Orbital Angular Momentum Formula: L = √(l(l+1)) h/2π

The orbital angular momentum formula for an electron in an atom is $L = \sqrt{l(l+1)}\,\frac{h}{2\pi}$ , often written $L = \sqrt{l(l+1)}\,\hbar$ , where l is the azimuthal quantum number of the orbital. An s electron (l = 0) has zero orbital angular momentum, a p electron has $\sqrt{2}\,\hbar$ , and so on up the subshells.

Physics students meet an older version first. Bohr's model of hydrogen assumes $mvr = \frac{nh}{2\pi}$ , so the angular momentum is a whole number multiple of ħ. Chemistry courses on atomic structure use the quantum mechanical form with l, and that is the one this page treats as the main result. Both appear in the modern physics formula sheet hub, while the classical idea behind them is on the angular momentum formula page.

Quick facts

Formula
$L = \sqrt{l(l+1)}\,\hbar$
Bohr's model form
$L = mvr = \frac{nh}{2\pi}$
SI unit
joule second (J s), the same as kg m²/s
Dimensional formula
[M¹ L² T⁻¹]
z component
$L_z = m_l\,\hbar$ , with $m_l = -l$ to $+l$
Type
Vector; only its size and one component can be fixed at once
Value of ħ
h/2π = 1.055 × 10⁻³⁴ J s

Symbols in the orbital angular momentum formula

Quantum numbers and constants used
Symbol	Name	Allowed values or unit
$L$	Size of the orbital angular momentum	J s
$l$	Azimuthal quantum number	0, 1, 2, ..., n − 1
$n$	Principal quantum number (shell)	1, 2, 3, ...
$m_l$	Magnetic quantum number	−l to +l, 2l + 1 values
$\hbar$	h divided by 2π	1.055 × 10⁻³⁴ J s

The shell number n limits l but does not fix it. In the n = 3 shell, l can be 0, 1 or 2, giving the 3s, 3p and 3d subshells, each with a different orbital angular momentum. Electrons also carry a separate spin angular momentum, $\sqrt{s(s+1)}\,\hbar$ with s = ½, which is not included in L.

Angular momentum formula in chemistry: subshell values

Orbital angular momentum by subshell
Subshell	l	$\sqrt{l(l+1)}$	L in units of h/2π
s	0	0	0
p	1	$\sqrt{2}$	1.414
d	2	$\sqrt{6}$	2.449
f	3	$\sqrt{12} = 2\sqrt{3}$	3.464

Chemistry questions often stop at an answer such as $\sqrt{6}\,\frac{h}{2\pi}$ and do not ask for a number in J s. Read the options: if they contain h/2π, leave the answer in that form. Note the zero for s orbitals. A 1s electron has no orbital angular momentum at all, which is one place where the quantum result and Bohr's model disagree.

Bohr's quantisation condition

Bohr's model of the hydrogen atom places the electron in circular orbits where

$mvr = \frac{nh}{2\pi}, \quad n = 1, 2, 3, \ldots$

This gives $L = n\hbar$ : ħ in the first orbit, 2ħ in the second. The condition, combined with the Coulomb attraction providing the centripetal force, makes the orbit radius grow as $n^2/Z$ from 0.529 Å, while the level energy is $-13.6\,Z^2/n^2$ eV. It predicts the hydrogen spectrum well, as the number of spectral lines formula page shows, but it gets the angular momentum of real orbitals wrong. Use nh/2π only when the question says Bohr orbit; use $\sqrt{l(l+1)}\,\hbar$ when it names an orbital or subshell.

Solved problems

Example 1: a 3d electron

Find the orbital angular momentum of an electron in a 3d orbital.

For d, l = 2. The value of n does not enter.

$L = \sqrt{2 \times 3}\,\hbar = \sqrt{6} \times 1.055 \times 10^{-34} = 2.58 \times 10^{-34}\ \text{J s}$

In the h/2π form the answer is √6 h/2π.

Example 2: second Bohr orbit

Find the angular momentum of the electron in the second Bohr orbit of hydrogen.

$L = \frac{nh}{2\pi} = \frac{2 \times 6.626 \times 10^{-34}}{2\pi} = \frac{h}{\pi} = 2.11 \times 10^{-34}\ \text{J s}$

Example 3: working backwards

An electron has orbital angular momentum $2\sqrt{3}\,\hbar$ . Which subshell is it in?

Set $l(l+1) = 12$ . Trying values, l = 3 gives 3 × 4 = 12.

So the electron is in an f subshell. The calculator above solves l the same way from L in J s.

Units and dimensions

Angular momentum is moment of inertia times angular velocity, or $mvr$ for a point mass. Multiply the units: kilograms times metres per second times metres gives kg m²/s. Planck's constant has exactly the same unit, J s = kg m² s⁻¹, which is why ħ can serve as the natural unit of angular momentum in atoms. The dimensional formula of angular momentum is therefore [M¹ L² T⁻¹], the same as for h. The square root factor $\sqrt{l(l+1)}$ is dimensionless, so it changes nothing here.

Traps in exam questions

Using n in the quantum formula. Put the azimuthal number l under the root, never the shell number.
Writing l² instead of l(l + 1). The magnitude is $\sqrt{l(l+1)}\,\hbar$ , always a little more than lħ.
Giving s electrons ħ. For l = 0 the orbital angular momentum is zero.
Confusing L with L_z. The z component $m_l\hbar$ can be zero or negative; the magnitude cannot be negative.
Dropping the 2π. h and ħ differ by a factor of 6.28, an easy slip in numerical answers.

Check: for any l above zero, √(l(l + 1)) lies between l and l + 1.

Frequently asked questions

What is the formula of orbital angular momentum?

For an electron in an orbital with azimuthal quantum number l, L = √(l(l + 1)) h/2π. In Bohr's model of hydrogen it is taken as L = nh/2π instead.

What is the angular momentum formula in chemistry?

Chemistry uses L = √(l(l + 1)) h/2π for orbital angular momentum. So s, p, d and f electrons have 0, √2, √6 and √12 units of h/2π.

What is the orbital angular momentum of a 2p electron?

For a p electron l = 1, so L = √2 h/2π, about 1.49 × 10⁻³⁴ J s. The shell number 2 does not change it.

What is the difference between orbital and spin angular momentum?

Orbital angular momentum comes from the electron's motion around the nucleus and depends on l. Spin is an intrinsic property, √(s(s + 1)) ħ with s = ½, and is the same for every electron.

What is the SI unit of orbital angular momentum?

Orbital angular momentum is measured in J s (joule seconds), equal to kg m²/s. Dimensionally it is written [M¹ L² T⁻¹].