Radial and Angular Nodes of Hydrogen-Like Orbitals

By Aniket Bhardwaj · 21 October 2026 · Updated 21 October 2026 · Physical Chemistry

Nodes are the places where an electron is never found. They decide the shape of every orbital, and they explain why a 2s orbital has a second, smaller region around the nucleus while a 2p orbital has none. Questions on nodes appear in IIT-JAM, GATE, CSIR-NET and CUET-PG as direct counting problems, and sometimes as a request to find the actual distance of a radial node. This page gives the counting rules, derives the node positions for a few hydrogen-like orbitals, and works five examples with the arithmetic checked.

The radial distribution of 2s has two maxima separated by a radial node at ρ = 2, while 2p has one maximum and no radial node. — 2s has two maxima and one radial node; 2p has one maximum and none.

The counting rules

Radial nodes = n − l − 1
Angular nodes = l
Total nodes = n − 1

n is the principal quantum number and l is the azimuthal (angular momentum) quantum number: l = 0, 1, 2, 3 for s, p, d, f. A radial node is a spherical surface at a fixed distance r from the nucleus where the wavefunction ψ is zero. An angular node is a plane or cone passing through the nucleus. The wavefunction changes sign when you cross a node.

The total number of nodes depends only on n. Each step up in n adds exactly one node, and the l value decides how many of the nodes are angular and how many are radial.

Table of nodes

Orbital	n	l	Radial nodes	Angular nodes	Total
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2
4f	4	3	0	3	3

Notice the pattern along each row of n: the first orbital of a given l has no radial node. The lowest p orbital is 2p, the lowest d is 3d and the lowest f is 4f; each has zero radial nodes.

Where the radial nodes lie

For a one-electron atom with nuclear charge Z, write ρ = Zr ÷ a₀, where a₀ is the Bohr radius, 52.9177 pm. The radial parts of the wavefunctions are a polynomial in ρ multiplied by a decaying exponential. A radial node is a root of the polynomial. The standard polynomials are:

Orbital	Polynomial in ρ	Roots (nodes)
2s	2 − ρ	ρ = 2
3s	27 − 18ρ + 2ρ²	ρ = 1.902 and 7.098
3p	ρ(6 − ρ)	ρ = 6 (the factor ρ only gives the zero at r = 0)

The node positions in a hydrogen-like atom are therefore r = ρ a₀ ÷ Z. A larger Z pulls the nodes in towards the nucleus. These forms follow from the hydrogen-like radial equation; the exponential factor never equals zero, so it does not create nodes.

Worked example 1: counting nodes

Question. Find the radial, angular and total nodes in a 4d orbital and in a 5f orbital.

4d. n = 4, l = 2. Radial = 4 − 2 − 1 = 1; angular = 2; total = 4 − 1 = 3. Check: 1 + 2 = 3.

5f. n = 5, l = 3. Radial = 5 − 3 − 1 = 1; angular = 3; total = 5 − 1 = 4. Check: 1 + 3 = 4.

Worked example 2: from nodes to the orbital

Question. An orbital has 2 radial nodes and 1 angular node. Name it.

l. Angular nodes = l, so l = 1 (a p orbital).

n. Radial nodes = n − l − 1 = 2, so n = 2 + 1 + 1 = 4.

Answer. It is a 4p orbital. Total nodes = 3 = n − 1, which confirms it.

Worked example 3: the 2s node in hydrogen

Question. Find the distance of the radial node of the 2s orbital of the hydrogen atom from the nucleus. Take a₀ = 52.9177 pm.

Condition. 2 − ρ = 0, so ρ = 2. For hydrogen Z = 1, so r = 2a₀.

Calculation. r = 2 × 52.9177 = 105.8 pm.

Worked example 4: the effect of nuclear charge

Question. Where is the 2s radial node in He⁺ (Z = 2) and in Li²⁺ (Z = 3)?

Rule. ρ = 2 for any hydrogen-like 2s orbital, and r = ρ a₀ ÷ Z.

He⁺. r = 2 × 52.9177 ÷ 2 = 52.9 pm.

Li²⁺. r = 2 × 52.9177 ÷ 3 = 35.3 pm.

The node moves inward as Z increases, because the stronger attraction shrinks the whole orbital.

Worked example 5: the two nodes of 3s

Question. Find the radial nodes of the 3s orbital of hydrogen.

Polynomial. 2ρ² − 18ρ + 27 = 0.

Quadratic formula. ρ = [18 ± √(18² − 4 × 2 × 27)] ÷ (2 × 2) = [18 ± √(324 − 216)] ÷ 4 = [18 ± √108] ÷ 4 = [18 ± 10.392] ÷ 4.

Roots. ρ₁ = 7.608 ÷ 4 = 1.902; ρ₂ = 28.392 ÷ 4 = 7.098.

Distances. r₁ = 1.902 × 52.9177 = 100.6 pm; r₂ = 7.098 × 52.9177 = 375.6 pm.

Check. Sum of roots = 18 ÷ 2 = 9 and 1.902 + 7.098 = 9.000. Product = 27 ÷ 2 = 13.5 and 1.902 × 7.098 = 13.50. The number of roots is 2, as n − l − 1 = 2 requires.

Nodes and the radial distribution curve

If you plot the radial distribution function 4πr²R²(r) against r, the curve touches zero at every radial node. It also touches zero at r = 0. A useful result is that this curve for a given orbital has (n − l) maxima. For example, 2s has two maxima, with the larger one outside the node, and 2p has one. The inner peak of 2s lies close to the nucleus, which is one reason a 2s electron is lowered in energy relative to 2p in many-electron atoms (penetration).

Common mistakes

Counting the nucleus. The zero of ψ at r = 0 for p, d and f orbitals is not a radial node. Radial nodes are at finite distances.
Using n − l instead of n − l − 1. n − l is the number of maxima in the radial distribution curve, one more than the number of nodes.
Forgetting angular nodes. A 2p orbital has one node, the nodal plane, even though it has no radial node.
Applying hydrogen-like roots to many-electron atoms. The positions r = ρ a₀ ÷ Z are for one-electron species only; the counting rules still hold for all atoms.
Mixing up total nodes and radial nodes. Total = n − 1 always.

Exam relevance

Question type	Method
Count nodes of a named orbital	Use n − l − 1, l and n − 1
Name the orbital from its nodes	l = angular nodes, then n = radial + l + 1
Find the distance of a node	Solve the radial polynomial for ρ, then r = ρ a₀ ÷ Z
Compare penetration	Use the number of maxima, n − l