Derive the expression for the energy of the electron in the nth orbit of the hydrogen atom and draw its energy level diagram.

1. 1 · Start from the force balance. mvₙ²/rₙ = Ze²/(4πε₀rₙ²), so K = ½mvₙ² = Ze²/(8πε₀rₙ).
2. 2 · Write the potential energy. U = −Ze²/(4πε₀rₙ) = −mvₙ² = −2K.
3. 3 · Add them. Eₙ = K + U = −½mvₙ².
4. 4 · Substitute the speed. vₙ = Ze²/(2ε₀nh) gives Eₙ = −mZ²e⁴/(8ε₀²n²h²).
5. 5 · Put in the numbers. For hydrogen, Eₙ = −13.6/n² eV. The minus sign means the electron is bound.
6. 6 · Draw the energy level diagram. [Diagram: lines at −13.6 (n = 1), −3.4 (n = 2), −1.51 (n = 3), −0.85 (n = 4) and 0 eV (n = ∞), crowding together near 0]
7. 7 · Real life. Visible photons (1.8 to 3.1 eV) cannot ionise hydrogen; ultraviolet light and X-rays can.
8. 8 · Close in one line. So, Eₙ ∝ −1/n², and 13.6 eV is the ionisation energy of hydrogen.

Scoring tip: Write K, U and E on separate lines before substituting. In the diagram, label each line with both n and its energy in eV.