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JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

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    An electron jumps from the 4th orbit to the 2nd orbit of hydrogen atom. Given the Rydberg's constant \[R={{10}^{5}}c{{m}^{-1}}\]. The frequency in Hz of the emitted radiation will be                                               [CPMT 1976]

    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 :

                       \[\frac{1}{\lambda }=R\,\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\Rightarrow \lambda =\frac{16}{3R}=\frac{16}{3}\times {{10}^{-5}}cm\]            Frequency \[n=\frac{c}{\lambda }=\frac{3\times {{10}^{10}}}{\frac{16}{3}\times {{10}^{-5}}}=\frac{9}{16}\times {{10}^{15}}Hz\]


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