2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-13

  • 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)  -54.4 eV                      

    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