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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    If the binding energy of the electron in a hydrogen atom is \[13.6\] eV, the energy required to remove the electron from the first excited state of \[L{{i}^{2+}}\] is     AIEEE  Solved  Paper-2003

    A) \[30.6\] eV

    B) \[13.6\] eV                         

    C) \[3.4\] eV                           

    D) \[122.4\] eV

    Correct Answer: A

    Solution :

                   Orbital energy, \[E=-{{Z}^{2}}\frac{13.6}{{{n}^{2}}}eV\] For first excited state,                     \[{{E}_{2}}=-{{3}^{2}}\times \frac{13.6}{4}\]                     = - 30.6 eV lonisation energy for first excited state of \[L{{i}^{2+}}\] is 30.6 eV.         


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