(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) \[n=4;\text{ }l=0\] |
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 \[V<I<III<II<IV\] \[\because \]\[(n+l)\,\,\,\,\,6\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,4\]You need to login to perform this action.
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