Tabulate the energy of radiation having a wavelength of 4000 A°
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Main articles: Radiation/Cyans, Cyan radiation, and Cyans
Planck's equation (colored curves) accurately describes black body radiation. Credit: .
The color box shows some of the variations of cyan. Credit: .
Aquamarine is a blue or turquoise variety of beryl. Credit: .
The turquoise gemstone is the namesake for the color. Credit: .
These are cyan colored fluorite crystals from Rogerley Mine, Frosterley, Weardale, North Pennines, Co. Durham, England, UK. Credit: Parent Géry.
Cyan is the color of clear water over a sandy beach. Credit: .
The image shows a blue sky, white clouds over a blue-green ocean on Earth. Credit: SKYLIGHTS.
The image shows blue-green algae collecting along the shores of the Madison Lakes in Wisconsin. Credit: Bryce Richter.
Cyan light has a wavelength of between 490 and 520 nanometers, between the wavelengths of blue and green.[45]
Planck's equation describes the amount of spectral radiance at a certain wavelength radiated by a black body in thermal equilibrium.
In terms of wavelength (λ), Planck's equation is written: as
{\displaystyle B_{\lambda }(T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}} {\displaystyle B_{\lambda }(T)={\frac {2hc^{2}}{\lambda ^{5}}}{\frac {1}{e^{\frac {hc}{\lambda k_{\mathrm {B} }T}}-1}}}
where B is the spectral radiance, T is the absolute temperature of the black body, kB is the Boltzmann constant, h is the Planck constant, and c is the speed of light.
This form of the equation contains several constants that are usually not subject to variation with wavelength. These are h, c, and kB. They may be represented by simple coefficients: c1 = 2h c2 and c2 = h c/kB.
By setting the first partial derivative of Planck's equation in wavelength form equal to zero, iterative calculations may be used to find pairs of (λ,T) that to some significant digits represent the peak wavelength for a given temperature and vice versa.
{\displaystyle {\frac {\partial B}{\partial \lambda }}={\frac {c1}{\lambda ^{6}}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}[{\frac {c2}{\lambda T}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}e^{\frac {c2}{\lambda T}}-5]=0.} {\displaystyle {\frac {\partial B}{\partial \lambda }}={\frac {c1}{\lambda ^{6}}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}[{\frac {c2}{\lambda T}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}e^{\frac {c2}{\lambda T}}-5]=0.}
Or,
{\displaystyle {\frac {c2}{\lambda T}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}e^{\frac {c2}{\lambda T}}-5=0.} {\displaystyle {\frac {c2}{\lambda T}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}e^{\frac {c2}{\lambda T}}-5=0.}
{\displaystyle {\frac {c2}{\lambda T}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}e^{\frac {c2}{\lambda T}}=5.} {\displaystyle {\frac {c2}{\lambda T}}{\frac {1}{e^{\frac {c2}{\lambda T}}-1}}e^{\frac {c2}{\lambda T}}=5.}
Use c2 = 1.438833 cm K.
For a star to have a peak in the cyan, iterative calculations using the last equation yield the pairs: approximately (476 nm, 6300 K) and (495 nm, 6100 K)