The Planck Function

In Planck and Blackbody Radiation you saw why quantized energies give the blackbody spectrum. Astrophysicists use the result as a working tool. It tells a star's temperature from its color, its power from its size, and it describes the oldest light in the universe.

Key idea

The Planck function gives the power emitted per unit area, per unit wavelength, per unit solid angle: $$B_\lambda(T) = \frac{2hc^2}{\lambda^5}\,\frac{1}{e^{hc/\lambda kT} - 1}.$$ Its peak gives Wien's law, $\lambda_{\max}T = 2.898 \times 10^{-3}$ m·K, and its area gives the flux $F = \pi\int_0^\infty B_\lambda\,d\lambda = \sigma T^4$.

Where the peak comes from

Why it works

Write $x = hc/(\lambda kT)$. Then $1/\lambda^5 = (kT/hc)^5x^5$, so at a fixed temperature $$B_\lambda = 2hc^2\left(\frac{kT}{hc}\right)^{5}\frac{x^5}{e^x - 1}.$$ Finding the peak means maximizing $x^5/(e^x - 1)$. Setting the derivative to zero gives $5(e^x - 1) = xe^x$, or $x = 5(1 - e^{-x})$. Start at $x = 5$ and iterate: $5(1 - e^{-5}) = 4.966$, then $4.965$. So the peak sits at $x = 4.965$, which means $\lambda_{\max}T = hc/(4.965\,k) = 2.898 \times 10^{-3}$ m·K. The peak wavelength is inversely proportional to temperature.

Where σT⁴ comes from

Why it works

Integrate over all wavelengths, substituting $\lambda = hc/(xkT)$: $$\int_0^\infty B_\lambda\,d\lambda = \frac{2k^4T^4}{h^3c^2}\int_0^\infty \frac{x^3}{e^x - 1}\,dx.$$ This is an improper integral, and it converges: near $0$ the integrand behaves like $x^2$, and for large $x$ it dies off like $x^3e^{-x}$. Its value is $\pi^4/15$. Multiplying by $\pi$ (to add up the light leaving a surface in every outward direction) gives $F = \sigma T^4$ with $\sigma = 2\pi^5k^4/(15h^3c^2) = 5.67 \times 10^{-8}$ W/(m²·K⁴). The temperature comes out as $T^4$ because the substitution pulled out $T^5$ from the curve's height and $1/T$ from its width.

Planck curves at 3,500 K, 5,772 K (the Sun) and 8,000 K. The curves never cross: the hotter body is brighter at every wavelength, and its peak moves toward the blue.

Worked examples

Worked example: The Sun's peak

Where does the Sun's spectrum ($T = 5772$ K) peak?

$\lambda_{\max} = (2.898 \times 10^{-3}\text{ m·K})/(5772\text{ K})$.
$= 5.02 \times 10^{-7}$ m.

Answer: About 502 nm, blue-green light. (Plotted per unit frequency instead, the peak lands elsewhere, so "the Sun's peak color" depends on how you plot it.)

Worked example: Hotter is brighter at every wavelength

At $\lambda = 500$ nm, compare $B_\lambda$ for 8,000 K and 4,000 K. Use $hc/k = 1.4388 \times 10^{-2}$ m·K.

The factor $2hc^2/\lambda^5$ cancels in the ratio; only $1/(e^x - 1)$ changes.
$$x_{4000} = \frac{1.4388 \times 10^{-2}}{(5.00 \times 10^{-7})(4000)} = 7.194,$$ and half that at 8,000 K: $x_{8000} = 3.597$.
Ratio $= (e^{7.194} - 1)/(e^{3.597} - 1) = 1330/35.5 \approx 37.5$.

Answer: Doubling the temperature raises the 500 nm output about 37 times, far more than 2 times.

Worked example: The long-wavelength limit

Repeat the comparison at $\lambda = 1$ cm, a radio wavelength.

Now $$x_{4000} = \frac{1.4388 \times 10^{-2}}{(0.01)(4000)} = 3.6 \times 10^{-4},$$ which is tiny.
For small $x$, $e^x - 1 \approx x$, so $B_\lambda \approx 2ckT/\lambda^4$ (the Rayleigh–Jeans law), proportional to $T$.

Answer: The ratio is $2.00$: at long wavelengths the hotter body is brighter only in proportion to its temperature.

Common mistake

Thinking a hotter blackbody is brighter only at short wavelengths, because its peak moves to the blue. In fact $B_\lambda$ grows with $T$ at every wavelength: raising $T$ lowers $x = hc/\lambda kT$, which lowers $e^x - 1$ and raises $B_\lambda$. The curves for different temperatures never cross. Check with the examples above: the hotter body wins at 500 nm and at 1 cm.

The Planck Function

Where the peak comes from

Where σT⁴ comes from

Worked examples

Practice

Builds on

Leads to

Sources and further reading