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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

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

    A)  16/3R                                  

    B)  2R/16

    C)  3R/16                                  

    D)  4R/16

    Correct Answer: C

    Solution :

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


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