A 600 K piece of metal would emit radiation at a peak wavelength of what?

1 Answer
Jan 17, 2016

lamda_"max" = "5000 nm"

Explanation:

For an object that emits black-body radiation, the peak wavelength is simply the wavelength at which the intensity of the emitted radiation is at its maximum value.

Now, black-body radiation is a type of electromagnetic radiation emitted by a black body that's held at a constant temperature. The important thing to realize about black-body radiation is that its intensity depends exclusively on the absolute temperature of the body.

The relationship between the maximum intensity of the black-body radiation emitted by an object and the wavelength at which it occurs is given by the Wien displacement law equation

color(blue)(lamda_"max" = b/T)" ", where

lamda_"max" - the wavelength at which the intensity of the radiation is maximum
b - the Wien displacement constant, usually given as 2.89777 * 10^(-3)"m K"
T - the absolute temperature of the object

In your case, the absolute temperature of the metal is said to be equal to "600 K". This means that its peak wavelength will be equal to

lamda_"max" = (2.89777 * 10^(-3) "m" color(red)(cancel(color(black)("K"))))/(600color(red)(cancel(color(black)("K")))) = 4.83 * 10^(-6)"m"

Expressed in nanometers and rounded to one sig fig, the number of sig figs you have for the temperature of the metal, the answer will be

lamda_"max" = color(green)("5000 nm")

This wavelength places the radiation in the infrared region of the electromagnetic spectrum. More specifically, this wavelength corresponds to the Far-IR region.

http://www.cosmosportal.org/view/article/140793/http://www.cosmosportal.org/view/article/140793/