Blackbody Radiation
A black body is one that absorbs all the EM radiation (light) that strikes it. To stay in thermal equilibrium, it must emit radiation at the same rate as it absorbs it so a black body also radiates well.
Radiation from a hot object is familiar to us. Objects around room temperature radiate mainly in the infrared as seen the the graph below.
If we heat an object up to about 1500 degrees we will begin to see a dull red glow and we say the object is red hot. If we heat something up to about 5000 degrees, near the temperature of the sun's surface, it radiates well throughout the visible spectrum and we say it is white hot.
By considering plates in thermal equilibrium it can be shown that the emissive power over the absorption coefficient must be the same as a function of wavelength, even for plates of different materials.
`(E_1(lamda,T))/(A_1(lamda))=(E_2(lamda,T))/(A_2(lamda))`
A black body is one that absorbs all radiation incident upon it.
`A_(BB)=1`
For a black body Wein's constant `b=0.288` cmK.
`lamda_m T= 0.288`
Radiation from a hot object is familiar to us. Objects around room temperature radiate mainly in the infrared as seen the the graph below.
If we heat an object up to about 1500 degrees we will begin to see a dull red glow and we say the object is red hot. If we heat something up to about 5000 degrees, near the temperature of the sun's surface, it radiates well throughout the visible spectrum and we say it is white hot.
By considering plates in thermal equilibrium it can be shown that the emissive power over the absorption coefficient must be the same as a function of wavelength, even for plates of different materials.
`(E_1(lamda,T))/(A_1(lamda))=(E_2(lamda,T))/(A_2(lamda))`
A black body is one that absorbs all radiation incident upon it.
`A_(BB)=1`
For a black body Wein's constant `b=0.288` cmK.
`lamda_m T= 0.288`