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NEET Physics Atomic Physics NEET PYQ-Atomic Physics

  • question_answer
    Energy E of a hydrogen atom with principal quantum number n is given by \[E=\frac{-13.6}{{{n}^{2}}}eV\]. The energy of a photon ejected when the electron jumps from \[n=3\] state to \[n=2\] state of hydrogen, is approximately:                                                                                                                                      [AIPMT (S) 2004]

    A)  1.5 eV             

    B)       0.85 eV

    C) 3.4 eV 

    D)  1.9 eV

    Correct Answer: D

    Solution :

    Given:  \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\]
    Energy of photon ejected when electron jumps
    from\[n=3\] state to \[n=2\] state is given by
    \[\Delta E={{E}_{3}}-{{E}_{2}}\]
    \[\therefore \]      \[{{E}_{3}}=-\frac{13.6}{{{(3)}^{2}}}eV=-\frac{13.6}{9}eV\]
    \[{{E}_{2}}=-\frac{13.6}{{{(2)}^{2}}}eV=-\frac{13.6}{4}eV\]
    So,       \[\Delta E={{E}_{3}}-{{E}_{2}}=-\frac{13.6}{9}-\left( -\frac{13.6}{4} \right)\]
    \[=1.9\,eV\] (approximately)


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