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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    The binding energy of the electron in a hydrogen atom is13.6 eV, the energy required to remove the electron from the first excited state of \[\text{L}{{\text{i}}^{\text{++}}}\] is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 122.4 eV                              

    B) 30.6 eV

    C) 13.6 eV                                

    D) 3.4 eV

    Correct Answer: B

    Solution :

                    For first excited state, n = 2 and for \[L{{i}^{++}}Z=3\] \[{{E}_{n}}=\frac{13.6}{{{n}^{2}}}\times {{Z}^{2}}=\frac{13.6}{4}\times 9=30.6eV\]


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