JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

  • question_answer
    An electron jumps from 5th orbit to 4th orbit of hydrogen atom. Taking the Rydberg constant as \[{{10}^{7}}\] per metre. What will be the frequency of radiation emitted [Pb. PMT 2001]

    A)            \[6.75\times {{10}^{12}}\,Hz\]                                         

    B)            \[6.75\times {{10}^{14}}\,Hz\]

    C)            \[6.75\times {{10}^{13}}\,Hz\]                                         

    D)            None of these

    Correct Answer: C

    Solution :

                       By using \[\nu =RC\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\]            \[\Rightarrow \nu ={{10}^{7}}\times (3\times {{10}^{8}})\,\left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{5}^{2}}} \right]\]= 6.75 ´ 1013 Hz


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