Question

Use the formula `lambda_(m)T=0.29` cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you ?

Answer 1

We know , every body at given temperature `T_(1)` emits radiations of all wavelengths in certain range .For a black body, the wavelength corresponding to maximum intensity of radiation at a given temperature .
`lambda_(m)T=0.29 cmk ` or `T=(0.29)/(lambda_(m))`
For `lambda_(m)=10^(-6)m =10^(-4) cm , T=(0.29)/(10^(-4))=2900 K`
Temperature for other wavelengths can be similarly found. These numbers tell us the temperature ranges required for obtained radiations in different parts of e.m spectrum . Thus to obtain visible radiation, say, `lambda_(m)=5xx10^(-5)` cm, the source should have temperature `T=(0.29)/(5xx10^(-5))=6000 K`
It is to be noted that , a body at lower temperature will also proudce this wavelength but not with maximum intensity.