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Which orbitals are degenerate in a multielectron atom, from quantum numbers

This type of question, asked 2 times: 14 Jun 2022, 28 Jan 2025. All 2 questions below, oldest first.

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  1. Question 1

    14 Jun 2022

    32
    Consider the following pairs of electrons (A) (a) n=3,l=1,ml=1,ms=+12n = 3,l = 1,m_{l}= 1,m_{s}= +\frac{1}{2} (b) n=3,l=2,ml=1,ms=+12n = 3,l = 2,m_{l}= 1,m_{s}= +\frac{1}{2} (B) (a) n=3,l=2,ml=−2,ms=−12n = 3,l = 2,m_{l}= - 2,m_{s}= -\frac{1}{2} (b) n=3,l=2,ml=−1,ms=−12n = 3,l = 2,m_{l}= - 1,m_{s}= -\frac{1}{2} (C) (a) n=4,l=2,ml=2,ms=+12n = 4,l = 2,m_{l}= 2,m_{s}= +\frac{1}{2} (b) n=3,l=2,ml=2,ms=+12n = 3,l = 2,m_{l}= 2,m_{s}= +\frac{1}{2} The pairs of electron present in degenerate orbitals is/are:

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  2. Question 2

    28 Jan 2025

    38
    In a multielectron atom, which of the following orbitals described by three quantum numbers will have the same energy in the absence of electric and magnetic fields? (A) n=1,l=0,ml=0n = 1,l = 0,m_{l}= 0 (B) n=2,l=0,ml=0n = 2,l = 0,m_{l}= 0 (C) n=2,l=1,ml=1n = 2,l = 1,m_{l}= 1 (D) n=3,l=2,ml=1n = 3,l = 2,m_{l}= 1 (E) n=3,l=2,ml=0n = 3,l = 2,m_{l}= 0 Choose the correct answer from the options given below :

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