Velocity and energy of the electron in the nth orbit
In the nth orbit, vₙ = e²/(2ε₀nh) falls as 1/n, and Eₙ = −13.6/n² eV. The negative energy shows the electron is bound to the nucleus.
Derivation: speed in the nth orbit
- 1From Bohr's quantum condition: vₙ = nh/(2πmrₙ).
- 2Put rₙ = n²h²ε₀/(πme²): vₙ = (nh/2πm) × (πme²/n²h²ε₀).
- 3Cancel π, m, n and h: vₙ = e²/(2ε₀nh).
- 4★So, vₙ = e²/(2ε₀nh), and vₙ ∝ 1/n: the electron moves more slowly in outer orbits.
★ key step
Derivation: energy in the nth orbit
- 1From the force balance mvₙ²/rₙ = e²/(4πε₀rₙ²): kinetic energy K = (1/2)mvₙ² = e²/(8πε₀rₙ).
- 2Potential energy: U = −e²/(4πε₀rₙ).
- 3Total energy: Eₙ = K + U = −e²/(8πε₀rₙ).
- 4Put rₙ = n²h²ε₀/(πme²): Eₙ = −(e²/8πε₀) × (πme²/n²h²ε₀).
- 5Simplify: Eₙ = −me⁴/(8ε₀²n²h²).
- 6Put in the exact values of m, e, ε₀ and h: Eₙ = −2.18 × 10⁻¹⁸/n² J = −13.6/n² eV.
- 7★So, Eₙ = −13.6/n² eV. Energy rises (becomes less negative) as n increases and reaches 0 at n = ∞, where the electron is free.
★ key step
Energy levels of the hydrogen atom
| 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 |
| ∞ | Ionised (electron free) | 0 | infinite |
How each quantity depends on n
| Quantity | Formula | Varies as |
|---|---|---|
| Radius rₙ | n²h²ε₀/(πme²) | n² |
| Speed vₙ | e²/(2ε₀nh) | 1/n |
| Angular momentum | nh/2π | n |
| Kinetic energy | 13.6/n² eV | 1/n² |
| Potential energy | −27.2/n² eV | −1/n² |
| Total energy | −13.6/n² eV | −1/n² |
| Time for one revolution | 2πrₙ/vₙ | n³ |
Worked example: energies in the n = 3 state
An electron in a hydrogen atom is in the n = 3 orbit. Find (a) its total energy, (b) its kinetic and potential energies, (c) the energy needed to ionise the atom from this state, and (d) the energy needed to excite it from the ground state to n = 3.
- Formula: Eₙ = −13.6/n² eV; K = −E; U = 2E.
- (a) E₃ = −13.6/9 eV = −1.51 eV.
- (b) K = +1.51 eV; U = 2 × (−1.51) eV = −3.02 eV.
- (c) To ionise, raise E from −1.51 eV to 0: energy needed = 1.51 eV.
- (d) Excitation energy = E₃ − E₁ = −1.51 eV − (−13.6 eV) = 12.09 eV.
(a) −1.51 eV; (b) K = 1.51 eV, U = −3.02 eV; (c) 1.51 eV; (d) 12.09 eV
Worked example: speed in the first orbit
Find the speed of the electron in the first orbit of hydrogen and compare it with the speed of light. (e = 1.6 × 10⁻¹⁹ C, ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s)
- Formula: v₁ = e²/(2ε₀h) (n = 1).
- Numerator: (1.6 × 10⁻¹⁹)² = 2.56 × 10⁻³⁸ C².
- Denominator: 2 × 8.85 × 10⁻¹² × 6.63 × 10⁻³⁴ = 1.174 × 10⁻⁴⁴.
- v₁ = 2.56 × 10⁻³⁸ ÷ 1.174 × 10⁻⁴⁴ ≈ 2.18 × 10⁶ m/s.
- c/v₁ = 3 × 10⁸ ÷ 2.18 × 10⁶ ≈ 137.
v₁ ≈ 2.18 × 10⁶ m/s, about 1/137 of the speed of light
A visible-light photon carries only about 1.8 to 3.1 eV, far less than the 13.6 eV needed to ionise hydrogen. That is why only ultraviolet light and X-rays can knock electrons out of hydrogen atoms.
3 marks: “Using Bohr's postulates, obtain the expression for the total energy of the electron in the nth orbit of hydrogen. Why is it negative?” K = e²/(8πε₀r), U = −e²/(4πε₀r), E = −e²/(8πε₀r). Put rₙ = n²h²ε₀/(πme²): Eₙ = −me⁴/(8ε₀²n²h²) = −13.6/n² eV. Negative: the electron is bound.
2 marks: “The total energy of an electron in the first excited state of hydrogen is −3.4 eV. What are its kinetic and potential energies?” K = −E = +3.4 eV; U = 2E = −6.8 eV.
3 marks: “Find the energy needed to excite a hydrogen atom from its ground state to n = 3, and the energy then needed to ionise it.” E₁ = −13.6 eV, E₃ = −1.51 eV. Excitation energy = 12.09 eV. Ionisation from n = 3 = 1.51 eV.
K is plus, U is double minus, E is minus K: K = 13.6/n², U = −27.2/n², E = −13.6/n² (in eV).