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NEET AIPMT SOLVED PAPER SCREENING 2004

  • 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:

    A)                                                                                                                                                                 1.5eV           

    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,                  \[\Delta E={{E}_{3}}-{{E}_{2}}=-\frac{13.6}{9}-\left( -\frac{13.6}{4} \right)\]          \[=1.9\,\text{eV}\]        (approximately)


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