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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

Q3(a). Most probable distance of the 1s electron of hydrogen

  • 10 marks
  • Answered
Question

The wave function of the hydrogen atom is given by R₁ₛ=(1)/(√(π))e^(-r) (in atomic units). What will be the most probable distance of an electron from the nucleus in its ground state (in unit of a₀, the Bohr radius and in Å unit)?

Answer in brief

The radial distribution function P(r)=4π r²ψ²=4r²e^(-2r) is largest at r=1 atomic unit: rₘₚ=1 a₀=0.529 Å.

Key points

  • Probability in a shell of radius r and thickness dr (volume 4π r²dr), since ψ₁ₛ is spherical: P(r) dr=ψ² 4π r² dr
  • Substitution: P(r)=4π r²·(1)/(π)e^(-2r)=4r²e^(-2r)
  • (dP)/(dr)=4(2r e^(-2r)-2r²e^(-2r))=8r e^(-2r)(1-r)=0. The roots r=0 and r→∞ are minima (P=0), so the maximum is at r=1.
  • in atomic units the unit of length is the Bohr radius: rₘₚ=1 a₀=0.529 Å (52.9 pm)

The full model answer

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

Reading: the function printed contains the angular factor, so it is the complete 1s wavefunction ψ₁ₛ=π^(-1/2)e^(-r) (the radial part alone is 2e^(-r)). The answer is the same either way.

  1. Probability in a shell of radius r and thickness dr (volume 4π r²dr), since ψ₁ₛ is spherical: P(r) dr=ψ² 4π r² dr
  2. Substitution: P(r)=4π r²·(1)/(π)e^(-2r)=4r²e^(-2r)
  3. Maximum: (dP)/(dr)=4(2r e^(-2r)-2r²e^(-2r))=8r e^(-2r)(1-r)=0. The roots r=0 and r→∞ are minima (P=0), so the maximum is at r=1.
  4. Result: in atomic units the unit of length is the Bohr radius: rₘₚ=1 a₀=0.529 Å (52.9 pm)
  5. ψ² against P(r): ψ² is largest at the nucleus, but the shell volume 4π r²dr is zero there; the product of a falling ψ² and a rising r² gives the peak at a₀.
  6. Agreement with Bohr: the most probable distance equals the radius of the first Bohr orbit; the average distance is larger, r=32a₀=0.79 Å, because the distribution has a long tail.

Application. the same method gives rₘₚ=a₀/Z for a hydrogen-like ion (0.265 Å for He⁺), the basis of the shrinking of orbitals with nuclear charge and of atomic radii in the periodic table.

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Model answer to Q3(a), 2020 Paper I: solution page 12
Solution page 12 of 45
Read the whole 2020 solution2020 question paper

All parts of Question 3