Indian Forest Service (Mains), Chemistry optional · 2022 · Paper I
Q1(a). Continuity of the wavefunction
Question
For an acceptable solution, the wavefunction φ(x) and its derivative dφ(x)/dx have to be continuous. Why is this condition imposed on the wavefunction?
Answer in brief
For a finite, nonsingular potential, both φ and φ′ are continuous, ensuring a well-defined probability amplitude and a Schrödinger-equation solution without unphysical singularities.
Key points
- The stationary equation is −ℏ²φ″/(2m) + Vφ = Eφ; for finite V and finite φ it requires a finite second derivative locally.
- Integrating across a boundary x₀ shows φ′(x₀+ε) − φ′(x₀−ε) → 0, so the first derivative is continuous even at a finite step in V.
- A discontinuity of φ would be inconsistent with the finite-potential equation; |φ|² must also be finite and normalisable.
- The wavefunction must additionally be single-valued and obey the system’s boundary conditions.
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.
Coverswell-behaved wavefunctioncontinuity of wavefunctionSchrödinger equationboundary conditionsnormalisation
All parts of Question 1
- 1(a) Continuity of the wavefunction8 marks
- 1(b) Dipole moments of ammonia and nitrogen trifluoride8 marks
- 1(c) MgO impurity in alumina8 marks
- 1(d) Neutral pH at body temperature8 marks
- 1(e) Activity coefficient of potassium ferricyanide8 marks