The Most Probable Radius and the Bohr Radius

Introduction

In Bohr's 1913 model the electron in hydrogen moved in a circular orbit of radius 52.9 pm. Quantum mechanics abolished such orbits, yet the same number survives: it is the most probable distance of the 1s electron from the nucleus. This page shows how that result follows from the radial distribution function by simple calculus, and why the old and new pictures happen to share one length while disagreeing about almost everything else.

Core explanation

The quantity to maximise. The probability density ψ² is largest at the nucleus, but the probability of finding the electron at a distance r is measured by the radial distribution function. For hydrogen 1s,

ψ₁ₛ = (1/√(πa₀³)) e^(−r/a₀)

P(r) = 4πr²ψ₁ₛ² = (4/a₀³) r² e^(−2r/a₀)

Differentiating. The maximum of P(r) is where dP/dr = 0. Using the product rule and ignoring the constant 4/a₀³:

d/dr [r² e^(−2r/a₀)] = 2r e^(−2r/a₀) − (2/a₀) r² e^(−2r/a₀) = 2r e^(−2r/a₀)(1 − r/a₀)

This is zero at r = 0 (a minimum, since P(0) = 0), as r → ∞ (P decays to zero) and at r = a₀ , which must be the maximum.

The value of a₀. The Bohr radius is built from fundamental constants:

a₀ = 4πε₀ħ²/(mₑe²) ≈ 5.29 × 10⁻¹¹ m = 52.9 pm

It balances two effects. A smaller ħ or a larger electron mass allows tighter confinement; a larger e (stronger attraction) pulls the electron in. The same combination sets the natural unit of length in atomic units.

Reduced mass. Strictly, the electron and proton both move about their common centre of mass, so mₑ should be replaced by the reduced mass μ = mₑmₚ/(mₑ + mₚ). For hydrogen μ ≈ 0.99946mₑ, making the true most probable radius about 0.05% larger than a₀. For positronium, where the two particles have equal masses, μ = mₑ/2 and the length scale doubles.

Other orbitals. For any orbital with l = n − 1 (1s, 2p, 3d, 4f) the radial function has no nodes and R ∝ r^(n−1)e^(−r/na₀), so P ∝ r^(2n)e^(−2r/na₀). Differentiating gives

r mp = n²a₀

so 2p peaks at 4a₀ and 3d at 9a₀. This reproduces the Bohr-model radii n²a₀, although Bohr's model gives these radii for every orbit rather than only for the nodeless orbitals. For a hydrogen-like ion of charge Z every distance is divided by Z: r mp(1s) = a₀/Z.

Most probable is not average. The mean radius ⟨r⟩ for 1s is 1.5a₀, larger than r mp, because P(r) has a long tail towards large r. Different measures of "size" give different numbers.

Formulae

For 1s: P(r) = (4/a₀³)r²e^(−2r/a₀), maximised at r mp = a₀. For l = n − 1: r mp = n²a₀/Z. a₀ = 4πε₀ħ²/(mₑe²) = 52.9 pm.

Step-by-step reasoning

To find a most probable radius:

1. Write the radial function R(r) for the orbital. 2. Form P(r) = r²R². 3. Differentiate (taking logarithms first often helps: d ln P/dr = 0). 4. Solve for r, discarding r = 0 and r → ∞. 5. If several roots remain (orbitals with radial nodes), compare the values of P to identify the largest peak.

Visual explanation

Plot ψ² and P(r) for 1s on the same graph. ψ² starts at its highest point and falls steadily. P(r) starts at zero, climbs steeply, peaks exactly at r = a₀ and then falls with a long tail. A vertical line at a₀ marks the "Bohr radius" as the summit of the probability-per-distance curve.

Real-world analogy

Imagine throwing darts at a board where the aim is very good, so the hits are densest at the bullseye. Count the darts in successive rings, however, and the busiest ring is not the bullseye itself, because it is tiny, but a ring slightly further out. The most probable radius is that busiest ring.

Real-world example

The Bohr radius sets the scale of atomic sizes and bond lengths. The H–H bond length in H₂ is 74 pm, about 1.4a₀, and the covalent radii of carbon, nitrogen and oxygen (about 66–77 pm) lie between one and one and a half Bohr radii. Scanning-tunnelling and electron-diffraction measurements confirm that atoms are a few tenths of a nanometre across, just as a₀ predicts.

Why?

Why do quantum mechanics and the Bohr model agree on this number? Both use the same balance between Coulomb attraction and the kinetic energy implied by quantisation, and both give the same energy −13.6 eV. The resulting length combination of ħ, mₑ, e and ε₀ is fixed by dimensional analysis, so both theories inevitably arrive at a₀.

Common misconception

"The electron in hydrogen orbits at 52.9 pm." Quantum mechanics has no orbit. The electron can be found at any distance; a₀ is only the single most likely distance. The probability of lying within a₀ of the nucleus is only about 32%.

Worked example

Question: Find the most probable radius of the 1s electron in Li²⁺ in picometres.

Reasoning: For a hydrogen-like ion, ψ₁ₛ ∝ e^(−Zr/a₀), so P ∝ r²e^(−2Zr/a₀). Setting the derivative to zero gives r = a₀/Z. With Z = 3: r = 52.9/3 pm.

Answer: r mp ≈ 17.6 pm.

Quick check

1. What is the most probable radius of a hydrogen 3d electron, expressed in units of the Bohr radius? Answer: 9a₀, because for l = n − 1 the most probable radius is n²a₀, and n = 3 gives 9a₀.

Exam focus

The derivation of r mp = a₀ by differentiating r²e^(−2r/a₀) is a standard examination question. Show the product rule clearly, identify all three stationary points and state which one is the maximum. Remember to distinguish r mp from ⟨r⟩.

Advanced insight

The probability of finding the 1s electron inside radius R is 1 − e^(−2x)(1 + 2x + 2x²), where x = R/a₀. At R = a₀ this gives about 0.32; the 90% boundary surface lies at about 2.66a₀. Such cumulative probabilities are what define boundary surfaces, and they show how diffuse the electron cloud really is compared with the sharp orbit of the Bohr picture.

Summary

The most probable radius is the maximum of P(r). For hydrogen 1s, differentiating (4/a₀³)r²e^(−2r/a₀) gives r mp = a₀ = 52.9 pm, the Bohr radius. Nodeless orbitals with l = n − 1 peak at n²a₀/Z. The Bohr radius survives into quantum mechanics as a most probable distance, not an orbit, and the mean radius is larger because of the tail of P(r).

Practice questions

1. Show that the 2p radial distribution P ∝ r⁴e^(−r/a₀) has its maximum at 4a₀. Answer: dP/dr ∝ (4r³ − r⁴/a₀)e^(−r/a₀) = r³(4 − r/a₀)e^(−r/a₀), which is zero at r = 4a₀. 2. Calculate the most probable radius of the 1s electron in He⁺. Answer: a₀/Z = 52.9/2 ≈ 26.5 pm. 3. Why is ⟨r⟩ larger than r mp for hydrogen 1s? Answer: P(r) is asymmetric with a long tail to large r, which pulls the average beyond the peak. 4. How would the most probable radius change if the electron were replaced by a muon, about 207 times heavier? Answer: a₀ is inversely proportional to the (reduced) mass, so the radius would shrink by a factor of roughly 186 when the reduced mass of the muon–proton system is used.