Lesson eLearning

An electron is accelerated from rest through a potential difference of 300 V. (i) Find its de Broglie wavelength in picometres. (ii) The same electron is confined in a one‑dimensional box of length L = 0.4 nm. Calculate the wavelength of the photon emitted when the electron makes a transition from the first excited state (n = 2) to the ground state (n = 1). Assume the photon energy equals the difference of the box energy levels. Give the wavelength in nanometres. Use h = 6.626×10⁻³⁴ J·s, mₑ = 9.11×10⁻³¹ kg, e = 1.602×10⁻¹⁹ C, and c = 3.00×10⁸ m/s.

hard 5 marks Class 12 · Physics · Atoms
Answer
(i) λ = 71 pm (ii) λ = 176 nm

The correct answer is (i) λ = 71 pm (ii) λ = 176 nm — Express in nanometres: 1.76×10⁻⁷ m = 176 nm.

Step-by-step solution

  1. (i) Compute the kinetic energy gained: K = eV = 1.602×10⁻¹⁹ C × 300 V = 4.806×10⁻¹⁷ J.
  2. Find the momentum: p = √(2mₑK) = √[2 × 9.11×10⁻³¹ kg × 4.806×10⁻¹⁷ J] = 9.36×10⁻²⁴ kg·m/s.
  3. Apply de Broglie’s relation: λ = h/p = 6.626×10⁻³⁴ J·s / 9.36×10⁻²⁴ kg·m/s = 7.08×10⁻¹¹ m.
  4. Convert to picometres: 7.08×10⁻¹¹ m = 70.8 pm ≈ 71 pm.
  5. (ii) For a particle in a 1‑D box, energy levels are E_n = n²h²/(8mₑL²).
  6. Calculate the common factor: h²/(8mₑL²) = (6.626×10⁻³⁴)² / [8 × 9.11×10⁻³¹ × (4.0×10⁻¹⁰)²] = 3.77×10⁻¹⁹ J.
  7. Find E₁ = 1² × factor = 3.77×10⁻¹⁹ J and E₂ = 2² × factor = 1.51×10⁻¹⁸ J.
  8. Energy of the emitted photon: ΔE = E₂ – E₁ = 3 × factor = 1.13×10⁻¹⁸ J.
  9. Convert ΔE to wavelength using λ = hc/ΔE: λ = (6.626×10⁻³⁴ × 3.00×10⁸) / 1.13×10⁻¹⁸ = 1.76×10⁻⁷ m.
  10. Express in nanometres: 1.76×10⁻⁷ m = 176 nm.
Common mistake: Using L = 0.4 nm as 0.4×10⁻⁹ m or forgetting the n² factor in E_n

Concept tested: de Broglie wavelength and particle‑in‑a‑box transitions

Published · How this solution was prepared and checked

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