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NEET Sample Paper NEET Sample Test Paper-56

  • question_answer
    If the electron falls from n = 3 to n = 2 in H-atom, then emitted energy is:

    A) 10.2ev              

    B) 12.09ev

    C) 1.9ev                           

    D) 0.65ev

    Correct Answer: C

    Solution :

    [c] \[1.9\,eV\] \[{{E}_{n}}=\frac{-\,13.6}{{{n}^{2}}}eV\] (energy of electron in nth orbit) \[{{E}_{3}}-{{E}_{2}}=13.6\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{3}^{2}}} \right)eV\] \[13.6\left( \frac{1}{4}-\frac{1}{9} \right)=13.6\left( \frac{9-4}{36} \right)1.9eV\]


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