AFMC AFMC Solved Paper-2010

  • question_answer
    An electron jumps from the 4th orbit to 2nd orbit of hydrogen atom. Given the Rydberg 's constant R = 105 cm-1 the frequency in hertz of the emitted radiation will be

    A) \[\frac{3}{16}\times {{10}^{5}}\]                              

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

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

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

    Correct Answer: C

    Solution :

    Frequency, \[v=\frac{c}{\lambda }=c.R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] \[=3\times {{10}^{8}}\times {{10}^{7}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\] \[=\frac{9}{16}\times {{10}^{5}}\text{Hz}\]


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