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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    An electron makes transition from\[n=3\]to the orbit \[n=2\]of a hydrogen atom (ionization potential 13.6 eV). The energy of the photon emitted in the process is

    A)  1.89 eV               

    B)  2.55 eV

    C)  12.09 eV             

    D)  12.75 eV

    Correct Answer: C

    Solution :

                     Energy released \[=13.6\left[ \frac{1}{{{(1)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right]\] \[=12.09\text{ }eV\]


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