JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    Consider 3rd orbit of \[H{{e}^{+}}\](Helium), using non-relativistic approach, the speed of electron in this orbit will be [given \[K=9\times {{10}^{9}}\]constant and h (Plank's Constant) \[6.6\times {{10}^{-34}}Js\]]

    A)  \[1.46\times {{10}^{6}}m/s\]

    B)  \[0.73\times {{10}^{6}}m/s\]

    C)  \[3.0\times {{10}^{8}}m/s\]

    D)  \[2.92\times {{10}^{6}}m/s\]

    Correct Answer: A

    Solution :

    [a] Speed of electron in nth orbit \[{{V}_{n}}=\frac{2\pi KZ{{e}^{2}}}{nh}\] \[V=\left( 2.19\times {{10}^{6}}m/s \right)\frac{Z}{n}\] \[V=(2.19\times {{10}^{6}})\frac{2}{3}\left( Z=2\And n=3 \right)\] \[V=1.46\times {{10}^{6}}m/s\]


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