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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2014

  • question_answer
    The wavelength of radiation emitted is \[\frac{27}{20}{{\lambda }_{0}}\] when an electron in hydrogen atom jumps from 3rd to 2nd orbit. If in the hydrogen atom itself, the electron jumps from fourth orbit to second orbit, then wavelength of emitted radiation will be

    A) \[\frac{20}{27}{{\lambda }_{0}}\]                             

    B) \[\frac{16}{25}{{\lambda }_{0}}\]   

    C) \[3\Omega \]                    

    D) \[4\Omega \]

    Correct Answer: C

    Solution :

                    From the relation \[\frac{2\omega }{3}\]              ... (i) Also,      \[\mu =\frac{3}{2}\] \[\mu =\frac{4}{3}\] From Eqs. (i) and (ii), we get \[{{\sin }^{-1}}\left( \frac{9}{8} \right)\] \[{{45}^{o}}\]


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