KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • question_answer
    The quantum number of four electrons are given below:
    (I) \[n=4\], \[l=2\], \[{{m}_{1}}=1,\]\[{{m}_{s}}=-\frac{1}{2}\]
    (II) \[n=3\], \[l=2\], \[{{m}_{1}}=1,\] \[{{m}_{s}}=+\frac{1}{2}\]
    (III) \[n=4\], \[l=1\], \[{{m}_{1}}=0,\]\[{{m}_{s}}=+\frac{1}{2}\]
    (IV) \[n=3\], \[l=1\], \[{{m}_{1}}=1,\] \[{{m}_{s}}=-\frac{1}{2}\]
    The Correct order of their increasing energies will be:

    A) \[I<III<II<IV\]  

    B) \[IV<II<III<I\]

    C) \[I<II<III<IV\]  

    D) \[IV<III<II<I\]  

    Correct Answer: B

    Solution :

    Higher-the value of\[(n+l)\]
    Higher will be energy of orbital. If \[(n+l)\] are equal, then higher the value n higher will be energy
    I. \[n+l=6\] \[n=4\]
    II. \[n+l=5\] \[n=3\]
    III. \[n+l=5\] \[n=4\]

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