JEE Main & Advanced Physics Atomic Physics Question Bank Topic Test - Atomic Physics

  • 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 radiations (R = Rydberg's constant) will be

    A) \[\frac{16}{3R}\]

    B) \[\frac{2R}{16}\]

    C) \[\frac{3R}{16}\]

    D) \[\frac{4R}{16}\]

    Correct Answer: C

    Solution :

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


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