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JCECE Medical JCECE Medical Solved Paper-2013

  • question_answer
    If the energy of hydrogen atom in the ground state is \[-13.6\text{ eV,}\] then energy of \[\text{H}{{\text{e}}^{+}}\] ion in first excited state will be

    A) \[6.8\,eV\]

    B)  \[-13.6\,eV\]

    C)  \[-27.2\,eV\]

    D)  \[-54.4\,eV\]

    Correct Answer: B

    Solution :

    \[{{E}_{H{{e}^{+}}}}=-\frac{{{Z}^{2}}}{{{n}^{2}}}\times 13.6\,eV\] For \[H{{e}^{+}}\] ion \[Z=2\] and for first excited state \[n=2\] \[\therefore \] \[{{E}_{H{{e}^{+}}}}=-\frac{{{2}^{2}}}{{{2}^{2}}}\times 13.6\,eV\] \[=-13.6\,eV\]


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