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The correct decreasing order of energy, for the orbitals having, following set of quantum numbers :
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1

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(B) > (D) > (C) > (A)

(D) > (B) > (C) > (A)

(B) > (C) > (D) > (A)

answer is A.

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Detailed Solution

$(A) n + l = 3 + 0 = 3$

$(B) n + l = 4 + 0 = 4$

$(C) n + l = 3 + 1 = 4$

$(D) n + l = 3 + 2 = 5$

Higher $n + l$ value, higher the energy & if same $n + l$ value, then higher n value, higher the energy.

Thus : D > B > C > A.

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