Compare the energies of following sets of quantum numbers for multielectron system. (A) n = 4, 1 = 1 (B) n = 4, l = 2 (C) n = 3, l = 1 (D) n = 3, l = 2 (E) n = 4, 1 = 0 Choose the correct answer from

Question

Compare the energies of following sets of quantum numbers for multielectron system. (A) n = 4, 1 = 1 (B) n = 4, l = 2 (C) n = 3, l = 1 (D) n = 3, l = 2 (E) n = 4, 1 = 0 Choose the correct answer from the options given below : (A) (B) > (A) > (C) > (E) > (D) (B) (E) > (C) < (D) < (A) < (B) (C) (E) > (C) > (A) > (D) > (B) (D) (C) < (E) < (D) < (A) < (B)

Prashant Kotian · Accepted Answer

Correct answer: D. Energy level can be determined by comparing (n + ) values (A) n = 4,  = 1  (n + ) = 5 (B) n = 4,  = 2  (n + ) = 6 (C) n = 3,  = 1  (n + ) = 4 (D) n = 3,  = 2  (n + ) = 5 (E) n = 4,  = 0  (n + ) = 4 For same value of (n + ), orbital having higher value of n, will have more energy. (B) > (A) > (D) > (E) > (C)

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