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JEE Main & Advanced Chemistry Structure of Atom / परमाणु संरचना Sample Paper Topic Test - Structure of Atom

  • question_answer
    The orbital angular momentum for an electron revolving in an orbit is given by \[\sqrt{l(l+1)}\frac{h}{2\pi }\]. This momentum for \[{{d}^{5}},\,{{d}^{6}},\,{{p}^{3}}\] electron will be respectively

    A) \[\sqrt{6}\,\frac{h}{2\pi },\,\sqrt{7}\,\frac{h}{2\pi },\,\sqrt{2}\,\frac{h}{2\pi }\]      

    B) \[0\,\frac{\sqrt{6}h}{2\pi },\,\frac{\sqrt{2}h}{2\pi }\]

    C) \[\sqrt{6}\,\frac{h}{2\pi },\,\sqrt{6}\frac{6}{2\pi },\,\sqrt{2}\frac{h}{2\pi }\]

    D) \[\sqrt{6}\,\frac{h}{2\pi },\,\sqrt{6}\frac{h}{2\pi },\,\sqrt{3}\frac{h}{2\pi }\]

    Correct Answer: C

    Solution :

    [c] For d-electron \[L=\sqrt{2(2+1)}\frac{h}{2\pi }=\sqrt{6}\frac{h}{2\pi }\]
    For p-electron \[L=\sqrt{1(1+1)}\frac{h}{2\pi }=\sqrt{2}.\frac{h}{2\pi }\]


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