BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    If an electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom, then the frequency of emitted radiation in the hertz will be (Take Rydbergs constant,\[R={{10}^{5}}c{{m}^{-1}}\])

    A) \[\frac{3}{4}\times {{10}^{15}}\]              

    B)         \[\frac{3}{16}\times {{10}^{15}}\]

    C)  \[\frac{3}{16}\times {{10}^{15}}\]           

    D)         \[\frac{9}{16}\times {{10}^{15}}\]                           

    Correct Answer: D

    Solution :

    Using the relation, \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[={{10}^{-5}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[={{10}^{-5}}\times \left( \frac{1}{4}-\frac{1}{16} \right)\] \[\lambda =\frac{16}{3}\times {{10}^{-5}}\,cm\]                 \[\therefore \]   Frequency \[\eta =\frac{c}{\lambda }\]                                 \[=\frac{3\times {{10}^{10}}}{\frac{16}{3}\times {{10}^{-5}}}\]                                 \[=\frac{9}{16}\times {{10}^{15}}\,Hz\] 


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