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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 · 2018 · Paper I

Q2(a)(iii). If Planck's constant were zero

  • 5 marks
  • Answered
Question

Would it be possible to determine the position and momentum of an electron simultaneously and accurately if the value of Plank's constant is zero?

Answer in brief

Yes. The uncertainty principle sets Δ x Δ p≥ h/4π; with h=0 the lower limit is zero, so position and momentum could both be known exactly, as in classical mechanics.

Key points

  • Δ x Δ pₓ≥(h)/(4π). The product of the uncertainties cannot be smaller than a fixed quantity proportional to h.
  • If h=0, the condition becomes Δ x Δ pₓ≥0, which any pair of values satisfies, including Δ x=0 and Δ pₓ=0 together. Position and momentum would be simultaneously exact.
  • The same limit removes all wave and quantum behaviour: λ=h/p→0 (no de Broglie wave), E=hν→0 (light carries no quanta), and Eₙ=n²h²/8mL²→0 (no quantisation, no zero-point energy).
  • an electron would follow a definite orbit. It would then radiate, lose energy and spiral into the nucleus, so stable atoms could not exist.

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. Heisenberg's uncertainty principle: Δ x Δ pₓ≥(h)/(4π). The product of the uncertainties cannot be smaller than a fixed quantity proportional to h.
  2. If h=0, the condition becomes Δ x Δ pₓ≥0, which any pair of values satisfies, including Δ x=0 and Δ pₓ=0 together. Position and momentum would be simultaneously exact.
  3. The same limit removes all wave and quantum behaviour: λ=h/p→0 (no de Broglie wave), E=hν→0 (light carries no quanta), and Eₙ=n²h²/8mL²→0 (no quantisation, no zero-point energy).
  4. The world would be classical: an electron would follow a definite orbit. It would then radiate, lose energy and spiral into the nucleus, so stable atoms could not exist.
  5. h=6.626×10⁻³⁴ J s is small but not zero. That is why the principle is negligible for a cricket ball and decisive for an electron.

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Model answer to Q2(a)(iii), 2018 Paper I: solution page 10
Solution page 10 of 55
Read the whole 2018 solution2018 question paper

All parts of Question 2