# Why 8–15 µm is considered “thermal infrared” if typical room temperature kT is 48 µm?

According to Wikipedia:

Long-wavelength infrared (8–15 µm, 20–37 THz, 83–155 meV): The "thermal imaging" region, in which sensors can obtain a completely passive image of objects only slightly higher in temperature than room temperature - for example, the human body - based on thermal emissions only and requiring no illumination such as the sun, moon, or infrared illuminator. This region is also called the "thermal infrared".

However, using $\frac{hc}{\lambda}=k_\mathrm B T$, the temperature range 288–308 K (15–35 °C) is equivalent to 50–46.7 µm, while 8–15 µm is equivalent to 1800–960 K (using the same equation).

$\lambda=\frac{b}{T}$
where Wien's displacement constant $b=2.8977729(17)×10^{−3} m K$. Put $T=288K,308K$ into that and you get:
$\lambda=9.4-10.0\mu m$, which as you'd expect is at the lower range of the thermal infra-red sensors.
Note that $b=\frac{hc}{xk_B}$ where $x$ can be determined from the finding the peak of the black body spectrum from Planck's law - which has to be done numerically. It turns out that $x=4.96$, as opposed the value of 1 you in effect used in your original estimate. Hence your values are around a factor of 5 too high.