Radius and speed of the electron in the nth orbit
Combining the Coulomb pull with mvr = nh/2π gives rₙ = n²h²ε₀/(πmZe²) and vₙ = Ze²/(2ε₀nh): outer orbits are bigger and slower.
Derivation: radius of the nth orbit
Take a nucleus of charge +Ze at rest. The electron (mass m, charge −e) moves in the nth orbit of radius rₙ with speed vₙ.
- 1Coulomb force = centripetal force: mvₙ²/rₙ = (1/4πε₀)(Ze²/rₙ²) ... (i)
- 2Multiply both sides by mrₙ³: m²vₙ²rₙ² = (1/4πε₀)mZe²rₙ ... (ii)
- 3Square Bohr's condition mvₙrₙ = nh/2π: m²vₙ²rₙ² = n²h²/4π² ... (iii)
- 4Equate (ii) and (iii): n²h²/4π² = mZe²rₙ/(4πε₀).
- 5Make rₙ the subject: rₙ = (n²h²/4π²) × (4πε₀/mZe²) = n²h²ε₀/(πmZe²).
- 6★So, rₙ = n²h²ε₀/(πmZe²), and rₙ ∝ n².
★ key step
Derivation: speed in the nth orbit
- 1From Bohr's condition: vₙ = nh/(2πmrₙ).
- 2Put in rₙ = n²h²ε₀/(πmZe²): vₙ = (nh/2πm) × (πmZe²/n²h²ε₀).
- 3Cancel π, m, one n and one h: vₙ = Ze²/(2ε₀nh).
- 4★So, vₙ = Ze²/(2ε₀nh), and vₙ ∝ 1/n: the electron moves more slowly in outer orbits.
★ key step
The Bohr radius and the spacing of orbits
For hydrogen (Z = 1) and n = 1, the radius is the Bohr radius a₀ = 5.3 × 10⁻¹¹ m, or 0.53 Å. Other orbits have radii 4a₀, 9a₀, 16a₀, and so on, so the radii are in the ratio 1 : 4 : 9 : 16. The gap between neighbouring orbits grows as n grows. The speed falls as 1/n, so the second orbit has half the speed of the first.
Worked example: radii of the third and fourth orbits
The radius of the first Bohr orbit of hydrogen is 0.53 Å. Find the radii of the third and fourth orbits. Which orbit has a radius of 13.25 Å?
- Formula: rₙ = n²a₀ for hydrogen.
- r₃ = 3² × 0.53 Å = 9 × 0.53 Å = 4.77 Å.
- r₄ = 4² × 0.53 Å = 16 × 0.53 Å = 8.48 Å.
- For 13.25 Å: n² = 13.25 ÷ 0.53 = 25, so n = 5.
r₃ = 4.77 Å and r₄ = 8.48 Å; 13.25 Å is the fifth orbit (n = 5)
Worked example: speed and revolutions in the first orbit
Find the speed of the electron in the first orbit of hydrogen. About how many times does it go round the nucleus in one second? (e = 1.6 × 10⁻¹⁹ C, ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², h = 6.63 × 10⁻³⁴ J s, a₀ = 5.3 × 10⁻¹¹ m)
- Formula: v₁ = e²/(2ε₀h), since Z = 1 and n = 1.
- Top: e² = (1.6 × 10⁻¹⁹)² = 2.56 × 10⁻³⁸ C².
- Bottom: 2 × 8.85 × 10⁻¹² × 6.63 × 10⁻³⁴ = 1.17 × 10⁻⁴⁴.
- v₁ = 2.56 × 10⁻³⁸ ÷ 1.17 × 10⁻⁴⁴ ≈ 2.18 × 10⁶ m/s.
- One trip round = 2πa₀ = 2 × 3.14 × 5.3 × 10⁻¹¹ m = 3.33 × 10⁻¹⁰ m.
- Revolutions per second = v₁ ÷ 2πa₀ = 2.18 × 10⁶ ÷ 3.33 × 10⁻¹⁰ ≈ 6.5 × 10¹⁵.
v₁ ≈ 2.18 × 10⁶ m/s; about 6.5 × 10¹⁵ revolutions every second
How to draw the orbit diagram
- Draw a small dot at the centre and label it nucleus (+Ze).
- Draw three circles round it for n = 1, 2, 3, with radii in the ratio 1 : 4 : 9, so the outer circles are much farther apart.
- Label the radii r₁ = a₀, r₂ = 4a₀, r₃ = 9a₀.
- On one orbit, mark the electron (−e), an arrow for its velocity v along the tangent, and an arrow for the force F pointing to the centre.
The Bohr radius sets the size of atoms. A sheet of notebook paper, about 0.1 mm thick, is nearly ten lakh hydrogen atoms thick.
3 marks: “Using Bohr's postulates, derive the radius of the nth orbit of the hydrogen atom. Show that rₙ ∝ n².” mv²/r = Ze²/(4πε₀r²) and mvr = nh/2π. Eliminate v: rₙ = n²h²ε₀/(πmZe²). For hydrogen everything except n is constant, so rₙ ∝ n²; r₁ = 0.53 Å.
2 marks: “Calculate the radii of the second and third orbits of hydrogen. (a₀ = 0.53 Å)” rₙ = n²a₀: r₂ = 4 × 0.53 = 2.12 Å; r₃ = 9 × 0.53 = 4.77 Å.
1 marks: “How does the speed of the electron change from the first to the second Bohr orbit of hydrogen?” vₙ ∝ 1/n, so it halves, to about 1.09 × 10⁶ m/s.
Radius rises as n², speed slides as 1/n: rₙ = n²a₀ and vₙ = v₁/n.