# Is Color Temperature and Wavelength the Same Thing?

I know that a higher color temperature is associated with cool light and a lower color temperature is associated with warm light (green and red)

However I'm trying to understand the correlation between color temperature and wavelength (or frequency) of light.

Most bulbs are rated with a color temperature in kelvin but contain no information regarding the wavelength of the light.

I need to produce a vast amount of light in the wavelength of 380 nm.

Is it possible to convert kelvin (color temp) into wavelength? If not how can I know the wavelength of a given bulb or light source? I suppose that light may be made up of multiple wavelengths not just one - if that is the case then how can I know which wavelength dominates or how much of each wavelength is emitted by a given source?

• Possibly related: physics.stackexchange.com/q/297533/26969 – Floris Sep 13 '17 at 16:01
• Have a look at Edgar Bonet's answer here. physics.stackexchange.com/questions/331899/… – Farcher Sep 13 '17 at 16:10
• FYI: The phrase "color temperature" comes from photography. Professional photographers sometimes use photo flood lamps in their studio. If the bulb's envelope is not colored blue (as some of them are) then the "color temperature" listed on the package of an incandescent photo flood bulb is the actual temperature of the tungsten filament (expressed in Kelvins) when the lamp operates from the recommended voltage. – Solomon Slow Sep 13 '17 at 17:14
• If you need "vast amounts of UV light" you might want to look at indoor tanning solutions. UVA is typically in the range of 320 nm to 400 nm. At least these are lamps that are designed to produce "reasonable amounts of" UV in the approximate range you are interested in. What exactly do you need 380 nm for? – Floris Sep 13 '17 at 20:30

Color temperature describes how a particular light source produces light (in the visible range) that "looks like" light from a black body radiator at a particular temperature.

Because it is only intended to help with understanding the apparent color, it will not (in general) help you with figuring out the amount of light in the UV, unless you are actually talking about an incandescent emitter, and the enclosure of the emitter (light bulb) has excellent transmission properties across the range of wavelengths of interest.

In general the spectrum of a black body radiator is given by the Planck Equation. This earlier answer may be helpful to give you some background and plots.

Colour temperature is related to black-body radiation, in theory the distribution of wavelengths of a light source of a certain colour temperature should resemble that of a black body of that temperature.

So colour temperature is not exactly related to a wavelength, but to a distribution of wavelengths. You can calculate the wavelength of the highest intensity using Wien's displacement law.

The application to fluorescent lamps is more or less a marketing trick (describing the colour the lamp is perceived to have), as the emitted spectrum contains discrete peaks from the mercury and materials within the used phosphor, together with more or less broad emission spectra by the phosphor, stimulated by the mercury's emissions, very unlike to the theoretical black body radiation.

If you need only one wavelength (or light in the range around a single wavelength), you could use an LED, or lasers, or maybe blacklight tubes.

I'm trying to understand the correlation between color temperature and wavelength (or frequency) of light.

You should probably be looking for a black light, which is a lamp specifically designed to produce UV. The Wiki article includes a table of different fluorescent phosphors used in blacklights, and their peak emission wavelengths. It looks like the closest to 380 nm is the $\rm SrB_{4}O_7, Eu$ type, with 370 nm center wavelength and 20 nm spectral width. Probably you want to look for a blacklight using this phosphor. From what's in Wiki, this type should be available from Osram, but might be hard to find in North America.