Bohr's model: the four postulates
Bohr kept the nucleus but added quantum rules: electrons move in stationary orbits with angular momentum nh/2π, and give out light only when they jump down.
Bohr's four postulates (1913)
- Circular orbits: the electron moves in a circle round the nucleus. The Coulomb attraction of the nucleus gives the centripetal force: mv²/r = (1/4πε₀)(Ze²/r²). Z is the number of protons (Z = 1 for hydrogen).
- Quantum condition: only orbits in which the angular momentum L = mvr is a whole-number multiple of h/2π are allowed: mvr = nh/2π, n = 1, 2, 3, .... The number n is the principal quantum number.
- Stationary states: an electron in an allowed orbit does not radiate energy. Its energy stays constant. These orbits are called stationary states.
- Frequency condition: the atom gives out energy only when the electron falls from a higher level to a lower one. The energy leaves as one photon: hν = E(higher) − E(lower). To jump up, the electron must absorb a photon of exactly that energy.
How the postulates solve Rutherford's problems
Niels Bohr, a Danish physicist, worked in Rutherford's laboratory. He kept the nucleus but used Planck's idea that energy comes in packets (quanta). Postulate 3 says allowed orbits do not radiate, so the atom does not collapse. Postulate 2 allows only some orbits, so the electron can have only certain energies. By postulate 4, only certain photon energies can come out. Fixed photon energies mean fixed frequencies, so the atom gives a line spectrum, as seen in experiments.
Classical and quantum parts of Bohr's model
Bohr quantised both energy and angular momentum. Planck had quantised only energy.
Worked example: angular momentum and its change
Find the angular momentum of an electron in the third orbit of hydrogen. By how much does it change when the electron jumps to the first orbit? (h = 6.63 × 10⁻³⁴ J s, π = 3.14)
- Formula: L = nh/2π.
- h/2π = 6.63 × 10⁻³⁴ ÷ 6.28 = 1.056 × 10⁻³⁴ J s.
- L₃ = 3 × 1.056 × 10⁻³⁴ = 3.17 × 10⁻³⁴ J s.
- L₁ = 1 × 1.056 × 10⁻³⁴ = 1.06 × 10⁻³⁴ J s.
- Change ΔL = L₃ − L₁ = 2 × h/2π = h/π = 2.11 × 10⁻³⁴ J s, a decrease.
L₃ ≈ 3.17 × 10⁻³⁴ J s; it falls by h/π ≈ 2.11 × 10⁻³⁴ J s
Worked example: frequency of the emitted photon
In a hydrogen atom, the electron falls from the n = 2 level (−3.4 eV) to the n = 1 level (−13.6 eV). Find the energy and the frequency of the photon given out. (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J)
- Formula: hν = E(higher) − E(lower).
- Photon energy = −3.4 eV − (−13.6 eV) = 10.2 eV.
- In joules: 10.2 × 1.6 × 10⁻¹⁹ = 1.632 × 10⁻¹⁸ J.
- ν = 1.632 × 10⁻¹⁸ J ÷ 6.63 × 10⁻³⁴ J s = 2.46 × 10¹⁵ Hz.
10.2 eV; ν ≈ 2.46 × 10¹⁵ Hz, in the ultraviolet
The orange glow of sodium vapour street lamps and the bright colours of Diwali sparklers both come from electrons falling between fixed energy levels, just as Bohr's fourth postulate says.
2 marks: “State Bohr's quantum condition. Find the angular momentum of an electron in the second orbit of hydrogen. (h = 6.63 × 10⁻³⁴ J s)” mvr = nh/2π. L₂ = 2h/2π = h/π = 6.63 × 10⁻³⁴ ÷ 3.14 ≈ 2.11 × 10⁻³⁴ J s.
3 marks: “State the postulates of Bohr's model of the atom. Which of them agrees with classical physics?” (1) Circular orbits with the Coulomb force as centripetal force; (2) mvr = nh/2π; (3) no radiation in stationary orbits; (4) hν = E(higher) − E(lower) on a jump. The first is classical; the other three are quantum.
1 marks: “According to Bohr, why does an atom not collapse?” An electron in an allowed (stationary) orbit does not radiate energy.
Circle, Quantise, Calm, Jump: Coulomb circle, mvr = nh/2π, no radiation in a stationary orbit, a photon hν on a jump.