UPSC Civil Services (Mains), Chemistry optional · 2024 · Paper I
Q1(b)(i). Separation factor for ²³⁵UF₆ and ²³⁸UF₆ in one effusion stage
Question
The ²³⁵U isotope undergoes fission when bombarded with neutrons. However, its natural abundance is only 0.72 percent. To separate it from the more abundant ²³⁸U isotope, U is first converted to UF₆, which is easily vaporized above room temperature. The mixture of ²³⁵UF₆ and ²³⁸UF₆ gases is then subjected to many stages of effusion. Calculate the separation factor, that is, the enrichment of ²³⁵U relative to ²³⁸U after one stage of effusion.
Answer in brief
Separation factor = √(352/349) = 1.0043, an enrichment of 0.43 % in one stage.
Key points
- M(²³⁵UF₆) = 235 + 6(19) = 349 g mol⁻¹; M(²³⁸UF₆) = 238 + 6(19) = 352 g mol⁻¹.
- At one temperature lighter molecules move faster, so the rate of effusion is proportional to 1/√M (Graham's law).
- Separation factor = (n₂₃₅/n₂₃₈)after / (n₂₃₅/n₂₃₈)before = √(352/349) = 1.0043.
- One stage raises the ²³⁵U/²³⁸U ratio only from 0.00725 to 0.00728; the gain multiplies as (1.0043)ⁿ.
- About 340 stages are needed to reach 3 % ²³⁵U, which is why many stages are used.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversGraham's law of effusionuranium enrichmentUF₆separation factorisotope separation
All parts of Question 1
- 1(a)(i) Most probable distance of the 1s electron5 marks
- 1(a)(ii) Lattice energies of NaBr and NaI above the calculated values5 marks
- 1(b)(i) Separation factor for ²³⁵UF₆ and ²³⁸UF₆ in one effusion stage5 marks
- 1(b)(ii) Unit cell and the cubic Bravais lattices5 marks
- 1(c) Normal boiling point of mercury10 marks
- 1(d)(i) Will CCl₂F₂ condense at 18 atm and 20 °C?5 marks
- 1(d)(ii) Overvoltage and its applications5 marks
- 1(e) Temperature at which the uncatalysed reaction matches catalase at 20 °C10 marks