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

  • question_answer
    When an electron in hydrogen atom is excited, from its 4th to 5th stationary orbit, the change in angular momentum of electron is (Planck’s constant: \[h=6.6\times {{10}^{-34}}J\text{-s}\]

    A) \[4.16\times {{10}^{-34}}\,J\text{-}s\]

    B) \[3.32\times {{10}^{-34}}\,J\text{-}s\]

    C) \[1.05\times {{10}^{-34}}\,J\text{-}s\]

    D) \[2.08\times {{10}^{-34}}\,J\text{-}s\]

    Correct Answer: C

    Solution :

    [c]         Change in the angular momentum
    \[\Delta L={{L}_{2}}-{{L}_{1}}=\frac{{{n}_{2}}h}{2\pi }-\frac{{{n}_{1}}h}{2\pi }\]\[\Rightarrow \Delta L=\frac{h}{2\pi }({{n}_{2}}-{{n}_{1}})\]
    \[=\frac{6.6\times {{10}^{-34}}}{2\times 3.14}(5-4)\]\[=1.05\times {{10}^{-34}}J\text{-}S\]


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