{1}}\] and \[{{n}_{2}}\] will be

    A)                 8 : 1                                       

    B)  4 : 1                      

    C)  2 : 1                      

    D)         1 : 2

    Correct Answer: C

    Solution :

                            Let T be the periodic time and r the radius of circular orbit, then \[{{T}^{2}}\propto {{r}^{3}}\] So,          \[{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2}}\propto {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}\] or            \[\frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{2/3}}\]                 \[={{(8)}^{2/3}}=4\] If n is principal quantum number, then \[r=\frac{{{\varepsilon }_{0}}{{h}^{2}}{{n}^{2}}}{\pi m{{e}^{2}}}\propto {{n}^{2}}\]                 So,          \[\frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{n}_{1}}}{{{n}_{2}}} \right)}^{2}}\] or            \[\frac{{{n}_{1}}}{{{n}_{2}}}=\sqrt{\frac{{{r}_{1}}}{{{r}_{2}}}}\]                 \[=\sqrt{\frac{4}{1}}=2:1\]          


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