Energy of the electron and the energy levels of hydrogen
The total energy in the nth orbit of hydrogen is Eₙ = −13.6/n² eV. It is negative because the electron is bound; zero means it is free.
Derivation: energy in the nth orbit
- 1Kinetic energy: from mvₙ²/rₙ = Ze²/(4πε₀rₙ²), K = ½mvₙ² = Ze²/(8πε₀rₙ).
- 2Potential energy of the electron (−e) at distance rₙ from the nucleus (+Ze): U = −Ze²/(4πε₀rₙ). It is zero at infinity.
- 3So U = −mvₙ² = −2K.
- 4Total energy: Eₙ = K + U = ½mvₙ² − mvₙ² = −½mvₙ².
- 5Put in vₙ = Ze²/(2ε₀nh): Eₙ = −½m × Z²e⁴/(4ε₀²n²h²).
- 6Simplify: Eₙ = −mZ²e⁴/(8ε₀²n²h²).
- 7★So, Eₙ = −mZ²e⁴/(8ε₀²n²h²). With standard values, for hydrogen (Z = 1), Eₙ = −13.6/n² eV.
★ key step
What the minus sign means
Energy is counted from zero when the electron is far away and free (n = ∞). A bound electron has less energy than a free one, so its energy is negative. The lowest level, n = 1, is the ground state, at −13.6 eV. Higher levels (n = 2, 3, ...) are excited states. As n grows, the energy rises towards 0 and the levels crowd together. Your NIOS lesson writes Eₙ = −R/n², with R = me⁴/(8ε₀²h²) = 2.17 × 10⁻¹⁸ J = 13.6 eV.
Energy levels of hydrogen
| n | Name | Energy Eₙ (eV) | Radius |
|---|---|---|---|
| 1 | Ground state | −13.6 | a₀ |
| 2 | First excited state | −3.40 | 4a₀ |
| 3 | Second excited state | −1.51 | 9a₀ |
| 4 | Third excited state | −0.85 | 16a₀ |
| 5 | Fourth excited state | −0.54 | 25a₀ |
| ∞ | Free electron (ionised) | 0 | very large |
Worked example: energies in the n = 4 state
The ionisation potential of hydrogen is 13.6 V. For a hydrogen atom in the n = 4 state, find (a) its total energy, (b) the kinetic and potential energies of the electron, (c) the energy needed to excite the atom from n = 1 to n = 4.
- An ionisation potential of 13.6 V means E₁ = −13.6 eV.
- (a) E₄ = −13.6/4² = −13.6/16 = −0.85 eV.
- (b) K = −E = +0.85 eV; U = 2E = −1.70 eV.
- (c) Excitation energy = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV.
(a) −0.85 eV; (b) K = 0.85 eV, U = −1.70 eV; (c) 12.75 eV
Worked example: comparing two levels
Find the ratio of the energy of a hydrogen atom in its first excited state to that in its third excited state. How much energy frees the electron from the first excited state?
- First excited state is n = 2: E₂ = −13.6/4 = −3.40 eV.
- Third excited state is n = 4: E₄ = −13.6/16 = −0.85 eV.
- Ratio E₂ : E₄ = (1/4) : (1/16) = 4 : 1.
- To free the electron from n = 2, raise its energy from −3.40 eV to 0: 3.40 eV.
E₂ : E₄ = 4 : 1; 3.4 eV frees the electron from the first excited state
How to draw the energy level diagram
- Draw a vertical arrow on the left labelled Energy (eV).
- Draw horizontal lines for n = 1 (−13.6), n = 2 (−3.40), n = 3 (−1.51), n = 4 (−0.85) and n = ∞ (0) at the top.
- Space them correctly: a big gap between n = 1 and n = 2, and lines crowding together near 0.
- Label n = 1 ground state, the others excited states, and the top line ionisation (electron free).
- Draw an upward arrow from n = 1 to n = ∞ and write 13.6 eV (ionisation energy) beside it.
A photon of visible light carries only 1.8 to 3.1 eV, far less than 13.6 eV. So the light from a tube light cannot ionise hydrogen; only ultraviolet light and X-rays can.
5 marks: “Derive an expression for the total energy of the electron in the nth orbit of hydrogen. Why is it negative? Draw the energy level diagram.” K = Ze²/(8πε₀r), U = −Ze²/(4πε₀r), so E = −½mv². With vₙ = Ze²/(2ε₀nh): Eₙ = −mZ²e⁴/(8ε₀²n²h²) = −13.6/n² eV. Negative: the electron is bound. Diagram: levels at −13.6, −3.4, −1.51, −0.85 and 0 eV.
2 marks: “What energy is needed to ionise a hydrogen atom in its first excited state?” First excited state is n = 2: E₂ = −13.6/4 = −3.4 eV. Energy needed to reach 0 = 3.4 eV.
1 marks: “The total energy of an electron in a Bohr orbit is negative. What does the minus sign tell us?” The electron is bound to the nucleus.
Minus thirteen point six over n squared: E₁ = −13.6, E₂ = −3.4, E₃ = −1.51 eV; and E = −K, U = 2E.