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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The electrons, identified by quantum numbers \[n\]and\[l,\]
    (I) \[n=3;\text{ }l=2\]                    
    (II)\[n=5;\text{ }l=0\]
    (III) \[n=4;\text{ }l=1\]                  
    (IV) \[n=4;\text{ }l=2\]
    (V) \[n=4;\text{ }l=0\]
    can be placed in order of increasing energy, as

    A) \[I<V<III<IV<II\]

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

    C) \[V<I<III<II<IV\]

    D) \[V<I<II<IIII<IV\]

    E) \[V<I<IV<III<II\]

    Correct Answer: C

    Solution :

    Higher the value of\[(n+l),\]higher will be the energy of electron. If value of\[(n+l)\]is same for any two or more electrons, the electron with higher value of n, has higher energy. Hence, the correct order of energy is     \[V<I<III<II<IV\] \[\because \]\[(n+l)\,\,\,\,\,6\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,4\]


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