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

(D) (C) $<$ (E) $<$ (D) $<$ (A) $<$ (B)

(A) $\text{(B)} > \text{(A)} > \text{(C)} > \text{(E)} > \text{(D)}$

(B) (E) $>$ (C) $<$ (D) $<$ (A) $<$ (B)

(C) (E) $>$ (C) $>$ (A) $>$ (D) $>$ (B)

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Atomic Structure

Compare the energies of following sets of quantum numbers for multielectron system.
(A)

n = 4, 1 = 1

(B)

n = 4, 1 = 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)✓ Correct

Answer: (D)

Solution

Order energies by (n+ℓ), and for equal (n+ℓ) the higher n wins. (A)4p:5, (B)4d:6, (C)3p:4, (D)3d:5, (E)4s:4. Increasing energy: (C) < (E) < (D) < (A) < (B). Option (4).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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