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MGIMS WARDHA MGIMS WARDHA Solved Paper-2007

  • question_answer
    The ground state energy of H-atom is 13.6 eV. The energy needed to ionize H-atom from its second excited state is

    A) 1.51 eV                                

    B) 3.4 eV

    C) 13.6 eV                                

    D) 12.1 eV

    Correct Answer: A

    Solution :

    Second excited state corresponds to\[n=3.\] \[\therefore \] \[E=-\frac{13.6}{{{n}^{2}}}\] \[\Rightarrow \]               \[E=-\frac{13.6}{{{(3)}^{2}}}eV\] \[\Rightarrow \]               \[E=-\frac{13.6}{9}eV=-1.51eV\]


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