9 min readConcept sheet3-mark derivation, often part of a 5-mark long answer with a numerical; also a NEET and JEE favourite

Radius of the nth Bohr orbit

Combining the electric pull as centripetal force with mvr = nh/2π gives rₙ = n²h²ε₀/(πme²), so the radius grows as n²: a₀, 4a₀, 9a₀.

Radius of the nth orbitrₙ = n²h²ε₀/(πme²) = n²a₀

Bohr radius a₀a₀ = h²ε₀/(πme²) = 5.29 × 10⁻¹¹ m ≈ 0.53 Å (radius of the n = 1 orbit)

Dependence on nrₙ ∝ n²: the radii of the n = 1, 2, 3 and 4 orbits are in the ratio 1 : 4 : 9 : 16

Symbolsh: Planck's constant; ε₀: permittivity of free space; m and e: mass and charge of the electron

Derivation: radius of the nth orbit

Assume a hydrogen atom with the nucleus at rest. The electron (mass m, charge −e) moves in the nth orbit of radius rₙ with speed vₙ.

1
Electrostatic attraction provides the centripetal force: mvₙ²/rₙ = (1/4πε₀)(e²/rₙ²) ... (i)
2
Multiply (i) by rₙ²: mvₙ²rₙ = e²/(4πε₀) ... (ii)
3
Bohr's quantum condition: mvₙrₙ = nh/2π, so vₙ = nh/(2πmrₙ) ... (iii)
4
Substitute (iii) in (ii): m rₙ × n²h²/(4π²m²rₙ²) = e²/(4πε₀).
5
Simplify the left side: n²h²/(4π²mrₙ) = e²/(4πε₀).
6
Make rₙ the subject: rₙ = (n²h²/4π²m) × (4πε₀/e²) = n²h²ε₀/(πme²).
7★
So, rₙ = n²h²ε₀/(πme²), which shows that rₙ ∝ n².

★ key step

The Bohr radius and the spacing of orbits

For n = 1 the radius is the Bohr radius a₀ = h²ε₀/(πme²) = 5.29 × 10⁻¹¹ m, about 0.53 Å. The next orbits have radii 4a₀, 9a₀, 16a₀, so the gap between neighbouring orbits grows as n increases. Extra reading (not in your textbook): for a hydrogen-like ion of nuclear charge Ze, such as He⁺ (Z = 2), e² becomes Ze² in step 1, so rₙ = n²a₀/Z and its orbits sit closer to the nucleus.

Worked example: radii of other orbits

Problem

The radius of the first Bohr orbit of hydrogen is 0.53 Å. (a) Find the radius of the third orbit. (b) Which orbit has a radius of 8.48 Å?

Working

Formula: rₙ = n²a₀ for hydrogen.
(a) r₃ = 3² × 0.53 Å = 9 × 0.53 Å = 4.77 Å.
(b) n² = 8.48 Å ÷ 0.53 Å = 16, so n = 4.

Answer

(a) 4.77 Å; (b) n = 4

How to draw the orbit diagram

Draw a small dot at the centre and label it nucleus (+e).
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.

In real life

The Bohr radius sets the size of atoms. A single strand of hair, about 0.07 mm thick, is roughly seven lakh hydrogen atoms across.

Exam angle

3 marks: “Using Bohr's postulates, derive an expression for the radius of the nth orbit of the hydrogen atom.” mv²/r = e²/(4πε₀r²) and mvr = nh/2π. Put v = nh/(2πmr) in the first: n²h²/(4π²mr) = e²/(4πε₀). So rₙ = n²h²ε₀/(πme²), and rₙ ∝ n².

2 marks: “The radius of the innermost electron orbit of hydrogen is 5.3 × 10⁻¹¹ m. Find the radii of the n = 2 and n = 3 orbits.” rₙ = n²r₁. r₂ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m; r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m.

1 marks: “How does the radius of a Bohr orbit depend on the principal quantum number n?” rₙ ∝ n².

Remember it

Radius Rises as n Squared: rₙ = n²a₀, with a₀ = 0.53 Å.