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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    An electron makes a transition from orbit\[n=4\]to be the orbit\[n=2\]of a hydrogen atom. The wave number of the emitted radiations (R = R vdberg's constant) will be

    A)  7/16 R           

    B)  3R/16

    C)  16/3 R           

    D)  9R/16

    Correct Answer: C

    Solution :

     \[\frac{1}{\lambda }=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right]=\frac{3R}{16}\] \[\therefore \] \[\lambda =\left( \frac{16}{3R} \right)\]


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