Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydbergs constant\[R={{10}^{5}}c{{m}^{-1}},\]the frequency in Hz of the emitted radiation will be

    A)  \[\frac{3}{10}\times {{10}^{5}}\]       

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

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

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

    Correct Answer: C

    Solution :

     From Bohfs theory, frequency of incident radiation \[v=Rc\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[={{10}^{5}}\times {{10}^{2}}\times 3\times {{10}^{9}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[=3\times {{10}^{15}}\left( \frac{1}{4}-\frac{1}{16} \right)\] \[=3\times {{10}^{15}}\times \frac{3}{16}\] \[=\frac{9}{16}\times {{10}^{15}}Hz\]


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