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 , 1 = 1 (D) n = 3 , 1 = 2 (E) n = 4 , 1 = 0 Choose the correct answer

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 , 1 = 1 (D) n = 3 , 1 = 2 (E) n = 4 , 1 = 0 Choose the correct answer from the options given below :

TestHub · Accepted Answer

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

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