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Velocity and energy of the electron in the nth orbit sheet
5-mark answer frame

Using Bohr's postulates, derive the expression for the total energy of the electron in the nth orbit of the hydrogen atom. Draw the energy level diagram for n = 1 to 4.

How to lay out the answer. Follow the frame, then write it in your own words.

  1. 1 · Start from the force balance. mvₙ²/rₙ = e²/(4πε₀rₙ²), so K = (1/2)mvₙ² = e²/(8πε₀rₙ).
  2. 2 · Write the potential energy. U = −e²/(4πε₀rₙ).
  3. 3 · Add to get the total energy. Eₙ = K + U = −e²/(8πε₀rₙ).
  4. 4 · Substitute the radius. With rₙ = n²h²ε₀/(πme²): Eₙ = −me⁴/(8ε₀²n²h²).
  5. 5 · Put in the numbers. Eₙ = −13.6/n² eV. The minus sign means the electron is bound.
  6. 6 · Draw the energy level diagram. [Diagram: horizontal lines at −13.6 eV (n = 1), −3.40 eV (n = 2), −1.51 eV (n = 3), −0.85 eV (n = 4) and 0 eV (n = ∞), getting closer together going up]
  7. 7 · Real life. Visible photons (1.8 to 3.1 eV) cannot ionise hydrogen; only UV 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 the radius formula before substituting it. Label every level in the diagram with both n and its energy in eV.