Wavelength and Wavenumber not matching from Spectracalc

Question

When I use Spectracalc.com to calculate the peak wavelength at 500 Kelvin, I am told the peak is at 5.79551 µm. When I switch the calculation result to wavenumber the peak is said to be at 980.506 cm-1. But if I were to simply convert wavelength to wavenumber I do not get this value.

1/(5.79551E-6) = 172,547.3 m-1 = 1725.47 cm-1

What is going on here?

Edit:

Several people have answered that this has to do with the differentials not being the same size. I don't think this answers my question. I get that a wavenumber increment is not the same size as a wavelength increment. That isn't the question at all.

The question is more about the physical meaning of what is measured. Let me restate it more clearly.

Say I measure with a highly accurate physical instrument and get a value of the peak at 5.8 µm. This is a physical quantity. It is the distance between two sequential peaks of the wave of the light measured as coming from the body at the temperature of 500 K. This physical quantity, that describes a physical thing, also can be described by looking at the number of times the wave cycles in a unit distance. The wave that has a wavelength of 5.8 µm cycles 1/5.8 µm times in a unit distance, giving it a wavenumber of 1724.1 1/cm. This is also a physically meaningful way a to describe the exact same wave I measured as the peak. They are interchangeable descriptions of the same physical thing.

Now, say I had used a different instrument to measure the exact same physical body in the exact same state, an instrument that instead measured wavenumber. According to the wavenumber form of Planck's law, I should measure a wavenumber of 980.5 1/cm when earlier I had measured a wave with a wavelength of 1724.1 1/cm. This is supposed to be the exact same physical thing that was measured. Yet the exact same physical thing cannot have two different values for the same measure!

So, which is it: Is one of the forms of Planck's law wrong, or are the two measures of the same wave not in fact interchangeable, i.e., is it false that a wave with a wavelength X has a wavenumber 1/X? If the peak is a real physical phenomenon that corresponds to a real physical wavelength of X, then this physical wave MUST have a wavenumber of 1/X, unless that relation is just false. But if that relation is not false, then how on earth can we get two different values for the same physical phenomenon? The hot body knows which instrument I am using and moves the peak around accordingly? Something is not right about this.

https://www.spectralcalc.com/blackbody_calculator/blackbody.php

Answer 1

Let's stick to your example of a black body with temperature $T = 500$ K.

According to Planck's law - Different forms the black-body spectrum in terms of wavelength $\lambda$ (i.e. the intensity $dI$ per wavelength range $d\lambda$) is $$ \frac{dI}{d\lambda} = B_\lambda(\lambda,T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda k_B T}-1} $$ For temperature $T = 500$ K it looks like this. Note especially the peak at wavelength $5.8\ \mu$m.

Likewise, the spectrum in terms of wavenumber $\tilde{\nu}$ (i.e. the intensity $dI$ per wavenumber range $d\tilde{\nu}$) is $$ \frac{dI}{d\tilde{\nu}} = B_{\tilde{\nu}}(\tilde{\nu},T) = 2hc^2\tilde{\nu}^3 \frac{1}{e^{hc\tilde{\nu}/k_B T}-1} $$ For temperature $T = 500$ K it looks like this. Note especially the peak at wavenumber $980$ cm$^{-1}$.

For further visualization I have highlighted some corresponding wavelength and wavenumber ranges.

• Red: $2\ \mu\text{m} < \lambda < 5\ \mu\text{m}$ corresponding to $5000\ \text{cm}^{-1} > \tilde{\nu} > 2000\ \text{cm}^{-1}$.
• Green: $10\ \mu\text{m} < \lambda < 25\ \mu\text{m}$ corresponding to $1000\ \text{cm}^{-1} > \tilde{\nu} > 400\ \text{cm}^{-1}$.

Note that both green areas have the same size, and both red areas have the same size. This is because it is the same spectral energy, regardless if you measure the spectrum in $\mu$m or cm$^{-1}$. But apart from that the change from wavelength to wavenumber heavily distorted the form of the spectral curve. With this distortion comes also a shift of the peak.

The important thing to understand is: In your experiment you don't measure the intensity of a single wave. You can only measure the intensity $\Delta I$ of a small wavelength range $\Delta\lambda$ or of a small wavenumber range $\Delta\tilde{\nu}$. And from that you then calculate $\frac{\Delta I}{\Delta\lambda}$, or $\frac{\Delta I}{\Delta\tilde{\nu}}$ respectively. Only this can be compared to the theoretical spectrum $\frac{dI}{d\lambda}$ shown above, or $\frac{dI}{d\tilde{\nu}}$ respectively.