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JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    The energy of \[H{{e}^{+}}\] in the ground state is -54.4 eV, then the energy of \[L{{i}^{++}}\] in the first excited state will be

    A)  -30.6 eV

    B)  27.2 eV

    C)  -13.6 eV

    D)  - 27.2 eV

    Correct Answer: A

    Solution :

    [a] Energy of electron in nth orbit is \[{{E}_{n}}=-\left( Rch \right)\frac{{{Z}^{2}}}{{{n}^{2}}}=-54.4eV\] For \[H{{e}^{+}}\] is ground state \[{{E}_{1}}=-(Rch)\frac{{{(2)}^{2}}}{{{(1)}^{2}}}=-\,54.4\,\Rightarrow \,Rch\,=13.6\] \[\therefore \] For \[L{{i}^{++}}\] in first excited state (n=2) \[E=-13.6\times \frac{{{\left( 3 \right)}^{2}}}{{{\left( 2 \right)}^{2}}}=-30.6eV\]


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