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

  • question_answer
    The energy of an electron in first Bohr orbit of H-atom is -13.6 eV. The energy value of electron in the excited state of\[L{{i}^{2+}}\]is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 27.2 eV                                

    B) 30.6 eV

    C) 30.6 eV                                

    D) 27.2 eV

    Correct Answer: C

    Solution :

                    For \[L{{i}^{2+}}\]ion \[E=-13.6\times \frac{{{Z}^{2}}}{{{n}^{2}}}\] \[=-13.6\times \frac{{{(3)}^{2}}}{{{(2)}^{2}}}\] \[=\frac{-13.6\times 9}{4}=-30.6eV\]


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