Bohr orbits: how r, v and E scale with n and Z.

Bohr orbits to scale for H, He⁺ or Li²⁺ with readouts and bars for r ∝ n²/Z, v ∝ Z/n, K, U and E = −13.6 Z²/n² eV.

Find the He⁺ orbit with the same energy as hydrogen's ground state (n = 2) and confirm U = 2E there.

Saying r ∝ n and v ∝ n, giving the bound electron positive total energy, or U = −E.

• Niels Bohr spent several months in Rutherford's laboratory in 1912 and accepted the nuclear model. In 1913 he concluded that classical electromagnetism, so successful on a large scale, cannot be applied to processes inside the atom, and he combined classical and early quantum ideas in three postulates.
• First postulate: an electron can revolve in certain stable orbits without radiating, contrary to classical theory. Each such stationary state of the atom has a definite total energy.
• Second postulate: the allowed orbits are those whose angular momentum is a whole multiple of h/2π, L = nh/2π (Eq. 12.5), where h is Planck's constant (NCERT quotes 6.6 × 10⁻³⁴ J s here). Angular momentum is quantised.
• Third postulate: an electron may jump from one stationary orbit to another of lower energy, emitting a photon whose energy equals the difference: hν = Ei − Ef (Eq. 12.6), with Ei > Ef.
• Combining L = nh/2π with the force balance gives the radius of the nth orbit, rn = (n²/m)(h/2π)²(4πε₀/e²) (Eq. 12.7), so rn ∝ n². For n = 1 this is the Bohr radius a₀ = 5.3 × 10⁻¹¹ m.
• Putting rn into E = −e²/(8πε₀r) gives En = −me⁴/(8n²ε₀²h²) (Eq. 12.8), which is −2.18 × 10⁻¹⁸ J/n² (Eq. 12.9), or En = −13.6 eV/n² (Eq. 12.10).
• Since v = nh/(2πmrn) and rn ∝ n², the speed goes as vn ∝ 1/n: 2.2 × 10⁶ m/s in the first orbit, half that in the second.
• The negative En means the electron is bound; energy must be supplied to take it infinitely far from the proton.

Teaching this idea? Put the simulation on your own page or LMS with this code.