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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2020 · Paper I

Q5(a). Degeneracy of the hydrogenic level with E=-R_H/36

  • 10 marks
  • Answered
Question

What do you mean by degeneracy of energy level? Find the number of degeneracy of hydrogenic orbital having energy -R_H/36. R_H = Rydberg constant.

Answer in brief

A level is degenerate when two or more different states (orbitals) have the same energy; the number of such states is its degeneracy. Eₙ=-R_H/n²=-R_H/36 gives n=6, whose degeneracy is n²=36 orbitals (72 spin-orbitals).

Key points

  • different wavefunctions (different sets of quantum numbers) that have the same energy. The degree of degeneracy g is the number of independent states at that energy. Example: the three 2p orbitals of H (g=3 within the subshell).
  • degeneracy reflects symmetry. In any atom the 2l+1 values of mₗ are degenerate (spherical symmetry); in hydrogen, with a pure 1/r potential, all l values of one n are degenerate too (accidental degeneracy).
  • Find n: -R_Hn²=-(R_H)/(36) ⇒ n²=36, n=6
  • l=0 to 5 (6s, 6p, 6d, 6f, 6g, 6h), each with 2l+1 values of mₗ: 4pt] g=1+3+5+7+9+11, g=n²=36 orbitals

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

  1. Degeneracy: different wavefunctions (different sets of quantum numbers) that have the same energy. The degree of degeneracy g is the number of independent states at that energy. Example: the three 2p orbitals of H (g=3 within the subshell).
  2. Origin: degeneracy reflects symmetry. In any atom the 2l+1 values of mₗ are degenerate (spherical symmetry); in hydrogen, with a pure 1/r potential, all l values of one n are degenerate too (accidental degeneracy).
  3. Find n: -R_Hn²=-(R_H)/(36) ⇒ n²=36, n=6
  4. Count the orbitals: l=0 to 5 (6s, 6p, 6d, 6f, 6g, 6h), each with 2l+1 values of mₗ: 4pt] g=1+3+5+7+9+11, g=n²=36 orbitals
  5. With spin (mₛ=±12 for each orbital): 2n²=72 states (spin-orbitals).
  6. Removal of degeneracy: in many-electron atoms, penetration makes the l levels of one n different (6s < 6p < 6d…); a magnetic field splits the mₗ levels (Zeeman effect).

Application. degeneracy is the statistical weight g in the Boltzmann distribution, Nᵢ/N₀=(gᵢ/g₀)e^(-Δ E/kT), which sets the populations and line intensities in atomic spectra.

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Model answer to Q5(a), 2020 Paper I: solution page 23
Solution page 23 of 45
Read the whole 2020 solution2020 question paper

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