Wien's law calculator

Locate the maximum of an ideal thermal spectrum, with a choice of wavelength-density or frequency-density peak.

Calculator

Blackbody temperature
300
Wavelength-density peak (nm)
9,659.23985062
Frequency-density peak
17,636,800,000,000
Wavelength displacement constant (m K)
0.00289777
Frequency displacement constant (Hz/K)
58,789,257,576.46825400
Scientific notation for small or large results
QuantityValue in the stated units (SI for dimensioned outputs)
Frequency-density peak1.7636777273e+13
Frequency displacement constant (Hz/K)5.8789257576e+10

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How this calculator works

Planck spectra have a peak whose location depends on the spectral coordinate. For energy per wavelength interval, peak wavelength times temperature is constant. For energy per frequency interval, peak frequency divided by temperature is a different constant. Converting one peak using c/lambda does not give the other maximum because the interval widths change. This page uses the energy spectrum, not photon-count spectrum. A wavelength-independent grey emissivity changes amplitude without moving the peak. Constants are derived from CODATA 2018 h, c and kB.

Formula

lambda_max T = b_lambda = hc/(4.965114231744276 kB); nu_max/T = b_nu = 2.821439372122079 kB/h. These maxima describe different spectral densities.

Worked example

At 300 K, the energy-per-wavelength peak is about 9659.239851 nm. The energy-per-frequency peak is about 1.763677727e13 Hz. Doubling temperature halves the first peak and doubles the second.

Frequently asked questions

Why is nu_max not c/lambda_max?

Changing coordinates changes the width of a spectral bin and therefore the location of the density maximum.

Can this give a star temperature?

It gives an approximate colour temperature for a near-blackbody spectrum. Absorption lines, dust and instrumental response affect observations.

Does emissivity move the peak?

Constant grey emissivity does not. Emissivity that varies with wavelength can shift the maximum.

Sources