What Is Planck’s Constant, Simply Explained?

Planck’s constant is one of the basic constants of nature. It tells you how energy is connected to frequency in quantum physics, and it helps set the scale at which quantum effects matter. Its small numerical value is why those effects are usually hard to notice in everyday objects.

The short version: a quantum scale

Planck’s constant is written $h$. In the International System of Units, its exact value is

$$h = 6.62607015 \times 10^{-34}\;\mathrm{J\,s}.$$

The unit joule-second (J s) means energy multiplied by time. The constant is not itself an amount of energy; it is a conversion factor that relates energy to frequency. For a photon—a quantum of light—the relation is

$$E = hf,$$

where $E$ is the photon’s energy and $f$ is its frequency. Higher-frequency light has higher-energy photons. This is one of the most useful ways to understand what $h$ does.

A worked example: energy of red light

Suppose light has wavelength $\lambda = 600\;\mathrm{nm}$. Its frequency is $f=c/\lambda$, where $c=2.998\times10^8\;\mathrm{m/s}$ is the speed of light. First convert the wavelength to meters: $600\;\mathrm{nm}=6.00\times10^{-7}\;\mathrm{m}$. Then

$$\begin{aligned} f &= \frac{c}{\lambda} \\ &= \frac{2.998\times10^8}{6.00\times10^{-7}}\;\mathrm{s^{-1}} \\ &\approx 5.00\times10^{14}\;\mathrm{Hz}. \end{aligned}$$

Use $E=hf$:

$$\begin{aligned} E &= (6.62607015\times10^{-34}\;\mathrm{J\,s})(5.00\times10^{14}\;\mathrm{s^{-1}}) \\ &\approx 3.31\times10^{-19}\;\mathrm{J} \\ &\approx 2.07\;\mathrm{eV}. \end{aligned}$$

The seconds cancel, leaving joules, as required for energy. This is the energy of one photon at that wavelength—not the total energy in a beam, which depends on how many photons it contains.

Why does a tiny constant matter?

The value of $h$ is tiny when expressed in joule-seconds, but its role is not to set an absolute cutoff between “quantum” and “classical.” Rather, it is a scale for comparing a system’s characteristic action—roughly, energy multiplied by time or momentum multiplied by distance—with the scale of quantum mechanics.

When the characteristic action is much larger than $h$, quantum behavior can often be approximated by classical physics. When it is comparable to $h$, quantum effects can become important. Atoms are a familiar example: their allowed energies and other properties cannot generally be described by simply treating their components as tiny classical objects.

Planck’s constant also appears in the de Broglie relation for matter waves:

$$\lambda = \frac{h}{p},$$

where $p$ is momentum. For an object with a large everyday momentum, this wavelength is extremely small and difficult to observe. For particles such as electrons, it can be significant.

What is the reduced Planck constant?

You will often see $\hbar$, pronounced “h-bar.” It is defined by

$$\hbar = \frac{h}{2\pi} \approx 1.05457\times10^{-34}\;\mathrm{J\,s}.$$

The two constants are not different physical quantities; they are related by the factor $2\pi$. The reduced form is convenient when equations use angular frequency $\omega$, measured in radians per second. Since $\omega=2\pi f$, the photon energy can be written either way:

$$E=hf=\hbar\omega.$$

Use $h$ with frequency in cycles per second (hertz), and $\hbar$ with angular frequency in radians per second. Many quantum-mechanics equations use $\hbar$ because rotations and waves are often expressed using angular quantities.

Common mistakes

• Treating $h$ as an energy. It is a constant with units J s. The product $hf$ has units of energy.
• Thinking all energy comes in identical chunks of size $h$. For a photon, the energy depends on its frequency: $E=hf$. In other systems, the allowed energies depend on the system and its conditions.
• Mixing up frequency and angular frequency. If you use $E=hf$, use $f$ in hertz. If you use $E=\hbar\omega$, use angular frequency $\omega$.
• Assuming the small value means quantum effects are negligible everywhere. The relevant comparison is between $h$ and the characteristic action of the system. At atomic scales, quantum effects are central.

Why is its value exact?

Since 2019, the SI definition of the kilogram fixes the numerical value of $h$ exactly at $6.62607015\times10^{-34}\;\mathrm{J\,s}$. This does not mean experiments cannot have uncertainty when measuring quantities involving $h$; it means the constant’s numerical value is part of the definition of the SI units.