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Given below are the quantum numbers for 4 electrons. A. n = 3, ℓ = 2, mℓ = 1, ms = +1/2 B. n = 4, ℓ = 1, mℓ = 0, ms = +1/2 C. n = 4, ℓ = 2, mℓ = −2, ms = −1/2 D. n = 3, ℓ = 1, mℓ = −1, ms = +1/2 The correct order of increasing energy is :
- D < B < A < C
- D < A < B < C
- B < D < A < C
- B < D < C < A
Correct answer: D < A < B < C
Solution
The correct order of increasing energy is determined by the principal quantum number (n) and the azimuthal quantum number (ℓ). Electrons in lower n and ℓ values have lower energy, so D (n=3, ℓ=1) has the lowest energy, followed by A (n=3, ℓ=2), B (n=4, ℓ=1), and C (n=4, ℓ=2) which has the highest energy due to its higher n and ℓ values.
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