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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength \[\lambda ,\]energy                                                [JEE Main Held on 12-4-2019 Morning] \[E=\frac{1240eV}{\lambda (in\,nm)}:\]

    A) n = 5              

    B) n = 4

    C) n = 6              

    D) n = 7

    Correct Answer: A

    Solution :

    \[\frac{1}{\lambda }=R\left( \frac{1}{{{m}^{2}}}-\frac{1}{{{n}^{2}}} \right){{z}^{2}}\]           \[\frac{1}{1085}=R\left( \frac{1}{{{m}^{2}}}-\frac{1}{{{n}^{2}}} \right){{2}^{2}}\]           \[\frac{1}{304}=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{m}^{2}}} \right){{2}^{2}}\]           \[\therefore m=2\]           \[\therefore n=5\]


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