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Manipal Medical Manipal Medical Solved Paper-2007

  • 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)  1.9eV

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


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