2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER SCREENING 2010

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

    A) -13.6 eV                              

    B) -27.2 eV

    C) -54.4eV                               

    D) - 6.8 eV

    Correct Answer: A

    Solution :

    Energy E of an atom with principal quantum number n is given by \[E=\frac{-13.6}{{{n}^{2}}}{{Z}^{2}}\] for first excited state n = 2 and for \[H{{e}^{+}}Z=2\] \[\Rightarrow \]                               \[E=\frac{-13.6\times {{(2)}^{2}}}{{{(2)}^{2}}}\]                                 \[=-13.6\,eV\]


You need to login to perform this action.
You will be redirected in 3 sec spinner