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BCECE Medical BCECE Medical Solved Papers-2010

  • question_answer
    What is the correct orbital designation of an electron with the quantum number, \[n=4,\,l=3,\,m=-2,\,\,s=\frac{1}{2}?\]

    A)  3s              

    B)  4f

    C)  5p              

    D)  6s

    Correct Answer: B

    Solution :

    The value of n decides the shell of electrons. The value of \[l\] decides the orbital; \[l=0\] is s, \[l=1\] is p, \[l=2\]is d and \[l=3\] is \[f\] Given, \[n=4,\,l=3,\,m-2,\,s=-\frac{1}{2}\] \[\therefore \] It is \[4f\] orbital/.


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