Assuming human skin is at 98.6 degrees Fahrenheit, what wavelength is the peak in the human thermal radiation spectrum? What type of waves are these?
2 Answers
Explanation:
Wien's displacement law states that the black-body radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature.
where b is Wien's displacement constant, equal to
plugging this into our equation, we get the peak wavelength of radiated light:
This is in the range of what is called infrared radiation.
Should be the infrared fingerprint region.
We have a relationship for this called Wien's Displacement Law:
#\mathbf(lambda_max = b/T)# where:
#b = 2.89777xx10^(-3)# #"m"cdot"K"# is a proportionality constant, probably experimentally determined.#T# is temperature in#"K"# .#lambda_max# is the wavelength that you observe at its largest spectral energy density.
The spectral energy density is depicted in the following diagram, with respect to wavelength in
![https://upload.wikimedia.org/]()
You can think of the spectral energy density as being proportional to the contribution of each wavelength range to some final observed color at a particular temperature. You can see that the peaks would correspond to
Converting temperature to
#(98.6^@ "F" - 32) xx 5/9 = 37^@ "C"#
#37 + 273.15 ~~ color(green)("310.15 K")#
And now we get a max wavelength of:
#lambda_max = (2.89777xx10^(-3) "m"cdotcancel"K")/("310.15" cancel"K")#
#= 9.343xx10^(-6) "m"#
Converting this to
#= 9.343xx10^(-6) cancel("m") xx (10^6 mu"m")/(1 cancel"m")#
#= color(blue)(9.343)# #color(blue)(mu"m")#

Being close to

