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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    Hydrogen atom from excited state comes to the ground stage by emitting a photon of wavelength \[\lambda \]. If R is the Rydberg constant, the principal quantum number n of the excited state

    A)  \[\sqrt{\frac{\lambda R}{\lambda R-1}}\]                            

    B) \[\sqrt{\frac{\lambda }{\lambda R-1}}\]

    C) \[\sqrt{\frac{\lambda {{R}^{2}}}{\lambda R-1}}\]              

    D)        \[\sqrt{\frac{\lambda R}{\lambda -1}}\]

    Correct Answer: A

    Solution :

    Here,     \[{{n}_{f}}=1,\,\,{{n}_{i}}=n\] \[\therefore \]\[\frac{1}{\lambda }=R\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{n}_{2}}} \right)\Rightarrow \frac{1}{\lambda }=R\left( 1-\frac{1}{{{n}^{2}}} \right)\]   ... (i) or            \[\frac{1}{\lambda R}=1-\frac{1}{{{n}^{2}}}\]or\[\frac{1}{{{n}^{2}}}=1-\frac{1}{\lambda R}\] or            \[n=\sqrt{\frac{\lambda R}{\lambda R-1}}\]


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